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Honors Chemistry
Class Policies and Grading
The students will receive a Unit Outline at the beginning of each Unit. It will
have information about the assignments that they will do, what it’s grade
classification will be, what action they will need to do to complete the
assignment and when it is due.
The students will receive a Weekly Memo of the activities they will be
responsible for that week. It will serve to inform the students of the learning
goal for the week. It will also give the students any special information
about that week.
The students will also receive daily lectures and assignments that are
designed to teach and re-enforce information related to the learning goal.
This will be time in which new material will be taught and reviewed and will
give the students the opportunity to ask questions regarding the concepts
being taught.
The students will work with a Lab partner and also be in a Lab group, but it
will be up to the individual student to do his or her part of all assignments
and the individual student will ultimately be responsible for all information
presented in the class.
The students will be required to follow all District and School Policies and to
follow all Lab Safety Procedures, which they will be given and will sign,
while performing labs. Students should come to class on time and with the
supplies needed for that class.
The following grading policy will be used.
Percent of Final Grade
Notebook 40%
Test/Projects 30%
Labs/Quizzes 20%
Work 10%
The students will be given a teacher generated Mid Term and a District
Final.
0

Unit 1
Measurement Lab
Separation of Mixtures Lab with Lab Write Up
Unit 2
Flame Test Lab
Nuclear Decay Lab
Element Marketing Project
Unit 3
Golden Penny Lab with Lab Write Up
Molecular Geometry
Research Presentation on a Chemical
Mid Term
Unit 4
Double Displacement Lab
Stoichiometry Lab with Lab Write Up
Mole Educational Demonstration Project
Unit 5
Gas Laws Lab with Lab Write Up
States of Matter Lab
Teach a Gas Law Project
Unit 6
Dilutions Lab
Titration Lab
District Final
1

Unit 1 (22 days)
Chapter 1 Introduction to Chemistry
Honors Chemistry
2016/2017 Syllabus
3 days
1.1 The Scope of Chemistry 1.3 Thinking Like a Scientist
1.2 Chemistry and You 1.4 Problem Solving in Chemistry
Chapter 2 Matter and Change
2.1 Properties of Matter 2.3 Elements and Compounds
2.2 Mixtures 2.4 Chemical Reactions
Chapter 3 Scientific Measurement
9 days
10 days
3.1 Using and Expressing Measurements 3.3 Solving Conversion Problems
3.2 Units of Measurement
Unit 2 (15 days)
Chapter 4 Atomic Structure
5 days
4.1 Defining the Atom 4.3 Distinguishing Among Atoms
4.2 Structure of the Nuclear Atom
Chapter 5 Electrons in Atoms
5 days
5.1 Revising the Atomic Model 5.2 Electron Arrangement in Atoms
5.3 Atomic Emission Spectrum and the Quantum Mechanical Model
Chapter 6 The Periodic Table
6.1 Organizing the Elements 6.3 Periodic Trends
6.2 Classifying Elements
Unit 3 (22 days)
Chapter 25 Nuclear Chemistry
25.1 Nuclear Radiation 25.3 Fission and Fusion
25.2 Nuclear Transformations 25.4 Radiation in Your Life
Chapter 7 Ionic and Metallic Bonding
7.1 Ions 7.3 Bonding in Metals
7.2 Ionic Bonds and Ionic Compounds
Chapter 8 Covalent Bonding
5 days
6 days
8 days
8 days
8.1 Molecular Compounds 8.3 Bonding Theories
8.2 The Nature of Covalent Bonding 8.4 Polar Bonds and Molecules
Unit 4 (14 days)
Chapter 9 Chemical Names and Formulas
6 days
9.1 Naming Ions 9.3 Naming & Writing Formulas Molecular Compounds
9.2 Naming and Writing Formulas for Ionic Compounds 9.4 Names for Acids and Bases
Chapter 22 Hydrocarbons Compounds
22.1 Hydrocarbons 22.4 Hydrocarbon Rings
Chapter 23 Functional Groups
4 days
4 days
23.1 Introduction to Functional Groups 23.4 Alcohols, Ethers, and Amines
2

