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Bryan’s Chemistry <strong>Notebook</strong>
Honors Chemistry<br />
Class Policies and Grading<br />
The students will receive a Unit Outline at the beginning of each Unit. It will<br />
have information about the assignments that they will do, what it’s grade<br />
classification will be, what action they will need to do to complete the<br />
assignment and when it is due.<br />
The students will receive a Weekly Memo of the activities they will be<br />
responsible for that week. It will serve to inform the students of the learning<br />
goal for the week. It will also give the students any special information<br />
about that week.<br />
The students will also receive daily lectures and assignments that are<br />
designed to teach and re-enforce information related to the learning goal.<br />
This will be time in which new material will be taught and reviewed and will<br />
give the students the opportunity to ask questions regarding the concepts<br />
being taught.<br />
The students will work with a Lab partner and also be in a Lab group, but it<br />
will be up to the individual student to do his or her part of all assignments<br />
and the individual student will ultimately be responsible for all information<br />
presented in the class.<br />
The students will be required to follow all District and School Policies and to<br />
follow all Lab Safety Procedures, which they will be given and will sign,<br />
while performing labs. Students should come to class on time and with the<br />
supplies needed for that class.<br />
The following grading policy will be used.<br />
Percent of Final Grade<br />
<strong>Notebook</strong> 40%<br />
Test/Projects 30%<br />
Labs/Quizzes 20%<br />
Work 10%<br />
The students will be given a teacher generated Mid Term and a District<br />
Final.<br />
0
Unit 1<br />
Measurement Lab<br />
Separation of Mixtures Lab with Lab Write Up<br />
Unit 2<br />
Flame Test Lab<br />
Nuclear Decay Lab<br />
Element Marketing Project<br />
Unit 3<br />
Golden Penny Lab with Lab Write Up<br />
Molecular Geometry<br />
Research Presentation on a Chemical<br />
Mid Term<br />
Unit 4<br />
Double Displacement Lab<br />
Stoichiometry Lab with Lab Write Up<br />
Mole Educational Demonstration Project<br />
Unit 5<br />
Gas Laws Lab with Lab Write Up<br />
States of Matter Lab<br />
Teach a Gas Law Project<br />
Unit 6<br />
Dilutions Lab<br />
Titration Lab<br />
District Final<br />
1
Unit 1 (22 days)<br />
Chapter 1 Introduction to Chemistry<br />
Honors Chemistry<br />
2016/2017 Syllabus<br />
3 days<br />
1.1 The Scope of Chemistry 1.3 Thinking Like a Scientist<br />
1.2 Chemistry and You 1.4 Problem Solving in Chemistry<br />
Chapter 2 Matter and Change<br />
2.1 Properties of Matter 2.3 Elements and Compounds<br />
2.2 Mixtures 2.4 Chemical Reactions<br />
Chapter 3 Scientific Measurement<br />
9 days<br />
10 days<br />
3.1 Using and Expressing Measurements 3.3 Solving Conversion Problems<br />
3.2 Units of Measurement<br />
Unit 2 (15 days)<br />
Chapter 4 Atomic Structure<br />
5 days<br />
4.1 Defining the Atom 4.3 Distinguishing Among Atoms<br />
4.2 Structure of the Nuclear Atom<br />
Chapter 5 Electrons in Atoms<br />
5 days<br />
5.1 Revising the Atomic Model 5.2 Electron Arrangement in Atoms<br />
5.3 Atomic Emission Spectrum and the Quantum Mechanical Model<br />
Chapter 6 The Periodic Table<br />
6.1 Organizing the Elements 6.3 Periodic Trends<br />
6.2 Classifying Elements<br />
Unit 3 (22 days)<br />
Chapter 25 Nuclear Chemistry<br />
25.1 Nuclear Radiation 25.3 Fission and Fusion<br />
25.2 Nuclear Transformations 25.4 Radiation in Your Life<br />
Chapter 7 Ionic and Metallic Bonding<br />
7.1 Ions 7.3 Bonding in Metals<br />
7.2 Ionic Bonds and Ionic Compounds<br />
Chapter 8 Covalent Bonding<br />
5 days<br />
6 days<br />
8 days<br />
8 days<br />
8.1 Molecular Compounds 8.3 Bonding Theories<br />
8.2 The Nature of Covalent Bonding 8.4 Polar Bonds and Molecules<br />
Unit 4 (14 days)<br />
Chapter 9 Chemical Names and Formulas<br />
6 days<br />
9.1 Naming Ions 9.3 Naming & Writing Formulas Molecular Compounds<br />
9.2 Naming and Writing Formulas for Ionic Compounds 9.4 Names for Acids and Bases<br />
Chapter 22 Hydrocarbons Compounds<br />
22.1 Hydrocarbons 22.4 Hydrocarbon Rings<br />
Chapter 23 Functional Groups<br />
4 days<br />
4 days<br />
23.1 Introduction to Functional Groups 23.4 Alcohols, Ethers, and Amines<br />
2
Unit 5 (28 days)<br />
Chapter 10 Chemical Quantities 8 days<br />
10.1 The Mole: A Measurement of Matter 10.3 % Composition & Chem. Formulas<br />
10.2 Mole-Mass and Mole-Volume Relationships<br />
Chapter 11 Chemical Reactions 8 days<br />
11.1 Describing Chemical Reactions 11.3 Reactions in Aqueous Solutions<br />
11.2 Types of Chemical Reactions<br />
Chapter 12 Stoichiometry 12 days<br />
12.1 The Arithmetic of Equations 12.3 Limiting Reagent and % Yield<br />
12.2 Chemical Calculations<br />
Unit 6 (22 days)<br />
Chapter 13 States of Matter 6 days<br />
13.1 The Nature of Gases 13.3 The Nature of Solids<br />
13.2 The Nature of Liquids 13.4 Changes in State<br />
Chapter 14 The Behavior of Gases 10 days<br />
14.1 Properties of Gases 14.3 Ideal Gases<br />
14.2 The Gas Laws 14.4 Gases: Mixtures and Movement<br />
Chapter 15 Water and Aqueous Systems 6 days<br />
15.1 Water and its Properties 15.3 Heterogeneous Aqueous Systems<br />
15.2 Homogeneous Aqueous Systems<br />
Unit 7 (18 days)<br />
Chapter 16 Solutions 8 days<br />
16.1 Properties of Solutions 16.3 Colligative Properties of Solutions<br />
16.2 Concentrations of Solutions 16.4 Calc. Involving Colligative Property<br />
Chapter 17 Thermochemistry 5 days<br />
17.1 The Flow of Energy 17.3 Heat in Changes of State<br />
17.2 Measuring and Expressing Enthalpy Change 17.4 Calculating Heats in Reactions<br />
Chapter 18 Reaction Rates and Equilibrium 5 days<br />
18.1 Rates of Reactions 18.3 Reversible Reaction & Equilibrium<br />
18.2 The Progress of Chemical Reactions 18.5 Free Energy and Entropy<br />
Unit 8 (14 days)<br />
Chapter 19 Acid and Bases 10 days<br />
19.1 Acid-Base Theories 19.4 Neutralization Reactions<br />
19.2 Hydrogen Ions and Acidity 19.5 Salts in Solutions<br />
19.3 Strengths of Acids and Bases<br />
Chapter 20 Oxidation-Reduction Reactions 4 days<br />
20.1 The Meaning of Oxidation and Reduction 20.3 Describing Redox Equations<br />
20.2 Oxidation Numbers<br />
3
Lorenzo Walker Technical High School<br />
MUSTANG LABORATORIES<br />
Chemistry Safety<br />
Safety in the MUSTANG LABORATORIES - Chemistry Laboratory<br />
Working in the chemistry laboratory is an interesting and rewarding experience. During your labs, you will be actively<br />
involved from beginning to end—from setting some change in motion to drawing some conclusion. In the laboratory, you<br />
will be working with equipment and materials that can cause injury if they are not handled properly.<br />
However, the laboratory is a safe place to work if you are careful. Accidents do not just happen; they are caused—by<br />
carelessness, haste, and disregard of safety rules and practices. Safety rules to be followed in the laboratory are listed<br />
below. Before beginning any lab work, read these rules, learn them, and follow them carefully.<br />
General<br />
1. Be prepared to work when you arrive at the lab. Familiarize yourself with the lab procedures before beginning the lab.<br />
2. Perform only those lab activities assigned by your teacher. Never do anything in the laboratory that is not called for in<br />
the laboratory procedure or by your teacher. Never work alone in the lab. Do not engage in any horseplay.<br />
3. Work areas should be kept clean and tidy at all times. Only lab manuals and notebooks should be brought to the work<br />
area. Other books, purses, brief cases, etc. should be left at your desk or placed in a designated storage area.<br />
4. Clothing should be appropriate for working in the lab. Jackets, ties, and other loose garments should be removed. Open<br />
shoes should not be worn.<br />
5. Long hair should be tied back or covered, especially in the vicinity of open flame.<br />
6. Jewelry that might present a safety hazard, such as dangling necklaces, chains, medallions, or bracelets should not be<br />
worn in the lab.<br />
7. Follow all instructions, both written and oral, carefully.<br />
8. Safety goggles and lab aprons should be worn at all times.<br />
9. Set up apparatus as described in the lab manual or by your teacher. Never use makeshift arrangements.<br />
10. Always use the prescribed instrument (tongs, test tube holder, forceps, etc.) for handling apparatus or equipment.<br />
11. Keep all combustible materials away from open flames.<br />
12. Never touch any substance in the lab unless specifically instructed to do so by your teacher.<br />
13. Never put your face near the mouth of a container that is holding chemicals.<br />
14. Never smell any chemicals unless instructed to do so by your teacher. When testing for odors, use a wafting motion to<br />
direct the odors to your nose.<br />
15. Any activity involving poisonous vapors should be conducted in the fume hood.<br />
16. Dispose of waste materials as instructed by your teacher.<br />
17. Clean up all spills immediately.<br />
18. Clean and wipe dry all work surfaces at the end of class. Wash your hands thoroughly.<br />
19. Know the location of emergency equipment (first aid kit, fire extinguisher, fire shower, fire blanket, etc.) and how to use them.<br />
20. Report all accidents to the teacher immediately.<br />
Handling Chemicals<br />
21. Read and double check labels on reagent bottles before removing any reagent. Take only as much reagent as you<br />
need.<br />
22. Do not return unused reagent to stock bottles.<br />
23. When transferring chemical reagents from one container to another, hold the containers out away from your body.<br />
24. When mixing an acid and water, always add the acid to the water.<br />
25. Avoid touching chemicals with your hands. If chemicals do come in contact with your hands, wash them immediately.<br />
26. Notify your teacher if you have any medical problems that might relate to lab work, such as allergies or asthma.<br />
27. If you will be working with chemicals in the lab, avoid wearing contact lenses. Change to glasses, if possible, or notify<br />
the teacher.<br />
Handling Glassware<br />
28. Glass tubing, especially long pieces, should be carried in a vertical position to minimize the likelihood of breakage and<br />
to avoid stabbing anyone.<br />
29. Never handle broken glass with your bare hands. Use a brush and dustpan to clean up broken glass. Dispose of the<br />
glass as directed by your teacher.<br />
4
30. Always lubricate glassware (tubing, thistle tubes, thermometers, etc.) with water or glycerin before attempting to insert<br />
it into a rubber stopper.<br />
31. Never apply force when inserting or removing glassware from a stopper. Use a twisting motion. If a piece of glassware<br />
becomes "frozen" in a stopper, take it to your teacher.<br />
32. Do not place hot glassware directly on the lab table. Always use an insulating pad of some sort.<br />
33. Allow plenty of time for hot glass to cool before touching it. Hot glass can cause painful burns. (Hot glass looks cool.)<br />
Heating Substances<br />
34. Exercise extreme caution when using a gas burner. Keep your head and clothing away from the flame.<br />
35. Always turn the burner off when it is not in use.<br />
36. Do not bring any substance into contact with a flame unless instructed to do so.<br />
37. Never heat anything without being instructed to do so.<br />
38. Never look into a container that is being heated.<br />
39. When heating a substance in a test tube, make sure that the mouth of the tube is not pointed at yourself or anyone<br />
else.<br />
40. Never leave unattended anything that is being heated or is visibly reacting.<br />
First Aid in the MUSTANG LABORATORIES - Chemistry Laboratory<br />
Accidents do not often happen in well-equipped chemistry laboratories if students understand safe laboratory procedures<br />
and are careful in following them. When an occasional accident does occur, it is likely to be a minor one.<br />
The instructor will assist in treating injuries such as minor cuts and burns. However, for some types of injuries, you must<br />
take action immediately. The following information will be helpful to you if an accident occurs.<br />
1. Shock. People who are suffering from any severe injury (for example, a bad burn or major loss of blood) may be in a<br />
state of shock. A person in shock is usually pale and faint. The person may be sweating, with cold, moist skin and a weak,<br />
rapid pulse. Shock is a serious medical condition. Do not allow a person in shock to walk anywhere—even to the campus<br />
security office. While emergency help is being summoned, place the victim face up in a horizontal position, with the feet<br />
raised about 30 centimeters. Loosen any tightly fitting clothing and keep him or her warm.<br />
2. Chemicals in the Eyes. Getting any kind of a chemical into the eyes is undesirable, but certain chemicals are<br />
especially harmful. They can destroy eyesight in a matter of seconds. Because you will be wearing safety goggles at all<br />
times in the lab, the likelihood of this kind of accident is remote. However, if it does happen, flush your eyes with water<br />
immediately. Do NOT attempt to go to the campus office before flushing your eyes. It is important that flushing with water<br />
be continued for a prolonged time—about 15 minutes.<br />
3. Clothing or Hair on Fire. A person whose clothing or hair catches on fire will often run around hysterically in an<br />
unsuccessful effort to get away from the fire. This only provides the fire with more oxygen and makes it burn faster. For<br />
clothing fires, throw yourself to the ground and roll around to extinguish the flames. For hair fires, use a fire blanket to<br />
smother the flames. Notify campus security immediately.<br />
4. Bleeding from a Cut. Most cuts that occur in the chemistry laboratory are minor. For minor cuts, apply pressure to the<br />
wound with a sterile gauze. Notify campus security of all injuries in the lab. If the victim is bleeding badly, raise the<br />
bleeding part, if possible, and apply pressure to the wound with a piece of sterile gauze. While first aid is being given,<br />
someone else should notify the campus security officer.<br />
5. Chemicals in the Mouth. Many chemicals are poisonous to varying degrees. Any chemical taken into the mouth<br />
should be spat out and the mouth rinsed thoroughly with water. Note the name of the chemical and notify the campus<br />
office immediately. If the victim swallows a chemical, note the name of the chemical and notify campus security<br />
immediately.<br />
If necessary, the campus security officer or administrator will contact the Poison Control Center, a hospital emergency<br />
room, or a physician for instructions.<br />
6. Acid or Base Spilled on the Skin.<br />
Flush the skin with water for about 15 minutes. Take the victim to the campus office to report the injury.<br />
7. Breathing Smoke or Chemical Fumes.<br />
All experiments that give off smoke or noxious gases should be conducted in a well-ventilated fume hood. This will make<br />
an accident of this kind unlikely. If smoke or chemical fumes are present in the laboratory, all persons—even those who<br />
do not feel ill—should leave the laboratory immediately. Make certain that all doors to the laboratory are closed after the<br />
last person has left. Since smoke rises, stay low while evacuating a smoke-filled room. Notify campus security<br />
immediately.<br />
5
MUSTANG LABORATORIES<br />
COMMITMENT TO SAFETY IN THE LABORATORY<br />
As a student enrolled in Chemistry at Lorenzo Walker Technical High<br />
School, I agree to use good laboratory safety practices at all times. I<br />
also agree that I will:<br />
1. Conduct myself in a professional manner, respecting both my personal safety and the safety of<br />
others in the laboratory.<br />
2. Wear proper and approved safety glasses or goggles in the laboratory at all times.<br />
3. Wear sensible clothing and tie back long hair in the laboratory. Understand that open-toed shoes<br />
pose a hazard during laboratory classes and that contact lenses are an added safety risk.<br />
4. Keep my lab area free of clutter during an experiment.<br />
5. Never bring food or drink into the laboratory, nor apply makeup within the laboratory.<br />
6. Be aware of the location of safety equipment such as the fire extinguisher, eye wash station, fire<br />
blanket, first aid kit. Know the location of the nearest telephone and exits.<br />
7. Read the assigned lab prior to coming to the laboratory.<br />
8. Carefully read all labels on all chemical containers before using their contents, remove a small<br />
amount of reagent properly if needed, do not pour back the unused chemicals into the original<br />
container.<br />
9. Dispose of chemicals as directed by the instructor only. At no time will I pour anything down the<br />
sink without prior instruction.<br />
10. Never inhale fumes emitted during an experiment. Use the fume hood when instructed to do so.<br />
11. Report any accident immediately to the instructor, including chemical spills.<br />
12. Dispose of broken glass and sharps only in the designated containers.<br />
13. Clean my work area and all glassware before leaving the laboratory.<br />
14. Wash my hands before leaving the laboratory.<br />
NAME __________________________<br />
PERIOD ________________________<br />
PARENT NAME ____________________________<br />
PARENT NUMBER _________________________<br />
SIGNATURE ____________________________<br />
DATE ____________________________________<br />
6
7
Chapter 1<br />
Unit 1<br />
Introduction to Chemistry<br />
The students will learn why and how to solve problems using<br />
chemistry.<br />
Identify what is science, what clearly is not science, and what superficially<br />
resembles science (but fails to meet the criteria for science).<br />
Students will identify a phenomenon as science or not science.<br />
Inference<br />
Hypothesis<br />
Science<br />
Observation<br />
Identify which questions can be answered through science and which<br />
questions are outside the boundaries of scientific investigation, such as<br />
questions addressed by other ways of knowing, such as art, philosophy, and<br />
religion.<br />
Students will differentiate between problems and/or phenomenon that can and<br />
those that cannot be explained or answered by science.<br />
Students will differentiate between problems and/or phenomenon that can and<br />
those that cannot be explained or answered by science.<br />
Observation<br />
Inference<br />
Hypothesis<br />
Theory<br />
Controlled experiment<br />
Describe how scientific inferences are drawn from scientific observations<br />
and provide examples from the content being studied.<br />
Students will conduct and record observations.<br />
Students will make inferences.<br />
Students will identify a statement as being either an observation or inference.<br />
Students will pose scientific questions and make predictions based on<br />
inferences.<br />
Inference<br />
Observation<br />
Hypothesis<br />
Controlled experiment<br />
Identify sources of information and assess their reliability according to the<br />
strict standards of scientific investigation.<br />
Students will compare and assess the validity of known scientific information<br />
from a variety of sources:<br />
8
Print vs. print<br />
Online vs. online<br />
Print vs. online<br />
Students will conduct an experiment using the scientific method and compare<br />
with other groups.<br />
Controlled experiment<br />
Investigation<br />
Peer Review<br />
Accuracy<br />
Precision<br />
Percentage Error<br />
Chapter 2<br />
Matter and Change<br />
The students will learn what properties are used to describe<br />
matter and how matter can change its form.<br />
Differentiate between physical and chemical properties and physical and<br />
chemical changes of matter.<br />
Students will be able to identify physical and chemical properties of various<br />
substances.<br />
Students will be able to identify indicators of physical and chemical changes.<br />
Students will be able to calculate density.<br />
mass<br />
physical property<br />
volume<br />
chemical property<br />
vapor<br />
extensive property<br />
Chapter 3<br />
mixture<br />
intensive property<br />
solution<br />
element<br />
compound<br />
Scientific Measurements<br />
The students will be able to solve conversion problems using<br />
measurements.<br />
Determine appropriate and consistent standards of measurement for the<br />
data to be collected in a survey or experiment.<br />
Students will participate in activities to collect data using standardized<br />
measurement.<br />
Students will be able to manipulate/convert data collected and apply the data<br />
to scientific situations.<br />
Scientific notation<br />
International System of Units (SI)<br />
Significant figures<br />
Accepted value<br />
Experimental value<br />
Percent error<br />
Dimensional analysis<br />
9
10<br />
Learning Goal: Bechmark
To use the Stair-Step method, find the prefix the original measurement starts with. (ex. milligram)<br />
If there is no prefix, then you are starting with a base unit.<br />
Find the step which you wish to make the conversion to. (ex. decigram)<br />
Count the number of steps you moved, and determine in which direction you moved (left or right).<br />
The decimal in your original measurement moves the same number of places as steps you moved and in the<br />
same direction. (ex. milligram to decigram is 2 steps to the left, so 40 milligrams = .40 decigrams)<br />
If the number of steps you move is larger than the number you have, you will have to add zeros to hold the<br />
places. (ex. kilometers to meters is three steps to the right, so 10 kilometers would be equal to 10,000 m)<br />
That’s all there is to it! You need to be able to count to 6, and know your left from your right!<br />
1) Write the equivalent<br />
a) 5 dm =_______m b) 4 mL = ______L c) 8 g = _______mg<br />
d) 9 mg =_______g e) 2 mL = ______L f) 6 kg = _____g<br />
g) 4 cm =_______m h) 12 mg = ______ g i) 6.5 cm 3 = _______L<br />
j) 7.02 mL =_____cm 3 k) .03 hg = _______ dg l) 6035 mm _____cm<br />
m) .32 m = _______cm n) 38.2 g = _____kg<br />
11
2. One cereal bar has a mass of 37 g. What is the mass of 6 cereal bars? Is that more than or less<br />
than 1 kg? Explain your answer.<br />
3. Wanda needs to move 110 kg of rocks. She can carry l0 hg each trip. How many trips must she<br />
make? Explain your answer.<br />
4. Dr. O is playing in her garden again She needs 1 kg of potting soil for her plants. She has 750 g.<br />
How much more does she need? Explain your answer.<br />
5. Weather satellites orbit Earth at an altitude of 1,400,000 meters. What is this altitude in kilometers?<br />
6. Which unit would you use to measure the capacity? Write milliliter or liter.<br />
a) a bucket __________<br />
b) a thimble __________<br />
c) a water storage tank__________<br />
d) a carton of juice__________<br />
7. Circle the more reasonable measure:<br />
a) length of an ant 5mm or 5cm<br />
b) length of an automobile 5 m or 50 m<br />
c) distance from NY to LA 450 km or 4,500 km<br />
d) height of a dining table 75 mm or 75 cm<br />
8. Will a tablecloth that is 155 cm long cover a table that is 1.6 m long? Explain your answer.<br />
9. A dollar bill is 15.6 cm long. If 200 dollar bills were laid end to end, how many meters long would<br />
the line be?<br />
10. The ceiling in Jan’s living room is 2.5 m high. She has a hanging lamp that hangs down 41 cm.<br />
Her husband is exactly 2 m tall. Will he hit his head on the hanging lamp? Why or why not?<br />
12
Using SI Units<br />
Match the terms in Column II with the descriptions in Column I. Write the letters of the correct term in<br />
the blank on the left.<br />
Column I Column II<br />
_____ 1. distance between two points<br />
a. time<br />
_____ 2. SI unit of length<br />
_____ 3. tool used to measure length<br />
_____ 4. units obtained by combining other units<br />
_____ 5. amount of space occupied by an object<br />
_____ 6. unit used to express volume<br />
_____ 7. SI unit of mass<br />
_____ 8. amount of matter in an object<br />
_____ 9. mass per unit of volume<br />
_____ 10. temperature scale of most laboratory thermometers<br />
_____ 11. instrument used to measure mass<br />
_____ 12. interval between two events<br />
_____ 13. SI unit of temperature<br />
_____ 14. SI unit of time<br />
_____ 15. instrument used to measure temperature<br />
b. volume<br />
c. mass<br />
d. density<br />
e. meter<br />
f. kilogram<br />
g. derived<br />
h. liter<br />
i. second<br />
j. Kelvin<br />
k. length<br />
1. balance<br />
m. meterstick<br />
n. thermometer<br />
o. Celsius<br />
Circle the two terms in each group that are related. Explain how the terms are related.<br />
16. Celsius degree, mass, Kelvin _____________________________________________________<br />
________________________________________________________________________________<br />
17. balance, second, mass __________________________________________________________<br />
________________________________________________________________________________<br />
18. kilogram, liter, cubic centimeter __________________________________________________<br />
________________________________________________________________________________<br />
19. time, second, distance __________________________________________________________<br />
________________________________________________________________________________<br />
20. decimeter, kilometer, Kelvin _____________________________________________________<br />
________________________________________________________________________________<br />
13
1. How many meters are in one kilometer? __________<br />
2. What part of a liter is one milliliter? __________<br />
3. How many grams are in two dekagrams? __________<br />
4. If one gram of water has a volume of one milliliter, what would the mass of one liter of water be in<br />
kilograms?__________<br />
5. What part of a meter is a decimeter? __________<br />
In the blank at the left, write the term that correctly completes each statement. Choose from the terms<br />
listed below.<br />
Metric SI standard ten<br />
prefixes ten tenth<br />
6. An exact quantity that people agree to use for comparison is a ______________ .<br />
7. The system of measurement used worldwide in science is _______________ .<br />
8. SI is based on units of _______________ .<br />
9. The system of measurement that was based on units of ten was the _______________ system.<br />
10. In SI, _______________ are used with the names of the base unit to indicate the multiple of ten<br />
that is being used with the base unit.<br />
11. The prefix deci- means _______________ .<br />
14
Standards of Measurement<br />
Fill in the missing information in the table below.<br />
S.I prefixes and their meanings<br />
Prefix<br />
Meaning<br />
0.001<br />
0.01<br />
deci- 0.1<br />
10<br />
hecto- 100<br />
1000<br />
Circle the larger unit in each pair of units.<br />
1. millimeter, kilometer 4. centimeter, millimeter<br />
2. decimeter, dekameter 5. hectogram, kilogram<br />
3. hectogram, decigram<br />
6. In SI, the base unit of length is the meter. Use this information to arrange the following units of<br />
measurement in the correct order from smallest to largest.<br />
Write the number 1 (smallest) through 7 - (largest) in the spaces provided.<br />
_____ a. kilometer<br />
_____ b. centimeter<br />
_____ c. meter<br />
_____ e. hectometer<br />
_____ f. millimeter<br />
_____ g. decimeter<br />
_____ d. dekameter<br />
Use your knowledge of the prefixes used in SI to answer the following questions in the spaces<br />
provided.<br />
7. One part of the Olympic games involves an activity called the decathlon. How many events do you<br />
think make up the decathlon?_____________________________________________________<br />
8. How many years make up a decade? _______________________________________________<br />
9. How many years make up a century? ______________________________________________<br />
10. What part of a second do you think a millisecond is? __________________________________<br />
15
The Learning Goal for this assignment is:<br />
Notes Section<br />
1. 7,485 6. 1.683<br />
2. 884.2 7. 3.622<br />
3. 0.00002887 8. 0.00001735<br />
4. 0.05893 9. 0.9736<br />
5. 0.006162 10. 0.08558<br />
11. 6.633 X 10−⁴ 16. 1.937 X 10⁴<br />
12. 4.445 X 10−⁴ 17. 3.457 X 10⁴<br />
13. 2.182 X 10−³ 18. 3.948 X 10−⁵<br />
14. 4.695 X 10² 19. 8.945 X 10⁵<br />
15. 7.274 X 10⁵ 20. 6.783 X 10²<br />
16
SCIENTIFIC NOTATION RULES<br />
How to Write Numbers in Scientific Notation<br />
Scientific notation is a standard way of writing very large and very small numbers so that they're<br />
easier to both compare and use in computations. To write in scientific notation, follow the form<br />
N X 10 ᴬ<br />
where N is a number between 1 and 10, but not 10 itself, and A is an integer (positive or negative<br />
number).<br />
RULE #1: Standard Scientific Notation is a number from 1 to 9 followed by a decimal and the<br />
remaining significant figures and an exponent of 10 to hold place value.<br />
Example:<br />
5.43 x 10 2 = 5.43 x 100 = 543<br />
8.65 x 10 – 3 = 8.65 x .001 = 0.00865<br />
****54.3 x 10 1 is not Standard Scientific Notation!!!<br />
RULE #2: When the decimal is moved to the Left the exponent gets Larger, but the value of the<br />
number stays the same. Each place the decimal moves Changes the exponent by one (1). If you<br />
move the decimal to the Right it makes the exponent smaller by one (1) for each place it is moved.<br />
Example:<br />
6000. x 10 0 = 600.0 x 10 1 = 60.00 x 10 2 = 6.000 x 10 3 = 6000<br />
(Note: 10 0 = 1)<br />
All the previous numbers are equal, but only 6.000 x 10 3 is in proper Scientific Notation.<br />
17
RULE #3: To add/subtract in scientific notation, the exponents must first be the same.<br />
Example:<br />
(3.0 x 10 2 ) + (6.4 x 10 3 ); since 6.4 x 10 3 is equal to 64. x 10 2 . Now add.<br />
(3.0 x 10 2 )<br />
+ (64. x 10 2 )<br />
67.0 x 10 2 = 6.70 x 10 3 = 6.7 x 10 3<br />
67.0 x 10 2 is mathematically correct, but a number in standard scientific notation can only<br />
have one number to the left of the decimal, so the decimal is moved to the left one place and<br />
one is added to the exponent.<br />
Following the rules for significant figures, the answer becomes 6.7 x 10 3 .<br />
RULE #4: To multiply, find the product of the numbers, then add the exponents.<br />
Example:<br />
(2.4 x 10 2 ) (5.5 x 10 –4 ) = ? [2.4 x 5.5 = 13.2]; [2 + -4 = -2], so<br />
(2.4 x 10 2 ) (5.5 x 10 –4 ) = 13.2 x 10 –2 = 1.3 x 10 – 1<br />
RULE #5: To divide, find the quotient of the number and subtract the exponents.<br />
Example:<br />
(3.3 x 10 – 6 ) / (9.1 x 10 – 8 ) = ? [3.3 / 9.1 = .36]; [-6 – (-8) = 2], so<br />
(3.3 x 10 – 6 ) / (9.1 x 10 – 8 ) = .36 x 10 2 = 3.6 x 10 1<br />
18
Convert each number from Scientific Notation to real numbers:<br />
1. 7.485 X 10³ 6. 1.683 X 10⁰<br />
2. 8.842 X 10² 7. 3.622 10⁰<br />
3. 2.887 X 10−⁵ 8. 1.735 X 10−⁵<br />
4. 5.893 X 10−² 9. 9.736 X 10−¹<br />
5. 6.162 X 10−³ 10. 8.558 X 10−²<br />
Convert each number from a real number to Scientific Notation:<br />
11. 0.0006633 16. 1,937,000<br />
12. 0.0004445 17. 34,570<br />
13. 0.002182 18. 0.00003948<br />
14. 469.5 19. 894,500<br />
15. 727,400 20. 678.3<br />
19
The Learning Goal for this assignment is:<br />
Notes Section:<br />
Question Sig Figs Question Add & Subtract Question Multiple & Divide<br />
1 4 1 55.36 1 20,000<br />
2 4 2 84.2 2 94<br />
3 3 3 115.4 3 300<br />
4 3 4 0.8 4 7<br />
5 4 5 245.53 5 62<br />
6 3 6 34.5 6 0.005<br />
7 3 7 74.0 7 4,000<br />
8 2 8 53.287 8 3,900,000<br />
9 2 9 54.876 9 2<br />
10 2 10 40.19 10 30,000,000<br />
11 3 11 7.7 11 1,200<br />
12 2 12 67.170 12 0.2<br />
13 3 13 81.0 13 0.87<br />
14 4 14 73.290 14 0.049<br />
15 4 15 29.789 15 2,000<br />
16 3 16 39.53 16 0.5<br />
17 4 17 70.58 17 1.9<br />
18 2 18 86.6 18 0.05<br />
19 2 19 64.990 19 230<br />
20 1 20 36.0 20 460,000<br />
20
Significant Figures Rules<br />
There are three rules on determining how many significant figures are in a<br />
number:<br />
1. Non-zero digits are always significant.<br />
2. Any zeros between two significant digits are significant.<br />
3. A final zero or trailing zeros in the DECIMAL PORTION ONLY are<br />
significant.<br />
Please remember that, in science, all numbers are based upon measurements (except for a very few<br />
that are defined). Since all measurements are uncertain, we must only use those numbers that are<br />
meaningful.<br />
Not all of the digits have meaning (significance) and, therefore, should not be written down. In<br />
science, only the numbers that have significance (derived from measurement) are written.<br />
Rule 1: Non-zero digits are always significant.<br />
If you measure something and the device you use (ruler, thermometer, triple-beam, balance, etc.)<br />
returns a number to you, then you have made a measurement decision and that ACT of measuring<br />
gives significance to that particular numeral (or digit) in the overall value you obtain.<br />
Hence a number like 46.78 would have four significant figures and 3.94 would have three.<br />
Rule 2: Any zeros between two significant digits are significant.<br />
Suppose you had a number like 409. By the first rule, the 4 and the 9 are significant. However, to<br />
make a measurement decision on the 4 (in the hundred's place) and the 9 (in the one's place), you<br />
HAD to have made a decision on the ten's place. The measurement scale for this number would have<br />
hundreds, tens, and ones marked.<br />
Like the following example:<br />
These are sometimes called "captured zeros."<br />
If a number has a decimal at the end (after the one’s place) then all digits (numbers) are significant<br />
and will be counted.<br />
In the following example the zeros are significant digits and highlighted in blue.<br />
960.<br />
70050.<br />
21
Rule 3: A final zero or trailing zeros in the decimal portion ONLY are<br />
significant.<br />
This rule causes the most confusion among students.<br />
In the following example the zeros are significant digits and highlighted in blue.<br />
0.07030<br />
0.00800<br />
Here are two more examples where the significant zeros are highlighted in blue.<br />
When Zeros are Not Significant Digits<br />
4.7 0 x 10−³<br />
6.5 0 0 x 10⁴<br />
Zero Type # 1 : Space holding zeros in numbers less than one.<br />
In the following example the zeros are NOT significant digits and highlighted in red.<br />
0.09060<br />
0.00400<br />
These zeros serve only as space holders. They are there to put the decimal point in its correct<br />
location.<br />
They DO NOT involve measurement decisions.<br />
Zero Type # 2 : Trailing zeros in a whole number.<br />
In the following example the zeros are NOT significant digits and highlighted in red.<br />
200<br />
25000<br />
For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal point)<br />
of the numbers ONLY. Here is what to do:<br />
1) Count the number of significant figures in the decimal portion of each number in the problem. (The<br />
digits to the left of the decimal place are not used to determine the number of decimal places in the<br />
final answer.)<br />
2) Add or subtract in the normal fashion.<br />
3) Round the answer to the LEAST number of places in the decimal portion of any number in the<br />
problem<br />
The following rule applies for multiplication and division:<br />
The LEAST number of significant figures in any number of the problem determines the number of<br />
significant figures in the answer.<br />
This means you MUST know how to recognize significant figures in order to use this rule.<br />
22
How Many Significant Digits for Each Number?<br />
Learning Goal: Benchmark<br />
1) 2359 = ______<br />
2) 2.445 x 10−⁵= ______<br />
3) 2.93 x 10⁴= ______<br />
4) 1.30 x 10−⁷= ______<br />
5) 2604 = ______<br />
6) 9160 = ______<br />
7) 0.0800 = ______<br />
8) 0.84 = ______<br />
9) 0.0080 = ______<br />
10) 0.00040 = ______<br />
11) 0.0520 = ______<br />
12) 0.060 = ______<br />
13) 6.90 x 10−¹= ______<br />
14) 7.200 x 10⁵= ______<br />
15) 5.566 x 10−²= ______<br />
16) 3.88 x 10⁸= ______<br />
.0000000388<br />
17) 3004 = ______<br />
18) 0.021 = ______<br />
19) 240 = ______<br />
20) 500 = ______<br />
23
For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal point) of the<br />
numbers ONLY. Here is what to do:<br />
1) Count the number of significant figures in the decimal portion of each number in the problem. (The<br />
digits to the left of the decimal place are not used to determine the number of decimal places in the<br />
final answer.)<br />
2) Add or subtract in the normal fashion.<br />
3) Round the answer to the LEAST number of places in the decimal portion of any number in the<br />
problem.<br />
Solve the Problems and Round Accordingly...<br />
Learning Goal: Benchmark<br />
1) 43.287 + 5.79 + 6.284 = _______<br />
2) 87.54 - 3.3 = _______<br />
3) 99.1498 + 6.5397 + 9.7 = _______<br />
4) 5.868 - 5.1 = _______<br />
5) 59.9233 + 86.21 + 99.396 = _______<br />
6) 7.7 + 26.756 = _______<br />
7) 66.8 + 2.3 + 4.8516 = _______<br />
8) 9.7419 + 43.545 = _______<br />
9) 4.8976 + 48.4644 + 1.514 = _______<br />
10) 4.335 + 35.85 = _______<br />
11) 9.448 - 1.7 = _______<br />
12) 75.826 - 8.6555 = _______<br />
13) 57.2 + 23.814 = _______<br />
14) 77.684 - 4.394 = _______ 73.29<br />
15) 26.4496 + 3.339 = _______ 29.79<br />
16) 9.6848 + 29.85 = _______ 39.53<br />
17) 63.11 + 2.5412 + 4.93 = _______ 70.6<br />
18) 11.2471 + 75.4 = _______ 86.6<br />
19) 73.745 - 8.755 = _______ 64.99<br />
20) 6.5238 + 1.7 + 27.79 = _______ 36<br />
24
The following rule applies for multiplication and division:<br />
The LEAST number of significant figures in any number of the problem determines the number of<br />
significant figures in the answer.<br />
This means you MUST know how to recognize significant figures in order to use this rule.<br />
Solve the Problems and Round Accordingly...<br />
1) 0.6 x 65.0 x 602 = __________<br />
2) 720 ÷ 7.7 = __________<br />
3) 929 x 0.3 = __________<br />
4) 300 ÷ 44.31 = __________<br />
5) 608 ÷ 9.8 = __________<br />
6) 0.06 x 0.079 = __________<br />
7) 0.008 x 72.91 x 7000 = __________<br />
8) 73.94 x 67 x 780 = __________<br />
9) 0.62 x 0.097 x 40 = __________<br />
10) 600 x 10 x 5030 = __________<br />
11) 5200 ÷ 4.46 = __________<br />
12) 0.0052 x 0.4 x 107 = __________<br />
13) 0.099 x 8.8 = __________<br />
14) 0.0095 x 5.2 = __________<br />
15) 8000 ÷ 4.62 = __________<br />
16) 0.6 x 0.8 = __________<br />
17) 2.84 x 0.66 = __________<br />
18) 0.5 x 0.09 = __________<br />
7 6.770480704<br />
62 62.04081633<br />
0.005 0.00474<br />
19) 8100 ÷ 34.84 = __________<br />
20) 8.24 x 6.9 x 8100 = __________<br />
4000 4082.96<br />
3900000 3864104.4<br />
2 2.4056<br />
30000000 30180000<br />
1200 1165.919283<br />
0.2 0.22256<br />
0.87 0.8712<br />
0.049 0.0494<br />
2000 1731.601732<br />
0.5 0.48<br />
1.9 1.8744<br />
0.05 0.045<br />
230 232.4913892<br />
460000 460533.6<br />
25
Dimensional Analysis<br />
This is a way to convert from one unit of a given substance to<br />
another unit using ratios or conversion units. What this video<br />
www.youtube.com/watch?v=aZ3J60GYo6U<br />
Let’ look at a couple of examples:<br />
1. Convert 2.6 qt to mL.<br />
First we need a ratio or conversion unit so that we can go from quarts to milliliters. 1.00 qt = 946 mL<br />
Next write down what you are starting with<br />
2.6 qt<br />
Then make you conversion tree<br />
2.6 qt<br />
Then fill in the units in your ratio so that you can cancel out the original unit and will be left with the<br />
unit you need for the answer. Cross out units, one at a time that are paired, and one on top one on<br />
the bottom.<br />
2.6 qt mL<br />
qt<br />
Now fill in the values from the ratio.<br />
2.6 qt 946 mL<br />
1.00 qt<br />
Now multiply all numbers on the top and multiply all numbers on the bottom and write them as a<br />
fraction.<br />
2.6 qt 946 mL = 2,459.6 mL<br />
1.00 qt 1.00<br />
Now divide the top number by the bottom number and write that number with the unit that was not<br />
crossed out.<br />
26
Learning Goal: Benchmark<br />
1qt=32 oz 1gal = 4qts 1.00 qt = 946 mL 1L = 1000mL<br />
2. Convert 8135.6 mL to quarts<br />
=<br />
3. Convert 115.2 oz to mL<br />
=<br />
4. Convert 2.3 g to Liters<br />
=<br />
5. Convert 8.42 L to oz<br />
=<br />
Go to http://science.widener.edu/svb/tutorial/ chose #7 “Converting Volume” and do 5 more in the<br />
space provided.<br />
1. Convert _________ to _________<br />
=<br />
2. Convert _________ to _________<br />
=<br />
3. Convert _________ to _________<br />
=<br />
4. Convert _________ to _________<br />
=<br />
5. Convert _________ to _________<br />
=<br />
27
Chapter 4<br />
Unit 2<br />
Atomic Structure<br />
The students will learn what makes up atoms and how are<br />
atoms of one element different from atoms of another element.<br />
Explore the scientific theory of atoms (also known as atomic theory) by<br />
describing changes in the atomic model over time and why those changes<br />
were necessitated by experimental evidence.<br />
<br />
<br />
<br />
Students will be able to draw/identify each atomic model.<br />
Students will be able to compare/contrast the different atomic models.<br />
Students will be able to describe how results of experimental evidence caused<br />
the atomic model to change.<br />
proton<br />
electron<br />
neutron<br />
nucleus<br />
electron cloud<br />
Explore the scientific theory of atoms (also known as atomic theory) by<br />
describing the structure of atoms in terms of protons, neutrons and<br />
electrons, and differentiate among these particles in terms of their mass,<br />
electrical charges and locations within the atom.<br />
<br />
Students will compare/contrast the characteristics of subatomic particles.<br />
atomic number<br />
mass number<br />
isotope<br />
atomic mass unit (amu)<br />
atomic mass
Chapter 5<br />
Electrons in Atoms<br />
The students will be able to describe the arrangement of<br />
electrons in atoms and predict what will happen when<br />
electrons in atoms absorb or release energy.<br />
Describe the quantization of energy at the atomic level.<br />
Students will participate in activities to view emission spectrums using a<br />
diffraction grating or a spectroscope.<br />
Students will be able to explain how the spectrum lines relate to electron motion.<br />
energy level<br />
atomic orbital<br />
quantum mechanical model<br />
Chapter 6<br />
The Periodic Table<br />
The student will learn what information the periodic table<br />
provides and how periodic trends can be explained.<br />
Relate properties of atoms and their position in the periodic table to the<br />
arrangement of their electrons.<br />
Students will be able to compare and contrast metals, nonmetals, and metalloids.<br />
Students will be able to describe the traits of various families on the periodic<br />
table.<br />
Students will be able to explain periodicity.<br />
Students will write/represent electron configuration of various elements.<br />
Students will be able to use a periodic table to calculate the number of p + , e - , and<br />
n 0 .<br />
Students will be able to calculate the average weight of mass.<br />
periodic law<br />
halogen<br />
metals<br />
noble gas<br />
nonmetals<br />
transition metal<br />
metalloid<br />
atomic radius<br />
alkali metal<br />
ionization energy<br />
alkaline earth metal<br />
electronegativity
The Learning Goal for this assignment is:<br />
Ch 5 The students will learn what makes up atoms and how are atoms of one element<br />
different from atoms of another element.<br />
Ch 6<br />
The student will learn what information the periodic table provides and how periodic<br />
trends can be explained.<br />
Notes Section<br />
http://www.learner.org/interactives/periodic/basics_interactive.html
Atoms Are Building Blocks<br />
Atoms are the basis of chemistry. They are the basis for everything in the Universe. You<br />
should start by remembering that matter is composed of atoms. Atoms and the study of<br />
atoms are a world unto themselves. We're going to cover basics like atomic structure<br />
and bonding between atoms.<br />
Smaller Than Atoms?<br />
Are there pieces of matter that are smaller than atoms?<br />
Sure there are. You'll soon be learning that atoms are<br />
composed of pieces like electrons, protons, and neutrons.<br />
But guess what? There are even smaller particles moving<br />
around in atoms. These super-small particles can be found<br />
inside the protons and neutrons. Scientists have many<br />
names for those pieces, but you may have heard of<br />
nucleons and quarks. Nuclear chemists and physicists<br />
work together at particle accelerators to discover the<br />
presence of these tiny, tiny, tiny pieces of matter.<br />
Even though super-tiny atomic particles exist, you only<br />
need to remember the three basic parts of an atom: electrons, protons, and neutrons.<br />
What are electrons, protons, and neutrons? A picture works best to show off the idea.<br />
You have a basic atom. There are three types of pieces in that atom: electrons, protons,<br />
and neutrons. That's all you have to remember. Three things! As you know, there are<br />
almost 120 known elements in the periodic table. Chemists and physicists haven't<br />
stopped there. They are trying to make new ones in labs every day. The thing that<br />
makes each of those elements different is the number of electrons, protons, and<br />
neutrons. The protons and neutrons are always in the center of the atom. Scientists call<br />
the center region of the atom the nucleus. The nucleus in<br />
a cell is a thing. The nucleus in an atom is a place where<br />
you find protons and neutrons. The electrons are always<br />
found whizzing around the center in areas called shells or<br />
orbitals.<br />
You can also see that each piece has either a "+", "-", or a<br />
"0." That symbol refers to the charge of the particle. Have<br />
you ever heard about getting a shock from a socket, static<br />
electricity, or lightning? Those are all different types of<br />
electric charges. Those charges are also found in tiny particles of matter. The electron<br />
always has a "-", or negative, charge. The proton always has a "+", or positive, charge. If<br />
the charge of an entire atom is "0", or neutral, there are equal numbers of positive and<br />
negative pieces. Neutral means there are equal numbers of electrons and protons. The<br />
third particle is the neutron. It has a neutral charge, also known as a charge of zero. All<br />
atoms have equal numbers of protons and electrons so that they are neutral. If there are<br />
more positive protons or negative electrons in an atom, you have a special atom called<br />
an ion.
Looking at Ions<br />
We haven’t talked about ions before, so let’s get down to basics. The<br />
atomic number of an element, also called a proton number, tells you the<br />
number of protons or positive particles in an atom. A normal atom has a<br />
neutral charge with equal numbers of positive and negative particles.<br />
That means an atom with a neutral charge is one where the number of<br />
electrons is equal to the atomic number. Ions are atoms with extra<br />
electrons or missing electrons. When you are missing an electron or<br />
two, you have a positive charge. When you have an extra electron<br />
or two, you have a negative charge.<br />
What do you do if you are a sodium (Na) atom? You have eleven<br />
electrons — one too many to have an entire shell filled. You need to<br />
find another element that will take that electron away from you. When you lose that<br />
electron, you will you’ll have full shells. Whenever an atom has full shells, we say it is<br />
"happy." Let's look at chlorine (Cl). Chlorine has seventeen electrons and only needs<br />
one more to fill its third shell and be "happy." Chlorine will take your extra sodium<br />
electron and leave you with 10 electrons inside of two filled shells. You are now a happy<br />
atom too. You are also an ion and missing one electron. That missing electron gives you<br />
a positive charge. You are still the element sodium, but you are now a sodium ion (Na + ).<br />
You have one less electron than your atomic number.<br />
Ion Characteristics<br />
So now you've become a sodium ion. You have ten electrons.<br />
That's the same number of electrons as neon (Ne). But you<br />
aren't neon. Since you're missing an electron, you aren't really<br />
a complete sodium atom either. As an ion you are now<br />
something completely new. Your whole goal as an atom was<br />
to become a "happy atom" with completely filled electron<br />
shells. Now you have those filled shells. You have a lower<br />
energy. You lost an electron and you are "happy." So what<br />
makes you interesting to other atoms? Now that you have<br />
given up the electron, you are quite electrically attractive.<br />
Other electrically charged atoms (ions) of the opposite charge<br />
(negative) are now looking at you and seeing a good partner to<br />
bond with. That's where the chlorine comes in. It's not only chlorine. Almost any ion with<br />
a negative charge will be interested in bonding with you.
Electrovalence<br />
Don't get worried about the big word. Electrovalence is just another word for something<br />
that has given up or taken electrons and become an ion. If you look at the periodic table,<br />
you might notice that elements on the left side usually become positively charged ions<br />
(cations) and elements on the right side get a negative charge (anions). That trend<br />
means that the left side has a positive valence and the right side has a negative<br />
valence. Valence is a measure of how much an atom wants to bond with other atoms. It<br />
is also a measure of how many electrons are excited about bonding with other atoms.<br />
There are two main types of bonding, covalent and electrovalent. You may have heard<br />
of the term "ionic bonds." Ionic bonds are electrovalent bonds. They are just groups of<br />
charged ions held together by electric forces. When in the presence of other ions, the<br />
electrovalent bonds are weaker because of outside electrical forces and attractions.<br />
Sodium and chlorine ions alone have a very strong bond, but as soon as you put those<br />
ions in a solution with H + (Hydrogen ion), OH - (Hydroxide), F - (Fluorine ion) or Mg ++<br />
(Magnesium ion), there are charged distractions that break the Na-Cl bond.<br />
Look at sodium chloride (NaCl) one more time. Salt is a very strong bond when it is<br />
sitting on your table. It would be nearly impossible to break those ionic/electrovalent<br />
bonds. However, if you put that salt into some water (H2O), the bonds break very<br />
quickly. It happens easily because of the electrical attraction of the water. Now you have<br />
sodium (Na + ) and chlorine (Cl - ) ions floating around the solution. You should remember<br />
that ionic bonds are normally strong, but they are very weak in water.
Neutron Madness<br />
We have already learned that ions are atoms that are<br />
either missing or have extra electrons. Let's say an atom<br />
is missing a neutron or has an extra neutron. That type of<br />
atom is called an isotope. An atom is still the same<br />
element if it is missing an electron. The same goes for<br />
isotopes. They are still the same element. They are just a<br />
little different from every other atom of the same element.<br />
For example, there are a lot of carbon (C) atoms in the<br />
Universe. The normal ones are carbon-12. Those atoms have 6 neutrons. There are a<br />
few straggler atoms that don't have 6. Those odd ones may have 7 or even 8 neutrons.<br />
As you learn more about chemistry, you will probably hear about carbon-14. Carbon-14<br />
actually has 8 neutrons (2 extra). C-14 is considered an isotope of the element carbon.<br />
Messing with the Mass<br />
If you have looked at a periodic table, you may have noticed that the atomic mass of<br />
an element is rarely an even number. That happens because of the isotopes. If you are<br />
an atom with an extra electron, it's no big deal. Electrons don't have much of a mass<br />
when compared to a neutron or proton.<br />
Atomic masses are calculated by figuring out the<br />
amounts of each type of atom and isotope there are in<br />
the Universe. For carbon, there are a lot of C-12, a<br />
couple of C-13, and a few C-14 atoms. When you<br />
average out all of the masses, you get a number that is a<br />
little bit higher than 12 (the weight of a C-12 atom). The<br />
average atomic mass for the element is actually 12.011.<br />
Since you never really know which carbon atom you are<br />
using in calculations, you should use the average mass<br />
of an atom.<br />
Bromine (Br), at atomic number 35, has a greater variety of isotopes. The atomic mass<br />
of bromine (Br) is 79.90. There are two main isotopes at 79 and 81, which average out<br />
to the 79.90amu value. The 79 has 44 neutrons and the 81 has 46 neutrons. While it<br />
won't change the average atomic mass, scientists have made bromine isotopes with<br />
masses from 68 to 97. It's all about the number of neutrons. As you move to higher<br />
atomic numbers in the periodic table, you will probably find even more isotopes for<br />
each element.
Summary
Electron Configuration<br />
Color the sublevel:<br />
s = Red<br />
d = Green<br />
p = Blue<br />
f = Orange<br />
s<br />
p<br />
d<br />
f<br />
1<br />
Write in sublevels<br />
1s 2<br />
1s 1<br />
2<br />
Write period, sublevel and super scripts.<br />
13 14 15 16 17<br />
7s 1 7s 2<br />
2s 1<br />
Ctrl Shift = gives you super scripts<br />
2s 2<br />
2p 1 2p 2 2p 3 2p 4 2p 5 2p 6<br />
3s 1 3s 2<br />
3 4 5 6 7 8 9 10 11 12 3p 1 3p 2 3p 3 3p 4 2p 5 2p 6<br />
4s 1 4s 2 3d 1 3d 2 3d 3 3d 4 3d 5 3d 6 3d 7 3d 8 3d 9 3d 10 4p 1 4p 2 4p 3 4p 4 4p 5 4p 6<br />
5s 1 5s 2 4d 1 4d 2 4d 3 4d 4 4d 5 4d 6 4d 7 4d 8 4d 9 4d 10 5p 1 5p 2 5p 3 5p 4 5p 5 5p 6<br />
6s 1 6s 2 5d 1 5d 2 5d 3 5d 4 5d 5 5d 6 5d 7 5d 8 5d 9 5d 10 6p 1 6p 2 6p 3 6p 4 6p 5 6p 6<br />
6d 1 6d 2 6d 3 6d 4 6d 5 6d 6 6d 7 6d 8 6d 9 6d 10<br />
18<br />
4f 1 4f 2 4f 3 4f 4 4f 5 4f 6 4f 7 4f 8 4f 9 4f 10 4f 11 4f 12 4f 13 4f 14<br />
5f 1 5f 2 5f 3 5f 4 5f 5 5f 6 5f 7 5f 8 5f 9 5f 10 5f 11 5f 12 5f 13 5f 14
The Learning Goal for this assignment is:<br />
Ch 6 The student will learn what information the periodic table provides and how periodic trends can be<br />
explained.<br />
Ch 5 The students will be able to describe the arrangement of electrons in atoms and predict what will happen<br />
when electrons in atoms absorb or release energy.<br />
www.youtube.com/watch?v=jtYzEzykFdg<br />
www.youtube.com/watch?<br />
annotation_id=annotation_2076&feature=iv&src_vid=jtYzEzykFdg&v=cOlac8ruD_0<br />
www.youtube.com/watch?<br />
annotation_id=annotation_570977&feature=iv&src_vid=cOlac8ruD_0&v=lR2vqHZWb5A<br />
Notes Section
Electron Configuration<br />
In order to write the electron configuration for an atom you must know the 3 rules of<br />
electron configurations.<br />
1. Aufbau<br />
Notation<br />
nO e<br />
where<br />
n is the energy level<br />
O is the orbital type (s, p, d, or f)<br />
e is the number of electrons in that orbital shell<br />
Principle<br />
electrons will first occupy orbitals of the lowest energy level<br />
2. Hund rule<br />
when electrons occupy orbitals of equal energy, one electron enters each orbital until<br />
all the orbitals contain one electron with the same spin.<br />
3. Pauli exclusion principle<br />
an orbital contains a maximum of 2 electrons and<br />
paired electrons will have opposite spin
In the space below, write the unabbreviated electron configurations of the following elements:<br />
1)<br />
1s<br />
sodium ________________________________________________<br />
,2s 2 ,2p 6 ,3s 1<br />
2) iron<br />
1s<br />
________________________________________________<br />
,2s 2 ,2p 6 ,3s 2 ,3p 6 ,4s 2 ,3d 6<br />
3) bromine ________________________________________________<br />
1s 2 ,2s 2 ,2p 6 ,3s 2 ,3p 6 ,4s 2 ,3d 10 ,4p 5<br />
4) barium ________________________________________________<br />
1s 2 ,2s 2 ,2p 6 ,3s 2 ,3p 6 ,4s 2 ,3d 10 ,4p 6 ,5s 2 ,4d 10 ,5p 6 ,6s 2<br />
5) neptunium 1s ________________________________________________<br />
,2s 2 ,2p 6 ,3s 2 ,3p 6 ,4s 2 ,3d 10 ,4p 6 ,5s 2 ,4d 10 ,5p 6 ,6s 2 ,4f 14 ,5d 10 ,6p 6 ,7s 2 ,5f 5<br />
In the space below, write the abbreviated electron configurations of the following elements:<br />
6) cobalt<br />
[Ar], 4s<br />
________________________________________________<br />
,3d 7<br />
7) silver<br />
[Kr],5s<br />
________________________________________________<br />
,4d 9<br />
8) tellurium<br />
[Kr],5s<br />
________________________________________________<br />
,4d 10 ,5p 4<br />
9) radium ________________________________________________<br />
[Rn],7s 2 [Rn],7s 2 ,5f 14 ,6d 1<br />
10) lawrencium ________________________________________________<br />
Determine what elements are denoted by the following electron configurations:<br />
11)<br />
Sulfur<br />
1s²s²2p⁶3s²3p⁴ ____________________<br />
12)<br />
Rubidium<br />
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹ ____________________<br />
13)<br />
Antimony<br />
[Kr] 5s²4d¹⁰5p³ ____________________<br />
14)<br />
Osmium<br />
[Xe] 6s²4f¹⁴5d⁶ ____________________<br />
15) [Rn] 7s²5f¹¹ ____________________<br />
Einsteinium<br />
Identify the element or determine that it is not a valid electron configuration:<br />
16)<br />
4d should be 3d<br />
1s²2s²2p⁶3s²3p⁶4s²4d¹⁰4p⁵ ____________________<br />
17)<br />
3p comes after 3s<br />
1s²2s²2p⁶3s³3d⁵ ____________________<br />
18)<br />
[Ra] is not a noble gas<br />
[Ra] 7s²5f⁸ ____________________<br />
19)<br />
This is valid<br />
[Kr] 5s²4d¹⁰5p⁵ ____________________<br />
20)<br />
An element can't be its own electron configuration<br />
[Xe] ____________________<br />
1)sodium 1s 2 2s 2 2p 6 3s 1 2)iron 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6<br />
3)bromine 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5 4)barium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2<br />
5)neptunium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 7s 2 5f 5 6)cobalt [Ar] 4s 2 3d 7<br />
7)silver [Kr] 5s 2 4d 9 8)tellurium[Kr] 5s 2 4d 10 5p 4<br />
9)radium [Rn] 7s 2 10)lawrencium[Rn] 7s 2 5f 14 6d 1<br />
1s 2 2s 2 2p 6 3s 2 3p 4 sulfur 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 rubidium<br />
[Kr] 5s 2 4d 10 5p 3 antimony [Xe] 6s 2 4f 14 5d 6 osmium<br />
[Rn] 7s 2 5f 11 einsteinium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 4d 10 4p 5 not valid (take a look at “4d”)<br />
1s 2 2s 2 2p 6 3s 3 3d 5 not valid (3p comes after 3s) [Ra] 7s 2 5f 8 not valid (radium isn’t a noble gas)<br />
[Kr] 5s 2 4d 10 5p 5 valid iodine<br />
20)[Xe] not valid (an element can’t be its own electron configuration)
Calcium -CA
Create groups for these Scientist and explain your groupings<br />
(use the information you got from your research)<br />
Radiation Discovery<br />
Glenn Seaborg<br />
Atoine Henri Becquerel<br />
Marie and Pierre Curie<br />
Electrons<br />
J.J Thomson<br />
Louis de Barogilie<br />
Eugene Goldstein<br />
Robert Millikan<br />
Alpha Particles<br />
James Chadwick<br />
Ernst Rutherford<br />
Periodic Table<br />
J.W Dobereiner<br />
Lothar Meyer<br />
Dmitri Mendeleev<br />
Models<br />
Erwin Shodinger<br />
Hantaro Nagaoka<br />
Niels Bohr
Research the Scientist and summarize their contributions to the Atomic Theory<br />
Antoine Henri Becquerel<br />
Antoine found strange radioactivity in his experiments with different substances which helped him invent the x-ray. He the<br />
found out about the deflection of beta particles.<br />
Niels Bohr<br />
Niels contributed to to the understanding of the structures to atoms.He then found out about the development of quantum<br />
mechanics.<br />
Louis de Barogilie<br />
Louis studied quantum theory. He then predicted the wave natures of electrons.<br />
Glenn Seaborg<br />
Glenn isolated radioactive elements.He then formulated the actinide concept.<br />
Hantaro Nagaoka<br />
Hanatro contributed to the atom model. He also suggested that the electrons were outside of the atom.<br />
Democritus<br />
Democritus created the atomic theory. Which then helped future scientists.<br />
Marie and Pierre Curie<br />
Marie and Pierre studied on radioactivity. In which they discovered 2 elements named radium and polonium<br />
Eugene Goldstein<br />
Eugene concluded that cathode rays negativity charged travel towards the positively charged anode. He also thought abou<br />
electrons doing the same thing.<br />
Dmitri Mendeleev<br />
Dmitri contributed to the periodic table by putting elements into rows and colums. He did it by atomic mass and chemical/ph<br />
propertys.<br />
J.J. Thomson<br />
J.J. discovered electrons when doing a experiment with cathode ray. He also discovered that cathode is negativity<br />
charged.<br />
James Chadwick<br />
James discovered neutrons. He did it by combining beryllium atoms with alpha particles.<br />
Erwin Shrodinger<br />
Erwin contributed to the wave theory. Which involved matter and quantum mechanics.<br />
John Dalton<br />
John contributed to the atomic theory. He did it by concluding when atoms combine in a ratio it must be presumed to be a<br />
binary.<br />
Lothar Meyer<br />
Lothar was one of the first to create the periodic table. The periodic table he had contained 28 elements.<br />
Robert Millikan<br />
Robert did a oil drop experiment which helped us find the quantity of an electron. After he did this it then helped us<br />
understand the structure of the atom.<br />
J.W. Dobereiner<br />
J.W. observed the similarities of elements. It then help contribute with the periodic table.<br />
Ernest Rutherford<br />
Ernest discovered alpha and beta rays. He then set created the law of radioactive decay.
The Learning Goal for this Assignment is<br />
Ch 5 The students will be able to describe the arrangement of electrons in atoms and predict what will<br />
happen when electrons in atoms absorb or release energy.<br />
Ch 6 The student will learn what information the periodic table provides and how periodic trends can be<br />
explained.<br />
Alkali Metals<br />
Alkali Earth Metals<br />
Transitional Metals<br />
Inter Transitional Metals<br />
Metals<br />
Metalloids<br />
Non Metals<br />
Noble Gases
Using Wikipedia, define the 8 categories of elements on the<br />
left page.<br />
Color your periodic table similar to the one on<br />
pages 168—169 of your book.<br />
alkali metals<br />
alkaline metals<br />
other metals<br />
transitional metals<br />
lanthanoids<br />
metalloids<br />
non metals<br />
halogens<br />
noble gases<br />
unknown elements<br />
actinoids
Define Atomic Size:<br />
Atomic Size<br />
The size of the atom is determined by the size of the nucleus and the<br />
number of energy levels<br />
I<br />
n<br />
c<br />
r<br />
e<br />
a<br />
s<br />
e<br />
Explanation:<br />
When looking at groups the size of the atom increases when going from top to bottom<br />
because we added more energy levels. When looking at a period the size increases from the<br />
right to the left we increase mass which increases gravity which pulls in electrons.
Ionization Energy<br />
Define Ionization Energy:<br />
The energy required to remove an electron from an atom.<br />
Explanation: The closer the electrons are to the nucleus of the atom, the higher the atom's<br />
ionization energy. Due to this you can predict what ions an element will form.<br />
The cation produced has a 1+ charge.
Electronegativity<br />
Define Electronegativity:<br />
The ability of an atom of an element to attract electrons when the<br />
atom is in a compound.<br />
Explanation:<br />
Metals at the far left of the periodic table have low values. Nonmetals at the far right have<br />
high values. The noble gases are omitted because they do not form many compounds.
Ion Size<br />
Define Ion Size:<br />
The size of the ion's atom in comparison to the size of the element atom.<br />
Cation<br />
Anions<br />
Explanation:<br />
Cations give up electrons so they go down the energy level so the larger mass means it is<br />
going to pull the energy levels. Since Calcium has little mass it has to give up the electrons.