Unit 5 (28 days)
Chapter 10 Chemical Quantities 8 days
10.1 The Mole: A Measurement of Matter 10.3 % Composition & Chem. Formulas
10.2 Mole-Mass and Mole-Volume Relationships
Chapter 11 Chemical Reactions 8 days
11.1 Describing Chemical Reactions 11.3 Reactions in Aqueous Solutions
11.2 Types of Chemical Reactions
Chapter 12 Stoichiometry 12 days
12.1 The Arithmetic of Equations 12.3 Limiting Reagent and % Yield
12.2 Chemical Calculations
Unit 6 (22 days)
Chapter 13 States of Matter 6 days
13.1 The Nature of Gases 13.3 The Nature of Solids
13.2 The Nature of Liquids 13.4 Changes in State
Chapter 14 The Behavior of Gases 10 days
14.1 Properties of Gases 14.3 Ideal Gases
14.2 The Gas Laws 14.4 Gases: Mixtures and Movement
Chapter 15 Water and Aqueous Systems 6 days
15.1 Water and its Properties 15.3 Heterogeneous Aqueous Systems
15.2 Homogeneous Aqueous Systems
Unit 7 (18 days)
Chapter 16 Solutions 8 days
16.1 Properties of Solutions 16.3 Colligative Properties of Solutions
16.2 Concentrations of Solutions 16.4 Calc. Involving Colligative Property
Chapter 17 Thermochemistry 5 days
17.1 The Flow of Energy 17.3 Heat in Changes of State
17.2 Measuring and Expressing Enthalpy Change 17.4 Calculating Heats in Reactions
Chapter 18 Reaction Rates and Equilibrium 5 days
18.1 Rates of Reactions 18.3 Reversible Reaction & Equilibrium
18.2 The Progress of Chemical Reactions 18.5 Free Energy and Entropy
Unit 8 (14 days)
Chapter 19 Acid and Bases 10 days
19.1 Acid-Base Theories 19.4 Neutralization Reactions
19.2 Hydrogen Ions and Acidity 19.5 Salts in Solutions
19.3 Strengths of Acids and Bases
Chapter 20 Oxidation-Reduction Reactions 4 days
20.1 The Meaning of Oxidation and Reduction 20.3 Describing Redox Equations
20.2 Oxidation Numbers
3