Unit 3<br />
Chapter 25 Nuclear Chemistry<br />
The students will learn what happens when an unstable<br />
nucleus decays and how nuclear chemistry affects their lives.<br />
Explore the theory of electromagnetism by comparing and contrasting the<br />
different parts of the electromagnetic spectrum in terms of wavelength,<br />
frequency, and energy, and relate them to phenomena and applications.<br />
Students will be able to compare and contrast the different parts of the<br />
electromagnetic spectrum.<br />
Students will be able to apply knowledge of the EMS to real world phenomena.<br />
Students will be able to quantitatively compare the relationship between energy,<br />
wavelength, and frequency of the EMS.<br />
amplitude<br />
wavelength<br />
frequency<br />
hertz<br />
electromagnetic radiation<br />
photon<br />
Planck’s constant<br />
Explain and compare nuclear reactions (radioactive decay, fission and<br />
fusion), the energy changes associated with them and their associated<br />
safety issues.<br />
Students will be able to compare and contrast fission and fusion reactions.<br />
Students will be able to complete nuclear decay equations to identify the type of<br />
decay.<br />
Students will participate in activities to calculate half-life.<br />
radioactivity<br />
nuclear radiation<br />
alpha particle<br />
beta particle<br />
gamma ray<br />
positron<br />
½ life<br />
transmutation<br />
fission<br />
fusion<br />
50
Chapter 7<br />
Ionic and Metallic Bonding<br />
The students will learn how ionic compounds form and how<br />
metallic bounding affects the properties of metals.<br />
Compare the magnitude and range of the four fundamental forces<br />
(gravitational, electromagnetic, weak nuclear, strong nuclear).<br />
Students will compare/contrast the characteristics of each fundamental force.<br />
gravity<br />
electromagnetic<br />
strong<br />
weak<br />
Distinguish between bonding forces holding compounds together and other<br />
attractive forces, including hydrogen bonding and van der Waals forces.<br />
Students will be able to compare/contrast traits of ionic and covalent bonds.<br />
Students will be able to compare/contrast basic attractive forces between<br />
molecules.<br />
Students will be able to predict the type of bond or attractive force between<br />
atoms or molecules.<br />
ionic bond<br />
covalent bond<br />
metallic bond<br />
polar covalent bond<br />
hydrogen bond<br />
van der Waals forces<br />
London dispersion forces<br />
Chapter 8<br />
Covalent Bonding<br />
The students will learn how molecular bonding is different<br />
than ionic bonding and electrons affect the shape of a<br />
molecule and its properties.<br />
Interpret formula representations of molecules and compounds in terms of<br />
composition and structure.<br />
Students will be able to interpret chemical formulas in terms of # of atoms.<br />
Students will be able to differentiate between ionic and molecular compounds.<br />
Students will be able to list various VSEPR shapes and identify examples of<br />
each.<br />
Students will be able to predict shapes of various compounds.<br />
Molecule<br />
empirical formula<br />
Atom<br />
Electron<br />
Element<br />
Compound<br />
51
Bryan Vega<br />
Name ____________________<br />
Go to the web site www.darvill.clara.net/emag<br />
1. Click on “How the waves fit into the spectrum” and fill in this table:<br />
>: look out for the<br />
RED words on the web site!<br />
frequency<br />
Low __________, Long wavelength<br />
High frequency, Short ______________<br />
wavelength<br />
Radio Waves<br />
Micro Waves<br />
Infra-Red<br />
Visible Light Ultra-Violent X-Rays<br />
Gamma rays<br />
2. Click on “Radio waves”. They are used for _______________________<br />
communications<br />
3. Click on “Microwaves”. They are used for cooking, mobile _________, phones speed<br />
_______ cameras and _________.<br />
wifi<br />
4. Click on “Infra-red”. These waves are given off by _____ hot _________. objects They are used for remote controls,<br />
cameras in police ____________ helicopters , and alarm systems.<br />
laser<br />
5. Click on “Visible Light”. This is used in ___ players and _______ printers, and for seeing where we’re going.<br />
DVD<br />
6. “UV” stands for “ ________ ultra ___________”. violent This can damage the _________ retina in your eyes, and cause<br />
sunburn and even _______ skin cancer. Its uses include detecting forged ______ bank _______. notes<br />
7. X-rays are used to see inside people, and for _________ airport security.<br />
8. Gamma rays are given off by some ________________<br />
radioactive<br />
substances. We can use them to kill ________<br />
cancer<br />
cells,<br />
which is called R_______________ adiotherapy .<br />
9. My Quiz score is ____%. 100<br />
52
10. Name ________________________________<br />
Go to the web site www.darvill.clara.net/emag<br />
Name How they’re made Uses Dangers<br />
They are made by stars They are used to kill living They are<br />
and radioactive<br />
cancer cells.<br />
dangerous cause<br />
substances.<br />
they damage cells .<br />
Gamma rays<br />
X-Rays<br />
Ultra-Violent<br />
Visible Light<br />
Infra-Red<br />
Micro waves<br />
Radio waves<br />
They are made by stars and They are used to see inside<br />
some types of nebula. people.<br />
They are made by special<br />
lamps.<br />
They are made by things<br />
hot enough to glow.<br />
Infra Red waves are made<br />
by stars, lamps, flames,<br />
yourself, and more<br />
Microwaves are made by<br />
frequency radio waves, and<br />
other types of transmitter.<br />
Radio waves are made by<br />
various types of transmitter<br />
UV light are used for<br />
getting a sun tan, detecting<br />
forged bank notes in shops.<br />
We use light to see things.<br />
They are used foe remote<br />
controls for TVs and video<br />
recorders, and more.<br />
We use microwaves to<br />
cook many types of food.<br />
They are used mainly for<br />
communications.<br />
They can cause<br />
cell damage<br />
and cancer.<br />
UV light can damage<br />
the retina in your<br />
eyes<br />
Too much light can<br />
damage the retina<br />
in your eyes.<br />
The danger is<br />
overheating.<br />
Microwaves is known<br />
to cause cataracts in<br />
your eyes<br />
Radio waves are known<br />
to cause cancer, and<br />
other disorders.<br />
_____ High Frequency _____ Low frequency,<br />
Short wavelength ______ High Wavelength<br />
53
Learning Goal for this section:<br />
Explain and compare nuclear reactions (radioactive decay, fission and<br />
fusion), the energy changes associated with them and their associated<br />
safety issues.<br />
Notes Section:<br />
Nucleus= protons+neutrons<br />
N-1amu-0 / (+,-)<br />
P-1amu-+<br />
E-none- -<br />
Element 10 has 20 isotopes<br />
All living things are made up of carbon<br />
Bottom of periodic table- the mass gets massive<br />
Isotopes are an element or atom that has a diferent mass than neutrons<br />
It becomes stable when the number of protons are equal or similar to the number of neutrons<br />
When the particles are connected to your cells it becomes cancerous<br />
Beta particles have a mass of an electron<br />
You are able to convert neutrons to protons<br />
Unit is AMU for protons and neutrons<br />
Gamma radiation can kill<br />
Gamma radiation is energy<br />
More dense the more the atoms are together<br />
Gamma radiation is associated with radioactive decay<br />
54
The Nucleus<br />
A typical model of the atom is called the Bohr Model, in<br />
honor of Niels Bohr who proposed the structure in 1913. The Bohr atom consists of a central nucleus<br />
composed of neutrons and protons, which is surrounded by electrons which “orbit” around the nucleus.<br />
Protons carry a positive charge of one and have a mass of about 1 atomic mass unit or amu (1 amu =1.7x10-<br />
27 kg, a very, very small number). Neutrons are electrically “neutral” and also have a mass of about 1 amu. In<br />
contrast electron carry a negative charge and have mass of only 0.00055 amu. The number of protons in a<br />
nucleus determines the element of the atom. For example, the number of protons in uranium is 92 and the<br />
number in neon is 10. The proton number is often referred to as Z.<br />
Atoms with different numbers of protons are called elements, and are arranged in the periodic table with<br />
increasing Z.<br />
Atoms in nature are electrically neutral so the number of electrons orbiting the nucleus equals the number of<br />
protons in the nucleus.<br />
Neutrons make up the remaining mass of the nucleus and provide a means to “glue” the protons in place.<br />
Without neutrons, the nucleus would split apart because the positive protons would repel each other. Elements<br />
can have nucleii with different numbers of neutrons in them. For example hydrogen, which normally only has<br />
one proton in the nucleus, can have a neutron added to its nucleus to from deuterium, ir have two neutrons<br />
added to create tritium, which is radioactive. Atoms of the same element which vary in neutron number are<br />
called isotopes. Some elements have many stable isotopes (tin has 10) while others have only one or two. We<br />
express isotopes with the nomenclature Neon-20 or 20 Ne 10, with twenty representing the total number of<br />
neutrons and protons in the atom, often referred to as A, and 10 representing the number of protons (Z).<br />
Alpha Particle<br />
Decay<br />
Alpha decay is a radioactive process in which a<br />
particle with two neutrons and two protons is<br />
ejected from the nucleus of a radioactive atom. The particle is identical to the nucleus of a helium atom.<br />
Alpha decay only occurs in very heavy elements such as uranium, thorium and radium. The nuclei of these<br />
atoms are very “neutron rich” (i.e. have a lot more neutrons in their nucleus than they do protons) which makes<br />
emission of the alpha particle possible.<br />
After an atom ejects an alpha particle, a new parent atom is formed which has two less neutrons and two less<br />
protons. Thus, when uranium-238 (which has a Z of 92) decays by alpha emission, thorium-234 is created<br />
(which has a Z of 90).<br />
Because alpha particles contain two protons, they have a positive charge of two. Further, alpha particles are<br />
very heavy and very energetic compared to other common types of radiation. These characteristics allow alpha<br />
particles to interact readily with materials they encounter, including air, causing many ionizations in a very short<br />
distance. Typical alpha particles will travel no more than a few centimeters in air and are stopped by a sheet of<br />
paper.<br />
55
Beta Particle Decay<br />
Beta decay is a radioactive process in which an electron is emitted from the nucleus of a radioactive<br />
atom Because this electron is from the nucleus of the atom, it is called a beta particle to distinguish it<br />
from the electrons which orbit the atom.<br />
Like alpha decay, beta decay occurs in isotopes which are “neutron rich” (i.e. have a lot more<br />
neutrons in their nucleus than they do protons). Atoms which undergo beta decay are located below<br />
the line of stable elements on the chart of the nuclides, and are typically produced in nuclear reactors.<br />
When a nucleus ejects a beta particle, one of the neutrons in the nucleus is transformed into a proton.<br />
Since the number of protons in the nucleus has changed, a new daughter atom is formed which has<br />
one less neutron but one more proton than the parent. For example, when rhenium-187 decays<br />
(which has a Z of 75) by beta decay, osmium-187 is created (which has a Z of 76). Beta particles<br />
have a single negative charge and weigh only a small fraction of a neutron or proton. As a result, beta<br />
particles interact less readily with material than alpha particles. Depending on the beta particles<br />
energy (which depends on the radioactive atom), beta particles will travel up to several meters in air,<br />
and are stopped by thin layers of metal or plastic.<br />
Positron emission or beta plus decay (β+ decay) is a subtype of radioactive decay called beta decay,<br />
in which a proton inside a radionuclide nucleus is converted into a neutron while releasing a positron<br />
and an electron neutrino (νe). Positron emission is mediated by the weak force.<br />
An example of positron emission (β+ decay) is shown with magnesium-23 decaying into sodium-23:<br />
23 Mg12 → 23 Na11 + e +<br />
Because positron emission decreases proton number relative to neutron number, positron decay<br />
happens typically in large "proton-rich" radionuclides. Positron decay results in nuclear transmutation,<br />
changing an atom of one chemical element into an atom of an element with an atomic number that is<br />
less by one unit.<br />
Positron emission should not be confused with electron emission or beta minus decay (β− decay),<br />
which occurs when a neutron turns into a proton and the nucleus emits an electron and an<br />
antineutrino.<br />
56
Gamma<br />
Radiation<br />
After a decay reaction, the nucleus is often in an<br />
“excited” state. This means that the decay has<br />
resulted in producing a nucleus which still has<br />
excess energy to get rid of. Rather than emitting another beta or alpha particle, this energy is lost by<br />
emitting a pulse of electromagnetic radiation called a gamma ray. The gamma ray is identical in<br />
nature to light or microwaves, but of very high energy.<br />
Like all forms of electromagnetic radiation, the gamma ray has no mass and no charge. Gamma rays<br />
interact with material by colliding with the electrons in the shells of atoms. They lose their energy<br />
slowly in material, being able to travel significant distances before stopping. Depending on their initial<br />
energy, gamma rays can travel from 1 to hundreds of meters in air and can easily go right through<br />
people.<br />
It is important to note that most alpha and beta emitters also emit gamma rays as part of their decay<br />
process. However, their is no such thing as a “pure” gamma emitter. Important gamma emitters<br />
including technetium-99m which is used in nuclear medicine, and cesium-137 which is used for<br />
calibration of nuclear instruments.<br />
Half Life<br />
Half-life is the time required for the quantity of a<br />
radioactive material to be reduced to one-half its<br />
original value.<br />
All radionuclides have a particular half-life, some<br />
of which a very long, while other are extremely<br />
short. For example, uranium-238 has such a<br />
long half life, 4.5x109 years, that only a small fraction has decayed since the earth was formed. In<br />
contrast, carbon-11 has a half-life of only 20 minutes. Since this nuclide has medical applications, it<br />
has to be created where it is being used so that enough will be present to conduct medical studies.<br />
57
The Learning Goal for this assignment is:<br />
Distinguish between bonding forces holding compounds together and other attractive<br />
forces, including hydrogen bonding and van der Waals forces<br />
Introduction to Ionic Compounds<br />
Those molecules that consist of charged ions with opposite charges are called IONIC. These ionic<br />
compounds are generally solids with high melting points and conduct electrical current. Ionic<br />
compounds are generally formed from metal and a non-metal elements. See Ionic Bonding below.<br />
Ionic Compound Example<br />
For example, you are familiar with the fairly benign unspectacular behavior of common white<br />
crystalline table salt (NaCl). Salt consists of positive sodium ions (Na + ) & negative chloride ions (Cl - ).<br />
On the other hand the element sodium is a silvery gray metal composed of neutral atoms which react<br />
vigorously with water or air. Chlorine as an element is a neutral greenish-yellow, poisonous, diatomic<br />
gas (Cl2).<br />
The main principle to remember is that ions are completely different in physical and chemical<br />
properties from the neutral atoms of the elements.<br />
The notation of the + and - charges on ions is very important as it conveys a definite meaning.<br />
Whereas elements are neutral in charge, IONS have either a positive or negative charge depending<br />
upon whether there is an excess of protons (positive ion) or excess of electrons (negative ion).<br />
Formation of Positive Ions<br />
Metals usually have 1-4 electrons in the outer energy level. The electron arrangement of a rare gas is<br />
most easily achieved by losing the few electrons in the newly started energy level. The number of<br />
electrons lost must bring the electron number "down to" that of a prior rare gas.<br />
How will sodium complete its octet?<br />
First examine the electron arrangement of the atom. The atomic number is eleven, therefore, there<br />
are eleven electrons and eleven protons on the neutral sodium atom. Here is the Bohr diagram and<br />
Lewis symbol for sodium:<br />
58
This analysis shows that sodium has only one electron in its outer level. The nearest rare gas is neon<br />
with 8 electron in the outer energy level. Therefore, this electron is lost so that there are now eight<br />
electrons in the outer energy level, and the Bohr diagrams and Lewis symbols for sodium ion and<br />
neon are identical. The octet rule is satisfied.<br />
Ion Charge?<br />
What is the charge on sodium ion as a result of losing one electron? A comparison of the atom and<br />
the ion will yield this answer.<br />
Sodium Atom<br />
Sodium Ion<br />
11 p+ to revert to 11 p + Protons are identical in<br />
12 n an octet 12 n<br />
the atom and ion.<br />
Positive charge is<br />
11 e- lose 1 electron 10 e-<br />
caused by lack of<br />
0 charge + 1 charge<br />
electrons.<br />
Formation of Negative Ions<br />
How will fluorine complete its octet?<br />
First examine the electron arrangement of the atom. The atomic number is nine, therefore, there are<br />
nine electrons and nine protons on the neutral fluorine atom. Here is the Bohr diagram and Lewis<br />
symbol for fluorine:<br />
This analysis shows that fluorine already has seven electrons in its outer level. The nearest rare gas<br />
is neon with 8 electron in the outer energy level. Therefore only one additional electron is needed to<br />
complete the octet in the fluorine atom to make the fluoride ion. If the one electron is added, the Bohr<br />
diagrams and Lewis symbols for fluorine and neon are identical. The octet rule is satisfied.<br />
59
Ion Charge?<br />
What is the charge on fluorine as a result of adding one electron? A comparison of the atom and the<br />
ion will yield this answer.<br />
Fluorine Atom Fluoride Ion *<br />
9 p+ to complete 9 p + Protons are identical in<br />
10 n octet 10 n<br />
9 e- add 1 electron 10 e-<br />
0 charge - 1 charge<br />
the atom and ion.<br />
Negative charge is<br />
caused by excess<br />
electrons<br />
* The "ide" ending in the name signifies a simple negative ion.<br />
Summary Principle of Ionic Compounds<br />
An ionic compound is formed by the complete transfer of electrons from a metal to a nonmetal and<br />
the resulting ions have achieved an octet. The protons do not change. Metal atoms in Groups 1-3<br />
lose electrons to non-metal atoms with 5-7 electrons missing in the outer level. Non-metals gain 1-4<br />
electrons to complete an octet.<br />
Octet Rule<br />
Elemental atoms generally lose, gain, or share electrons with other atoms in order to achieve the<br />
same electron structure as the nearest rare gas with eight electrons in the outer level.<br />
The proper application of the Octet Rule provides valuable assistance in predicting and explaining<br />
various aspects of chemical formulas.<br />
Introduction to Ionic Bonding<br />
Ionic bonding is best treated using a simple<br />
electrostatic model. The electrostatic model<br />
is simply an application of the charge<br />
principles that opposite charges attract and<br />
similar charges repel. An ionic compound<br />
results from the interaction of a positive and<br />
negative ion, such as sodium and chloride in<br />
common salt.<br />
The IONIC BOND results as a balance<br />
between the force of attraction between<br />
opposite plus and minus charges of the ions<br />
and the force of repulsion between similar<br />
negative charges in the electron clouds. In<br />
crystalline compounds this net balance of<br />
forces is called the LATTICE ENERGY.<br />
Lattice energy is the energy released in the<br />
formation of an ionic compound.<br />
60<br />
DEFINITION: The formation of an IONIC<br />
BOND is the result of the transfer of one or<br />
more electrons from a metal onto a nonmetal.
Metals, with only a few electrons in the outer energy level, tend to lose electrons most readily. The<br />
energy required to remove an electron from a neutral atom is called the IONIZATION POTENTIAL.<br />
Energy + Metal Atom ---> Metal (+) ion + e-<br />
Non-metals, which lack only one or two electrons in the outer energy level have little tendency to lose<br />
electrons - the ionization potential would be very high. Instead non-metals have a tendency to gain<br />
electrons. The ELECTRON AFFINITY is the energy given off by an atom when it gains electrons.<br />
Non-metal Atom + e- --- Non-metal (-) ion + energy<br />
The energy required to produce positive ions (ionization potential) is roughly balanced by the energy<br />
given off to produce negative ions (electron affinity). The energy released by the net force of<br />
attraction by the ions provides the overall stabilizing energy of the compound.<br />
Notes Section:<br />
Ion is a charged particle<br />
Caton on the left and antons on the right on the periodic table<br />
Ionic compounds are generally solids with high melting points and conduct electrical current<br />
crystalline compounds this net balance of forces is called the lattice energy<br />
Energy + Metal Atom ---> Metal (+) ion + e-<br />
Non-metal Atom + e- --- Non-metal (-) ion + energy<br />
An ionic compound is formed by the complete transfer of electrons from a metal to a nonmetal and the<br />
resulting ions have achieved an octet.<br />
61
The Learning Goal for this assignment is:<br />
The students will learn how molecular<br />
bonding is different than ionic bonding and electrons affect the shape of a molecule and its properties.<br />
Introduction to Covalent Bonding:<br />
Bonding between non-metals consists of two electrons shared between two atoms. Using the Wave<br />
Theory, the covalent bond involves an overlap of the electron clouds from each atom. The electrons<br />
are concentrated in the region between the two atoms. In covalent bonding, the two electrons shared<br />
by the atoms are attracted to the nucleus of both atoms. Neither atom completely loses or gains<br />
electrons as in ionic bonding.<br />
There are two types of covalent bonding:<br />
1. Non-polar bonding with an equal sharing of electrons.<br />
2. Polar bonding with an unequal sharing of electrons. The number of shared electrons depends on<br />
the number of electrons needed to complete the octet.<br />
NON-POLAR BONDING results when two identical non-metals equally share electrons between<br />
them. One well known exception to the identical atom rule is the combination of carbon and hydrogen<br />
in all organic compounds.<br />
Hydrogen<br />
The simplest non-polar covalent molecule is hydrogen. Each hydrogen<br />
atom has one electron and needs two to complete its first energy level.<br />
Since both hydrogen atoms are identical, neither atom will be able to<br />
dominate in the control of the electrons. The electrons are therefore<br />
shared equally. The hydrogen covalent bond can be represented in a<br />
variety of ways as shown here:<br />
The "octet" for hydrogen is only 2 electrons since the nearest rare gas is<br />
He. The diatomic molecule is formed because individual hydrogen atoms<br />
containing only a single electron are unstable. Since both atoms are<br />
identical a complete transfer of electrons as in ionic bonding is<br />
impossible.<br />
Instead the two hydrogen atoms SHARE both electrons equally.<br />
Oxygen<br />
Molecules of oxygen, present in about 20% concentration in air are<br />
also covalent molecules. See the graphic on the left of the Lewis Dot<br />
Structure.<br />
There are 6 electrons in the outer shell, therefore, 2 electrons are<br />
needed to complete the octet. The two oxygen atoms share a total of<br />
four electrons in two separate bonds, called double bonds.<br />
The two oxygen atoms equally share the four electrons.<br />
62
POLAR BONDING results when two different non-metals unequally share electrons between them.<br />
One well known exception to the identical atom rule is the combination of carbon and hydrogen in all<br />
organic compounds.<br />
The non-metal closer to fluorine in the Periodic Table has a greater tendency to keep its own electron<br />
and also draw away the other atom's electron. It is NOT completely successful. As a result, only<br />
partial charges are established. One atom becomes partially positive since it has lost control of its<br />
electron some of the time. The other atom becomes partially negative since it gains electron some of<br />
the time.<br />
Hydrogen Chloride<br />
Hydrogen Chloride forms a polar covalent molecule. The graphic<br />
on the left shows that chlorine has 7 electrons in the outer shell.<br />
Hydrogen has one electron in its outer energy shell. Since 8<br />
electrons are needed for an octet, they share the electrons.<br />
However, chlorine gets an unequal share of the two electrons,<br />
although the electrons are still shared (not transferred as in ionic<br />
bonding), the sharing is unequal. The electrons spends more of the<br />
time closer to chlorine. As a result, the chlorine acquires a "partial"<br />
negative charge. At the same time, since hydrogen loses the<br />
electron most - but not all of the time, it acquires a "partial" charge.<br />
The partial charge is denoted with a small Greek symbol for delta.<br />
Water<br />
Water, the most universal compound on all of the earth, has the property of<br />
being a polar molecule. As a result of this property, the physical and<br />
chemical properties of the compound are fairly unique.<br />
Dihydrogen Oxide or water forms a polar covalent molecule. The graphic on<br />
the left shows that oxygen has 6 electrons in the outer shell. Hydrogen has<br />
one electron in its outer energy shell. Since 8 electrons are needed for an<br />
octet, they share the electrons.<br />
Notes Section:<br />
1 Count the valence electrons<br />
2 Find the central atom and bond the other atoms to it.<br />
Subtract the number of electrons in the bonds from the total.<br />
Add lone pairs to the terminal atoms. Add lone pairs to the<br />
central atom or double or triple bonds<br />
3 Find the formal charges. Try to get the charges as close to<br />
zero as possible by moving electrons and bonds.<br />
63
C 2 H 6 O Ethanol CH 3 CH 2 O<br />
Step 1<br />
Find valence e- for all atoms. Add them together.<br />
C: 4 x 2 = 8<br />
H: 1 x 6 = 6<br />
O: 6<br />
Total = 20<br />
Step 2<br />
Find octet e- for each atom and add them together.<br />
C: 8 x 2 = 16<br />
H: 2 x 6 = 12<br />
O: 8<br />
Total = 36<br />
Step 3<br />
Subtract Step 1 total from Step 2.<br />
Gives you bonding e-.<br />
36 – 20 = 16e-<br />
Step 4<br />
Find number of bonds by diving the number in step 3 by 2<br />
(because each bond is made of 2 e-)<br />
16e- / 2 = 8 bond pairs<br />
These can be single, double or triple bonds.<br />
Step 5<br />
Determine which is the central atom<br />
Find the one that is the least electronegative.<br />
Use the periodic table and find the one farthest<br />
away from Fluorine or<br />
The one that only has 1 atom.<br />
64
Step 6<br />
Put the atoms in the structure that you think it will<br />
have and bond them together.<br />
Put Single bonds between atoms.<br />
Step 7<br />
Find the number of nonbonding (lone pairs) e-.<br />
Subtract step 3 number from step 1.<br />
20 – 16 = 4e- = 2 lone pairs<br />
Step 8<br />
<strong>Complete</strong> the Octet Rule by adding the lone<br />
pairs.<br />
Then, if needed, use any lone pairs to make<br />
double and triple bonds so that all atoms meet<br />
the Octet Rule.<br />
See Step 4 for total number of bonds.<br />
65
Linear<br />
Molecular Geometry<br />
Orbital Equation Lone Pairs Angle<br />
sp AX 2 None 180<br />
BeCl 2<br />
Cl<br />
Be<br />
Cl<br />
Beryllium dichloride<br />
66<br />
element bond lone pair<br />
C
Trigonal Planar<br />
Molecular Geometry<br />
Orbital Equation Lone Pairs Angle<br />
sp 2 AX 3 None 120<br />
BF 3<br />
Boron<br />
F<br />
Trifloride<br />
B<br />
F<br />
F<br />
element bond lone pair<br />
C<br />
67
Bent<br />
Molecular Geometry<br />
Orbital Equation Lone Pairs Angle<br />
sp 2 AX 2 E 1 116<br />
O 3<br />
O<br />
O<br />
O<br />
Trioxide<br />
element bond lone pair<br />
C<br />
68
Tetrahedral<br />
Molecular Geometry<br />
Orbital Equation Lone Pairs Angle<br />
sp 3 AX 4 None 109.5<br />
PO 4<br />
3-<br />
C<br />
-3<br />
O<br />
Posphate<br />
P<br />
O<br />
O<br />
Tetraoxide<br />
element bond lone pair<br />
C<br />
69
Trigonal Pyramid<br />
Molecular Geometry<br />
Orbital Equation Lone Pairs Angle<br />
Sp 3 Ax 3 E 1 107<br />
PH 3<br />
H<br />
Phophorus<br />
Trihydride<br />
P<br />
H<br />
H<br />
element bond lone pair<br />
C<br />
70
Bent<br />
Molecular Geometry<br />
Orbital Equation Lone Pairs Angle<br />
Sp 3 AX 2 E 2 2 104.5<br />
H 2 O<br />
H<br />
O<br />
H<br />
Dihydrogen Oxide<br />
element bond lone pair<br />
C<br />
71
Trigonal Bi Pyramid<br />
Molecular Geometry<br />
Orbital Equation Lone Pairs Angle<br />
Sp 3 d AX 5 None 120/90<br />
PCl 5<br />
Cl<br />
Cl<br />
Phosporus<br />
Pentachloride<br />
P<br />
Cl Cl Cl<br />
element bond lone pair<br />
C<br />
72
T-Shaped<br />
Molecular Geometry<br />
Orbital Equation Lone Pairs Angle<br />
Sp 3 d Ax 3 E 2 2 90<br />
ClF 3<br />
F<br />
Carbon<br />
Iodine<br />
Trifloride<br />
Cl<br />
F<br />
F<br />
element bond lone pair<br />
C<br />
73
Octahedral<br />
Molecular Geometry<br />
Orbital Equation Lone Pairs Angle<br />
Sp 3 d 2 Ax 6 None 90<br />
SF 6<br />
F<br />
F<br />
S<br />
F<br />
F<br />
Sulfur<br />
Hexafloride<br />
F<br />
F<br />
element bond lone pair<br />
C<br />
74
Square Planar<br />
Molecular Geometry<br />
Orbital Equation Lone Pairs Angle<br />
Sp 3 d 2 Ax 4 E 2 2 90<br />
ICl 4<br />
-<br />
Cl<br />
-<br />
Cl<br />
I<br />
Cl<br />
Iodine<br />
Tetrachloride<br />
Cl<br />
el bond lone pair<br />
C<br />
75
Orbitals Equation Lone Pairs Angle<br />
sp AX2 None 180<br />
sp 2 AX 3 None 120<br />
sp 2 AX2E 1 116<br />
sp 3 AX4 None 109.5<br />
sp 3 AX3E 1 107<br />
sp 3 AX2E2 2 104.5<br />
sp 3 d AX5 None 120/90<br />
sp 3 d AX3E2 2 90<br />
sp 3 d 2 AX6 None 90<br />
sp 3 d 2 AX4E2 2 90<br />
Name<br />
Linear<br />
Trig. Planar<br />
Bent<br />
Tetrahedral<br />
Trig. Pyramid<br />
Bent<br />
Trig. BiPyramid<br />
T-Shaped<br />
Octahedral<br />
Square Planar<br />
76
Name Formula Charge<br />
Dichromate Cr₂O₇ 2-<br />
Sulfate SO₄ 2-<br />
Hydrogen Carbonate HCO₃ 1-<br />
Hypochlorite ClO 1-<br />
Phosphate PO₄ 3-<br />
Nitrite NO₂ 1-<br />
Chlorite ClO₂ 1-<br />
Dihydrogen phosphate H₂PO₄ 1-<br />
Chromate CrO₄ 2-<br />
Carbonate CO₃ 2-<br />
Hydroxide OH 1-<br />
Hydrogen phosphate HPO₄ 2-<br />
Ammonium NH₄ 1+<br />
Acetate C₂H₃O₂ 1-<br />
Perchlorate ClO₄ 1-<br />
Permanganate MnO₄ 1-<br />
Chlorate ClO₃ 1-<br />
Hydrogen Sulfate HSO₄ 1-<br />
Phosphite PO₃ 3-<br />
Sulfite SO₃ 2-<br />
Silicate SiO₃ 2-<br />
Nitrate NO₃ 1-<br />
Hydrogen Sulfite HSO₃ 1-<br />
Oxalate C₂O₄ 2-<br />
Cyanide CN 1-<br />
Hydronium H₃O 1+<br />
Thiosulfate S₂O₃ 2-<br />
77
Chapter 9<br />
Unit 4<br />
Chemical Names and Formulas<br />
The students will learn how the periodic table helps them<br />
determine the names and formulas of ions and compounds.<br />
Chapter 22 Hydrocarbon Compounds<br />
The student will learn how Hydrocarbons are named and the<br />
general properties of Hydrocarbons.<br />
Describe how different natural resources are produced and how their rates<br />
of use and renewal limit availability.<br />
Students will explore local, national, and global renewable and nonrenewable<br />
resources.<br />
Students will explain the environmental costs of the use of renewable and<br />
nonrenewable resources.<br />
Students will explain the benefits of renewable and nonrenewable resources.<br />
Nuclear reactors<br />
Natural gas<br />
Petroleum<br />
Refining<br />
Coal<br />
78
Chapter 23 Functional Groups<br />
The student will learn what effects functional groups have on<br />
organic compounds and how chemical reactions are used in<br />
organic compounds.<br />
Describe the properties of the carbon atom that make the diversity of carbon<br />
compounds possible.<br />
Identify selected functional groups and relate how they contribute to<br />
properties of carbon compounds.<br />
Students will identify examples of important carbon based molecules.<br />
Students will create 2D or 3D models of carbon molecules and explain why this<br />
molecule is important to life.<br />
covalent bond<br />
single bond<br />
double bond<br />
triple bond<br />
monomer<br />
polymer<br />
79
http://www.bbc.co.uk/education/guides/zm9hvcw/revision<br />
Learning Goal:<br />
Describe the properties of the carbon atom that<br />
make the diversity of carbon compounds possible.<br />
Identify selected functional groups and relate how they<br />
contribute to properties of carbon compounds.<br />
Addition of bromine water can be used<br />
as a test for unsaturation.<br />
Alkenes quickly turn bromine water colourless,<br />
while alkanes and cycloalkanes do not.<br />
Indicate the position of the branches with a<br />
number, numbering from the end nearest<br />
the functional group.<br />
The alkanes don't contain a functional group<br />
and so the branches are numbered from<br />
the end that gives the lowest set of<br />
position numbers for the branches.<br />
80
Notes:<br />
The test<br />
that you<br />
gave us<br />
was easy<br />
because<br />
I didn't<br />
have Ronny<br />
and instead<br />
I had a<br />
paper.<br />
81
Unit 5<br />
Chapter 10 Chemical Quantities<br />
The student will learn why the mole is important and how the<br />
molecular formula of a compound can be determined<br />
experimentally.<br />
Chapter 11 Chemical Reactions<br />
The students will learn how chemical reactions obey the law of<br />
conservation of mass and how they can predict the products<br />
of a chemical reaction.<br />
Characterize types of chemical reactions, for example: redox, acid-base,<br />
synthesis, and single and double replacement reactions.<br />
<br />
<br />
<br />
<br />
Students will be able to identify the type of chemical reaction that occurs.<br />
Students will be able to compare/contrast reactants and products of various<br />
types of chemical reactions.<br />
Students will be able to predict the product of various reactants.<br />
Students will be able to write balanced chemical equations for each type of<br />
reaction.<br />
Decomposition<br />
Combustion<br />
Redox<br />
Acid-Base<br />
Synthesis<br />
single-replacement<br />
double-replacement<br />
Differentiate between chemical and nuclear reactions.<br />
Students will compare/contrast chemical and nuclear reactions.<br />
fission<br />
fusion<br />
Pg82
Chapter 12 Stoichiometry<br />
The students will learn how balanced chemical equations are<br />
used in stoichiometric calculations and how to calculate<br />
amounts of reactants and products in a chemical equation.<br />
Apply the mole concept and the law of conservation of mass to calculate<br />
quantities of chemicals participating in reactions.<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
Students will be able to use a balanced equation to determine mole ratios.<br />
Students will be able to apply law of conservation of mass to chemical equations.<br />
Students will be able to calculate empirical and molecular formulas.<br />
Students will be able to calculate the % composition of a compound.<br />
Students will be able to calculate theoretical yield.<br />
Students will be able to calculate % error.<br />
Students will be able to calculate molar mass.<br />
Students will be able to perform stoichiometric calculations, including limiting<br />
reagents.<br />
mole<br />
Avogadro’ s number<br />
molar mass<br />
gram formula mass<br />
Pg83
The Mole<br />
LG:The students will learn how balanced chemical<br />
www.youtube.com/watch?v=AsqEkF7hcII equations are used in stoichiometric calculations and how<br />
www.youtube.com/watch?v=tEn0N4R2dqA to calculate amounts of reactants and products in a<br />
www.youtube.com/watch?v=Pft2CASl0M0 chemical equation. Part<br />
www.youtube.com/watch?v=rwhJklbK8R0<br />
_________<br />
NO<br />
N<br />
O<br />
14.01 16.00<br />
H20m<br />
H2x1=2<br />
01x16=16<br />
18amu=H20<br />
Percent<br />
x <strong>Whole</strong><br />
18gH 2 O=1 move H 2 O = 6.02x10 23 molecules H 2 O<br />
H 2 + O=H 2 O<br />
%= Part divided by whole<br />
The mole(mol)-Sl Base Unit used to measure the amount of a substance<br />
The # of particles -carbon atomsin exactly 12g of pure (-12)<br />
Avogadro's number 6.02 x 10 23<br />
602 000 000 000 000 000 000 000!!!!!!!!!!!!!!!!!!!!!!!!!!!!<br />
3.5dozen roses =?????<br />
2.5mol of roses =?????<br />
90340 23 molecules of H 2 O =????? moles of H 2 O<br />
=????? moles of H atoms<br />
l<br />
3.5 dozen roses l<br />
12 roses<br />
------------------------------------------------<br />
l<br />
l 1 dozen roses<br />
l<br />
= 42 roses<br />
Pg84
www.youtube.com/watch?v=BTRm8PwcZ3U<br />
www.youtube.com/watch?v=F9NkYSKJifs<br />
www.youtube.com/watch?v=xPdqEX_WMjo<br />
Molar Mass<br />
LG:The students will learn how balanced chemical<br />
equations are used in stoichiometric calculations and<br />
how to calculate amounts of reactants and products in a<br />
chemical equation.<br />
H<br />
Elements<br />
Check Periodic table<br />
Mna = 22.99 g/ml<br />
Molecules<br />
1.008<br />
O<br />
16.00<br />
S<br />
32.07<br />
Na<br />
22.99<br />
Add up masses of each atom<br />
H 2 SO 4 :<br />
M H2So4 = 2 x 1.008 + 32.07 + 4 x 16.00<br />
= 98.09 g/ml<br />
M o2 = 2 x 16.00 = 32.00 g/mol<br />
Pg85
LG:The students will learn how chemical reactions obey the law of<br />
conservation of mass and how they can predict the products<br />
of a chemical reaction Categories of Reactions<br />
All chemical reactions can be placed into one of six categories. Here they are, in no<br />
particular order:<br />
1) Synthesis: A synthesis reaction is when two or more simple compounds combine to form a<br />
more complicated one. These reactions come in the general form of:<br />
A+B--->AB<br />
H2+02--->H20<br />
One example of a synthesis reaction is the combination of iron and sulfur to form iron (II) sulfide:<br />
8 Fe + S8 ---> 8 FeS<br />
If two elements or very simple molecules combine with each other, it’s probably a synthesis reaction.<br />
The products will probably be predictable using the octet rule to find charges.<br />
2) Decomposition: A decomposition reaction is the opposite of a synthesis reaction - a<br />
complex molecule breaks down to make simpler ones. These reactions come in the general form:<br />
AB-->A+B<br />
H2O-->H2+O2<br />
One example of a decomposition reaction is the electrolysis of water to make oxygen and hydrogen<br />
gas:<br />
2 H2O ---> 2 H2 + O2<br />
If one compound has an arrow coming off of it, it’s probably a decomposition reaction. The products<br />
will either be a couple of very simple molecules, or some elements, or both.<br />
3) Single displacement: This is when one element trades places with another element in a<br />
compound. These reactions come in the general form of:<br />
A+BC-->AC+B<br />
One example of a single displacement reaction is when magnesium replaces hydrogen in water to<br />
make magnesium hydroxide and hydrogen gas:<br />
Mg + 2 H2O ---> Mg(OH)2 + H2<br />
If a pure element reacts with another compound (usually, but not always, ionic), it’s probably a single<br />
displacement reaction. The products will be the compounds formed when the pure element switches<br />
places with another element in the other compound.<br />
Important note: these reactions will only occur if the pure element on the reactant side of the equation<br />
is higher on the activity series than the element it replaces.<br />
Pg86
4) Double displacement: This is when the anions and cations of two different molecules<br />
switch places, forming two entirely different compounds. These reactions are in the general form:<br />
AB+CD-->AD+CB<br />
One example of a double displacement reaction is the reaction of lead (II) nitrate with potassium<br />
iodide to form lead (II) iodide and potassium nitrate:<br />
Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3<br />
If two ionic compounds combine, it’s probably a double displacement reaction. Switch the cations<br />
and balance out the charges to figure out what will be made.<br />
Important note: These reactions will only occur if both reactants are soluble in water and only one<br />
product is soluble in water.<br />
5) Acid-base: This is a special kind of double displacement reaction that takes place when an<br />
acid and base react with each other. The H + ion in the acid reacts with the OH - ion in the base,<br />
causing the formation of water. Generally, the product of this reaction is some ionic salt and water:<br />
HA+BOH-->AB+H2O<br />
One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium<br />
hydroxide:<br />
HBr + NaOH ---> NaBr + H2O<br />
If an acid and a base combine, it’s an acid-base reaction. The products will be an ionic compound<br />
and water.<br />
6) Combustion: A combustion reaction is when oxygen combines with another compound to<br />
form water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An<br />
example of this kind of reaction is the burning of napthalene:<br />
C10H8 + 12 O2 ---> 10 CO2 + 4 H2O<br />
If something that has carbon and hydrogen reacts with oxygen, it’s probably a combustion reaction.<br />
The products will be CO2 and H2O. CH+O2-->H2O + CO<br />
Follow this series of questions. When you can answer "yes" to a question, then<br />
stop!<br />
1) Does your reaction have two (or more) chemicals combining to form one chemical? If yes, then it's<br />
a synthesis reaction<br />
2) Does your reaction have one large molecule falling apart to make several small ones? If yes, then<br />
it's a decomposition reaction<br />
3) Does your reaction have any molecules that contain only one element? If yes, then it's a single<br />
displacement reaction<br />
4) Does your reaction have water as one of the products? If yes, then it's an acid-base reaction<br />
5) Does your reaction have oxygen as one of it's reactants and carbon dioxide and water as<br />
products? If yes, then it's a combustion reaction<br />
6) If you haven't answered "yes" to any of the questions above, then you've got a double<br />
displacement reaction.<br />
Pg 87
List what type the following reactions are:<br />
1) NaOH + KNO3 --> NaNO3 + KOH<br />
Double Displacement<br />
2) CH4 + 2 O2 --> CO2 + 2 H2O<br />
Combustion<br />
3) 2 Fe + 6 NaBr --> 2 FeBr3 + 6 Na<br />
Single Displacement<br />
4) CaSO4 + Mg(OH)2 --> Ca(OH)2 + MgSO4<br />
Double Displacement<br />
5) NH4OH + HBr --> H2O + NH4Br<br />
Acid-Base<br />
6) Pb + O2 --> PbO2<br />
Synthesis<br />
7) Na2CO3 --> Na2O + CO2<br />
Decomposition<br />
Pg88
Determine the Type of Reaction for each equation.<br />
Then predict the products of each of the following chemical reactions. If a reaction will not occur,<br />
explain why not.<br />
Then Balance the equation.<br />
1) __Ag2SO4 + __NaNO3 2 →2AgNO 3 + Na 2 SO4<br />
2) __NaI 2 + __CaSO4 →<br />
3) __HNO3 + __Ca(OH)2 →<br />
Na 2 SO 4 + CaI 2<br />
Na 1 2<br />
I 1 2<br />
Ca 1<br />
S 1<br />
O 4<br />
2H 2 O+Ca(NO 3 ) 2<br />
H 2 4<br />
N 1 2<br />
Ca 1<br />
O 5 8<br />
Ag 2<br />
S1<br />
O 7 10<br />
Na 1 2<br />
N 1 2<br />
4 2 H<br />
2 N<br />
1 Ca<br />
8 7 O<br />
2 Na<br />
2 I<br />
1 Ca<br />
1 S<br />
4 O<br />
2 1 Ag<br />
1 S<br />
10 7 O<br />
Na 2<br />
2 1 N<br />
4) __CaCO3 →<br />
CaO + CO 2<br />
Ca 1<br />
C 2<br />
O 3<br />
1 Ca<br />
2 C<br />
3 O<br />
5) __AlCl3 + __(NH4)PO4 → AlPO 4 + 3NH 4 Cl Al 1<br />
Cl 3<br />
N 3<br />
H 12<br />
P 1<br />
O 4<br />
6) __Pb + __Fe(NO3)3 → No reaction<br />
Al 1<br />
Cl 1 3<br />
N 1 3<br />
H 4 12<br />
P 1<br />
O 4<br />
7) 2 __C3H6 + 9 __O2 →<br />
6H 2 O + 6CO 2<br />
C 3<br />
H 6<br />
O 2 9<br />
C 1 3<br />
H 2 6<br />
O 3 7 9<br />
8) __Na 2 + __CaSO4 →<br />
Ca + Na 2 SO 4<br />
Na 1 2<br />
Ca 1<br />
S 1<br />
O 4<br />
Na 2<br />
Ca 1<br />
S 1<br />
O 4<br />
Pg89
LG:The students will learn how chemical reactions obey the law of<br />
conservation of mass and how they can predict the products<br />
of a chemical reaction<br />
How to Balance Chemical Equations<br />
A chemical equation is a theoretical or written representation of what happens during a chemical<br />
reaction. The law of conservation of mass states that no atoms can be created or destroyed in a<br />
chemical reaction, so the number of atoms that are present in the reactants has to balance the<br />
number of atoms that are present in the products. Follow this guide to learn how to balance chemical<br />
equations.<br />
Step 1<br />
Write down your given equation. For this example, we will use:<br />
C3H8 + O2 --> H2O + CO2<br />
Step 2<br />
Write down the number of atoms that you have on each side of the equation. Look at the subscripts<br />
next to each atom to find the number of atoms in the equation.<br />
Left side: 3 carbon, 8 hydrogen and 2 oxygen<br />
Right side: 1 carbon, 2 hydrogen and 3 oxygen<br />
Pg90
Step 3<br />
Always leave hydrogen and oxygen for last. This means that you will need to balance the carbon<br />
atoms first.<br />
Step 4<br />
Add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon<br />
atoms on the left of the equation.<br />
C3H8 + O2 --> H2O + 3CO2<br />
The coefficient 3 in front of carbon on the right side indicates 3 carbon atoms just as the subscript 3<br />
on the left side indicates 3 carbon atoms.<br />
In a chemical equation, you can change coefficients, but you should never alter the subscripts.<br />
Pg91
Step 5<br />
Balance the hydrogen atoms next. You have 8 on the left side, so you'll need 8 on the right side.<br />
C3H8 + O2 --> 4H2O + 3CO2<br />
On the right side, we added a 4 as the coefficient because the subscript showed that we already<br />
had 2 hydrogen atoms.<br />
When you multiply the coefficient 4 times the subscript 2, you end up with 8.<br />
Step 6<br />
Finish by balancing the oxygen atoms.<br />
Because we've added coefficients to the molecules on the right side of the equation, the number of<br />
oxygen atoms has changed. We now have 4 oxygen atoms in the water molecule and 6 oxygen<br />
atoms in the carbon dioxide molecule. That makes a total of 10 oxygen atoms.<br />
Add a coefficient of 5 to the oxygen molecule on the left side of the equation. You now have 10<br />
oxygen molecules on each side.<br />
C3H8 + 5O2 --> 4H2O + 3CO2.<br />
The carbon, hydrogen and oxygen atoms are balanced. Your equation is complete.<br />
Pg92
1) ___ 2 NaNO3 + ___ PbO ___ Pb(NO3)2 + ___ Na2O<br />
Na 1<br />
O 3<br />
Pb 1<br />
N 1<br />
x 2 =2<br />
x 2=6+1=7<br />
x 2=2<br />
Na 2<br />
O 7<br />
Pb 1<br />
N 2<br />
2) ___ 6 AgI + ___ Fe2(CO3)3 ___ 2 FeI3 + ___ 3 Ag2CO3<br />
Ag 1x 6 =6<br />
I 1x6 =6<br />
Fe 2<br />
C 3<br />
O 9<br />
Ag 2<br />
I 3x2=6<br />
Fe 1x2=2<br />
C 1x3=3<br />
O3x3=9<br />
3) ___ C2H4O2 + ___ 2 O2 ___ 2 CO2 + ___ 2 H2O<br />
C2<br />
H4<br />
O4x2=8<br />
C1x2=2<br />
H2x2=4<br />
O2x4=8<br />
4) ___ ZnSO4 + ___ Li2CO3 ___ ZnCO3 + ___ Li2SO4<br />
IT'S GOOD ;)<br />
5) ___ V2O5 + ___ 5 CaS ___ 5 CaO + ___ V2S5<br />
V2<br />
O5<br />
Ca1x5=5<br />
S1x5=5<br />
V2<br />
O1x5=5<br />
Ca 1x5=5<br />
S5<br />
Pg93
6) ___ Mn(NO2)2 + ___ BeCl2 ___ Be(NO2)2 + ___ MnCl2<br />
IT'S GOOD ;)<br />
7) ___ 3 AgBr + ___ GaPO4 ___ Ag3PO4 + ___ GaBr3<br />
Ag1x3=3<br />
Br1x3=3<br />
Ga1<br />
P1<br />
O4<br />
Ag3<br />
Br3<br />
Ga1<br />
P1<br />
O4<br />
8) ___ 3 H2SO4 + ___ 2 B(OH)3 __ B2(SO4)3 + ___ 6 H2O<br />
H5x2=10+2=12<br />
S1x3=3<br />
O7x2+2=18<br />
B1x2=2<br />
H2x6=12<br />
S3<br />
O13-18<br />
B2<br />
9) ___ S8 + ___ 8 O2 ___ 8 SO2<br />
S8<br />
O2x8=16<br />
S1x8=8<br />
O2x8=16<br />
10) ___ Fe + ___ 2 AgNO3 ___ Fe(NO3)2 + ___ 2 Ag<br />
Fe1<br />
Ag1x2=2<br />
N1x=2=2<br />
O3x2=6<br />
Fe1<br />
Ag1x2=2<br />
N2<br />
O6<br />
Pg94
1) 2 NaNO3 + PbO Pb(NO3)2 + Na2O<br />
2) 6 AgI + Fe2(CO3)3 2 FeI3 + 3 Ag2CO3<br />
3) C2H4O2 + 2 O2 2 CO2 + 2 H2O<br />
4) ZnSO4 + Li2CO3 ZnCO3 + Li2SO4<br />
5) V2O5 + 5 CaS 5 CaO + V2S5<br />
6) Mn(NO2)2 + BeCl2 Be(NO2)2 + MnCl2<br />
7) 3 AgBr + GaPO4 Ag3PO4 + GaBr3<br />
8) 3 H2SO4 + 2 B(OH)3 B2(SO4)3 + 6 H2O<br />
9) S8 + 8 O2 8 SO2<br />
10) Fe + 2 AgNO3 Fe(NO3)2 + 2 Ag<br />
Additional Notes:<br />
Make sure you multiply instead of divide<br />
Don't look at the answer until you finish<br />
Always double check your work<br />
Always make sure the elements on both sides are the same number<br />
Write out the equation when you are done<br />
Pick the right partner at the beginning of the year :)<br />
Pg 95
The Learning Goal for this assignment is: Apply the mole concept and the law of conservation of mass<br />
to calculate quantities of chemicals participating of chemicals<br />
participating in reactions<br />
Stoichiometry and Balancing Reactions<br />
Stoichiometry is a section of chemistry that involves using relationships between reactants and/or<br />
products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means<br />
element and metron means measure, so stoichiometry literally translated means the measure of<br />
elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to<br />
first understand the relationships that exist between products and reactants and why they exist, which<br />
require understanding how to balanced reactions.<br />
Balancing<br />
In chemistry, chemical reactions are frequently written as an equation, using chemical symbols. The<br />
reactants are displayed on the left side of the equation and the products are shown on the right, with<br />
the separation of either a single or double arrow that signifies the direction of the reaction. The<br />
significance of single and double arrow is important when discussing solubility constants, but we will<br />
not go into detail about it in this module. To balance an equation, it is necessary that there are the<br />
same number of atoms on the left side of the equation as the right. One can do this by raising the<br />
coefficients.<br />
Reactants to Products<br />
A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a<br />
chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as<br />
well as their states, and the proportion for how much of each particle is create relative to one another,<br />
through the stoichiometric coefficient. The following equation demonstrates the typical format of a<br />
chemical equation:<br />
2Na(s) + 2HCl(aq) → 2NaCl(aq) + H2(g)<br />
In the above equation, the elements present in the reaction are represented by their chemical<br />
symbols. Based on the Law of Conservation of Mass, which states that matter is neither created nor<br />
destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and<br />
products, though the elements they are paired up with often change in a reaction. In this reaction,<br />
sodium (Na), hydrogen (H), and chloride (Cl) are the elements present in both reactants, so based on<br />
the law of conservation of mass, they are also present on the product side of the equations.<br />
Displaying each element is important when using the chemical equation to convert between<br />
elements.<br />
Stoichiometric Coefficients<br />
In a balanced reaction, both sides of the equation have the same number of elements. The<br />
stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical<br />
reaction to balance the number of each element on both the reactant and product sides of the<br />
equation. These stoichiometric coefficients are useful since they establish the mole ratio between<br />
reactants and products. In the balanced equation:<br />
2Na(s)+2HCl(aq)→2NaCl(aq)+H2(g)<br />
Pg96
we can determine that 2 moles of HCl will react with 2 moles of Na(s) to form 2 moles of NaCl(aq) and 1<br />
mole of H2(g). If we know how many moles of Na we start out with, we can use the ratio of 2 moles<br />
of NaCl to 2 moles of Na to determine how many moles of NaCl were produced or we can use the<br />
ration of 1 mole of H2 to 2 moles of Na to convert to NaCl. This is known as the coefficient factor. The<br />
balanced equation makes it possible to convert information about one reactant or product to<br />
quantitative data about another element. Understanding this is essential to solving stoichiometric<br />
problems.<br />
Example 1<br />
Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.<br />
Solution<br />
___Pb(OH)4 +___H2SO4→___Pb(SO4)2 +___H2O<br />
Start by counting the number of atoms of each element.<br />
Unbalanced<br />
Pb 1 1 Pb<br />
O 8 9 O<br />
H 6 2 H<br />
S 1 2 S<br />
The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does<br />
not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the<br />
equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a<br />
coefficient of 2 should be added in front of H2SO4to have an equal number of sulfur on both sides of<br />
the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4<br />
coefficient should be added in front of H2O where there is a deficiency of oxygen. Count the number<br />
of elements now present on either side of the equation. Since the numbers are the same, the<br />
equation is now balanced.<br />
Pb(OH)4 + 2H2SO4→ Pb(SO4)2 + 4H2O<br />
Balanced<br />
Pb 1 1 Pb<br />
O 8 12 12 9 O<br />
H 6 8 8 2 H<br />
S 1 2 2 2 S<br />
Balancing reactions involves finding least common multiples between numbers of elements present<br />
on both sides of the equation. In general, when applying coefficients, add coefficients to the<br />
molecules or unpaired elements last.<br />
A balanced equation ultimately has to satisfy two conditions.<br />
1. The numbers of each element on the left and right side of the equation must be equal.<br />
2. The charge on both sides of the equation must be equal. It is especially important to pay<br />
attention to charge when balancing redox reactions.<br />
Pg97
Stoichiometry and Balanced Equations<br />
In stoichiometry, balanced equations make it possible to compare different elements through the<br />
stoichiometric factor discussed earlier. This is the mole ratio between two factors in a chemical<br />
reaction found through the ratio of stoichiometric coefficients. Here is a real world example to show<br />
how stoichiometric factors are useful.<br />
Example 2<br />
There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How<br />
many party invitations can be sent?<br />
Solution<br />
The equation for this can be written as<br />
I+2S→IS2<br />
where<br />
I represent invitations,<br />
S represents stamps, and<br />
IS 2 represents the sent party invitations consisting of one invitation and two stamps.<br />
<br />
Based on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced equation.<br />
Pg98<br />
Invitations Stamps Party Invitations Sent<br />
In this example are all the reactants (stamps and invitations) used up? No, and this is normally the<br />
case with chemical reactions. There is often excess of one of the reactants. The limiting reagent, the<br />
one that runs out first, prevents the reaction from continuing and determines the maximum amount of<br />
product that can be formed.<br />
Example 3<br />
What is the limiting reagent in this example?<br />
Solution<br />
Stamps, because there was only enough to send out invitations, whereas there were enough<br />
invitations for 12 complete party invitations. Aside from just looking at the problem, the problem can<br />
be solved using stoichiometric factors.<br />
12 I x 1IS2 = 12 IS2 possible<br />
1I<br />
20 S x 1IS2 = 10 IS2 possible<br />
2S
When there is no limiting reagent because the ratio of all the reactants caused them to run out at the<br />
same time, it is known as stoichiometric proportions.<br />
Types of Reactions<br />
There are 6 basic types of reactions.<br />
Combustion: Combustion is the formation of CO2 and H2O from the reaction of a chemical<br />
and O2<br />
Combination (synthesis): Combination is the addition of 2 or more simple reactants to form a<br />
complex product.<br />
Decomposition: Decomposition is when complex reactants are broken down into simpler<br />
products.<br />
Single Displacement: Single displacement is when an element from on reactant switches with<br />
an element of the other to form two new reactants.<br />
Double Displacement: Double displacement is when two elements from on reactants<br />
switched with two elements of the other to form two new reactants.<br />
Acid-Base: Acid- base reactions are when two reactants form salts and water.<br />
Molar Mass<br />
Before applying stoichiometric factors to chemical equations, you need to understand molar mass.<br />
Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual<br />
element as listed in the periodic table established this relationship for atoms or ions. For compounds<br />
or molecules, you have to take the sum of the atomic mass times the number of each atom in order to<br />
determine the molar mass.<br />
Example 4<br />
What is the molar mass of H2O?<br />
Solution<br />
Molar mass = 2 × (1.00g/mol) + 1×(16.0g/mol) = 18.0g/mol<br />
Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of<br />
products or vice versa.<br />
Example 5: Combustion of Propane<br />
Propane (C3H8) burns in this reaction:<br />
C3H8 + 5O2 → 4H2O + 3CO2<br />
If 200 g of propane is burned, how many g of H2Ois produced?<br />
Solution<br />
Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products<br />
you must convert from grams of C3H8 to moles of C3H8 then from moles of C3H8 to moles of H2O.<br />
Then convert from moles of H2O to grams of H2O.<br />
Pg99
Step 1: 200g C3H8 is equal to 4.54 mol C3H8.<br />
Step 2: Since there is a ratio of 4:1 H2O to C3H8, for every 4.54 mol C3H8 there are 18.18 mol<br />
H2O.<br />
Step 3: Convert 18.18 mol H2O to g H2O 18.18 mol H2O is equal to 327.27 g H2O.<br />
Variation in Stoichiometric Equations<br />
Almost every quantitative relationship can be converted into a ratio that can be useful in data<br />
analysis.<br />
Density<br />
Density (ρ) is calculated as mass/volume. This ratio can be useful in determining the volume of a<br />
solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse<br />
relationship would be used.<br />
Volume x (Mass/Volume) = Mass<br />
Mass x (Volume/Mass) = Volume<br />
Percent Mass<br />
Percent establishes a relationship as well. A percent mass states how many grams of a mixture are of<br />
a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams<br />
are of the stated element or compound. This is useful in determining mass of a desired substance in<br />
a molecule.<br />
Example 6<br />
A substance is 5% carbon by mass. If the total mass of the substance is 10.0 grams, what is the<br />
mass of carbon in the sample? How many moles of carbon are there?<br />
Solution<br />
10 g sample x 5 g carbon = 0.5 g carbon<br />
100 g sample<br />
0.5g carbon x 1 mol carbon = 0.0416 mol carbon<br />
12.0g carbon<br />
Molarity<br />
Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it<br />
is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical<br />
equations and dilutions.<br />
Pg100
Example 7<br />
How much 5M stock solution is needed to prepare 100 mL of 2M solution?<br />
Solution<br />
100 mL of dilute solution (1 L/1000 mL) (2 mol/1L solution) (1 L stock solution/5 mol solution) (1000<br />
ml stock solution/1L stock solution) = 40 mL stock solution.<br />
These ratios of molarity, density, and mass percent are useful in complex examples ahead.<br />
Determining Empirical Formulas<br />
An empirical formula can be determined through chemical stoichiometry by determining which<br />
elements are present in the molecule and in what ratio. The ratio of elements is determined by<br />
comparing the number of moles of each element present.<br />
Example 8<br />
1. Find the molar mass of the empirical formula CH2O.<br />
12.0g C + (1.00g H) * (2H) + 16.0g O = 30.0 g/mol CH2O<br />
2. Determine the molecular mass experimentally. For our compound, it is 120.0 g/mol.<br />
3. Divide the experimentally determined molecular mass by the mass of the empirical formula.<br />
(120.0 g/mol) / (30.0 g/mol) = 3.9984<br />
4. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a<br />
slight error in the experimentally determined molecular mass. If the answer is not close to a whole<br />
number, there was either an error in the calculation of the empirical formula or a large error in the<br />
determination of the molecular mass.<br />
5. Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular<br />
formula.<br />
CH2O * 4 =?<br />
C: 1 * 4 = 4<br />
H: 2 * 4 = 8<br />
O 1 * 4 = 4<br />
CH2O * 4 = C4H8O4<br />
6. Check your result by calculating the molar mass of the molecular formula and comparing it to the<br />
experimentally determined mass.<br />
molar mass of C4H8O4= 120.104 g/mol<br />
experimentally determined mass = 120.056 g/mol<br />
% error = | theoretical - experimental | / theoretical * 100%<br />
% error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100%<br />
% error = 0.040 %<br />
Pg101
Stoichiometry and balanced equations make it possible to use one piece of information to calculate<br />
another. There are countless ways stoichiometry can be used in chemistry and everyday life. Try and<br />
see if you can use what you learned to solve the following problems.<br />
Problem 1<br />
Why are the following equations not considered balanced?<br />
a. H2O(l)→H2(g)+O2(g)<br />
a)If listing each molecule it will be seen that on the on<br />
b. Zn(s)+Au + (aq) →Zn 2+ (aq) +Ag(s) the right there are 2 oxygens while the left has only 1<br />
b)The equation is not changes, it is the same on both<br />
product and rectant<br />
Problem 2<br />
Hydrochloric acid reacts with a solid chunk of aluminum to produce hydrogen gas and aluminum ions.<br />
Write the balanced chemical equation for this reaction.<br />
2Al + 6HCl ---> 3H 2 + AlCl 3<br />
Problem 3<br />
Given a 10.1M stock solution, how many mL must be added to water to produce 200 mL of 5M<br />
solution?<br />
10.1m 1m<br />
5m<br />
200mL<br />
1000mL<br />
=<br />
2020<br />
5000<br />
=.404mL<br />
Problem 4<br />
If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide and water, what is the<br />
limiting reagent and how many moles of water are produced? The unbalanced equation is provided<br />
below.<br />
CH4(g)+O2(g)→CO2(g)+H2O(l)<br />
.502gCH 4<br />
1mCH 4<br />
1mCO 2<br />
44gCO 2<br />
1.38<br />
16gCH 4<br />
1mCH 4<br />
1mCO 2<br />
Limiting reagent is O 2 with product of .37g<br />
.27gO 2<br />
32gO 2<br />
1mO 2<br />
1mO 2<br />
1mCO 2<br />
1mCO 2<br />
44gCO 2<br />
.37125<br />
Pg102
Theoretical and Actual Yields<br />
Key Terms<br />
<br />
<br />
<br />
(Excess reagent, limiting reagent)<br />
Theoretical and actual yields<br />
Percentage or actual yield<br />
Skills to Develop<br />
Use stoichiometric calculation to determine excess and limiting reagents in a chemical reaction<br />
<br />
<br />
and explain why.<br />
Calculate theoretical yields of products formed in reactions that involve limiting reagents.<br />
Evaluate percentage or actual yields from known amounts of reactants<br />
Theoretical and Actual Yields<br />
Reactants not completely used up are called excess reagents, and the reactant that completely<br />
reacts is called the limiting reagent. This concept has been illustrated for the reaction:<br />
2Na+Cl2 →2NaCl<br />
Amounts of products calculated from the complete reaction of the limiting reagent are called<br />
theoretical yields, whereas the amount actually produced of a product is the actual yield. The ratio of<br />
actual yield to theoretical yield expressed in percentage is called the percentage yield.<br />
percent yield = actual yield / theoretical yield ×100<br />
Chemical reaction equations give the ideal stoichiometric relationship among reactants and products.<br />
Thus, the theoretical yield can be calculated from reaction stoichiometry. For many chemical<br />
reactions, the actual yield is usually less than the theoretical yield, understandably due to loss in the<br />
process or inefficiency of the chemical reaction.<br />
Example 1<br />
Methyl alcohol can be produced in a high-pressure reaction<br />
CO(g) + 2H2(g) →CH3OH(l)<br />
If 6.1 metric tons of methyl alcohol is obtained from 1.2 metric tons of hydrogen reacting with excess<br />
amount of CO, estimate the theoretical and the percentage yield?<br />
Hint:<br />
To calculate the theoretical yield, consider the reaction<br />
CO(g) + 2H2(g) → CH3OH(l)<br />
28.0 + 4.0 = 32.0 (stoichiometric masses in, g, kg, or tons)<br />
1.2 tons H2 × 32.0 CH3OH = 9.6 tons CH3OH<br />
4.0 H2<br />
Thus, the theoretical yield from 1.2 metric tons (1.2x10 6 g) of hydrogen gas is 9.6 tons. The actual<br />
yield is stated in the problem, 6.1 metric tons. Thus, the percentage yield is<br />
Pg103
%yield = 6.1 tons × 100 = 64%<br />
9.6tons<br />
Discussion<br />
Due to chemical equilibrium or the mass action law, the limiting reagent may not be completely<br />
consumed. Thus, a lower yield is expected in some cases. Losses during the recovery process of the<br />
product will cause an even lower actual yield.<br />
Example 2<br />
A solution containing silver ion, Ag + , has been treated with excess of chloride ions Cl − . When dried,<br />
0.1234 g of AgCl was recovered. Assuming the percentage yield to be 98.7%, how many grams of<br />
silver ions were present in the solution?<br />
Hint:<br />
The reaction and relative masses of reagents and product are:<br />
The calculation,<br />
Ag + (aq) + Cl − (aq) → AgCl(s)<br />
107.868 + 35.453 = 143.321<br />
0.1234 g AgCl ×107.868 g Ag + =0.09287 g Ag +<br />
143.321g AgCl<br />
shows that 0.1234 g dry AgCl comes from 0.09287g Ag + ions. Since the actual yield is only 98.7%,<br />
the actual amount of Ag + ions present is therefore<br />
0.09287 g Ag + = 0.09409 g Ag +<br />
0.987<br />
Discussion<br />
One can also calculate the theoretical yield of AgCl from the percentage yield of 98.7% to be<br />
0.1234 g AgCl =0.1250 g AgCl<br />
0.987<br />
From 0.1250 g AgCl, the amount of Ag + present is also 0.09409 g.<br />
Stoichiometry - A Review<br />
Skills Taught<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
evaluate molecular weight for a given formula<br />
evaluate weight (mass) percentages of elements for a given formula<br />
evaluate amounts (in mass and mole units) produced in a chemical reaction from given<br />
conditions<br />
classify reactions by types: combination, combustion, displacement, formation, etc<br />
determine the chemical formula when weight percentages are given and then evaluate the<br />
mole percentages of elements in the formula<br />
determine the chemical formula when weight percentages are given and molecular weight is<br />
known<br />
determine the amount produced, the actual yield, and other stoichiometry quantities for a given<br />
reaction<br />
Pg104
Review Purposes<br />
<br />
<br />
<br />
<br />
<br />
To get an overall view of stoichiometry.<br />
Apply skills learned to perform quantitative chemical analysis.<br />
Apply theories and rules of chemistry to solve problems.<br />
Assess areas of strength and weakness for review purposes.<br />
Improve problem solving strategy and learning efficiency.<br />
Stoichiometry<br />
STOICHIOMETRY is the quantitative relationship of reactants and products. This unit has been<br />
divided into the following objects. A brief review is given here so that you can get a birds'-eye or<br />
overall view of stoichiometry.<br />
1.Amounts of substances<br />
Express amounts of substance in mass units of g, kg, tons, and convert them to moles,<br />
kilomoles, or millimoles.<br />
2.Chemical formulas<br />
Represent a substance with a formula that reflects its chemical composition, structure, and<br />
bonding; evaluate weight and mole percentages of elements in a substance; and determine<br />
chemical formula by elemental analysis.<br />
3.Reaction features<br />
Define some common features of chemical reactions; classify chemical reactions by common<br />
features such as combination, combustion, decomposition, displacement, and redox reactions.<br />
4.Reaction equations<br />
Express quantitative relationships using chemical reaction equations; evaluate quantities of<br />
reactants and products in a chemical reaction; and solve reaction stoichiometry problems.<br />
5.Excess and limiting reagents<br />
Define excess and limiting reagents; determine excess and limiting reagents in a reaction<br />
mixture; and determine quantities produced in a chemical reaction.<br />
6.Yields<br />
Define theoretical and actual yields due to limiting reagent; apply the concept of limiting<br />
reagent to evaluate theoretical yield; convert actual yield to percentage yields.<br />
Pg105
Use the space provided to write out the steps you take to solve different types of problems.<br />
Use any additional space for notes. These 2 pages should be full when you turn in your notebook.<br />
Steps to solve problems<br />
Mole to Mole:<br />
(1) In the equation: _H^2+_O^2 -> _H^2O , If you are given 6<br />
moLs of H^2, how many moLs of O^2 will react with it?<br />
1. Balance the equation: _2_H^2+ _1_O^2 -> _2_H^2O<br />
2. Take what you are given and convert to what you need it to<br />
be through the balanced equation<br />
3. Multiply across the bottom and top<br />
4. Match significant figures: The answer is: 3 moLs of O^2<br />
(1) For the balanced equation shown below, how many moles<br />
of H^2 will be produced by 53.5grams of Al?<br />
_2_Al+_3_H^2SO^4=>_1_Al^2(SO^4)^3+_3_H^2<br />
1. Take what you are given and convert to what you need it to be<br />
through mole:mole<br />
and mole:mass ratios<br />
2. Take the grams and figure out the molar mass of the compound, so<br />
you know how much is 1 moL<br />
Al = 27amu, therefore, 27g of Al = 1 moL Al<br />
3. Convert the given mass to moLs through the balanced equation<br />
4. Now convert to the unit that you want to end with (in this case,<br />
moLs of H^2)<br />
5. Multiply across the top and the bottom and then divide to find the<br />
amount you're looking for in moLs<br />
6. Apply significant figures<br />
The answer would be: 2.97 moLs of H^2<br />
Pg106
Moles to Mass: (*in a balanced equation, everything on both<br />
sides is equal to each other as a ratio)<br />
(1) For the balanced equation shown below, how many grams<br />
of S reacted, if 0.147moles of Al^2S^3 are produced?<br />
_2_Al+_3_S=>_1_Al^2S^3<br />
1. Take the amount you are given and convert to moLs of the<br />
desired element<br />
2. Take the moLs of the substance you've converted to, and<br />
turn those into grams<br />
3.Multiply across the top and the bottom<br />
4.Apply significant figures<br />
The answer is : 14.1 g of S<br />
(1) For the balanced equation shown below, how many grams<br />
of Al will react with 12.4 grams of Al2O3?_2_Al<br />
+_3_MnO=>_1_Al^2O^3+_3_Mn<br />
1. Take the amount you are given and convert to the moLs of<br />
the compound<br />
to the amount of the element you have<br />
2.Convert to the desired element/compound through mole:mole<br />
ratios (in this case, to moLs of Al through the balanced<br />
equation)<br />
3.Find the mass of the desired element through conversion of<br />
the mole<br />
4.Multiply across the top and the bottom, then divide.<br />
Pg107
Unit 6<br />
Chapter 13 States of Matter<br />
The students will learn what are the factors that determine and<br />
characteristics that distinguish gases liquids and solids and<br />
how substances change from one state to another.<br />
Differentiate among the four states of matter.<br />
Students will measure the physical characteristics of matter such as temperature<br />
and density.<br />
Students will compare and contrast the physical characteristics of the 4 states of<br />
matter.<br />
solid<br />
liquid<br />
gas<br />
plasma<br />
Relate temperature to the average molecular kinetic energy.<br />
Students will be able to compare and contrast the motion of particles of a sample<br />
at various temperatures.<br />
Kinetic energy<br />
Kinetic theory<br />
Temperature<br />
Describe phase transitions in terms of kinetic molecular theory.<br />
Students will be able to identify and describe phase changes.<br />
Students will be able to compare and contrast the change in particle motion<br />
for phase changes.<br />
Students will be able to interpret heating/cooling curves and phase diagrams.<br />
melting point<br />
freezing point<br />
boiling point<br />
condensation<br />
sublimation<br />
phase diagram<br />
kinetic molecular theory<br />
Pg108
Chapter 14 The Behavior of Gases<br />
The students will learn how gases respond to changes in<br />
pressure, volume, and temperature and why the ideal gas law<br />
is useful even though ideal gases do not exist.<br />
Interpret the behavior of ideal gases in terms of kinetic molecular theory.<br />
Students will be able to describe the behavior of an ideal gas.<br />
Students will participate in activities to apply the Ideal Gas Law and its<br />
component laws to predict gas behavior.<br />
Students will be able to perform temperature/pressure conversions.<br />
Compressibility<br />
Boyle’s Law<br />
Charles’s Law<br />
Gay-Lussac’s Law<br />
Combine Gas Law<br />
Ideal Gas Law<br />
Partial pressure<br />
Dalton’s Law of partial pressure<br />
Diffusion<br />
Effusion<br />
Graham’s Law of effusion<br />
Chapter 15 Water and Aqueous Systems<br />
The students will learn how the interactions between water<br />
molecules account for the unique properties of water and how<br />
aqueous solutions form.<br />
Discuss the special properties of water that contribute to Earth's suitability<br />
as an environment for life: cohesive behavior, ability to moderate<br />
temperature, expansion upon freezing, and versatility as a solvent.<br />
Students will be able to prepare a solution of known molarity<br />
Students will participate in activities to calculate molarity<br />
Surface tension<br />
Surfactant<br />
Aqueous solutionSolvent<br />
Solute<br />
Pg109
The Learning Goal for this assignment is: Differentiate among the 4 states<br />
of matter<br />
Take note over the following chapter. Use the Headings provided to organize your notes. Define and<br />
number all highlighted vocabulary (total 23 ) as well as summarize the sections. You may add pictures<br />
where needed. The pictures should be an appropriate size. Use Arial 12 for all text. This document<br />
should be 3 pages and should be saved as a pdf before you submit it into Angel.<br />
Chapter 13 States of Matter<br />
Pages 420 - 439<br />
13.1 The Nature of Gases<br />
Kinetic Theory and a Model for Gases<br />