Lorenzo Walker Technical High School
MUSTANG LABORATORIES
Chemistry Safety
Safety in the MUSTANG LABORATORIES - Chemistry Laboratory
Working in the chemistry laboratory is an interesting and rewarding experience. During your labs, you will be actively
involved from beginning to end—from setting some change in motion to drawing some conclusion. In the laboratory, you
will be working with equipment and materials that can cause injury if they are not handled properly.
However, the laboratory is a safe place to work if you are careful. Accidents do not just happen; they are caused—by
carelessness, haste, and disregard of safety rules and practices. Safety rules to be followed in the laboratory are listed
below. Before beginning any lab work, read these rules, learn them, and follow them carefully.
General
1. Be prepared to work when you arrive at the lab. Familiarize yourself with the lab procedures before beginning the lab.
2. Perform only those lab activities assigned by your teacher. Never do anything in the laboratory that is not called for in
the laboratory procedure or by your teacher. Never work alone in the lab. Do not engage in any horseplay.
3. Work areas should be kept clean and tidy at all times. Only lab manuals and notebooks should be brought to the work
area. Other books, purses, brief cases, etc. should be left at your desk or placed in a designated storage area.
4. Clothing should be appropriate for working in the lab. Jackets, ties, and other loose garments should be removed. Open
shoes should not be worn.
5. Long hair should be tied back or covered, especially in the vicinity of open flame.
6. Jewelry that might present a safety hazard, such as dangling necklaces, chains, medallions, or bracelets should not be
worn in the lab.
7. Follow all instructions, both written and oral, carefully.
8. Safety goggles and lab aprons should be worn at all times.
9. Set up apparatus as described in the lab manual or by your teacher. Never use makeshift arrangements.
10. Always use the prescribed instrument (tongs, test tube holder, forceps, etc.) for handling apparatus or equipment.
11. Keep all combustible materials away from open flames.
12. Never touch any substance in the lab unless specifically instructed to do so by your teacher.
13. Never put your face near the mouth of a container that is holding chemicals.
14. Never smell any chemicals unless instructed to do so by your teacher. When testing for odors, use a wafting motion to
direct the odors to your nose.
15. Any activity involving poisonous vapors should be conducted in the fume hood.
16. Dispose of waste materials as instructed by your teacher.
17. Clean up all spills immediately.
18. Clean and wipe dry all work surfaces at the end of class. Wash your hands thoroughly.
19. Know the location of emergency equipment (first aid kit, fire extinguisher, fire shower, fire blanket, etc.) and how to use them.
20. Report all accidents to the teacher immediately.
Handling Chemicals
21. Read and double check labels on reagent bottles before removing any reagent. Take only as much reagent as you
need.
22. Do not return unused reagent to stock bottles.
23. When transferring chemical reagents from one container to another, hold the containers out away from your body.
24. When mixing an acid and water, always add the acid to the water.
25. Avoid touching chemicals with your hands. If chemicals do come in contact with your hands, wash them immediately.
26. Notify your teacher if you have any medical problems that might relate to lab work, such as allergies or asthma.
27. If you will be working with chemicals in the lab, avoid wearing contact lenses. Change to glasses, if possible, or notify
the teacher.
Handling Glassware
28. Glass tubing, especially long pieces, should be carried in a vertical position to minimize the likelihood of breakage and
to avoid stabbing anyone.
29. Never handle broken glass with your bare hands. Use a brush and dustpan to clean up broken glass. Dispose of the
glass as directed by your teacher.
4

30. Always lubricate glassware (tubing, thistle tubes, thermometers, etc.) with water or glycerin before attempting to insert
it into a rubber stopper.
31. Never apply force when inserting or removing glassware from a stopper. Use a twisting motion. If a piece of glassware
becomes "frozen" in a stopper, take it to your teacher.
32. Do not place hot glassware directly on the lab table. Always use an insulating pad of some sort.
33. Allow plenty of time for hot glass to cool before touching it. Hot glass can cause painful burns. (Hot glass looks cool.)
Heating Substances
34. Exercise extreme caution when using a gas burner. Keep your head and clothing away from the flame.
35. Always turn the burner off when it is not in use.
36. Do not bring any substance into contact with a flame unless instructed to do so.
37. Never heat anything without being instructed to do so.
38. Never look into a container that is being heated.
39. When heating a substance in a test tube, make sure that the mouth of the tube is not pointed at yourself or anyone
else.
40. Never leave unattended anything that is being heated or is visibly reacting.
First Aid in the MUSTANG LABORATORIES - Chemistry Laboratory
Accidents do not often happen in well-equipped chemistry laboratories if students understand safe laboratory procedures
and are careful in following them. When an occasional accident does occur, it is likely to be a minor one.
The instructor will assist in treating injuries such as minor cuts and burns. However, for some types of injuries, you must
take action immediately. The following information will be helpful to you if an accident occurs.
1. Shock. People who are suffering from any severe injury (for example, a bad burn or major loss of blood) may be in a
state of shock. A person in shock is usually pale and faint. The person may be sweating, with cold, moist skin and a weak,
rapid pulse. Shock is a serious medical condition. Do not allow a person in shock to walk anywhere—even to the campus
security office. While emergency help is being summoned, place the victim face up in a horizontal position, with the feet
raised about 30 centimeters. Loosen any tightly fitting clothing and keep him or her warm.
2. Chemicals in the Eyes. Getting any kind of a chemical into the eyes is undesirable, but certain chemicals are
especially harmful. They can destroy eyesight in a matter of seconds. Because you will be wearing safety goggles at all
times in the lab, the likelihood of this kind of accident is remote. However, if it does happen, flush your eyes with water
immediately. Do NOT attempt to go to the campus office before flushing your eyes. It is important that flushing with water
be continued for a prolonged time—about 15 minutes.
3. Clothing or Hair on Fire. A person whose clothing or hair catches on fire will often run around hysterically in an
unsuccessful effort to get away from the fire. This only provides the fire with more oxygen and makes it burn faster. For
clothing fires, throw yourself to the ground and roll around to extinguish the flames. For hair fires, use a fire blanket to
smother the flames. Notify campus security immediately.
4. Bleeding from a Cut. Most cuts that occur in the chemistry laboratory are minor. For minor cuts, apply pressure to the
wound with a sterile gauze. Notify campus security of all injuries in the lab. If the victim is bleeding badly, raise the
bleeding part, if possible, and apply pressure to the wound with a piece of sterile gauze. While first aid is being given,
someone else should notify the campus security officer.
5. Chemicals in the Mouth. Many chemicals are poisonous to varying degrees. Any chemical taken into the mouth
should be spat out and the mouth rinsed thoroughly with water. Note the name of the chemical and notify the campus
office immediately. If the victim swallows a chemical, note the name of the chemical and notify campus security
immediately.
If necessary, the campus security officer or administrator will contact the Poison Control Center, a hospital emergency
room, or a physician for instructions.
6. Acid or Base Spilled on the Skin.
Flush the skin with water for about 15 minutes. Take the victim to the campus office to report the injury.
7. Breathing Smoke or Chemical Fumes.
All experiments that give off smoke or noxious gases should be conducted in a well-ventilated fume hood. This will make
an accident of this kind unlikely. If smoke or chemical fumes are present in the laboratory, all persons—even those who
do not feel ill—should leave the laboratory immediately. Make certain that all doors to the laboratory are closed after the
last person has left. Since smoke rises, stay low while evacuating a smoke-filled room. Notify campus security
immediately.
5