The energy an object has because of its motion is called<br />
kinetic energy 1 . According to the kinetic theory 2 , all matter<br />
consists of tiny particles that are in constant motion. The<br />
particles in a gas are considered to be small, hard spheres with an insignificant volume. Within a gas,<br />
the particles are relatively far apart compared with the distance between particles in a liquid or solid.<br />
The motion of the particles in a gas is rapid, constant, and random. As a result, gases fill their<br />
containers regardless of the shape and volume of the containers. All collisions between particles in a<br />
gas are perfectly elastic. During an elastic collision, kinetic energy is transferred without loss from one<br />
particle to another.<br />
Gas Pressure<br />
Gas pressure 3 results from the force exerted by a gas per unit surface area of an object. An empty<br />
space with no particles and no pressure is called a vacuum 4 . The collisions of atoms and molecules<br />
in air with objects results in atmospheric pressure 5 . A barometer 6 is a device that is used to measure<br />
atmospheric pressure. The SI unit of pressure is the pascal(Pa) 7 . One standard atmosphere(atm) 8 is<br />
the pressure required to support 760 mm of mercury in a mercury barometer at 25 o C. Moving bodies<br />
exert a force when they collide with other bodies. Although a single particle in gas is a moving body,<br />
the force it exerts is extremely small. Gas pressure is the result of billions of rapidly moving particles<br />
in a gas simultaneously colliding with an object. If no particles are present, no collisions can occur.<br />
Consequently, there is no pressure.<br />
Kinetic Energy and Temperature<br />
The particles in any collection of atoms or molecules at a given temperature have a wide range of<br />
kinetic energies. Most of the particles have kinetic energies somewhere in the middle of this range.<br />
Therefore, we use average kinetic energy when discussing the kinetic energy of a collection of<br />
particles in a substance. The Kelvin temperature scale reflects the relationship between temperature<br />
and kinetic energy. The Kelvin temperature of a substance is directly proportional to the average<br />
kinetic energy of the particles of the substance. For example, the particle in helium gas at 200 K have<br />
twice the average kinetic energy as the particles in helium gas at 100K. The effects of temperature on<br />
particle motion in liquids and solids are more complex than in gases.<br />
Pg110
13.2 The Nature of Liquids<br />
A Model of Liquids<br />
These intermolecular attractions keep the particles in a liquid<br />
close together, which is why liquids have a definite volume. The<br />
interplay between the disruptive motions of particles in a liquid<br />
and the attractions among the particles determines the physical<br />
properties of liquids. Intermolecular attractions reduce the<br />
amount of space between the particles in a liquid. Thus liquids<br />
are much denser than gases.<br />
Evaporation<br />
The conversion of a liquid to a gas or vapor is called vaporization 9 . When this conversion occurs at<br />
the surface of a liquid that is not boiling, the process is called evaporation 10 . Most of the molecules in<br />
liquid don’t have enough kinetic energy to overcome the attractive forces and escape into the<br />
gaseous state. During evaporation, only those molecules with a certain minimum kinetic energy can<br />
escape from the surface of the liquid. Even some of the particles that do escape collide with<br />
molecules in the air and rebound back into the liquid. Heating the liquid increases the average kinetic<br />
energy of its particle.<br />
Vapor Pressure<br />
A measure of the force exerted by a gas above a liquid is called vapor pressure 11 . In a system at<br />
constant vapor pressure, a dynamic equilibrium exists between the vapor and the liquid. The system<br />
is in equilibrium because the rate of evaporation of liquid equals the rate of condensation of vapor. At<br />
equilibrium, the particles in the system continue to evaporate and condense, but no net charge occurs<br />
in the number of particles in the liquid or vapor. Eventually, the number of particles condensing will<br />
equal the number of particles vaporizing. The vapor pressure will then remain constant.<br />
Boiling Point<br />
The boiling point 12 (bp) is the temperature at which the vapor pressure of the liquid is just equal to the<br />
external pressure on the liquid. The normal boiling point 13 is defined as the boiling point of a liquid at a<br />
pressure of 101.3 kPa. When a liquid is heated to a temperature at which particles throughout the<br />
liquid have enough kinetic energy to vaporize, the liquid begins to boil. Heating allows a freater<br />
number of particles at the liquid’s surface to overcome the attractive forces that keep them in the<br />
liquid state. The remaining particles in the liquid move faster and faster as they absorb the added<br />
energy. Thus, the average kinetic energy of the particles in the liquid increases and the temperature<br />
of the liquid rises.<br />
Pg111
13.3 The Nature of Solids<br />
A Model of Solids<br />
The melting point 14 (mp) is the temperature at which a<br />
solid changes into a liquid. The freezing point 15 (fp) is<br />
the temperature at which a liquid changes into a solid.<br />
The general properties of solids reflect the orderly<br />
arrangement of their particles and the fixed locations of<br />
their particles. The particles in liquids are relatively free<br />
to move. The particles in solids, however, are not.<br />
These solids are dense and not easy to compress.<br />
Because the particles in solids tend to vibrate about<br />
fixed points, solids do not flow. When you heat a solid,<br />
its particles vibrate more rapidly as their kinetic energy<br />
increases.<br />
Crystal Structure and Unit Cell<br />
In a crystal 16 , the particles are arranged in an orderly, repeating, 3-dimensional pattern called a<br />
crystal lattice. The smallest group of particles within a crystal that retains the geometric shape of the<br />
crystal is known as unit cell 17 . Allotropes 18 are 2 or more different molecular forms of the same<br />
element in the same physical state. An amorphous solid 19 lacks an order internal structure. A glass 20<br />
is a transparent fusion product of inorganic substances that have cooled to a rigid state without<br />
crystallizing. The shape of a crystal reflects the arrangement of the particles within the solid. A crystal<br />
has sides, or faces.<br />
13.4 Changes of State<br />
Sublimation<br />
Ice changes directly to water vapor without melting and passing through the liquid state. The change<br />
of a substance from a solid to a vapor without passing through the liquid state is called sublimation 21 .<br />
Sublimation can occur because solids, like liquids, have a vapor pressure. Sublimation occurs in<br />
solids with vapor pressures that exceed atmospheric pressure at or near room temperature. Iodine is<br />
an example of a substance that undergoes sublimation. Sublimations can occur because solids, like<br />
liquids, have a vapor pressure. The violet-black solid changes into a purple vapor without passing<br />
through a liquid state.<br />
Phase Diagrams<br />
A phase diagram 22 gives the<br />
conditions of temperature and<br />
pressure at which a substance exists<br />
as solid, liquid, or gas (vapor). The<br />
triple point 23 describes the only set of<br />
conditions at which all 3 phases can<br />
exist in equilibrium with one another.<br />
For water, the triple point is a<br />
temperature of 0,0160 o C and a<br />
pressure of 0. 61lPa.The relationships<br />
among the solid, liquid, and vapor<br />
states of a substance in a sealed<br />
container can be represented in a<br />
single graph. The conditions of<br />
pressure and temperature at which 2<br />
phase exist in equilibrium are<br />
indicated on a phase diagram by a line separating the 2 regions representing the phases.<br />
Pg112
LG:The students will learn what are the factors that determine and<br />
characteristics that distinguish gases liquids and solids and<br />
how substances change from one state to another.<br />
Name: Bryan Vega<br />
Name: Ronny Sosa<br />
Grade:<br />
States of Matter Project<br />
You and you lab partner are going to create a study aid in the form of a game for the<br />
information in Chapter 13 States of Matter.<br />
First, each of you, independently from each other, will summarize the chapter on 3<br />
pages of a pdf which will be submitted in Angel by the end of class on Wednesday Feb<br />
22.<br />
Second, you and you lab partner will be given a game platform which you will use for<br />
your questions and answers, either Jeopardy or Kahoot.<br />
Third, you will fill in the information at the bottom of this page with your username,<br />
passwords and/or websites so that you do not forget this and I have a copy in case<br />
anything gets misplaced. This page will be submitted into Angel as a Word Document<br />
on Wednesday February 22 during class.<br />
Fourth, you will use your notes to generate the questions and answers.<br />
Finally you will give me access to your game by putting the website or Game Number<br />
on this page adding this page to your 3 pages of notes and resubmitting it in Angel as<br />
a pdf by the end of the class on Friday Feb 24.<br />
This page is due by the end of class on Wednesday February 22.<br />
This Project is due by the end of class on Friday February 24.<br />
Jeopardy (https://jeopardylabs.com)<br />
Password:<br />
Edit Link:<br />
Play Link:<br />
Kahoot (https://getkahoot.com)<br />
Username:bryanwithay<br />
Email:bryanvegalwths@gmail.com<br />
Password:boss304421<br />
Game PIN: 917416<br />
Pg113
LG:The students will learn how gases respond to changes in<br />
pressure, volume, and temperature and why the ideal gas law<br />
is useful even though ideal gases do not exist.<br />
Temperature<br />
Average Kinetic Energy<br />
Pressure<br />
Volume<br />
Gas laws are there to help predict how changing one of these variables will effect the other 2<br />
If we hold the third one constant it changes the other 2<br />
3 Gas laws-If we change 2 it affects the other one<br />
Pg114
We are hold mass constant and temperature constant the only things changing is pressure and volume<br />
As the volume increases, the pressure decreases<br />
When squeezing a balloon it decreases the volume and the pressure increases<br />
Pg115
When the volume increases, the temperature increases<br />
As the temperature increases so does the volume as long as the pressure stays the same<br />
We have to make an equality<br />
Pg116
The pressure and the temperature increase as the volume is still constant When<br />
the temperature increases so does the pressure as well<br />
This used to be used for pressure cooking<br />
If the pressure is 8 and the temperature is 8 and the pressure on the other side is<br />
4 that means the temperature as too be 4<br />
Pg117
Pg118
Boyle's Law<br />
-<br />
A gas system has initial pressure and volume of 1980 torr and 7730 ml If the<br />
pressure changes to 2.95 atm, what will the resultant volume be in ml?<br />
A gas system has an initial volume of 5140 ml with the pressure unknown.<br />
When the volume changes to 8.79 l the pressure is found to be 2070 torr.<br />
What was the initial pressure in atm? #1<br />
Pg119
Water freezes at<br />
-0 o C<br />
Kelvin at 237 o<br />
K----->C -273<br />
C----K>=+273 o<br />
Pg120
Pg121
Combined gas Law- When we put Boyle's law, Charles' law, and Gay-<br />
Lussac's law together, we come up with the combined gas law<br />
Equation<br />
Question-A closed gas system initially has volume and temperature of 2.49L and<br />
28.00C with the pressure unknown. If the same closed system has values of<br />
0.662atm, 9.75L and 178.0C, what was the initial pressure in atm?<br />
Pg122
Ideal gas laws- A law that describes the relationships between measurable<br />
properties of an ideal gas. The law states that P × V = n × (R) × T, where P is<br />
pressure, V is volume, n is the number of moles of molecules, T is the absolute<br />
temperature, and R is the gas constant (8.314 joules per degree Kelvin or 1.985<br />
calories per degree Celsius).<br />
Equation<br />
Question- A gas system has pressure, volume and moles of 1.63atm,6.15L and<br />
0.206moles, respectively. What is the temperature in K?<br />
Pg123
The Learning Goal for this assignment is: The students will learn how the interactions<br />
between water molecules account for the unique properties of water and how aqueous<br />
solutions form.<br />
Take note over the following chapter. Use the Headings provided to organize your notes. Define and<br />
number all highlighted vocabulary (total 22 ) as well as summarize and take notes over the sections.<br />
You may add pictures where needed. The pictures should be an appropriate size. Use Arial 12 for<br />
all text. This document should be 2 pages and should be saved as a pdf before you submit it into<br />
Angel.<br />
Chapter 15 Water and Aqueous Systems<br />
Pages 488 - 507<br />
15.1 Water and Its Properties<br />
Water in the Liquid State-The inward force, or pull, that tends to minimize the surface area of a<br />
liquid is called surface tension 1 . All liquids have a surface tension, but water’s surface tension is<br />
higher than most. This is why, on some surfaces, water tends to bead up rather than spread out. A<br />
surfactant 2 is any substance that interferes with the hydrogen bonding between water molecules and<br />
thereby reduces surface tension. Soaps and detergents are surfactant.<br />
Water in the Solid State- As water begins to cool, it behaves initially like a typical liquid. It<br />
contracts slightly and its density gradually increases. At 4 o C, the density of water is at its maximum of<br />
1.0000g/cm 3 . When the temperature of the water falls below 4 o C the density of water actually starts to<br />
decrease. Below 4 o C, water no longer behaves like a typical liquid. Ice, which forms at 0 o C. The<br />
structure of ice is a regular open framework of water molecules in a hexagonal arrangement. When<br />
ice melts, the framework collapses.<br />
15.2 Homogeneous Aqueous Systems<br />
Solutions- An aqueous solution 3 is water that contains dissolved substances. Even tap water that<br />
we drink is a solution that contains varying amounts of dissolved minerals and gases. In a solution,<br />
the dissolving medium is the solvent 4 . The dissolved particles in a solution are the solute 5 . A solvent<br />
dissolves the solute, and the solute becomes dispersed in the solvent. Solvents and solutes may be<br />
gases, liquids, or solids. Recall that solutions are homogeneous mixtures. They are also stable<br />
mixtures. Substances that dissolve most readily in water include ionic compounds and polar covalent<br />
compounds. The process by which the positive and negative ions of an ionic solid become<br />
surrounded by solvent molecules is called solvation 6 .<br />
124
Electrolytes and Nonelectrolytes- An electrolyte 7 is a compound that conducts an electric<br />
current when it is an aqueous solution or in the molten core. Conduction of an electric current<br />
requires ions that are mobile and, thus, able to carry charges through a liquid. All ionic compounds<br />
are electrolytes because they dissociate into ions. Barium sulfate is an ionic compound that cannot<br />
conduct an electric current in aqueous solution because it is insoluble, but it can conduct in the<br />
molten core. A nonelectrolyte 8 is a compound that does not conduct an electric current in either an<br />
aqueous solution or the molten state. Many molecular compounds are nonelectrolytes because they<br />
are not composed of ions. Most compounds of carbon, such as table sugar and the alcohol in rubbing<br />
alcohol, are nonelectrolytes. In a solution that contains a strong electrolyte 9 , all or nearly all of the<br />
solute exists as ions. The ions move in solution and conduct an electric current. Most soluble salts,<br />
inorganic acids, and inorganic bases are strong electrolytes. A weak electrolyte 10 conducts an<br />
electric current poorly because only a fraction of the solute in the solution exists as ions. Organic<br />
acids and bases are also examples of weak electrolytes.<br />
Hydrates- The water contained in a crystal is called the water of hydration 11 or water of<br />
crystallization. A compound that contains water of hydration is called a hydrate 12 . A substance that is<br />
anhydrous 13 does not contain water. If a hydrate has a pressure higher than the pressure of water<br />
vapor in the air, the hydrate will lose its water of hydration or effloresce 14 . These hydrates and other<br />
compounds that remove moisture from air called hygroscopic 15 . A desiccant 16 is a substance used to<br />
absorb moisture from the air and create a dry atmosphere. These compounds are deliquescent 17 ,<br />
which means that they remove sufficient water from air to dissolve completely and form solutions.<br />
The forces holding the water molecules in hydrates are not very strong, so the water is easily lost and<br />
regained.<br />
15.3 Heterogeneous Aqueous Systems<br />
Suspensions- In contrast, heterogeneous mixtures are not solutions. A suspension 18 is a mixture<br />
from which particles settle out upon standing. A suspension differs from a solution because the<br />
particles of a suspension are much larger and do not stay suspended indefinitely. Suspensions are<br />
heterogeneous because at least two substances can be clearly identified.<br />
Colloids- A colloid 19 is a heterogeneous mixture containing particles that range in size from 1 nm to<br />
1000nm. The particles are spread, or dispersed, throughout the dispersion medium, which can be<br />
solid, liquid, or gas. The scattering of visible light by colloidal particles is called the Tyndall effect 20 .<br />
The chaotic movement of colloidal particles, which was first observed by the Scottish botanist Robert<br />
Brown (1773-1858), is called Brownian motion 21 . An emulsion 22 is a colloidal dispersion of a liquid in<br />
a liquid. Colloids have particles smaller than those in suspensions and larger than lose in solutions.<br />
Many colloids are cloudy or milky in appearance, like suspensions, when they are concentrated.<br />
Colloids may look clear or almost clear, like solutions, when they are diluted. The important difference<br />
between colloids and solutions and suspensions is in the size of the particles. The first substances to<br />
be identified as a colloid was glue. Other substances that were also colloids were paint, aerosol<br />
spray, and smoke.<br />
125
126<br />
Unit 7<br />
Chapter 16 Solutions<br />
The students will learn what properties are used to describe<br />
the nature of solutions and how to quantify the concentration<br />
of a solution.<br />
Chapter 17 Thermochemistry<br />
The student will learn how energy is converted in a chemical<br />
or physical process and how to determine the amount of<br />
energy is absorbed or released in that process.<br />
Differentiate among the various forms of energy and recognize that they can<br />
be transformed from one form to others.<br />
Students will participate in activities to investigate and describe the<br />
transformation of energy from one form to another (i.e. batteries, food, fuels,<br />
etc.)<br />
Explore the Law of Conservation of Energy by differentiating among open,<br />
closed, and isolated systems and explain that the total energy in an isolated<br />
system is a conserved quantity.<br />
<br />
Students will be able to calculate various energy changes:<br />
o q = mc∆t<br />
o ∆Hfus<br />
o ∆Hmelt<br />
Thermochemistry<br />
Heat<br />
System<br />
Surrounding<br />
Law of conservation of energy<br />
Bond Making is exothermic<br />
Bond Breaking is endothermic<br />
Heat capacity<br />
Specific heat<br />
Calorimetry<br />
Enthalpy<br />
Thermochemical equation<br />
Molar heat of (fusion, solidification,<br />
vaporization, condensation, solution)<br />
Distinguish between endothermic and exothermic chemical processes.<br />
Students will be able to recognize exothermic and endothermic reactions through<br />
experimentation.<br />
Students will participate in activities (Pasco) to create exothermic and<br />
endothermic graphs.<br />
Endothermic<br />
Exothermic
Create and interpret potential energy diagrams, for example: chemical<br />
reactions, orbits around a central body, motion of a pendulum<br />
Students will participate in activities (Pasco) to create exothermic and<br />
endothermic graphs.<br />
Students will be able to interpret exothermic and endothermic reaction graphs.<br />
Potential energy diagram<br />
Thermochemical equations<br />
Chapter 18 Reaction Rates and Equilibrium<br />
The student will learn how the rate of a chemical reaction can<br />
be controlled, what the role of energy is and why some<br />
reactions occur naturally and others do not.<br />
Explain how various factors, such as concentration, temperature, and<br />
presence of a catalyst affect the rate of a chemical reaction.<br />
Students will be able to describe how each factor may affect the rate of a<br />
chemical reaction.<br />
Students will be able to compare the relative effect of each factor on the rate of a<br />
chemical reaction.<br />
Rate<br />
Collision theory<br />
Activation energy<br />
Catalyst<br />
Activated complex<br />
Inhibitor<br />
Explain the concept of dynamic equilibrium in terms of reversible processes<br />
occurring at the same rates.<br />
Students will be able to describe a system in dynamic equilibrium.<br />
Students will be able to describe how factors may affect the equilibrium of a<br />
reaction.<br />
Reversible reaction<br />
Chemical equilibrium<br />
Le Chatelier principle<br />
Explain entropy’s role in determining the efficiency of processes that convert<br />
energy to work.<br />
Students will be able to describe the change in entropy of a reaction.<br />
Students will be able to determine if a reaction is spontaneous<br />
Entropy<br />
Law of disorder<br />
Spontaneous/nonspontaneous reaction<br />
127
The Learning Goal for this assignment is:<br />
The students will learn what properties are used to describe<br />
the nature of solutions and how to quantify the concentration<br />
of a solution.<br />
Defining Concentration<br />
Measures of Concentration<br />
Concentration is defined as the amount of dissolved solute in a given amount of solvent or<br />
solution. There are several terms that describe concentration. Some of these terms are relative;<br />
that is, they can be used only to compare the concentration of one solution to another. Dilute and<br />
concentrated are two such terms. A dilute solution contains less dissolved solute than a<br />
concentrated solution (in equal volumes of solution).<br />
The terms saturated, unsaturated, and supersaturated are terms that describe concentration more<br />
precisely.<br />
<br />
<br />
<br />
<br />
<br />
<br />
Saturated: The maximum amount of solute is dissolved in a given amount of solvent at a<br />
particular temperature. Such solutions are stable.<br />
Unsaturated: Less than the maximum amount of solute is dissolved in a given amount of<br />
solvent at a particular temperature. Such solutions are stable.<br />
Supersaturated: More than the maximum amount of solute is dissolved in a given amount of<br />
solvent at a particular temperature. Such solutions are unstable.<br />
Look at the solubility curve shown below:<br />
128<br />
The solubility of NaNO3 is 86.0 g/100 mL H2O (at 20 °C). If you<br />
prepare a solution of 50.0 g NaNO3 dissolved in 100 mL H2O, an unsaturated solution results<br />
(Point A on the graph).<br />
continue adding NaNO to the solution until 86.0 g are dissolved, a saturated solution results<br />
(Point B).<br />
heat the solution to 50 °C, 113 grams NaNO3 can be dissolved (Point C). When the solution<br />
cools back down to 20 °C, it will be supersaturated (Point D).<br />
Quantifying Concentration<br />
To describe the concentration of a solution even more precisely, various measures of concentration<br />
can be used. Some of the ways concentration can be quantified include calculating the<br />
Mass of solute per solution mass (expressed as a percent or parts per million)<br />
Moles of solute per kilogram solvent (molality)<br />
Mass of solute per liter of solution (grams/liter)<br />
Moles of solute per liter of solution (molarity)
Part 1: Mass Percent<br />
Mass percent (also called percent by mass, weight percent, or percent by weight) compares the<br />
mass of the solute to the entire mass of the solution.<br />
Notes:<br />
Mass percentage is one way of representing the concentration of an element in a compound or a component in a mixture.<br />
Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%.<br />
Part 2: Parts per Million<br />
Parts per million (ppm) is another measure of concentration. It is similar to mass percent. But<br />
mass percent indicates the number of grams of solute per 100 g solution. Parts per million<br />
indicates the number of grams of solute per 1,000,000 g solution. This measure of concentration is<br />
often used to express the concentrations of very dilute solutions.<br />
Notes:<br />
This is a way of expressing very dilute concentrations of substances. Just as per cent means out of a hundred,<br />
so parts per million or ppm means out of a million.<br />
Part 3: Molality<br />
Molality (m) is the ratio of the moles of solute to the kilograms of solvent. Note: this is the first<br />
measure of concentration that is concerned with the mass of the solvent, not the mass of the<br />
solution as a whole.<br />
Notes:<br />
The molal unit is not used nearly as frequently as the molar unit. A molality is the number of moles of solute<br />
dissolved in one kilogram of solvent.<br />
Part 4: Grams per Liter<br />
To express the concentration of a solution in grams per liter, you must know the mass of the solute<br />
and the volume of the solution, not just the volume of the solvent.<br />
Notes:<br />
A gram per litre or gram per liter (g/L or g/l) is a unit of measurement of mass concentration that shows how many<br />
grams of a certain substance are present in one litre of a usually liquid or gaseous mixture.<br />
Suppose you wanted to know what the concentration would be before making the solution. Could that<br />
be done? In order to relate the volume or mass of solvent to the volume of solution, you would have<br />
to know the density of the solution. You will see how solution density can be used to calculate<br />
molarity in the next section.<br />
129
Part 5: Molarity<br />
Molarity (M) is the most common measure of concentration. The concentration of most solutions<br />
you see in the lab are expressed in terms of molarity. Just like g/L, molarity calculations require<br />
that you know either the volume or density of the resulting solution.<br />
Notes:<br />
Molarity is the concentration of a solution expressed as the number of moles of solute per litre of solution.<br />
To get the molarity, you divide the moles of solute by the litres of solution.<br />
Using Density to Calculate Molarity<br />
If the volume of the resulting solution is not known, molarity is calculated as follows:<br />
Convert the volume of solvent to grams. (The simulation does this step for you.)<br />
Determine the total mass of the solution (mass of solute + mass of solvent).<br />
Convert the solution mass to volume in milliliters, using its density (volume = mass / density).<br />
Convert the solution volume to liters (divide by 1000).<br />
Convert solute grams to moles.<br />
Calculate the molarity (moles solute / L solution).<br />
Reinforcing What You've Learned<br />
1. 35.0 g potassium dichromate are dissolved in 354.0 g distilled water. What is the concentration<br />
of the resulting solution, expressed as percent by mass? 9.00%<br />
2. A chemist discovers that 2587.0 g distilled water are contaminated with 13.0 g NaNO3. What is<br />
the concentration of NaNO3, expressed in ppm? 5.00 X 10 3<br />
3. 175.00 g H2SO4 are added to 61.49 g distilled H2O. What is the molality of the resulting solution?<br />
29.04m<br />
4. 225 g NaBr are dissolved in 525 g distilled water. If the volume of the resulting solution is<br />
584 mL, what is its concentration when expressed in g/L? 385g/L<br />
5. 470.0 g Na2CO3 are dissolved in 4230.0 g distilled water. What is the molarity of the resulting<br />
solution (D = 1.1029 g/mL)? 1.1561 M<br />
Applying What You've Learned<br />
6. Think about the equations for mass percent and parts/million. Consider how you might convert<br />
between these two measures of concentration.<br />
Mass percent = ppm ÷ 10,000 ppm = mass percent X .0001<br />
7. Think about the equations for grams/liter and molarity. Consider how you might convert<br />
between these two measures of concentration.<br />
grams / L = molarity X molar mass<br />
molarity = grams / L ÷ molar mass<br />
130
Glossary<br />
Concentration: Amount of dissolved solute in a given amount of solvent or solution.<br />
Density: Amount of matter per unit volume. Density is calculated by dividing an object's mass by<br />
its volume.<br />
Mass: Amount of matter an object contains or, more scientifically, the measure of an object's<br />
resistance to changes in motion. The SI (Systéme International) unit for mass is the kilogram. In<br />
the lab, mass is often measured in grams.<br />
Mass percent: Also referred to as percent by mass and, occasionally, weight percent or percent<br />
by weight. Mathematically, mass % = (mass of solute / mass of entire solution) x 100.<br />
Molar mass: Mass of a compound, calculated by adding up the individual masses for its component<br />
atoms, which are obtained from the periodic table of the elements. Molar mass is expressed in<br />
grams/mole and is sometimes referred to as molecular mass (for molecular compounds) or formula<br />
mass (for ionic compounds).<br />
Molality (m): Moles of solute per kilogram of solvent. Mathematically, m = moles of solute / kilogram of<br />
solvent.<br />
Molarity (M): Moles of solute per liter of solution. Mathematically, M = moles of solute / liter of<br />
solution.<br />
Mole (mol): Counting unit used to express the large numbers of particles, such as atoms or<br />
molecules, that are involved in chemical processes. One mole of particles contains 6.02 x 10<br />
particles. The mass of one mole of an element, in grams, is equivalent to the atomic mass for that<br />
element, as indicated on the periodic table.<br />
Parts per million (ppm): Measure of concentration often used for dilute solutions. Mathematically,<br />
ppm = (mass of solute / mass of solution) x 10.<br />
Solubility: Measure of the maximum amount of solute that can be dissolved in a given amount of<br />
solvent at a given temperature, forming a stable solution.<br />
Solute: Dissolved substance in a solution. The solute is generally the solution component present<br />
in the lesser amount.<br />
Solution: Homogeneous mixture in which one substance has been dissolved in another.<br />
Solvent: Substance in which a solute is dissolved to form a solution. The solvent is generally the<br />
solution component present in the greater amount.<br />
Volume: Amount of space an object occupies. The SI (Systéme International) unit for volume is<br />
the cubic meter. In the lab, volume is often measured in cubic centimeters, milliliters, or liters.<br />
SAS Curriculum Pathways VLab #866<br />
131
The Learning Goal for this assignment is:<br />
The student will learn how energy is converted in a chemical or physical process and how to<br />
determine the amount of energy is absorbed or released in that process.<br />
The System and the Surroundings in Chemistry<br />
Thermochemistry<br />
The system is the part of the universe we wish to focus our attention on. In the world of chemistry, the<br />
system is the chemical reaction. For example:<br />
2H2 + O2 ---> 2H2O<br />
The system consists of those molecules which are reacting.<br />
The surroundings are everything else; the rest of the universe. For example, say the above reaction is<br />
happening in gas phase; then the walls of the container are part of the surroundings.<br />
There are two important issues:<br />
1. a great majority of our studies will focus on the change in the amount of energy, not the<br />
absolute amount of energy in the system or the surroundings.<br />
2. regarding the direction of energy flow, we have a "sign convention."<br />
Two possibilities exist concerning the flow of energy between system and surroundings:<br />
1. The system can have energy added to it, which increases its amount and lessens the energy<br />
amount in the surroundings.<br />
2. The system can have energy removed from it, thereby lowering its amount and increasing the<br />
amount in the surroundings.<br />
We will signify an increase in energy with a positive sign and a loss of energy with a negative sign.<br />
Also, we will take the point-of-view from the system. Consequently:<br />
1. When energy (heat or work) flow out of the system, the system decreases in its amount. This<br />
is assigned a negative sign and is called exothermic.<br />
2. When energy (heat or work) flows into the system, the system increases its energy amount.<br />
This is assigned a positive sign and is called endothermic.<br />
We do not discuss chemical reactions from the surrounding's point-of-view. Only from the system's.<br />
Notes:<br />
When a reaction occurs, it may either give out energy, in the form of heat, or take in energy.<br />
The amount of energy that is given out or taken in can be measured as the Enthalpy Change of a reaction.<br />
132
Specific Heat<br />
Here is the definition of specific heat:<br />
the amount of heat necessary for 1.00 gram of a substance to change 1.00 °C<br />
Note the two important factors:<br />
1. It's 1.00 gram of a substance<br />
2. and it changes 1.00 °C<br />
Keep in mind the fact that this is a very specific value. It is only for one gram going one degree. The<br />
specific heat is an important part of energy calculations since it tells you how much energy is needed<br />
to move each gram of the substance one degree.<br />
Every substance has its own specific heat and each phase has its own distinct value. In fact, the<br />
specific heat value of a substance changes from degree to degree, but we will ignore that.<br />
The units are often Joules per gram-degree Celsius (J/g*°C). Sometimes the unit J/kg K is also used.<br />
This last unit is technically the most correct unit to use, but since the first one is quite common, you<br />
will need to know both.<br />
I will ignore calorie-based units almost entirely.<br />
Here are the specific heat values for water:<br />
Phase J g¯1 °C¯1 J kg¯1<br />
K¯1<br />
Gas 2.02 2.02 x 10 3<br />
Liquid 4.184 4.184 x 10 3<br />
Solid 2.06 2.06 x 10 3<br />
Notice that one set of values is simply 1000 times bigger than the other. That's to offset the influence<br />
of going from grams to kilograms in the denominator of the unit.<br />
Notice that the change from Celsius to Kelvin does not affect the value. That is because the specific<br />
heat is measured on the basis of one degree. In both scales (Celsius and Kelvin) the jump from one<br />
degree to the next are the same "distance." Sometimes a student will think that 273 must be involved<br />
somewhere. Not in this case.<br />
Specific heat values can be looked up in reference books. Typically, in the classroom, you will not be<br />
asked to memorize any specific heat values. However, you may be asked to memorize the values for<br />
the three phases of water.<br />
As you go about the Internet, you will find different values cited for specific heats of a given<br />
substance. For example, I have seen 4.186 and 4.187 used in place of 4.184 for liquid water. None of<br />
the values are wrong, it's just that specific heat values literally change from degree to degree. What<br />
happens is that an author will settle on one particular value and use it. Often, the one particular value<br />
used is what the author used as a student.<br />
Hence, 4.184.<br />
133
The Time-Temperature Graph<br />
We are going to heat a container that has 72.0 grams of ice (no liquid water yet!) in it. To make the<br />
illustration simple, please consider that 100% of the heat applied goes into the water. There is no loss<br />
of heat into heating the container and no heat is lost to the air.<br />
Let us suppose the ice starts at -10.0 °C and that the pressure is always one atmosphere. We will<br />
end the example with steam at 120.0 °C.<br />
There are five major steps to discuss in turn before this problem is completely solved. Here they are:<br />
1. the ice rises in temperature from -10.0 to 0.00 °C.<br />
2. the ice melts at 0.00 °C.<br />
3. the liquid water then rises in temperature from zero to 100.0 °C.<br />
4. the liquid water then boils at 100.0 °C.<br />
5. the steam then rises in temperature from 100.0 to 120.0 °C<br />
Each one of these steps will have a calculation associated with it. WARNING: many homework and<br />
test questions can be written which use less than the five steps. For example, suppose the water in<br />
the problem above started at 10.0 °C. Then, only steps 3, 4, and 5 would be required for solution.<br />
To the right is the type of graph which is typically used to<br />
show this process over time.<br />
You can figure out that the five numbered sections on the<br />