MUSTANG LABORATORIES
COMMITMENT TO SAFETY IN THE LABORATORY
As a student enrolled in Chemistry at Lorenzo Walker Technical High
School, I agree to use good laboratory safety practices at all times. I
also agree that I will:
1. Conduct myself in a professional manner, respecting both my personal safety and the safety of
others in the laboratory.
2. Wear proper and approved safety glasses or goggles in the laboratory at all times.
3. Wear sensible clothing and tie back long hair in the laboratory. Understand that open-toed shoes
pose a hazard during laboratory classes and that contact lenses are an added safety risk.
4. Keep my lab area free of clutter during an experiment.
5. Never bring food or drink into the laboratory, nor apply makeup within the laboratory.
6. Be aware of the location of safety equipment such as the fire extinguisher, eye wash station, fire
blanket, first aid kit. Know the location of the nearest telephone and exits.
7. Read the assigned lab prior to coming to the laboratory.
8. Carefully read all labels on all chemical containers before using their contents, remove a small
amount of reagent properly if needed, do not pour back the unused chemicals into the original
container.
9. Dispose of chemicals as directed by the instructor only. At no time will I pour anything down the
sink without prior instruction.
10. Never inhale fumes emitted during an experiment. Use the fume hood when instructed to do so.
11. Report any accident immediately to the instructor, including chemical spills.
12. Dispose of broken glass and sharps only in the designated containers.
13. Clean my work area and all glassware before leaving the laboratory.
14. Wash my hands before leaving the laboratory.
NAME _________________________
Juan Roque
PARENT NAME ____________________________
Yvonne Roque
PERIOD ________________________
1 PARENT NUMBER _________________________
239-913-8612
SIGNATURE ____________________________
DATE ____________________________________
8/25/16
6