graph relate to the five numbered parts of the list just above<br />
the graph.<br />
Also, note that numbers 2 and 4 are phases changes: solid<br />
to liquid in #2 and liquid to gas in #4.<br />
Q=mcΔT<br />
where ΔT is (Tf – Ti)<br />
Here are some symbols that will be used, A LOT!!<br />
Δt = the change in temperature from start to finish in degrees Celsius (°C)<br />
m = mass of substance in grams<br />
c = the specific heat. Its unit is Joules per gram X degree Celsius (J / g °C is one way to write<br />
the unit; J g¯1 °C¯1 is another)<br />
q = the amount of heat involved, measured in Joules or kilojoules (symbols = J and kJ)<br />
mol = moles of substance.<br />
ΔH is the symbol for the molar heat of fusion and ΔH is the symbol for the molar heat of<br />
vaporization.<br />
We will also require the molar mass of the substance. In this example it is water, so the molar mass is<br />
18.0 g/mol.<br />
Notes:<br />
When a system adjusts due to a temperature change, there are no sudden changes in<br />
concentration of any species, so there are no vertical lines on the graph.<br />
134
Step One: solid ice rises in temperature<br />
As we apply heat, the ice will rise in temperature until it<br />
arrives at its normal melting point of zero Celsius.<br />
Once it arrives at zero, the Δt equals 10.0 °C.<br />
Here is an important point: THE ICE HAS NOT MELTED<br />
YET.<br />
At the end of this step we have SOLID ice at zero<br />
degrees. It has not melted yet. That's an important point.<br />
Each gram of water requires a constant amount of energy<br />
to go up each degree Celsius. This amount of energy is<br />
called specific heat and has the symbol c.<br />
72.0 grams of ice (no liquid water yet!) has changed 10.0 °C. We need to calculate the energy<br />
needed to do this.<br />
This summarizes the information needed:<br />
Δt = 10 °C<br />
The mass = 72.0 g<br />
c = 2.06 Joules per gram-degree Celsius<br />
The calculation needed, using words & symbols is:<br />
q = (mass) (Δt) (c)<br />
Why is this equation the way it is?<br />
Think about one gram going one degree. The ice needs 2.06 J for that. Now go the second degree.<br />
Another 2.06 J. Go the third degree and use another 2.06 J. So one gram going 10 degrees needs<br />
2.06 x 10 = 20.6 J. Now we have 72 grams, so gram #2 also needs 20.6, gram #3 needs 20.6 and so<br />
on until 72 grams.<br />
With the numbers in place, we have:<br />
q = (72.0 g) (10 °C) (2.06 J/g °C)<br />
So we calculate and get 1483.2 J. We won't bother to round off right now since there are four more<br />
calculations to go. Maybe you can see that we will have to do five calculations and then sum them all<br />
up.<br />
One warning before going on: three of the calculations will yield J as the unit on the answer and two<br />
will give kJ. When you add the five values together, you MUST have them all be the same unit.<br />
In the context of this problem, kJ is the preferred unit. You might want to think about what 1483.2 J is<br />
in kJ.<br />
Notes:<br />
As we apply heat, the ice will rise in temperature until it arrives at its normal melting point of zero<br />
Celsius.Each gram of water requires a constant amount of energy to go up each degree Celsius. This<br />
amount of energy is called specific heat and has the symbol c.<br />
135
Step Two: solid ice melts<br />
Now, we continue to add energy and the ice begins to<br />
melt.<br />
However, the temperature DOES NOT CHANGE. It<br />
remains at zero during the time the ice melts.<br />
Each mole of water will require a constant amount of<br />
energy to melt. That amount is named the molar heat of<br />
fusion and its symbol is ΔHf. The molar heat of fusion is<br />
the energy required to melt one mole of a substance at its<br />
normal melting point. One mole of solid water, one mole<br />
of solid benzene, one mole of solid lead. It does not<br />
matter. Each substance has its own value.<br />
During this time, the energy is being used to overcome water molecules' attraction for each other,<br />
destroying the three-dimensional structure of the ice.<br />
The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion<br />
between calories and Joules is 4.184 J = 1.000 cal.<br />
Sometimes you also see this number expressed "per gram" rather than "per mole." For example,<br />
water's molar heat of fusion is 6.02 kJ/mol. Expressed per gram, it is 334.16 J/g.<br />
Typically, the term "heat of fusion" is used with the "per gram" value.<br />
72.0 grams of solid water is 0.0 °C. It is going to melt AND stay at zero degrees. This is an important<br />
point. While the ice melts, its temperature will remain the same. We need to calculate the energy<br />
needed to do this.<br />
This summarizes the information needed:<br />
ΔHf = 6.02 kJ/mol<br />
The mass = 72.0 g<br />
The molar mass of H2O = 18.0 gram/mol<br />
The calculation needed, using words & symbols is:<br />
q = (moles of water) (ΔHf)<br />
We can rewrite the moles of water portion and make the equation like this:<br />
q = (grams water / molar mass of water) (ΔHf)<br />
Why is this equation the way it is?<br />
Think about one mole of ice. That amount of ice (one mole or 18.0 grams) needs 6.02 kilojoules of<br />
energy to melt. Each mole of ice needs 6.02 kilojoules. So the (grams water / molar mass of water) in<br />
the above equation calculates the amount of moles.<br />
With the numbers in place, we have:<br />
q = (72.0 g / 18.0 g mol¯1 ) (6.02 kJ / mol)<br />
So we calculate and get 24.08 kJ. We won't bother to round off right now since there are three more<br />
calculations to go. We're doing the second step now. When all five are done, we'll sum them all up.<br />
136
Step Three: liquid water rises in temperature<br />
Once the ice is totally melted, the temperature can now<br />
begin to rise again.<br />
It continues to go up until it reaches its normal boiling<br />
point of 100.0 °C.<br />
Since the temperature went from zero to 100, the Δt is<br />
100.<br />
Here is an important point: THE LIQUID HAS NOT<br />
BOILED YET.<br />
At the end of this step we have liquid water at 100 degrees. It has not turned to steam yet.<br />
Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount<br />
of energy is called specific heat and has the symbol c. There will be a different value needed,<br />
depending on the substance being in the solid, liquid or gas phase.<br />
72.0 grams of liquid water is 0.0 °C. It is going to warm up to 100.0 °C, but at that temperature, the<br />
water WILL NOT BOIL. We need to calculate the energy needed to do this.<br />
This summarizes the information needed:<br />
Δt = 100.0 °C (100.0 °C – 0.0 °C)<br />
The mass = 72.0 g<br />
c = 4.184 Joules per gram-degree Celsius<br />
The calculation needed, using words & symbols is:<br />
q = (mass) (Δt) (c)<br />
Why is this equation the way it is?<br />
Think about one gram going one degree. The liquid water needs 4.184 J for that. Now go the second<br />
degree. Another 4.184 J. Go the third degree and use another 4.184 J. So one gram going 100<br />
degrees needs 4.184 x 100 = 418.4 J. Now we have 72 grams, so gram #2 also needs 418.4, gram<br />
#3 needs 418.4 and so on until 72 grams.<br />
With the numbers in place, we have:<br />
q = (72.0 g) (100.0 °C) (4.184 J/g °C)<br />
So we calculate and get 30124.8 J. We won't bother to round off right now since there are two more<br />
calculations to go. We will have to do five calculations and then sum them all up.<br />
Notes:<br />
THE LIQUID HAS NOT BOILED YET.Each gram of water requires a constant amount of energy to go up<br />
each degree Celsius. This amount of energy is called specific heat and has the symbol c. There will be a<br />
different value needed, depending on the substance being in the solid, liquid or gas phase.<br />
137
Step Four: liquid water boils<br />
Now, we continue to add energy and the water begins to<br />
boil.<br />
However, the temperature DOES NOT CHANGE. It<br />
remains at 100 during the time the water boils.<br />
Each mole of water will require a constant amount of<br />
energy to boil. That amount is named the molar heat of<br />
vaporization and its symbol is ΔH. The molar heat of<br />
vaporization is the energy required to boil one mole of a<br />
substance at its normal boiling point. One mole of liquid water, one mole of liquid benzene, one mole<br />
of liquid lead. It does not matter. Each substance has its own value.<br />
During this time, the energy is being used to overcome water molecules' attraction for each other,<br />
allowing them to move from close together (liquid) to quite far apart (the gas state).<br />
The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion<br />
between calories and Joules is 4.184 J = 1.000 cal.<br />
Typically, the term "heat of vaporization" is used with the "per gram" value.<br />
72.0 grams of liquid water is at 100.0 °C. It is going to boil AND stay at 100 degrees. This is an<br />
important point. While the water boils, its temperature will remain the same. We need to calculate the<br />
energy needed to do this.<br />
This summarizes the information needed:<br />
ΔH = 40.7 kJ/mol<br />
The mass = 72.0 g<br />
The molar mass of H2O = 18.0 gram/mol<br />
The calculation needed, using words & symbols is:<br />
q = (moles of water) (ΔH)<br />
We can rewrite the moles of water portion and make the equation like this:<br />
q = (grams water / molar mass of water) (ΔH)<br />
Why is this equation the way it is?<br />
Think about one mole of liquid water. That amount of water (one mole or 18.0 grams) needs 40.7<br />
kilojoules of energy to boil. Each mole of liquid water needs 40.7 kilojoules to boil. So the (grams<br />
water / molar mass of water) in the above equation calculates the amount of moles.<br />
With the numbers in place, we have:<br />
q = (72.0 g / 18.0 g mol¯1 ) (40.7 kJ / mol)<br />
So we calculate and get 162.8 kJ. We won't bother to round off right now since there is one more<br />
calculation to go. We're doing the fourth step now. When all five are done, we'll sum them all up.<br />
138
Step Five: steam rises in temperature<br />
Once the water is completely changed to steam, the<br />
temperature can now begin to rise again.<br />
It continues to go up until we stop adding energy. In this<br />
case, let the temperature rise to 120 °C.<br />
Since the temperature went from 100 °C to 120°C, the Δt<br />
is 20°C.<br />
Each gram of water requires a constant amount of energy<br />
to go up each degree Celsius. This amount of energy is called specific heat and has the symbol c.<br />
There will be a different value needed, depending on the substance being in the solid, liquid or gas<br />
phase.<br />
72.0 grams of steam is 100.0 °C. It is going to warm up to 120.0 °C. We need to calculate the energy<br />
needed to do this.<br />
This summarizes the information needed:<br />
Δt = 20 °C<br />
The mass = 72.0 g<br />
c = 2.02 Joules per gram-degree Celsius<br />
The calculation needed, using words & symbols is:<br />
q = (mass) (Δt) (c)<br />
Why is this equation the way it is?<br />
Think about one gram going one degree. The liquid water needs 2.02 J for that. Now go the second<br />
degree. Another 2.02 J. Go the third degree and use another 2.02 J. So one gram going 20 degress<br />
needs 2.02 x 20 = 44 J. Now we have 72 grams, so gram #2 also needs 44, gram #3 needs 44 and<br />
so on until 72 grams.<br />
I hope that helped.<br />
With the numbers in place, we have:<br />
q = (72.0 g) (20 °C) (2.02 J/g °C)<br />
So we calculate and get 2908.8 J. We won't bother to round off right now since we still need to sum<br />
up all five values.<br />
Notes:<br />
Once the water is completely changed to steam, the temperature can now begin to rise again.<br />
Each gram of water requires a constant amount of energy to go up each degree Celsius. This<br />
amount of energy is called specific heat and has the symbol c. There will be a different value<br />
needed, depending on the substance being in the solid, liquid or gas phase.<br />
139
The following table summarizes the five steps and their results. Each step number is a link back to<br />
the explanation of the calculation.<br />
Converting to kJ gives us this:<br />
1.4832 kJ<br />
24.08 kJ<br />
30.1248 kJ<br />
162.8 kJ<br />
2.9088 kJ<br />
Step q 72.0 g of H2O<br />
1 1483.2 J Δt = 10 (solid)<br />
2 24.08 kJ melting<br />
3 30124.8 J Δt = 100 (liquid)<br />
4 162.8 kJ boiling<br />
5 2908.8 J Δt = 20 (gas)<br />
Summing up gives 221.3968 kJ and proper significant digits gives us 221.4 kJ for the answer.<br />
Notice how all units were converted to kJ before continuing on. Joules is a perfectly fine unit; it's just<br />
that 221,396.8 J is an awkward number to work with. Usually Joules is used for values under 1000,<br />
otherwise kJ is used.<br />
By the way, on other sites you may see kj used for kilojoules. I've also seen Kj used. Both of these<br />
are wrong symbols. kJ is the only correct symbol.<br />
Enthalpy<br />
When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal<br />
to the change in enthalpy. Enthalpy (H) is the sum of the internal energy (U) and the product of<br />
pressure and volume (PV) given by the equation:<br />
H=U+PV<br />
When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal<br />
to the change in enthalpy. Enthalpy is a state function which depends entirely on the state<br />
functions T, P and U. Enthalpy is usually expressed as the change in enthalpy (ΔH) for a process<br />
between initial and final states:<br />
ΔH=ΔU+ΔPVΔ<br />
If temperature and pressure remain constant through the process and the work is limited to pressurevolume<br />
work, then the enthalpy change is given by the equation:<br />
ΔH=ΔU+PΔV<br />
Also at constant pressure the heat flow (q) for the process is equal to the change in enthalpy defined<br />
by the equation:<br />
ΔH=q<br />
By looking at whether q is exothermic or endothermic we can determine a relationship between ΔH<br />
and q. If the reaction absorbs heat it is endothermic meaning the reaction consumes heat from the<br />
surroundings so q>0 (positive). Therefore, at constant temperature and pressure, by the equation<br />
above, if q is positive then ΔH is also positive. And the same goes for if the reaction releases heat,<br />
140
then it is exothermic, meaning the system gives off heat to its surroundings, so q
The Learning Goal for this section is:<br />
The student will learn how the rate of a chemical reaction can be controlled, what the role of energy is<br />
and why some reactions occur naturally and others do not.<br />
You are going to answer these 15 questions first in the order they are given to you. This will be a<br />
quiz grade. You are then going to explain the chemical concepts that are used in each question. You<br />
can do this in the order given or group them by concepts. You can use your book or information<br />
found on the internet but all information must be written in your own word. The font needs to be Arial<br />
12. This is due on Monday April 24 by midnight in the drop box. This document should be a 6-<br />
page pdf.<br />
1.B 6.B 11.A<br />
2.C 7.B 12.D<br />
3.C 8.D 13.C<br />
4.D 9.A 14.D<br />
5.A 10.B 15.D<br />
1 This graph represents the change in energy for two laboratory trials of the same reaction.<br />
Which factor could explain the energy difference between the trials?<br />
A Heat was added to trial #2.<br />
B A catalyst was added to trial #2.<br />
C Trial #1 was stirred.<br />
D Trial #1 was cooled.<br />
Explanation: IN the first trial, the amount of energy required for the reaction to take place is more than<br />
the amount in the second trial. A catalyst reduces the activation energy for the reaction to take place.<br />
Collisions only result in a reaction if the particles collide with a certain minimum energy called the<br />
activation energy for the reaction. To increase the rate of a reaction you need to increase the number<br />
of successful collisions. One possible way of doing this is to provide an alternative way for the<br />
reaction to happen which has a lower activation energy.<br />
2 Consider this balanced chemical equation:<br />
Which will increase the rate of the reaction?<br />
A increasing pressure on the reaction<br />
B decreasing concentration of the reactants<br />
C adding a catalyst to the reaction<br />
D decreasing the temperature of the reaction<br />
142<br />
2H2O2 (aq) → 2H2O(l) + O2 (g)
Explanation: With the addition of the catalyst it will allow the amount of energy required to be less<br />
than what it was originally was. Resulting in an increase rate of the reaction.<br />
3 For the reaction<br />
A + (aq) + B — (aq) → AB (s)<br />
Increasing the temperature increases the rate of the reaction. Which is the best explanation for this<br />
happening?<br />
A The pressure increases, which in turn increases the production of products.<br />
B The concentration of reactants increases with an increase in temperature.<br />
C The average kinetic energy increases, so the likelihood of more effective collisions between ions<br />
increases.<br />
D Systems are more stable at high temperatures.<br />
Explanation: When you increase the surface area, more of the substances is exposed to particles<br />
which in turn will increase the collision of particles that cause the reaction. Therefore, this will result in<br />
an increase in the speed of some reactions. Particles can only react when they collide. If you heat a<br />
substance, the particles move faster and so collide more frequently. That will speed up the rate of<br />
reaction. Collisions only result in a reaction if the particles collide with enough energy to get the<br />
reaction started. This minimum energy required is called the activation energy for the reaction.<br />
4 Which statement explains why the speed of some reactions is increased when the surface area of<br />
one or all the reactants is increased?<br />
A increasing surface area changes the electronegativity of the reactant particles<br />
B increasing surface area changes the concentration of the reactant particles<br />
C increasing surface area changes the conductivity of reactant particles<br />
D increasing surface area enables more reactant particles to collide<br />
Explanation: The more finely divided the solid is, the faster the reaction happens. A powdered solid<br />
will normally produce a faster reaction than if the same mass is present as a single lump. The<br />
powdered solid has a greater surface area than the single lump. You are only going to get a reaction<br />
if the particles in the gas or liquid collide with the particles in the solid. Increasing the surface area of<br />
the solid increases the chances of collision taking place. Increasing the number of collisions per<br />
second increases the rate of reaction.<br />
catalyst<br />
C6H6 + Br2 → C6H5Br + HBr<br />
5 Which of the following changes will cause an increase in the rate of the above reaction?<br />
A increasing the concentration of Br2<br />
B decreasing the concentration of C6H6<br />
C increasing the concentration of HBr<br />
D decreasing the temperature<br />
Explanation: The same argument applies whether the reaction involves collision between two<br />
different particles or two of the same particle. In order for any reaction to happen, those particles<br />
must first collide. This is true whether both particles are in solution, or whether one is in solution and<br />
the other a solid. If the concentration is higher, the chances of collision are greater. For many<br />
reactions involving liquids or gases, increasing the concentration of the reactants increases the rate<br />
of reaction. In a few cases, increasing the concentration of one of the reactants may have little<br />
noticeable effect of the rate.<br />
143
2CO + O2 → 2CO2<br />
6 If the above reaction takes place inside a sealed reaction chamber, then which of these procedures<br />
will cause a decrease in the rate of reaction?<br />
A raising the temperature of the reaction chamber<br />
B increasing the volume inside the reaction chamber<br />
C removing the CO2 as it is formed<br />
D adding more CO to the reaction chamber<br />
Explanation: Increasing the pressure of a gas is exactly the same as increasing its concentration. If<br />
you have a given mass of gas, the way you increase its pressure is to squeeze it into a smaller<br />
volume. If you have the same mass in a smaller volume, then its concentration is higher.<br />
Because "RT" is constant as long as the temperature is constant, this shows that the pressure is<br />
directly proportional to the concentration. If you double one, you will also double the other. Increasing<br />
the pressure on a reaction involving reacting gases increases the rate of reaction. Changing the<br />
pressure on a reaction which involves only solids or liquids has no effect on the rate.<br />
7 A catalyst can speed up the rate of a given chemical reaction by<br />
A increasing the equilibrium constant in favor of products.<br />
B lowering the activation energy required for the reaction to occur.<br />
C raising the temperature at which the reaction occurs.<br />
D increasing the pressure of reactants, thus favoring products.<br />
Explanation: Catalysts increase the rate of reaction without being used up. They do this by lowering<br />
the activation energy needed. With a catalyst, more collisions result in a reaction, so<br />
the rate of reaction increases. Different reactions need different catalysts. Another important idea<br />
about catalysts is that they are selective. That is the catalyst doesn't just speed up all reactions, but<br />
only a very particular reaction. This is the key to many chemical transformations. When you only want<br />
to perform a particular chemical change, you look for a catalyst that will speed up that specific<br />
reaction but not others. Enzymes are remarkable in this way. Living biological systems require a<br />
myriad of specific chemical transformations and there is a unique enzyme to catalyze each of them.<br />
8 Which reaction diagram shows the effect of using the appropriate catalyst in a chemical reaction?<br />
A<br />
C<br />
B<br />
D<br />
144
Explanation: Simple energy level diagrams only show the energy levels at the beginning and end of a<br />
reaction. Energy levels change gradually during a reaction, and this can be shown using a curve<br />
between the reactant and product energy levels. This is an exothermic reaction because the energy<br />
level of the reactants is higher than the energy level of the products. However, the energy curve goes<br />
up from the reactants’ energy level to begin with, then drops to the products’ energy level. This is<br />
because many reactions need an input of energy to start the reaction off. This is energy is called<br />
the activation energy. It is represented on an energy level diagram as the difference between the<br />
reactants’ energy level and the top of the curve.<br />
9 H2O2, hydrogen peroxide, naturally breaks down into H2O and O2 over time. MnO2, manganese<br />
dioxide, can be used to lower the energy of activation needed for this reaction to take place and, thus,<br />
increase the rate of reaction. What type of substance is MnO2?<br />
A a catalyst<br />
B an enhancer<br />
C an inhibitor<br />
D a reactant<br />
Explanation: Manganese(IV) oxide is the inorganic compound with the formula MnO2. This blackish<br />
or brown solid occurs naturally as the mineral pyrolite, which is the main ore of manganese and a<br />
component of manganese nodules. The principal use for MnO2 is for dry-cell batteries, such as<br />
the alkaline battery and the zinc-carbon battery. MnO2 is also used as a pigment and as a precursor<br />
to other manganese compounds, such as link=potassium permanganateKMnO4. It is used as<br />
a reagent in organic synthesis, for example, for the oxidation of allylic alcohols. MnO2 in the α<br />
polymorph can incorporate a variety of atoms in the "tunnels" or "channels" between the magnesium<br />
oxide octahedral.<br />
10 When a reaction is at equilibrium and more reactant is added, which of the following changes is<br />
the immediate result?<br />
A The reverse reaction rate remains the same.<br />
B The forward reaction rate increases.<br />
C The reverse reaction rate decreases.<br />
D The forward reaction rate remains the same.<br />
Explanation: Equilibrium means it wants to stay balanced. If you add reactant, the system is<br />
unbalanced. To undo that, the reaction goes in the forward direction to create more product and use<br />
up the reactant - puts the systems back in balance/equilibrium.<br />
11 In which of the following reactions involving gases would the forward reaction be favored by an<br />
increase in pressure?<br />
A A + B ⇄AB<br />
B A + B ⇄ C + D<br />
C 2A + B ⇄ C + 2D<br />
D AC ⇄ A + C<br />
Explanation: A + b ↔ ab -when dealing with pressure, you need to count the number of moles on<br />
both side of the equation. If there is an increase in pressure, the reaction favors the direction with less<br />
number of moles. If there is a decrease in pressure, the reaction will favor the direction with more<br />
moles of molecules. If the pressure is increased, the reaction will favor the side with less number of<br />
moles of molecules, which in this case is the forward reaction.<br />
4HCl(g) + O2(g) ⇄ 2H2O(l) + 2Cl2(g) + 113 kJ<br />
145
12 Which action will drive the reaction to the right?<br />
A heating the equilibrium mixture<br />
B adding water to the system<br />
C decreasing the oxygen concentration<br />
D increasing the system’s pressure<br />
Explanation: If one or more of the reactants is a gas then increasing pressure will effectively increase<br />
the concentration of the reactant molecules and speed up the reaction. So, for gaseous reactants<br />
only, pressure is essentially a concentration factor. Increasing pressure has virtually no effect on<br />
solids or solutions engaged in a chemical reaction. The particles are, therefore on average, closer<br />
together and collisions between the particles will occur more frequently.<br />
NO2(g) + CO(g) ⇄ NO(g) + CO2(g)<br />
13 The reaction shown above occurs inside a closed flask. What action will shift the reaction to the<br />
left?<br />
A pumping CO gas into the closed flask<br />
B raising the total pressure inside the flask<br />
C increasing the NO concentration in the flask<br />
D venting some CO2 gas from the flask<br />
<br />
Explanation: The reactants are favored which causes them to shift to left because there are no<br />
gases present which means no change. The equilibrium will move in such a way that the<br />
pressure increases again. It can do that by producing more gaseous molecules. In this case, the<br />
position of equilibrium will move towards the left-hand side of the reaction.<br />
NH4Cl(s) + heat ⇄ NH3(g) + HCl(g)<br />
14 What kind of change will shift the reaction above to the right to form more products?<br />
A a decrease in total pressure<br />
B an increase in the concentration of HCl<br />
C an increase in the pressure of NH3<br />
D a decrease in temperature<br />
Explanation: Increasing the temperature increases reaction rates because of the disproportionately<br />
large increase in the number of high energy collisions. It is only these collisions (possessing at least<br />
the activation energy for the reaction) which result in a reaction. When you raise the temperature of a<br />
system, the molecules bounce around a lot more. They have more energy. When they bounce around<br />
more, they are more likely to collide. That fact means they are also more likely to combine. When you<br />
lower the temperature, the molecules are slower and collide less. That temperature drop lowers the<br />
rate of the reaction.<br />
15 In a sealed bottle that is half full of water, equilibrium will be attained when water molecules<br />
A cease to evaporate.<br />
B begin to condense.<br />
C are equal in number for both the liquid and the gas phase.<br />
D evaporate and condense at equal rates.<br />
Explanation: At equilibrium, the rate of forward reaction is equal to the rate of backward reaction.<br />
Therefore, when the rate of condensation will become equal to the rate of evaporation equilibrium is<br />
146
attained. When a liquid is in dynamic equilibrium with its vapor phase, the rates of evaporation and<br />
condensation are exactly equal to each other. The vapor pressure of a liquid is the equilibrium<br />
pressure of a vapor above its liquid (or solid); that is, the pressure of the vapor resulting from<br />
evaporation of a liquid (or solid) above a sample of the liquid (or solid) in a closed container.<br />
147
Unit 8<br />
Chapter 19 Acid and Bases<br />
The student will learn what are the different ways chemists<br />
define aids and bases, what the pH of a solution means and<br />
how chemist use acid-base reactions.<br />
Relate acidity and basicity to hydronium and hydroxyl ion concentration and<br />
pH.<br />
<br />
<br />
<br />
Students will be able to use a pH scale to identify substances as acids or bases.<br />
Students will be able to use various equipment (probeware, universal pH, etc.) to<br />
identify the pH of substances.<br />
Students will be able to calculate H3O+ and OH- concentration of various<br />
substances.<br />
pH scale<br />
Hydronium ion<br />
Arrhenius acid/base<br />
Lewis acid/base<br />
Bronsted-Lowry acid/base<br />
Strong acid/base<br />
Weak acid/base<br />
Neutralization reaction<br />
Titration<br />
Chapter 20 Oxidation-Reduction Reactions<br />
The student will learn what happens during oxidation and<br />
reduction and how to balance redox equations.<br />
Describe oxidation-reduction reactions in living and non-living systems.<br />
<br />
<br />
<br />
Students will be able to compare and contrast redox reactions.<br />
Students will be able to assign oxidation numbers to redox reactions.<br />
Students will be able to write half reactions<br />
148<br />
Oxidation
Reduction<br />
Oxidation reduction reaction<br />
Oxidation number<br />
Half reaction<br />
Electrochemical process<br />
Battery<br />
Cathode<br />
Anode<br />
Electrolysis<br />
149
The Learning Goal for this section is:<br />
The student will learn what are the different ways chemists define aids and bases, what the<br />
pH of a solution means and how chemist use acid-base reactions.<br />
Acids and Bases<br />
The Observable Properties of Acids and Bases<br />
The words acid and alkaline (an older word for base) are derived from direct sensory experience.<br />
Acid Property #1:<br />
The word acid comes from the Latin word acere, which means "sour." All acids taste sour. Well<br />
known from ancient times were vinegar, sour milk and lemon juice. Aspirin (scientific name:<br />
acetylsalicylic acid) tastes sour if you don't swallow it fast enough. Other languages derive their word<br />
for acid from the meaning of sour. So, in France, we have acide. In Germany, we have säure from<br />
saure and in Russia, kislota from kisly.<br />
Base Property #1:<br />
The word "base" has a more complex history (see below) and its name is not related to taste. All<br />
bases taste bitter. For example, mustard is a base. It tastes bitter. Many medicines, because they are<br />
bases, taste bitter. This is the reason cough syrups are advertised as having a "great grape taste."<br />
The taste is added in order to cover the bitterness of the active ingredient in cough syrup.<br />
Acid Property #2:<br />
Acids make a blue vegetable dye called litmus turn red.<br />
Base Property #2:<br />
Bases are substances which will restore the original blue color of litmus after having been reddened<br />
by an acid.<br />
Acid Property #3:<br />
Acids destroy the chemical properties of bases.<br />
Base Property #3:<br />
Bases destroy the chemical properties of acids.<br />
Neutralization is the name for this type of reaction.<br />
Acid Property #4:<br />
Acids conduct an electric current.<br />
Base Property #4:<br />
Bases conduct an electric current.<br />
This is a common property shared with salts. Acids, bases and salts are grouped together into a<br />
category called electrolytes, meaning that a water solution of the given substance will conduct an<br />
electric current.<br />
Non-electrolyte solutions cannot conduct a current. The most common example of this is sugar<br />
dissolved in water.<br />
150
So far, the properties have an obvious relationship: taste, color change, mutual destruction, and<br />
response to electric current. This last property is related, but in a less obvious way. The property<br />
below identifies a unique chemical reaction that acids and bases engage in.<br />
Acid Property #5:<br />
Upon chemically reacting with an active metal, acids will evolve hydrogen gas (H2). The key word, of<br />
course, is active. Some metals, like gold, silver or platinum, are rather unreactive and it takes rather<br />
extreme conditions to get these "unreactive" metals to react. Not so with the metals in this property.<br />
They include the alkali metals (Group I, Li to Rb), the alkaline earth metals (Group II, Be to Ra), as<br />
well as zinc and aluminum. Just bring the acid and the metal together at anything close to room<br />
temperature and you get a reaction. Here's a sample reaction:<br />
Zn + 2 HCl(aq) ---> ZnCl2 + H2<br />
Another common acid reaction some sources mention is that acids react with carbonates (and<br />
bicarbonates) to give carbon dioxide gas:<br />
HCl + NaCO3 ---> CO2 + H2O + NaCl<br />
Base Property #5:<br />
Bases feel slippery, sometimes people say soapy. This is because they dissolve the fatty acids and<br />
oils from your skin and this cuts down on the friction between your fingers as you rub them together.<br />
In essence, the base is making soap out of you. Yes, bases are involved in the production of soap! In<br />
the early years of soap making, the soaps were very harsh on the skin and clothes due to the high<br />
base content. Even today, people with very sensitive skin must sometimes use a non-soap-based<br />
product for bathing.<br />
It was not until more modern times that the chemical nature (as opposed to observable properties) of<br />
acids and bases began to be explored. That leads to this property that is not directly observable by<br />
the senses.<br />
Acid Property #6:<br />
Acids produce hydrogen ion (H + ) in solution. A more correct formula for what is produced is that of the<br />
hydronium ion, H3O + . Both formulas are used interchangeably.<br />
Acid base theories: Svante Arrhenius<br />
I. Introduction<br />
The basic idea is that certain substances remain ionized in solution all the time. Today, everyone<br />
accepts this without question, but it was the subject of much dissention and disagreement in 1884,<br />
when a twenty-five-year-old Arrhenius presented and defended his dissertation.<br />
II. The Acid Base Theory<br />
Acid - any substance which delivers hydrogen ion (H + ) to the solution.<br />
Base - any substance which delivers hydroxide ion (OH¯) to the solution.<br />
Here is a generic acid dissociating, according to Arrhenius:<br />
HA ---> H + + A¯<br />
151
This would be a generic base:<br />
152<br />
XOH ---> X + + OH¯<br />
When acids and bases react according to this theory, they neutralize each other, forming water and a<br />
salt:<br />
HA + XOH ---> H2O + XA<br />
Keeping in mind that the acid, the base and the salt all ionize, we can write this:<br />
Finally, we can drop all spectator ions, to get this:<br />
H + + A¯ + X + + OH¯ ---> H2O + X + + A¯<br />
H + + OH¯ ---> H2O<br />
These ideas covered all of the known acids at the time (the usual suspects like hydrochloric acid,<br />
acetic acid, and so on) and most of the bases (sodium hydroxide, potassium hydroxide, calcium<br />
hydroxide and so on). HOWEVER, and it is a big however, the theory did not explain why ammonia<br />
(NH3) was a base. There are other problems with the theory also.<br />
III. Problems with Arrhenius' Theory<br />
1. The solvent has no role to play in Arrhenius' theory. An acid is expected to be an acid in any<br />
solvent. This was found to not be the case. For example, HCl is an acid in water, behaving in<br />
the manner Arrhenius expected. However, if HCl is dissolved in benzene, there is no<br />
dissociation, the HCl remaining as un-dissociated molecules. The nature of the solvent plays a<br />
critical role in acid-base properties of substances.<br />
2. All salts in Arrhenius' theory should produce solutions that are neither acidic or basic. This is<br />
not the case. If equal amounts of HCl and ammonia react, the solution is slightly acidic. If equal<br />
amounts of acetic acid and sodium hydroxide are reacted, the resulting solution is basic.<br />
Arrhenius had no explanation for this.<br />
3. The need for hydroxide as the base led Arrhenius to propose the formula NH4OH as the<br />
formula for ammonia in water. This led to the misconception that NH4OH is the actual base,<br />
not NH3.<br />
In fact, by 1896, several years before Arrhenius announced his theory, it had been recognized that<br />
characteristic base properties where just as evident in such solvents as aniline, where no hydroxide<br />
ions were possible.<br />
4. H + , a bare proton, does not exist for very long in water. The proton affinity of H2O is about 799<br />
kJ/mol. Consequently, this reaction:<br />
H2O + H + ---> H3O +<br />
happens to a very great degree. The "concentration" of free protons in water has been estimated to<br />
be 10¯130 M. A rather preposterous value, indeed.<br />
The Arrhenius theory of acids and bases will be fully supplanted by the theory proposed<br />
independently by Johannes and Thomas Lowry in 1923.
The acid base theory of Brønsted and Lowry<br />
I. Introduction<br />
In 1923, within several months of each other, Johannes Nicolaus Brønsted (Denmark) and Thomas<br />
Martin Lowry (England) published essentially the same theory about how acids and bases behave.<br />
Since they came to their conclusions independently of each other, both names have been used for<br />
the theory name.<br />
II. The Acid Base Theory<br />
Using the words of Brønsted:<br />
". . . acids and bases are substances that are capable of splitting off or taking up hydrogen ions,<br />
respectively."<br />
Or an acid-base reaction consists of the transfer of a proton from an acid to a base. KEEP THIS<br />
THOUGHT IN MIND!!<br />
Here is a more recent way to say the same thing:<br />
An acid is a substance from which a proton can be removed.<br />
A base is a substance that can remove a proton from an acid.<br />
Remember: proton, hydrogen ion and H + all mean the same thing<br />
Very common in the chemistry world is this definition set:<br />
An acid is a "proton donor."<br />
A base is a "proton acceptor."<br />
In an acid, the hydrogen ion is bonded to the rest of the molecule. It takes energy (sometimes a little,<br />
sometimes a lot) to break that bond. So the acid molecule does not "give" or "donate" the proton, it<br />
has it taken away. In the same sense, you do not donate your wallet to the pickpocket, you have it<br />
removed from you.<br />
The base is a molecule with a built-in "drive" to collect protons. As soon as the base approaches the<br />
acid, it will (if it is strong enough) rip the proton off the acid molecule and add it to itself.<br />
Now this is where all the fun stuff comes in that you get to learn. You see, some bases are stronger<br />
than others, meaning some have a large "desire" for protons, while other bases have a weaker drive.<br />
It's the same way with acids, some have very weak bonds and the proton is easy to pick off, while<br />
other acids have stronger bonds, making it harder to "get the proton."<br />
One important contribution coming from Lowry has to do with the state of the hydrogen ion in solution.<br />
In Brønsted's announcement of the theory, he used H + . Lowry, in his paper (actually a long letter to<br />
the editor) used the H3O + that is commonly used today.<br />
III. Sample Equations written in the Brønsted-Lowry Style<br />
A. Reactions that proceed to a large extent:<br />
153
HCl + H2O ⇌ H3O + + Cl¯<br />
HCl - this is an acid, because it has a proton available to be transferred.<br />
H2O - this is a base, since it gets the proton that the acid lost.<br />
Now, here comes an interesting idea:<br />
H3O + - this is an acid, because it can give a proton.<br />
Cl¯ - this is a base, since it has the capacity to receive a proton.<br />
Notice that each pair (HCl and Cl¯ as well as H2O and H3O + differ by one proton (symbol = H + ). These<br />
pairs are called conjugate pairs.<br />
HNO3 + H2O ⇌ H3O + + NO3¯<br />
The acids are HNO3 and H3O + and the bases are H2O and NO3¯.<br />
Remember that an acid-base reaction is a competition between two bases (think about it!) for a<br />
proton. If the stronger of the two acids and the stronger of the two bases are reactants (appear on the<br />
left side of the equation), the reaction is said to proceed to a large extent.<br />
Here are some more conjugate acid-base pairs to look for:<br />
H2O and OH¯<br />
HCO3¯ and CO3 2¯<br />
H2PO4¯ and HPO4 2¯<br />
HSO4¯ and SO4 2¯<br />
NH4 + and NH3<br />
CH3NH3 + and CH3NH2<br />
HC2H3O2 and C2H3O2¯<br />
B. Reactions that proceed to a small extent:<br />
If the weaker of the two acids and the weaker of the two bases are reactants (appear on the left side<br />
of the equation), the reaction is said to proceed to only a small extent:<br />
HC2H3O2 + H2O ⇌ H3O + + C2H3O2¯<br />
NH3 + H2O ⇌ NH4 + + OH¯<br />
Identify the conjugate acid base pairs in each reaction.<br />
HC 2H 3O 2 and C 2H 3O 2¯<br />
is one conjugate pair.<br />
H 2O and H 3O + is the other.<br />
NH 3 and NH 4<br />
+<br />
is one pair.<br />
H 2O and OH¯ is the other.<br />
Notice that H 2O in the first equation is acting as a base and in the second equation is acting as an acid.<br />
154
IV. Problems with the Theory<br />
This theory works very nicely in all protic solvents (water, ammonia, acetic acid, etc.), but fails to<br />
explain acid base behavior in aprotic solvents such as benzene and dioxane. That job will be left for a<br />
more general theory, such as the Lewis Theory of Acids and Bases.<br />
The Lewis theory of acids and bases<br />
I. Introduction<br />
Lewis gives his definition of an acid and a base:<br />
"We are inclined to think of substances as possessing acid or basic properties, without having a<br />
particular solvent in mind. It seems to me that with complete generality we may say that a basic<br />
substance is one which has a lone pair of electrons which may be used to complete the stable group<br />
of another atom, and that an acid is one which can employ a lone pair from another molecule in<br />
completing the stable group of one of its own atoms."<br />
"In other words, the basic substance furnishes a pair of electrons for a chemical bond, the acid<br />
substance accepts such a pair."<br />
It is important to make two points here:<br />
1. NO hydrogen ion need be involved.<br />
2. NO solvent need be involved.<br />
The Lewis theory of acids and bases is more general than the "one sided" nature of the Bronsted-<br />
Lowry theory. Keep in mind that Bronsted-Lowry, which defines an acid as a proton donor and a base<br />
as a proton acceptor, REQUIRES the presence of a solvent, specifically a protic solvent, of which<br />
water is the usual example. Since almost all chemistry is done in water, the fact that this limits the<br />
Bronsted-Lowry definition is of little practical consequence.<br />
The Lewis definitions of acid and base do not have the constraints that the Bronsted-Lowry theory<br />
does and, as we shall see, many more reactions were seen to be acid base in nature using the Lewis<br />
definition than when using the Bronsted-Lowry definitions.<br />
II. The Acid Base Theory<br />
The modern way to define a Lewis acid and base is a bit more concise than above:<br />
Acid: an electron acceptor.<br />
Base: an electron donor.<br />
A "Lewis acid" is any atom, ion, or molecule which can accept electrons and a "Lewis base" is any<br />
atom, ion, or molecule capable of donating electrons. However, a warning: many textbooks will say<br />
"electron pair" where I have only written "electron." The truth is that it sometimes is an electron pair<br />
and sometimes it is not.<br />
It turns out that it may be more accurate to say that "Lewis acids" are substances which are electrondeficient<br />
(or low electron density) and "Lewis bases" are substances which are electron-rich (or high<br />
electron density).<br />
155
Several categories of substances can be considered Lewis acids:<br />
1. positive ions<br />
2. having less than a full octet in the valence shell<br />
3. polar double bonds (one end)<br />
4. expandable valence shells<br />
Several categories of substances can be considered Lewis bases:<br />
1. negative ions<br />
2. one of more unshared pairs in the valence shell<br />
3. polar double bonds (the other end)<br />
4. the presence of a double bond<br />
Sören Sörenson and the pH scale<br />
I. Short Historical Introduction<br />
In the late 1880's, Svante Arrhenius proposed that acids were substances that delivered hydrogen ion<br />
to the solution. He has also pointed out that the law of mass action could be applied to ionic<br />
reactions, such as an acid dissociating into hydrogen ion and a negatively charged anion.<br />
This idea was followed up by Wilhelm Ostwald, who calculated the dissociation constants (the<br />
modern symbol is Ka) of many weak acids. Ostwald also showed that the size of the constant is a<br />
measure of an acid's strength.<br />
By 1894, the dissociation constant of water (today called Kw) was measured to the modern value of<br />
1 x 10¯14 .<br />
In 1904, H. Friedenthal recommended that the hydrogen ion concentration be used to characterize<br />
solutions. He also pointed out that alkaline (modern word = basic) solutions could also be<br />
characterized this way since the hydroxyl concentration was always 1 x 10¯14 ÷ the hydrogen ion<br />
concentration. Many consider this to be the real introduction of the pH scale.<br />
III. The Introduction of pH<br />
Sörenson defined pH as the negative logarithm of the hydrogen ion concentration.<br />
pH = - log [H + ]<br />
Remember that sometimes H3O + is written, so<br />
pH = - log [H3O + ]<br />
means the same thing.<br />
So let's try a simple problem: The [H + ] in a solution is measured to be 0.010 M. What is the pH?<br />
The solution is pretty straightforward. Plug the [H + ] into the pH definition:<br />
pH = - log 0.010<br />
An alternate way to write this is:<br />
pH = - log 10¯2<br />
Since the log of 10¯2 is -2, we have:<br />
pH = - (- 2)<br />
Which, of course, is 2.<br />
156
Let's discuss significant figures and pH.<br />
Another sample problem: Calculate the pH of a solution in which the [H3O + ] is 1.20 x 10¯3 M.<br />
For the solution, we have:<br />
pH = - log 1.20 x 10¯3<br />
This problem can be done very easily using your calculator. However, be warned about putting<br />
numbers into the calculator.<br />
So you enter (-), log, 1.20, X10 n , (-), 3, enter.<br />
The answer, to the proper number of significant digits is: 2.921.<br />
III. Significant Figures in pH<br />
Here is the example problem: Calculate the pH of a solution where the [H + ] is 0.00100 M. (This could<br />
also be a pOH problem. The point being made is the same.)<br />
OK, you say, that's pretty easy, the answer is 3. After all, 0.00100 is 10¯3 and the negative log of 10¯3<br />
is 3.<br />
You would be graded wrong!! Why? Because the pH is not written to reflect the number of significant<br />
figures in the concentration.<br />
Notice that there are three sig figs in 0.00100. (Hopefully you remember significant figures, since you<br />
probably studied them months ago before getting to acid base stuff. THEY ARE STILL IMPORTANT!)<br />
So, our pH value should also reflect three significant figures.<br />
However, there is a special rule to remember with pH (and pOH) values. The whole number portion<br />
DOES NOT COUNT when figuring out how many digits to write down.<br />
Let's phrase that another way: in a pH (and a pOH), the only place where significant figures are<br />
contained is in the decimal portion.<br />
So, the correct answer to the above problem is 3.000. Three sig figs and they are all in the decimal<br />
portion, NOT (I repeat NOT) in the whole number portion.<br />
Practice Problems<br />
Convert each hydrogen ion concentration into a pH. Identify each as an acidic pH or a basic pH.<br />
1. 0.0015<br />
2. 5.0 x 10¯9<br />
3. 1.0<br />
4. 3.27 x 10¯4<br />
5. 1.00 x 10¯12<br />
6. 0.00010<br />
2.82 acidic pH<br />
8.30 basic pH<br />
0.00 acidic pH<br />
3.485 acidic pH<br />
12.000 basic pH<br />
4.00 acidic pH<br />
157
1. 2.82<br />
2. 8.30<br />
3. 0.00<br />
4. 3.485<br />
5. 12.000<br />
6. 4.00<br />
Sörenson also just mentions the reverse direction. That is, suppose you know the pH and you want to<br />
get to the hydrogen ion concentration ([H + ])?<br />
Here is the equation for that:<br />
[H + ] = 10¯pH<br />
That's right, ten to the minus pH gets you back to the [H + ] (called the hydrogen ion concentration).<br />
This is actually pretty easy to do with the calculator. Here's the sample problem: calculate the [H + ]<br />
from a pH of 2.45.<br />
This problem can be done very easily using your calculator. However, be warned about putting<br />
numbers into the calculator.<br />
So you enter 2nd, 10 x , (-), 2.45, enter.<br />
The answer, to the proper number of significant digits is: .00355.<br />
The pH of an acidic pond is 5. What is the hydrogen ion concentration (moles per liter)?<br />
The answer is:<br />
pH = -log (hydrogen ion concentration)<br />
The answer was .00001. Thus, 5 = -log (.00001).<br />
We'll take the formula that you started with (pH = -log([H+])) and work to the answer (solve for [H+]).<br />
pH = - log ([H+]) Given.<br />
pH = log ([H+] (-1) ) Since logarithms are like exponents, when you multiply a log by<br />
something, you can just move it to the inside of log as an exponent.<br />
10 pH = 10 log ([H+] (-1)) Take each side to tenth power.<br />
10 pH = [H+] (-1) Since "log" is just another notation for "log base 10", when you<br />
raise a log to the tenth power, the log cancels out.<br />
[H+] = 10 (-pH)<br />
Take the reciprocal of both sides.<br />
That is the general form. To answer the specific question,<br />
5 = - log ([H+])<br />
5 = log ([H+] (-1) )<br />
10 5 = [H+] (-1)<br />
10 (-5) = [H+]<br />
[H+]<br />
= .00001 mol/L<br />
158
On your calculator you would input 10, ^, (-), 5 and you would get 0.00001.<br />
This is also the way to find the amount of OH + that are present in a base.<br />
To find the pH: -log(concentration)<br />
To find the concentration: 10 -pH<br />
Define these terms:<br />
pH scale<br />
The pH scale measures how acidic or basic a substance is. The pH scale ranges from 0<br />
to 14.<br />
Hydronium ion<br />
The hydrogen ion bonded to a molecule of water, H 3 O + , the form in which hydrogen ions<br />
are found in aqueous solution.<br />
Arrhenius acid/base<br />
As defined by Arrhenius, acid-base reactions are characterized by acids, which dissociate in<br />
aqueous solution to form hydrogen ions and bases, which form hydroxide ions. Acids are<br />
defined as a compound or element that releases hydrogen ions into the solution.<br />
Lewis acid/base<br />
In the Lewis theory of acid-base reactions, bases donate pairs of electrons and acids<br />
accept pairs of electrons. A Lewis acid is therefore any substance, such as the H + ion,<br />
that can accept a pair of nonbonding electrons.<br />
Bronsted-Lowry acid/base<br />
In this theory, acids are defined as proton donors; whereas bases are defined as proton<br />
acceptors.<br />
Strong acid/base<br />
Acids and bases that are completely ionized when dissolved in water are called strong<br />
acids and strong bases<br />
Weak acid/base<br />
Weak acids and bases are only partially ionized in their solutions<br />
Neutralization reaction<br />
A neutralization reaction is when an acid and a base react to form water and a salt and<br />
involves the combination of H+ ions and OH- ions to generate water. The neutralization<br />
of a strong acid and strong base has a pH equal to 7.<br />
Titration<br />
Titration is the slow addition of one solution of a known concentration to a known<br />
volume of another solution of unknown concentration until the reaction reaches<br />
neutralization, which is often indicated by a color change.<br />
159
The Learning Goal for this assignment is:<br />
The student will learn what happens during oxidation and<br />
reduction and how to balance redox equations.<br />
1. Oxidation Numbers<br />
Redox Reactions<br />
Oxidation and reduction<br />
Every atom, ion or polyatomic ion has a formal oxidation number associated with it. This value<br />
compares the number of protons in an atom (positive charge) and the number of electrons assigned<br />
to that atom (negative charge).<br />
In many cases, the oxidation number reflects the actual charge on the atom, but there are many<br />
cases where it does not. Think of oxidation numbers as a bookkeeping exercise simply to keep track<br />
of where electrons go.<br />
2. Reduction<br />
Reduction means what it says: the oxidation number is reduced in reduction.<br />
This is accomplished by adding electrons. The electrons, being negative, reduce the overall oxidation<br />
number of the atom receiving the electrons.<br />
3. Oxidation<br />
Oxidation is the reverse process: the oxidation number of an atom is increased during oxidation.<br />
This is done by removing electrons. The electrons, being negative, make the atom that lost them<br />
more positive.<br />
I use this mnemonic to help me remember which is which: LEO the lion says GER.<br />
LEO = Loss of Electrons is Oxidation<br />
GER = Gain of Electrons is Reduction<br />
Another well-known mnemonic is this: OIL RIG<br />
OIL = Oxidation Is Loss (of Electrons)<br />
RIG = Reduction Is Gain (of Electrons)<br />
Another way is to simply remember that reduction is to reduce the oxidation number. Therefore,<br />
oxidation must increase the value.<br />
4. Reduction-Oxidation Reactions<br />
There are many chemical reactions in which one substance gets reduced in oxidation number<br />
(reduction) while another participating substance gets increased in oxidation number (oxidation).<br />
Such a reaction is called called a REDOX reaction. The RED, of course, comes from REDuction and<br />
OX from OXidation. However, it is pronounced re-dox and not red-ox.<br />
Here is a simple example of a redox reaction:<br />
160
Ag+ + Cu ---> Ag + Cu2+<br />
I have deliberately not balanced it and I have also written it in net ionic form. I have found that kids<br />
studying redox get confused by net ionic form and how to change a full equation into net ionic form.<br />
Redox equations need to be balanced but, except for the simplest ones, it cannot be done by<br />
inspection (also called trial and error). I take that back, complex ones can be done by trial and error. It<br />
typically takes quite a bit of work, especially when compared to how long it takes when the proper<br />
technique is used.<br />
There is a technique used to balance redox reactions. It is called "balancing by half-reactions." The<br />
basic plan will be to split the full equation into two simpler parts (called half-reactions), balance them<br />
following several standard steps, then recombine the balanced half-reactions into the final answer.<br />
This is another technique called the "ion-electron method." I plan to ignore it.<br />
Notes:<br />
Redox equations need to be balanced out. There is a technique used to balance redox<br />
reactions<br />
5. Some Definitions<br />
Oxidizing Agent - that substance which oxidizes somebody else. It is reduced in the process.<br />
Reducing Agent - that substance which reduces somebody else. It is oxidized in the process.<br />
It helps me to remember these definitions by the opposite nature of what happens. By that, I mean<br />
the oxidizing agent gets reduced and the reducing agent gets oxidized.<br />
6. Rules for Assigning Oxidation Numbers<br />
The Oxidation Number of an element corresponds to the number of electrons, e - , that an atom loses,<br />
gains, or appears to use when joining with other atoms in compounds. In determining the Oxidation<br />
Number of an atom, there are seven guidelines to follow:<br />
1. The Oxidation Number of an individual atom is 0. This includes diatomic elements such as<br />
O2 or others like P4 and S8<br />
2. The total Oxidation Number of all atoms in: a neutral species is 0 and in an ion is equal to the<br />
ion charge.<br />
3. Group 1 metals have an Oxidation Number of +1 and Group 2 an Oxidation Number of +2<br />
4. The Oxidation Number of fluorine is -1 in compounds<br />
5. Hydrogen generally has an Oxidation Number of +1 in compounds, except hydrides<br />
6. Oxygen generally has an Oxidation Number of -2 in compounds, except peroxides<br />
7. In binary metal compounds, Group 17 elements have an Oxidation Number of -1, Group 16 of -<br />
2, and Group 15 of -3.<br />
Note: The sum of the Oxidation Number s is equal to zero for neutral compounds and equal to the<br />
charge for polyatomic ion species.<br />
Now, some examples:<br />
1. What is the oxidation number of Cl in HCl?<br />
Since H = +1, the Cl must be -1 (minus one).<br />
161
2. What is the oxidation number of Na in Na2O?<br />
Since O = -2, the two Na must each be +1.<br />
3. What is the oxidation number of Cl in ClO¯?<br />
The O is -2, but since a -1 must be left over, then the Cl is +1.<br />
4. What is the oxidation number for each element in KMnO4?<br />
K = +1 because KCl exists. We know the Cl = -1 because HCl exists.<br />
O = -2 by definition<br />
Mn = +7. There are 4 oxygens for a total of -8, K is +1, so Mn must be the rest.<br />
5. What is the oxidation number of S in SO4 2¯<br />
O = -2. There are four oxygens for -8 total. Since -2 must be left over, the S must = +6.<br />
Please note that, if there is no charge indicated on a formula, the total charge is taken to be zero.<br />
Practice Problems<br />
Find oxidation numbers<br />
1. N in NO3¯<br />
2. C in CO3 2¯<br />
+5<br />
+4<br />
3. Cr in CrO4 2¯<br />
+6<br />
4. Cr in Cr2O7 2¯ +6<br />
5. Fe in Fe2O3<br />
6. Pb in PbOH +<br />
+3<br />
+2<br />
7. V in VO2 +<br />
8. V in VO 2+<br />
+5<br />
+4<br />
9. Mn in MnO4¯<br />
+7<br />
10. Mn in MnO4 2¯<br />
Notes:<br />
+6<br />
Then you have a - at the end you subtract the other oxidation number<br />
When you have a positive you add more to the other oxidation<br />
number<br />
162
7. Half Reactions<br />
A half-reaction is simply one which shows either reduction OR oxidation, but not both. Here is the<br />
example redox reaction used in a different file:<br />
Ag + + Cu → Ag + Cu 2+<br />
It has BOTH a reduction and an oxidation in it. That is why we call it a redox reaction, from REDuction<br />
and OXidation.<br />
What you must be able to do is look at a redox reaction and separate out the two half-reactions in it.<br />
To do that, identify the atoms which get reduced and get oxidized. Here are the two half-reactions<br />
from the above example:<br />
Ag+ → Ag<br />
Cu → Cu 2+<br />
The silver is being reduced, its oxidation number going from +1 to zero. The copper's oxidation<br />
number went from zero to +2, so it was oxidized in the reaction. In order to figure out the halfreactions,<br />
you MUST be able to calculate the oxidation number of an atom.<br />
Keep in mind that a half-reaction shows only one of the two behaviors we are studying. A single halfreaction<br />
will show ONLY reduction or ONLY oxidation, never both in the same equation.<br />
Also, notice that the reaction is read from left to right to determine if it is reduction or oxidation. If you<br />
read the reaction in the opposite direction (from right to left) it then becomes the other of our two<br />
choices (reduction or oxidation). For example, the silver half-reaction above is a reduction, but in the<br />
reverse direction it is an oxidation, going from zero on the right to +1 on the left.<br />
There will be times when you want to switch a half-reaction from one of the two types to the other. In<br />
that case, rewrite the entire equation and swap sides for everything involved. If I needed the silver<br />
half-reaction to be oxidation, I'd write Ag → Ag+ rather than just doing it mentally.<br />
The next step is that both half-reactions must be balanced. However, there is a twist. When you<br />
learned about balancing equation, you made equal the number of atoms of each element on each<br />
side of the arrow. That still applies, but there is one more thing: the total amount of charge on each<br />
side of the half-reaction MUST be the same.<br />
When you look at the two half-reactions above, you will see they are already balanced for atoms with<br />
one Ag on each side and one Cu on each side. So, all we need to do is balance the charge. To do<br />
this you add electrons to the more positive side. You add enough to make the total charge on each<br />
side become EQUAL.<br />
To the silver half-reaction, we add one electron:<br />
To the copper half-reaction, we add two electrons:<br />
Ag+ + e¯ ---> Ag<br />
Cu ---> Cu 2+ + 2e¯<br />
163
One point of concern: notice that each half-reaction wound up with a total charge of zero on each<br />
side. This is not always the case. You need to strive to get the total charge on each side EQUAL, not<br />
zero.<br />
One more point to make before wrapping this up. A half-reaction is a "fake" chemical reaction. It's just<br />
a bookkeeping exercise. Half-reactions NEVER occur alone. If a reduction half-reaction is actually<br />
happening (say in a beaker in front of you), then an oxidation reaction is also occurring. The two halfreactions<br />
can be in separate containers, but they do have to have some type of "chemical<br />
connection" between them.<br />
Half-Reaction Practice Problems<br />
Balance each half-reaction for atoms and charge:<br />
1) Cl2 → Cl¯<br />
2) Sn → Sn 2+<br />
3) Fe 2+ → Fe 3+<br />
4) I3¯ → I¯<br />
5) ICl2¯ → I¯ (I'm being mean on this one. Hint: the iodine is the only thing reduced or oxidized.)<br />
ICl 2¯ + 2e¯ →I¯ + 2Cl¯<br />
Separate each of these redox reactions into their two half-reactions (but do not balance):<br />
6) Sn + NO3¯ →SnO2 + NO2<br />
7) HClO + Co →Cl2 + Co 2+<br />
8) NO2 →NO3¯ + NO<br />
Here are the two half-reactions to be combined:<br />
Here is the rule to follow:<br />
Ag+ + e¯ → Ag<br />
Cu → Cu 2+ + 2e¯<br />
the total electrons MUST cancel when the two half-reactions are added.<br />
Another way to say it:<br />
Cl 2 + 2e¯ →2Cl¯<br />
Sn →Sn 2+ + 2e¯<br />
Fe 2+ →Fe 3+ + e¯<br />
I 3¯ → 3I¯ + 2e¯<br />
Sn →SnO 2 and NO 3¯ →NO 2<br />
HClO →Cl 2 and Co ---> Co 2+<br />
NO 2 →NO 3¯ and NO 2 →NO<br />
the number of electrons in each half-reaction MUST be equal when the two half-reactions are<br />
added.<br />
What that means is that one (or both) equations must be multiplied through by a factor. The value of<br />
the factor is selected so as to make the number of electrons equal.<br />
In our example problem, the top reaction (the one with silver) must be multiplied by two, like this:<br />
164<br />
2Ag+ + 2e¯ → 2Ag
Notice that each separate substance is increased by the factor amount. Occasionally, a student will<br />
multiply ONLY the electrons by the factor. That is incorrect.<br />
When the two half-reactions are added, we get:<br />
2Ag+ + 2e¯ + Cu → 2Ag + Cu 2+ + 2e¯<br />
With two electrons on each side, they may be canceled, resulting in:<br />
2Ag+ + Cu → 2Ag + Cu 2+<br />
This is the correct answer. Notice that there are two silvers on each side and one copper. Notice also<br />
that the total charge on each side is +2. It is balanced for both atoms and charge. Sometimes, I am<br />
asked if the order matters, if the Cu could be first on the left-hand side. The answer is that the order<br />
does not matter. There happen to be certain styles about where particular substances are put in the<br />
final answer, but these are only styles. They do not affect the chemical correctness of the answer.<br />
Notes:<br />
The total number electrons must cancel when the two half-reactions are<br />
added.<br />
Practice Problems<br />
Separate into half-reactions, balance them and then recombine.<br />
1) Sn + Cl2 ---> Sn 2+ + Cl¯<br />
Sn 0 -->Sn 2+ +2e - e - +Cl 0 -->Cl 1- --> Sn 2+ +2Cl -<br />
2) Fe 2+ + I3¯ ---> Fe 3+ + I¯<br />
???????????????????????????????<br />
1. N in NO 3¯<br />
The O is -2 and three of them makes -6. Since -1 is left over, the N must be +5<br />
2. C in CO 3<br />
2¯<br />
The O is -2 and three of them makes -6. Since -2 is left over, the C must be +4<br />
3. Cr in CrO 4<br />
2¯<br />
The O is -2 and four of them makes -8. Since -2 is left over, the Cr must be +6<br />
4. Cr in Cr 2 O 7<br />
2¯<br />
The O is -2 and seven of them makes -14. Since -2 is left over, the two Cr must be +12<br />
What follows is an important point. Each Cr is +6. Each one. We do not speak of Cr 2 being +12. Each Cr atom is considered individually.<br />
5. Fe in Fe 2 O 3<br />
The O is -2 and three of them makes -6. Each Fe must then be +3<br />
6. +2 9. +7<br />
7. +5 10. +6<br />
8. +4<br />
1) Cl 2 + 2e¯ →2Cl¯ 5) ICl 2¯<br />
+ 2e¯ →I¯ + 2Cl¯<br />
2) Sn →Sn 2+ + 2e¯ 6) Sn →SnO 2 and NO 3¯<br />
→NO 2<br />
3) Fe 2+ →Fe 3+ + e¯ 7) HClO →Cl 2 and Co ---> Co 2+<br />
4) I 3¯ → 3I¯ + 2e¯ 8) NO 2 →NO 3¯<br />
and NO 2 →NO<br />
Note that in this last redox equation the NO 2 (actually just the N) is both being reduced AND being oxidized. In the first half-reaction, the N goes from +4<br />
to +5 (oxidation) and in the second, the N goes from +4 to +2 (reduction).<br />