7

Chapter 1
Unit 1
Introduction to Chemistry
The students will learn why and how to solve problems using
chemistry.
Identify what is science, what clearly is not science, and what superficially
resembles science (but fails to meet the criteria for science).
Students will identify a phenomenon as science or not science.
Science
Observation
Inference
Hypothesis
Identify which questions can be answered through science and which
questions are outside the boundaries of scientific investigation, such as
questions addressed by other ways of knowing, such as art, philosophy, and
religion.
Students will differentiate between problems and/or phenomenon that can and
those that cannot be explained or answered by science.
Students will differentiate between problems and/or phenomenon that can and
those that cannot be explained or answered by science.
Observation
Inference
Hypothesis
Theory
Controlled experiment
Describe how scientific inferences are drawn from scientific observations
and provide examples from the content being studied.
Students will conduct and record observations.
Students will make inferences.
Students will identify a statement as being either an observation or inference.
Students will pose scientific questions and make predictions based on
inferences.
Inference
Observation
Hypothesis
Controlled experiment
Identify sources of information and assess their reliability according to the
strict standards of scientific investigation.
Students will compare and assess the validity of known scientific information
from a variety of sources:
8

Print vs. print
Online vs. online
Print vs. online
Students will conduct an experiment using the scientific method and compare
with other groups.
Controlled experiment
Investigation
Peer Review
Accuracy
Precision
Percentage Error
Chapter 2
Matter and Change
The students will learn what properties are used to describe
matter and how matter can change its form.
Differentiate between physical and chemical properties and physical and
chemical changes of matter.
Students will be able to identify physical and chemical properties of various
substances.
Students will be able to identify indicators of physical and chemical changes.
Students will be able to calculate density.
mass
physical property
volume
chemical property
vapor
extensive property
Chapter 3
mixture
intensive property
solution
element
compound
Scientific Measurements
The students will be able to solve conversion problems using
measurements.
Determine appropriate and consistent standards of measurement for the
data to be collected in a survey or experiment.
Students will participate in activities to collect data using standardized
measurement.
Students will be able to manipulate/convert data collected and apply the data
to scientific situations.
Scientific notation
International System of Units (SI)
Significant figures
Accepted value
Experimental value
Percent error
Dimensional analysis
9

K H D B D C M
KING HENRY DIED DRINKING CHOCOLATE MILK
0.001
.01
Kilo hecto deca base unit deci centi milli
.1
10
10

To use the Stair-Step method, find the prefix the original measurement starts with. (ex. milligram)
If there is no prefix, then you are starting with a base unit.
Find the step which you wish to make the conversion to. (ex. decigram)
Count the number of steps you moved, and determine in which direction you moved (left or right).
The decimal in your original measurement moves the same number of places as steps you moved and in the
same direction. (ex. milligram to decigram is 2 steps to the left, so 40 milligrams = .40 decigrams)
If the number of steps you move is larger than the number you have, you will have to add zeros to hold the
places. (ex. kilometers to meters is three steps to the right, so 10 kilometers would be equal to 10,000 m)
That’s all there is to it! You need to be able to count to 6, and know your left from your right!
1) Write the equivalent
a) 5 dm =_______m .5 b) 4 mL = ______L .004 c) 8 g = _______mg 8000
d) 9 mg =_______g .009 e) 2 mL = ______L .002 f) 6 kg = 6000 _____g
g) 4 cm =_______m .04 cm^3= mL h) 12 mg = ______ .0012 g i) 6.5 cm 3 = _______L .0065
j) 7.02 mL =_____cm 7.02 3 k) .03 hg = _______ 30 dg l) 6035 mm .06035 _____cm
m) .32 m = _______cm 32 n) 38.2 g = .00382 _____kg
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
than 1 kg? Explain your answer.
1k=1000g
222g
less 1 kilo is equal to 1k grams
3. Wanda needs to move 110 kg of rocks. She can carry l0 hg each trip. How many trips must she
make? Explain your answer.
1100
4. Dr. O is playing in her garden again She needs 1 kg of potting soil for her plants. She has 750 g.
How much more does she need? Explain your answer.
she needs 250 more grams 1,000-750=
5. Weather satellites orbit Earth at an altitude of 1,400,000 meters. What is this altitude in kilometers?
6. Which unit would you use to measure the capacity? Write milliliter or liter.
a) a bucket __________ liter
b) a thimble __________ milliliter
c) a water storage tank__________ liter
d) a carton of juice__________ liter
7. Circle the more reasonable measure:
a) length of an ant 5mm or 5cm
b) length of an automobile 5 m or 50 m
c) distance from NY to LA 450 km or 4,500 km
d) height of a dining table 75 mm or 75 cm
8. Will a tablecloth that is 155 cm long cover a table that is 1.6 m long? Explain your answer.
no 1.6 meters is longer 5.25 ft 5.09 ft
9. A dollar bill is 15.6 cm long. If 200 dollar bills were laid end to end, how many meters long would
the line be?
3,120 cm long
10. The ceiling in Jan’s living room is 2.5 m high. She has a hanging lamp that hangs down 41 cm.
Her husband is exactly 2 m tall. Will he hit his head on the hanging lamp? Why or why not?
yes he is way taller than the lamp and could run into it.
12