By the way, we could flip the reaction so that NO 3¯<br />
and NO are reacting together to produce only one product, the NO 2 . In that case, the two halfreactions<br />
would be reversed.<br />
165
Reference Tables for Physical Setting/CHEMISTRY<br />
Table A<br />
Standard Temperature and Pressure<br />
Name Value Unit<br />
Standard Pressure 101.3 kPa kilopascal<br />
1 atm atmosphere<br />
Standard Temperature 273 K kelvin<br />
0°C degree Celsius<br />
Table D<br />
Selected Units<br />
Symbol Name Quantity<br />
m meter length<br />
g gram mass<br />
Pa pascal pressure<br />
K kelvin temperature<br />
Table B<br />
Physical Constants for Water<br />
mol<br />
J<br />
mole<br />
joule<br />
amount of<br />
substance<br />
energy, work,<br />
quantity of heat<br />
Heat of Fusion<br />
Heat of Vaporization<br />
Specific Heat Capacity of H 2<br />
O()<br />
Specific Heat Capacity of H 2<br />
O(s)<br />
Specific Heat Capacity of H 2<br />
O(g)<br />
Table C<br />
Selected Prefixes<br />
Factor Prefix Symbol<br />
10 3 kilo- k<br />
334 J/g<br />
2260 J/g<br />
4.18 J/g•K<br />
2.10 J/g•K<br />
2.01 J/g•K<br />
s second time<br />
min minute time<br />
h hour time<br />
d day time<br />
y year time<br />
L liter volume<br />
ppm parts per million concentration<br />
M<br />
molarity<br />
solution<br />
concentration<br />
u atomic mass unit atomic mass<br />
10 –1 deci- d<br />
10 –2 centi- c<br />
10 –3 milli- m<br />
10 –6 micro- μ<br />
10 –9 nano- n<br />
10 –12 pico- p<br />
R1
Table E<br />
Selected Polyatomic Ions<br />
Formula Name Formula Name<br />
H 3<br />
O +<br />
hydronium<br />
CrO 4<br />
2–<br />
chromate<br />
Hg 2<br />
2+<br />
mercury(I)<br />
Cr 2<br />
O 7<br />
2–<br />
dichromate<br />
NH 4<br />
+<br />
C 2<br />
H 3<br />
O<br />
–<br />
2 –}<br />
CH 3<br />
COO<br />
CN –<br />
CO 3<br />
2–<br />
HCO<br />
–<br />
3<br />
C 2<br />
O<br />
2–<br />
4<br />
ClO –<br />
ammonium<br />
acetate<br />
cyanide<br />
carbonate<br />
hydrogen<br />
carbonate<br />
oxalate<br />
hypochlorite<br />
MnO 4<br />
–<br />
NO<br />
–<br />
2<br />
NO<br />
–<br />
3<br />
O<br />
2–<br />
2<br />
OH –<br />
PO 4<br />
3–<br />
SCN –<br />
SO 3<br />
2–<br />
permanganate<br />
nitrite<br />
nitrate<br />
peroxide<br />
hydroxide<br />
phosphate<br />
thiocyanate<br />
sulfite<br />
ClO 2<br />
–<br />
chlorite<br />
SO 4<br />
2–<br />
sulfate<br />
ClO 3<br />
–<br />
chlorate<br />
HSO 4<br />
–<br />
hydrogen sulfate<br />
ClO 4<br />
–<br />
perchlorate<br />
S 2<br />
O 3<br />
2–<br />
thiosulfate<br />
Table F<br />
Solubility Guidelines for Aqueous Solutions<br />
Ions That Form<br />
Soluble Compounds<br />
Group 1 ions<br />
(Li + , Na + , etc.)<br />
ammonium (NH 4 + )<br />
nitrate (NO 3 – )<br />
acetate (C 2<br />
H 3<br />
O 2 – or<br />
CH 3<br />
COO – )<br />
hydrogen carbonate<br />
(HCO 3 – )<br />
chlorate (ClO 3 – )<br />
halides (Cl – , Br – , I – )<br />
Exceptions<br />
when combined with<br />
Ag + , Pb 2+ , or Hg 2<br />
2+<br />
sulfates (SO 4 2– ) when combined with Ag + ,<br />
Ca 2+ , Sr 2+ , Ba 2+ , or Pb 2+<br />
Ions That Form<br />
Insoluble Compounds*<br />
Exceptions<br />
carbonate (CO 3 2– ) when combined with Group 1<br />
ions or ammonium (NH 4 + )<br />
chromate (CrO 4 2– ) when combined with Group 1<br />
ions, Ca 2+ , Mg 2+ , or<br />
ammonium (NH 4 + )<br />
phosphate (PO 4 3– ) when combined with Group 1<br />
ions or ammonium (NH 4 + )<br />
sulfide (S 2– ) when combined with Group 1<br />
ions or ammonium (NH 4 + )<br />
hydroxide (OH – ) when combined with Group 1<br />
ions, Ca 2+ , Ba 2+ , Sr 2+ , or<br />
ammonium (NH 4 + )<br />
*compounds having very low solubility in H 2 O<br />
R2
150.<br />
140.<br />
Table G<br />
Solubility Curves at Standard Pressure<br />
KI<br />
NaNO 3<br />
130.<br />
120.<br />
KNO 3<br />
110.<br />
100.<br />
Solubility (g solute/100. g H 2<br />
O)<br />
90.<br />
80.<br />
70.<br />
60.<br />
HCl<br />
NH 4<br />
Cl<br />
KCl<br />
50.<br />
40.<br />
30.<br />
NaCl<br />
KClO 3<br />
NH 3<br />
20.<br />
10.<br />
SO 2<br />
0<br />
0 10. 20. 30. 40. 50. 60. 70. 80. 90. 100.<br />
Temperature (°C)<br />
R3
Table H<br />
Vapor Pressure of Four Liquids<br />
200.<br />
propanone<br />
ethanol<br />
150.<br />
water<br />
Vapor Pressure (kPa)<br />
100.<br />
101.3 kPa<br />
ethanoic<br />
acid<br />
50.<br />
0<br />
0 25 50. 75 100. 125<br />
R4
Table I<br />
Heats of Reaction at 101.3 kPa and 298 K<br />
Reaction<br />
ΔH (kJ)*<br />
CH 4<br />
(g) + 2O 2<br />
(g) CO 2<br />
(g) + 2H 2<br />
O() –890.4<br />
C 3<br />
H 8<br />
(g) + 5O 2<br />
(g) 3CO 2<br />
(g) + 4H 2<br />
O() –2219.2<br />
2C 8<br />
H 18<br />
() + 25O 2<br />
(g) 16CO 2<br />
(g) + 18H 2<br />
O() –10943<br />
2CH 3<br />
OH() + 3O 2<br />
(g) 2CO 2<br />
(g) + 4H 2<br />
O() –1452<br />
C 2<br />
H 5<br />
OH() + 3O 2<br />
(g) 2CO 2<br />
(g) + 3H 2<br />
O() –1367<br />
C 6<br />
H 12<br />
O 6<br />
(s) + 6O 2<br />
(g) 6CO 2<br />
(g) + 6H 2<br />
O() –2804<br />
2CO(g) + O 2<br />
(g) 2CO 2<br />
(g) –566.0<br />
C(s) + O 2<br />
(g) CO 2<br />
(g) –393.5<br />
4Al(s) + 3O 2<br />
(g) 2Al 2<br />
O 3<br />
(s) –3351<br />
N 2<br />
(g) + O 2<br />
(g) 2NO(g) +182.6<br />
N 2<br />
(g) + 2O 2<br />
(g) 2NO 2<br />
(g) +66.4<br />
2H 2<br />
(g) + O 2<br />
(g) 2H 2<br />
O(g) –483.6<br />
2H 2<br />
(g) + O 2<br />
(g) 2H 2<br />
O() –571.6<br />
N 2<br />
(g) + 3H 2<br />
(g) 2NH 3<br />
(g) –91.8<br />
2C(s) + 3H 2<br />
(g) C 2<br />
H 6<br />
(g) –84.0<br />
2C(s) + 2H 2<br />
(g) C 2<br />
H 4<br />
(g) +52.4<br />
2C(s) + H 2<br />
(g) C 2<br />
H 2<br />
(g) +227.4<br />
H 2<br />
(g) + I 2<br />
(g) 2HI(g) +53.0<br />
KNO 3<br />
(s) H 2 O K + (aq) + NO 3 – (aq) +34.89<br />
NaOH(s) H 2 O Na + (aq) + OH – (aq) –44.51<br />
NH 4<br />
Cl(s) H 2 O NH 4 + (aq) + Cl – (aq) +14.78<br />
NH 4<br />
NO 3<br />
(s) H 2 O NH 4 + (aq) + NO 3 – (aq) +25.69<br />
NaCl(s) H 2 O Na + (aq) + Cl – (aq) +3.88<br />
LiBr(s) H 2 O Li + (aq) + Br – (aq) –48.83<br />
H + (aq) + OH – (aq) H 2<br />
O() –55.8<br />
*The ΔH values are based on molar quantities represented in the equations.<br />
A minus sign indicates an exothermic reaction.<br />
Most<br />
Active<br />
Least<br />
Active<br />
Table J<br />
Activity Series**<br />
Metals Nonmetals Most<br />
Active<br />
Li<br />
Rb<br />
K<br />
Cs<br />
Ba<br />
Sr<br />
Ca<br />
Na<br />
Mg<br />
Al<br />
Ti<br />
Mn<br />
Zn<br />
Cr<br />
Fe<br />
Co<br />
Ni<br />
Sn<br />
Pb<br />
H 2<br />
Cu<br />
Ag<br />
Au<br />
F 2<br />
Cl 2<br />
Br 2<br />
I 2<br />
**Activity Series is based on the hydrogen<br />
standard. H 2 is not a metal.<br />
Least<br />
Active<br />
R5
Table K<br />
Common Acids<br />
Table N<br />
Selected Radioisotopes<br />
HCl(aq)<br />
Formula<br />
HNO 2<br />
(aq)<br />
HNO 3<br />
(aq)<br />
H 2<br />
SO 3<br />
(aq)<br />
H 2<br />
SO 4<br />
(aq)<br />
H 3<br />
PO 4<br />
(aq)<br />
H 2<br />
CO 3<br />
(aq)<br />
or<br />
CO 2<br />
(aq)<br />
CH 3<br />
COOH(aq)<br />
or<br />
HC 2<br />
H 3<br />
O 2<br />
(aq)<br />
Name<br />
hydrochloric acid<br />
nitrous acid<br />
nitric acid<br />
sulfurous acid<br />
sulfuric acid<br />
phosphoric acid<br />
carbonic acid<br />
ethanoic acid<br />
(acetic acid)<br />
Nuclide Half-Life Decay<br />
Mode<br />
Nuclide<br />
Name<br />
198 Au 2.695 d β – gold-198<br />
14 C 5715 y β – carbon-14<br />
37 Ca 182 ms β + calcium-37<br />
60 Co 5.271 y β – cobalt-60<br />
137 Cs 30.2 y β – cesium-137<br />
53 Fe 8.51 min β + iron-53<br />
220 Fr 27.4 s α francium-220<br />
3 H 12.31 y β – hydrogen-3<br />
131 I 8.021 d β – iodine-131<br />
37 K 1.23 s β + potassium-37<br />
42 K 12.36 h β – potassium-42<br />
Table L<br />
Common Bases<br />
85 Kr 10.73 y β – krypton-85<br />
16 N 7.13 s β – nitrogen-16<br />
Formula<br />
NaOH(aq)<br />
KOH(aq)<br />
Ca(OH) 2<br />
(aq)<br />
NH 3<br />
(aq)<br />
Name<br />
sodium hydroxide<br />
potassium hydroxide<br />
calcium hydroxide<br />
aqueous ammonia<br />
19 Ne 17.22 s β + neon-19<br />
32 P 14.28 d β – phosphorus-32<br />
239 Pu 2.410 × 10 4 y α plutonium-239<br />
226 Ra 1599 y α radium-226<br />
222 Rn 3.823 d α radon-222<br />
90 Sr 29.1 y β – strontium-90<br />
Table M<br />
Common Acid–Base Indicators<br />
Approximate<br />
Indicator pH Range Color<br />
for Color Change<br />
Change<br />
methyl orange 3.1–4.4 red to yellow<br />
bromthymol blue 6.0–7.6 yellow to blue<br />
phenolphthalein 8–9 colorless to pink<br />
litmus 4.5–8.3 red to blue<br />
bromcresol green 3.8–5.4 yellow to blue<br />
thymol blue 8.0–9.6 yellow to blue<br />
99 Tc 2.13 × 10 5 y β – technetium-99<br />
232 Th 1.40 × 10 10 y α thorium-232<br />
233 U 1.592 × 10 5 y α uranium-233<br />
235 U 7.04 × 10 8 y α uranium-235<br />
238 U 4.47 × 10 9 y α uranium-238<br />
Source: CRC Handbook of Chemistry and Physics, 91 st ed., 2010–2011,<br />
CRC Press<br />
Source: The Merck Index, 14 th ed., 2006, Merck Publishing Group<br />
R6
Table O<br />
Symbols Used in Nuclear Chemistry<br />
Name Notation Symbol<br />
alpha particle<br />
4<br />
2<br />
He or 4 2 α α<br />
beta particle<br />
0<br />
–1<br />
e or 0<br />
–1 β β–<br />
gamma radiation<br />
0<br />
0<br />
γ γ<br />
neutron<br />
1<br />
0<br />
n n<br />
proton<br />
1<br />
1<br />
H or 1 1 p p<br />
positron<br />
0<br />
+1<br />
e or 0<br />
+1 β β+<br />
Table P<br />
Organic Prefixes<br />
Prefix<br />
meth- 1<br />
eth- 2<br />
prop- 3<br />
but- 4<br />
pent- 5<br />
hex- 6<br />
hept- 7<br />
oct- 8<br />
non- 9<br />
dec- 10<br />
Number of<br />
Carbon Atoms<br />
Table Q<br />
Homologous Series of Hydrocarbons<br />
Name General Examples<br />
Formula Name Structural Formula<br />
R7<br />
alkanes C n<br />
H 2n+2<br />
ethane<br />
alkenes C n<br />
H 2n<br />
ethene<br />
alkynes C n<br />
H 2n–2<br />
ethyne<br />
Note: n = number of carbon atoms<br />
H H<br />
H C C H<br />
H H<br />
H<br />
H<br />
C C<br />
H<br />
H<br />
H C C H
Table R<br />
Organic Functional Groups<br />
Class of<br />
Compound<br />
Functional<br />
Group<br />
General<br />
Formula<br />
Example<br />
halide<br />
(halocarbon)<br />
F (fluoro-)<br />
Cl (chloro-)<br />
Br (bromo-)<br />
I (iodo-)<br />
R X<br />
(X represents<br />
any halogen)<br />
CH 3<br />
CHClCH 3<br />
2-chloropropane<br />
alcohol<br />
OH<br />
R<br />
OH<br />
CH 3<br />
CH 2<br />
CH 2<br />
OH<br />
1-propanol<br />
ether<br />
O<br />
R O R′<br />
CH 3<br />
OCH 2<br />
CH 3<br />
methyl ethyl ether<br />
aldehyde<br />
O<br />
C H<br />
R<br />
O<br />
C H<br />
O<br />
CH 3<br />
CH 2<br />
C H<br />
propanal<br />
ketone<br />
O<br />
C<br />
O<br />
R C R′<br />
O<br />
CH 3<br />
CCH 2<br />
CH 2<br />
CH 3<br />
2-pentanone<br />
organic acid<br />
O<br />
C OH<br />
R<br />
O<br />
C OH<br />
O<br />
CH 3<br />
CH 2<br />
C OH<br />
propanoic acid<br />
ester<br />
O<br />
C O<br />
O<br />
R C O R′<br />
O<br />
CH 3<br />
CH 2<br />
COCH 3<br />
methyl propanoate<br />
amine<br />
N<br />
R<br />
R′<br />
N R′′<br />
CH 3<br />
CH 2<br />
CH 2<br />
NH 2<br />
1-propanamine<br />
amide<br />
O<br />
C NH<br />
R<br />
O R′<br />
C NH<br />
O<br />
CH 3<br />
CH 2<br />
C NH 2<br />
propanamide<br />
Note: R represents a bonded atom or group of atoms.<br />
R8
0<br />
6.941<br />
+1<br />
Li<br />
3<br />
2-1<br />
Na<br />
39.0983<br />
K +1<br />
19<br />
2-8-8-1<br />
85.4678 +1<br />
Rb<br />
Cs<br />
(223)<br />
Fr<br />
87<br />
-18-32-18-8-1<br />
+1<br />
Ra<br />
88<br />
-18-32-18-8-2<br />
39<br />
138.9055<br />
La<br />
57<br />
2-8-18-18-9-2<br />
+2 (227)<br />
Ac<br />
89<br />
-18-32-18-9-2<br />
47.867<br />
Ti<br />
22<br />
2-8-10-2<br />
91.224<br />
Zr<br />
40<br />
2-8-18-10-2<br />
+3 178.49<br />
Hf<br />
72<br />
*18-32-10-2<br />
+3 (261)<br />
Rf<br />
104<br />
+2<br />
+3<br />
+4<br />
+4<br />
+4<br />
50.9415<br />
V<br />
23<br />
2-8-11-2<br />
+2<br />
+3<br />
+4<br />
+5<br />
51.996<br />
Cr<br />
24<br />
2-8-13-1<br />
95.94<br />
Mo<br />
42<br />
2-8-18-13-1<br />
183.84<br />
W<br />
74<br />
-18-32-12-2<br />
+2<br />
+3<br />
+6<br />
+6<br />
+6<br />
54.9380<br />
Mn<br />
25<br />
2-8-13-2<br />
+2<br />
+3<br />
+4<br />
+7<br />
55.845<br />
Fe<br />
26<br />
2-8-14-2<br />
+2<br />
+3 58.9332<br />
Co<br />
27<br />
2-8-15-2<br />
+2<br />
+3<br />
58.693<br />
Ni<br />
28<br />
2-8-16-2<br />
+2<br />
+3 63.546 Cu<br />
2-8-18-1<br />
107.868<br />
Ag<br />
47<br />
2-8-18-18-1<br />
79<br />
+1<br />
+2<br />
+1<br />
65.409<br />
Zn<br />
30<br />
2-8-18-2<br />
10.81<br />
+3 12.011<br />
B<br />
5<br />
2-3<br />
26.98154<br />
Al<br />
13<br />
2-8-3<br />
+2 69.723<br />
Ga<br />
31<br />
2-8-18-3<br />
+3<br />
+3<br />
–4<br />
+2<br />
+4<br />
C<br />
6<br />
2-4<br />
28.0855<br />
Si<br />
14<br />
2-8-4<br />
72.64<br />
Ge<br />
32<br />
2-8-18-4<br />
Pb<br />
–4<br />
+2<br />
+4<br />
+2<br />
+4<br />
74.9216<br />
As<br />
33<br />
2-8-18-5<br />
Sb<br />
–3<br />
+3<br />
15.9994 O<br />
–2 18.9984<br />
8<br />
2-6 2-7<br />
78.96<br />
Se<br />
34<br />
2-8-18-6<br />
127.60<br />
Te<br />
52<br />
2-8-18-18-6<br />
(209)<br />
Po<br />
84<br />
-18-32-18-6<br />
–2<br />
+4<br />
+6<br />
–2<br />
+4<br />
+6<br />
+2<br />
+4<br />
F<br />
79.904<br />
Br<br />
35<br />
2-8-18-7<br />
126.904<br />
l<br />
53<br />
2-8-18-18-7<br />
(210)<br />
At<br />
85<br />
-18-32-18-7<br />
( ? )<br />
Uus<br />
117<br />
4.00260 0<br />
He<br />
2<br />
2<br />
–1 20.180<br />
Ne<br />
10<br />
2-8<br />
0<br />
22.98977<br />
11<br />
2-8-1<br />
1<br />
1.00794 +1<br />
–1<br />
H<br />
1<br />
1<br />
1<br />
37<br />
2-8-18-8-1<br />
–1<br />
+1<br />
+5<br />
–1<br />
+1<br />
+5<br />
+7<br />
83.798<br />
Kr<br />
36<br />
2-8-18-8<br />
131.29<br />
Xe<br />
54<br />
2-8-18-18-8<br />
(222)<br />
Rn<br />
86<br />
-18-32-18-8<br />
0<br />
+2<br />
0<br />
+2<br />
+4<br />
+6<br />
0<br />
132.905<br />
55<br />
2-8-18-18-8-1<br />
Symbol<br />
Relative atomic masses are based<br />
Group on 12 C = 12 (exact)<br />
Group<br />
2<br />
13 14 15 16 17 18<br />
Atomic Number<br />
+1<br />
+1<br />
9.01218 +2<br />
Be<br />
4<br />
2-2<br />
24.305<br />
Mg<br />
12<br />
2-8-2<br />
40.08<br />
Ca<br />
20<br />
2-8-8-2<br />
87.62<br />
Sr<br />
38<br />
2-8-18-8-2<br />
137.33<br />
Ba<br />
56<br />
2-8-18-18-8-2<br />
(226)<br />
+2<br />
+2<br />
+2<br />
+2<br />
3<br />
44.9559<br />
Sc<br />
21<br />
2-8-9-2<br />
88.9059<br />
Y<br />
2-8-18-9-2<br />
+3<br />
+3<br />
4<br />
KEY<br />
92.9064<br />
Nb +3<br />
+5<br />
41<br />
2-8-18-12-1<br />
180.948<br />
Ta<br />
73<br />
-18-32-11-2<br />
(262)<br />
105<br />
5<br />
Periodic Table of the Elements<br />
Atomic Mass<br />
Electron Configuration<br />
+4<br />
Db<br />
+5<br />
6<br />
(266)<br />
Sg<br />
106<br />
12.011 2-4<br />
–4<br />
6<br />
C<br />
+2<br />
+4<br />
(98)<br />
Tc<br />
43<br />
2-8-18-13-2<br />
186.207<br />
Re<br />
75<br />
-18-32-13-2<br />
(272)<br />
Bh<br />
107<br />
7<br />
Group<br />
+4<br />
+6<br />
+7<br />
+4<br />
+6<br />
+7<br />
8<br />
101.07<br />
Ru<br />
44<br />
2-8-18-15-1<br />
190.23<br />
Os<br />
76<br />
-18-32-14-2<br />
(277)<br />
Hs<br />
108<br />
+3<br />
+3<br />
+4<br />
Selected Oxidation States<br />
Note: Numbers in parentheses<br />
are mass numbers of the most<br />
stable or common isotope.<br />
9<br />
102.906<br />
Rh<br />
45<br />
2-8-18-16-1<br />
192.217<br />
Ir<br />
77<br />
-18-32-15-2<br />
(276)<br />
Mt<br />
109<br />
+3<br />
+3<br />
+4<br />
106.42<br />
Pd<br />
46<br />
2-8-18-18<br />
195.08<br />
Pt<br />
78<br />
-18-32-17-1<br />
+2<br />
+4<br />
+2<br />
+4<br />
196.967<br />
Au<br />
-18-32-18-1<br />
(281)<br />
Ds (280) Rg<br />
110<br />
10<br />
29<br />
111<br />
11 12<br />
+1<br />
+3<br />
112.41<br />
Cd<br />
48<br />
2-8-18-18-2<br />
200.59<br />
Hg<br />
80<br />
-18-32-18-2<br />
(285)<br />
Cn<br />
112<br />
+2 114.818<br />
In<br />
+1<br />
+2<br />
49<br />
2-8-18-18-3<br />
204.383<br />
Tl<br />
81<br />
-18-32-18-3<br />
(284)<br />
Uut<br />
113**<br />
+3<br />
+1<br />
+3<br />
118.71<br />
Sn<br />
50<br />
2-8-18-18-4<br />
207.2<br />
82<br />
-18-32-18-4<br />
(289)<br />
Uuq<br />
114<br />
+2<br />
+4<br />
+2<br />
+4<br />
14.0067 –3<br />
–2<br />
N<br />
–1<br />
7<br />
2-5<br />
30.97376<br />
P<br />
15<br />
2-8-5<br />
121.760<br />
51<br />
2-8-18-18-5<br />
208.980<br />
Bi<br />
83<br />
-18-32-18-5<br />
(288)<br />
Uup<br />
115<br />
+1<br />
+2<br />
+3<br />
+4<br />
+5<br />
–3<br />
+3<br />
+5<br />
+5<br />
–3<br />
+3<br />
+5<br />
+3<br />
+5<br />
32.065<br />
S<br />
16<br />
2-8-6<br />
(292)<br />
Uuh<br />
116<br />
–2<br />
+4<br />
+6<br />
35.453<br />
Cl<br />
17<br />
2-8-7<br />
–1<br />
+1<br />
+5<br />
+7<br />
39.948<br />
Ar<br />
18<br />
2-8-8<br />
18<br />
(294)<br />
Uuo<br />
118<br />
140.116<br />
Ce<br />
58<br />
232.038<br />
Th<br />
90<br />
+3<br />
+4<br />
140.908<br />
Pr +3<br />
59<br />
144.24<br />
Nd<br />
60<br />
+4 231.036<br />
Pa +4 238.029 +5<br />
U +3<br />
+4<br />
+5<br />
+6<br />
91<br />
92<br />
+3<br />
(145)<br />
Pm<br />
61<br />
+3<br />
150.36<br />
Sm<br />
62<br />
+2<br />
+3<br />
151.964<br />
Eu<br />
63<br />
+2<br />
+3<br />
157.25<br />
Gd<br />
64<br />
+3<br />
158.925<br />
(237)Np (244) Pu (243) Am (247) Cm +3 (247) Bk +3<br />
+3<br />
+4<br />
+5<br />
+6<br />
93 94<br />
+3<br />
+4<br />
+5<br />
+6<br />
65<br />
+3<br />
+4<br />
+5<br />
+6<br />
95 96 97<br />
Tb<br />
+3<br />
+4<br />
162.500<br />
Dy<br />
66<br />
(251)<br />
+3<br />
164.930<br />
Ho<br />
67<br />
+3<br />
167.259<br />
Er<br />
68<br />
Cf +3 (252) Es (257) Fm<br />
100<br />
98 99<br />
+3<br />
+3<br />
+3<br />
168.934<br />
Tm +3<br />
69<br />
(258)<br />
Md<br />
101<br />
+2<br />
+3<br />
173.04<br />
Yb<br />
70<br />
(259)<br />
No<br />
102<br />
+2<br />
+3<br />
+2<br />
+3<br />
174.9668<br />
Lu<br />
71<br />
(262)<br />
Lr<br />
103<br />
+3<br />
+3<br />
*denotes the presence of (2-8-) for elements 72 and above<br />
**The systematic names and symbols for elements of atomic numbers 113 and above<br />
will be used until the approval of trivial names by IUPAC.<br />
Source: CRC Handbook of Chemistry and Physics, 91 st ed., 2010–2011, CRC Press<br />
9<br />
R9<br />
Period<br />
1<br />
2<br />
3<br />
4<br />
5<br />
6<br />
7
Table S<br />
Properties of Selected Elements<br />
First<br />
Atomic Symbol Name Ionization<br />
Electro- Melting Boiling* Density** Atomic<br />
Number Energy negativity Point Point (g/cm 3 ) Radius<br />
(kJ/mol) (K) (K) (pm)<br />
1 H hydrogen 1312 2.2 14 20. 0.000082 32<br />
2 He helium 2372 — — 4 0.000164 37<br />
3 Li lithium 520. 1.0 454 1615 0.534 130.<br />
4 Be beryllium 900. 1.6 1560. 2744 1.85 99<br />
5 B boron 801 2.0 2348 4273 2.34 84<br />
6 C carbon 1086 2.6 — — .— 75<br />
7 N nitrogen 1402 3.0 63 77 0.001145 71<br />
8 O oxygen 1314 3.4 54 90. 0.001308 64<br />
9 F fluorine 1681 4.0 53 85 0.001553 60.<br />
10 Ne neon 2081 — 24 27 0.000825 62<br />
11 Na sodium 496 0.9 371 1156 0.97 160.<br />
12 Mg magnesium 738 1.3 923 1363 1.74 140.<br />
13 Al aluminum 578 1.6 933 2792 2.70 124<br />
14 Si silicon 787 1.9 1687 3538 2.3296 114<br />
15 P phosphorus (white) 1012 2.2 317 554 1.823 109<br />
16 S sulfur (monoclinic) 1000. 2.6 388 718 2.00 104<br />
17 Cl chlorine 1251 3.2 172 239 0.002898 100.<br />
18 Ar argon 1521 — 84 87 0.001633 101<br />
19 K potassium 419 0.8 337 1032 0.89 200.<br />
20 Ca calcium 590. 1.0 1115 1757 1.54 174<br />
21 Sc scandium 633 1.4 1814 3109 2.99 159<br />
22 Ti titanium 659 1.5 1941 3560. 4.506 148<br />
23 V vanadium 651 1.6 2183 3680. 6.0 144<br />
24 Cr chromium 653 1.7 2180. 2944 7.15 130.<br />
25 Mn manganese 717 1.6 1519 2334 7.3 129<br />
26 Fe iron 762 1.8 1811 3134 7.87 124<br />
27 Co cobalt 760. 1.9 1768 3200. 8.86 118<br />
28 Ni nickel 737 1.9 1728 3186 8.90 117<br />
29 Cu copper 745 1.9 1358 2835 8.96 122<br />
30 Zn zinc 906 1.7 693 1180. 7.134 120.<br />
31 Ga gallium 579 1.8 303 2477 5.91 123<br />
32 Ge germanium 762 2.0 1211 3106 5.3234 120.<br />
33 As arsenic (gray) 944 2.2 1090. — 5.75 120.<br />
34 Se selenium (gray) 941 2.6 494 958 4.809 118<br />
35 Br bromine 1140. 3.0 266 332 3.1028 117<br />
36 Kr krypton 1351 — 116 120. 0.003425 116<br />
37 Rb rubidium 403 0.8 312 961 1.53 215<br />
38 Sr strontium 549 1.0 1050. 1655 2.64 190.<br />
39 Y yttrium 600. 1.2 1795 3618 4.47 176<br />
40 Zr zirconium 640. 1.3 2128 4682 6.52 164<br />
R10
First<br />
Atomic Symbol Name Ionization<br />
Electro- Melting Boiling* Density** Atomic<br />
Number Energy negativity Point Point (g/cm 3 ) Radius<br />
(kJ/mol) (K) (K) (pm)<br />
41 Nb niobium 652 1.6 2750. 5017 8.57 156<br />
42 Mo molybdenum 684 2.2 2896 4912 10.2 146<br />
43 Tc technetium 702 2.1 2430. 4538 11 138<br />
44 Ru ruthenium 710. 2.2 2606 4423 12.1 136<br />
45 Rh rhodium 720. 2.3 2237 3968 12.4 134<br />
46 Pd palladium 804 2.2 1828 3236 12.0 130.<br />
47 Ag silver 731 1.9 1235 2435 10.5 136<br />
48 Cd cadmium 868 1.7 594 1040. 8.69 140.<br />
49 In indium 558 1.8 430. 2345 7.31 142<br />
50 Sn tin (white) 709 2.0 505 2875 7.287 140.<br />
51 Sb antimony (gray) 831 2.1 904 1860. 6.68 140.<br />
52 Te tellurium 869 2.1 723 1261 6.232 137<br />
53 I iodine 1008 2.7 387 457 4.933 136<br />
54 Xe xenon 1170. 2.6 161 165 0.005366 136<br />
55 Cs cesium 376 0.8 302 944 1.873 238<br />
56 Ba barium 503 0.9 1000. 2170. 3.62 206<br />
57 La lanthanum 538 1.1 1193 3737 6.15 194<br />
Elements 58–71 have been omitted.<br />
72 Hf hafnium 659 1.3 2506 4876 13.3 164<br />
73 Ta tantalum 728 1.5 3290. 5731 16.4 158<br />
74 W tungsten 759 1.7 3695 5828 19.3 150.<br />
75 Re rhenium 756 1.9 3458 5869 20.8 141<br />
76 Os osmium 814 2.2 3306 5285 22.587 136<br />
77 Ir iridium 865 2.2 2719 4701 22.562 132<br />
78 Pt platinum 864 2.2 2041 4098 21.5 130.<br />
79 Au gold 890. 2.4 1337 3129 19.3 130.<br />
80 Hg mercury 1007 1.9 234 630. 13.5336 132<br />
81 Tl thallium 589 1.8 577 1746 11.8 144<br />
82 Pb lead 716 1.8 600. 2022 11.3 145<br />
83 Bi bismuth 703 1.9 544 1837 9.79 150.<br />
84 Po polonium 812 2.0 527 1235 9.20 142<br />
85 At astatine — 2.2 575 — — 148<br />
86 Rn radon 1037 — 202 211 0.009074 146<br />
87 Fr francium 393 0.7 300. — — 242<br />
88 Ra radium 509 0.9 969 — 5 211<br />
89 Ac actinium 499 1.1 1323 3471 10. 201<br />
Elements 90 and above have been omitted.<br />
*boiling point at standard pressure<br />
**density of solids and liquids at room temperature and density of gases at 298 K and 101.3 kPa<br />
— no data available<br />
Source: CRC Handbook for Chemistry and Physics, 91 st ed., 2010–2011, CRC Press<br />
R11
Table T<br />
Important Formulas and Equations<br />
d = density<br />
m<br />
Density d = m = mass<br />
V<br />
V = volume<br />
Mole Calculations number of moles =<br />
given mass<br />
gram-formula mass<br />
measured value – accepted value<br />
Percent Error % error = × 100<br />
accepted value<br />
mass of part<br />
Percent Composition % composition by mass = × 100<br />
mass of whole<br />
mass of solute<br />
parts per million = × 1000000<br />
mass of solution<br />
Concentration<br />
molarity =<br />
moles of solute<br />
liter of solution<br />
Combined Gas Law<br />
P<br />
P = pressure<br />
1<br />
V 1<br />
P<br />
= 2<br />
V 2<br />
V = volume<br />
T 1<br />
T 2 T = temperature<br />
M A<br />
= molarity of H + M B<br />
= molarity of OH –<br />
Titration M A<br />
V A<br />
= M B<br />
V B<br />
V A<br />
= volume of acid V B<br />
= volume of base<br />
q = mCΔT q = heat H f<br />
= heat of fusion<br />
Heat q = mH f<br />
m = mass H v<br />
= heat of vaporization<br />
q = mH v<br />
C=specific heat capacity<br />
ΔT = change in temperature<br />
Temperature<br />
K = °C + 273<br />
K = kelvin<br />
°C = degree Celsius<br />
R12
Collier County CHEMISTRY EXAM FORMULA AND RESOURCE PACKET<br />
GENERAL<br />
D m V<br />
[ ExperimentalValue AcceptedVa lue]<br />
% error <br />
x100<br />
AcceptedVa lue<br />
% yield <br />
ExperimentalYield<br />
TheoreticalYield<br />
x100<br />
CONCENTRATIONS<br />
moles of solute<br />
M = Molarity <br />
liters of solution<br />
KEY<br />
P = pressure<br />
V = volume<br />
T = temperature<br />
n = number of moles<br />
m = mass<br />
M = molar mass (grams/mole)<br />
D = density<br />
KE = kinetic energy<br />
Avogadro’s Number = 6.02 x 10 23<br />
GASES, LIQUIDS, SOLUTIONS<br />
m = Molality <br />
M1V1 M2V2<br />
S1<br />
P1<br />
S 2<br />
P 2<br />
ACID/BASE<br />
pH = - log[H + ]<br />
[H + ]=10 -pH<br />
moles of solute<br />
kilograms of solvent<br />
<br />
Gas constant<br />
R 8.314 L kPa L atm L mmHg<br />
0.0821 62.4<br />
K mol K mol K mol<br />
1 atm = 760 mmHg = 760 torr = 101.3 kPa<br />
K = o C + 273<br />
o C = K - 273<br />
STP = Standard Temperature and Pressure = 0 o C<br />
and 1 atm<br />
P V<br />
1 1<br />
<br />
P2V<br />
2<br />
pOH = - log [OH - ]<br />
[OH - ]= 10 -pOH<br />
pH + pOH = 14<br />
Kw = [H + ] x [OH - ] = 1.0 x 10 -14 M 2<br />
V<br />
T<br />
1<br />
1<br />
P<br />
T<br />
1<br />
1<br />
V<br />
<br />
T<br />
P<br />
<br />
T<br />
2<br />
2<br />
2<br />
2<br />
Or V1T2 = V2T1<br />
Or P1T2 = P2T1<br />
THERMOCHEMISTRY<br />
ΔH= mCΔT, where ΔT = T f - T<br />
P1V<br />
1<br />
T<br />
1<br />
<br />
P2V<br />
2<br />
T<br />
2<br />
Or<br />
P1V1T2=P2V2T1<br />
q = mCΔT<br />
PV<br />
nRT<br />
Specific Heat of Water = 4.18 J/g*˚C or 1.0 cal/g*˚C<br />
Specific Heat of Ice = 2.1 J/g*˚C or 0.5 cal/g*˚C<br />
Specific Heat of Steam = 2.0 J/g*˚C or 0.4 cal/g*˚C<br />
P<br />
Total<br />
P<br />
1<br />
P<br />
2<br />
Rate A<br />
Rate B<br />
...<br />
<br />
Molar MassB<br />
Molar MassA<br />
R13
CHEMISTRY EXAM FORMULA AND RESOURCE PACKET<br />
Solubility of Compounds at 25 o C and 1 atm<br />
acetate<br />
bromide<br />
carbonate<br />
chlorate<br />
chloride<br />
hydroxide<br />
iodide<br />
nitrate<br />
oxide<br />
perchlorate<br />
phosphate<br />
sulfate<br />
sulfide<br />
aluminum S S - S S I S S I S I S d<br />
ammonium S S S S S S S S - S S S S<br />
barium S S I S S S S S sS S I I d<br />
calcium S S I S S S S S sS S I sS I<br />
copper(II) S S - S S I S S I S I S I<br />
iron(II) S S I S S I S S I S I S I<br />
iron(III) S S - S S I S S I S I sS d<br />
lithium S S sS S S S S S S S sS S S<br />
magnesium S S I S S I S S I S I S d<br />
potassium S S S S S S S S S S S S S<br />
silver sS I I S I - I S I S I sS I<br />
sodium S S S S S S S S S S S S S<br />
strontium S S I S S S S S S S I I I<br />
zinc S S I S S I S S I S I S I<br />
S=soluble<br />
sS = slightly soluble in water<br />
I = insoluble in water<br />
d = decomposes in water<br />
- = no such compound<br />
R14
CHEMISTRY EXAM FORMULA AND RESOURCE PACKET<br />
Common Polyatomic Ions<br />
1- Charge 2- Charge 3- Charge<br />
Formula Name Formula Name Formula Name<br />
H₂PO₄<br />
Dihydrogen<br />
Hydrogen<br />
HPO₄<br />
Phosphate<br />
phosphate PO₃ Phosphite<br />
C₂H₃O₂ Acetate C₂O₄ Oxalate PO₄ Phosphate<br />
HSO₃ Hydrogen<br />
sulfite<br />
SO₃ Sulfite<br />
HSO₄<br />
HCO₃<br />
Hydrogen<br />
SO₄<br />
sulfate<br />
Sulfate<br />
Hydrogen<br />
carbonate<br />
CO₃<br />
Carbonate<br />
NO₂ Nitrite CrO₄ Chromate 1+ Charge<br />
NO₃ Nitrate Cr₂O₇ Dichromate Formula Name<br />
CN Cyanide SiO₃ Silicate NH₄ Ammonium<br />
OH<br />
MnO₄<br />
ClO<br />
Hydroxide<br />
Permanganate<br />
Hypochlorite<br />
S₂O₃ Thiosulfate H₃O Hydronium<br />
ClO₂ Chlorite<br />
ClO₃<br />
Chlorate<br />
ClO₄ Perchlorate<br />
R15
Activity Series of Metals<br />
Name<br />
Symbol<br />
D<br />
Lithium<br />
Li<br />
e<br />
Potassium<br />
K<br />
c<br />
r<br />
Barium<br />
Ba<br />
e<br />
Calcium<br />
Ca<br />
a<br />
Sodium<br />
Na<br />
s<br />
i<br />
Magnesium<br />
Mg<br />
n<br />
Aluminum<br />
Al<br />
g<br />
Zinc<br />
Zn<br />
Iron<br />
Fe<br />
A<br />
c<br />
Nickel<br />
Ni<br />
t<br />
Tin<br />
Sn<br />
i<br />
v<br />
Lead<br />
Pb<br />
i<br />
(Hydrogen)<br />
(H)*<br />
t<br />
Copper<br />
Cu<br />
y<br />
Mercury<br />
Hg<br />
Silver<br />
Ag<br />
Gold<br />
Au<br />
*Metals from Li to Na will replace H from acids and water; from Mg to<br />
Pb they will replace H from acids only.<br />
Decreasing<br />
Activity<br />
Activity Series of Nonmetal (Halogens)<br />
Name<br />
Symbol<br />
Fluorine F 2<br />
Chlorine Cl 2<br />
Bromine Br 2<br />
Iodine I 2<br />
R16