Using SI Units
Match the terms in Column II with the descriptions in Column I. Write the letters of the correct term in
the blank on the left.
Column I Column II
_____ k 1. distance between two points
a. time
_____ e 2. SI unit of length
_____ m 3. tool used to measure length
_____ g 4. units obtained by combining other units
_____ b 5. amount of space occupied by an object
_____ d 6. unit used to express volume
_____ f 7. SI unit of mass
_____ d 8. amount of matter in an object
_____ k 9. mass per unit of volume
_____ j 10. temperature scale of most laboratory thermometers
_____ 1 11. instrument used to measure mass
_____ k 12. interval between two events
_____ o 13. SI unit of temperature
_____ i 14. SI unit of time
_____ n 15. instrument used to measure temperature
b. volume
c. mass
d. density
e. meter
f. kilogram
g. derived
h. liter
i. second
j. Kelvin
k. length
1. balance
m. meterstick
n. thermometer
o. Celsius
Circle the two terms in each group that are related. Explain how the terms are related.
16. Celsius degree, mass, Kelvin _____________________________________________________
both are temperatures used to scale most laboratory thermometers.
________________________________________________________________________________
17. balance, second, mass __________________________________________________________
instrument used to measure mass.\
________________________________________________________________________________
18. kilogram, liter, cubic centimeter __________________________________________________
________________________________________________________________________________
19. time, second, distance __________________________________________________________
unit of measure used to tell time
________________________________________________________________________________
20. decimeter, kilometer, Kelvin _____________________________________________________
both units of measure are used to tell an items specific weight.
________________________________________________________________________________
13

1. How many meters are in one kilometer? __________
1,000
2. What part of a liter is one milliliter? __________ l milli per part
3. How many grams are in two dekagrams? __________ 0.2
4. If one gram of water has a volume of one milliliter, what would the mass of one liter of water be in
kilograms?__________
.001
5. What part of a meter is a decimeter? __________ 10 deci squared parts
In the blank at the left, write the term that correctly completes each statement. Choose from the terms
listed below.
Metric SI standard ten
prefixes ten tenth
6. An exact quantity that people agree to use for comparison is a ______________ ten
.
7. The system of measurement used worldwide in science is _______________ metric .
8. SI is based on units of _______________ SI .
9. The system of measurement that was based on units of ten was the _______________ standar ten system.
10. In SI, _______________ prefixes are used with the names of the base unit to indicate the multiple of ten
that is being used with the base unit.
11. The prefix deci- means _______________ tenth .
14

Standards of Measurement
Fill in the missing information in the table below.
S.I prefixes and their meanings
Prefix
Meaning
0.001
0.01
deci- 0.1
10
hecto- 100
1000
Circle the larger unit in each pair of units.
1. millimeter, kilometer 4. centimeter, millimeter
2. decimeter, dekameter 5. hectogram, kilogram
3. hectogram, decigram
6. In SI, the base unit of length is the meter. Use this information to arrange the following units of
measurement in the correct order from smallest to largest.
Write the number 1 (smallest) through 7 - (largest) in the spaces provided.
6_____ a. kilometer
_____ 2 b. centimeter
_____ 4 c. meter
_____ e. hectometer
_____ 1 f. millimeter
_____ 3 g. decimeter
_____ 5 d. dekameter
Use your knowledge of the prefixes used in SI to answer the following questions in the spaces
provided.
7. One part of the Olympic games involves an activity called the decathlon. How many events do you
think make up the decathlon?_____________________________________________________
10
8. How many years make up a decade? _______________________________________________
10
9. How many years make up a century? ______________________________________________
100
10. What part of a second do you think a millisecond is? __________________________________
1,00th
15

The Learning Goal for this assignment is:
students will learn why and how to solve problems using chemistry.
Notes Section
if the decimal is on the right like 56000. then you have to move it left till its between the 5 and 6.
=5.6x10^4
1. 7,485 6. 1.683
2. 884.2 7. 3.622
3. 0.00002887 8. 0.00001735
4. 0.05893 9. 0.9736
5. 0.006162 10. 0.08558
11. 6.633 X 10−⁴ 16. 1.937 X 10⁴
12. 4.445 X 10−⁴ 17. 3.457 X 10⁴
13. 2.182 X 10−³ 18. 3.948 X 10−⁵
14. 4.695 X 10² 19. 8.945 X 10⁵
15. 7.274 X 10⁵ 20. 6.783 X 10²
16

17
SCIENTIFIC NOTATION RULES
How to Write Numbers in Scientific Notation
Scientific notation is a standard way of writing very large and very small numbers so that they're
easier to both compare and use in computations. To write in scientific notation, follow the form
N X 10 ᴬ
where N is a number between 1 and 10, but not 10 itself, and A is an integer (positive or negative
number).
RULE #1: Standard Scientific Notation is a number from 1 to 9 followed by a decimal and the
remaining significant figures and an exponent of 10 to hold place value.
Example:
5.43 x 10 2 = 5.43 x 100 = 543
8.65 x 10 – 3 = 8.65 x .001 = 0.00865
****54.3 x 10 1 is not Standard Scientific Notation!!!
RULE #2: When the decimal is moved to the Left the exponent gets Larger, but the value of the
number stays the same. Each place the decimal moves Changes the exponent by one (1). If you
move the decimal to the Right it makes the exponent smaller by one (1) for each place it is moved.
Example:
6000. x 10 0 = 600.0 x 10 1 = 60.00 x 10 2 = 6.000 x 10 3 = 6000
(Note: 10 0 = 1)
All the previous numbers are equal, but only 6.000 x 10 3 is in proper Scientific Notation.

RULE #3: To add/subtract in scientific notation, the exponents must first be the same.
Example:
(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.
(3.0 x 10 2 )
+ (64. x 10 2 )
67.0 x 10 2 = 6.70 x 10 3 = 6.7 x 10 3
67.0 x 10 2 is mathematically correct, but a number in standard scientific notation can only
have one number to the left of the decimal, so the decimal is moved to the left one place and
one is added to the exponent.
Following the rules for significant figures, the answer becomes 6.7 x 10 3 .
RULE #4: To multiply, find the product of the numbers, then add the exponents.
Example:
(2.4 x 10 2 ) (5.5 x 10 –4 ) = ? [2.4 x 5.5 = 13.2]; [2 + -4 = -2], so
(2.4 x 10 2 ) (5.5 x 10 –4 ) = 13.2 x 10 –2 = 1.3 x 10 – 1
RULE #5: To divide, find the quotient of the number and subtract the exponents.
Example:
(3.3 x 10 – 6 ) / (9.1 x 10 – 8 ) = ? [3.3 / 9.1 = .36]; [-6 – (-8) = 2], so
(3.3 x 10 – 6 ) / (9.1 x 10 – 8 ) = .36 x 10 2 = 3.6 x 10 1
18

Convert each number from Scientific Notation to real numbers:
1. 7.485 X 10³ 6. 1.683 X 10⁰
7,485 1.683
2. 8.842 X 10² 7. 3.622 10⁰
884.2
3.622
3. 2.887 X 10−⁵ 8. 1.735 X 10−⁵
0.00002887
0.0001735
4. 5.893 X 10−² 9. 9.736 X 10−¹
0.0583
0.9736
5. 6.162 X 10−³ 10. 8.558 X 10−²
0.0061612
0.08558
Convert each number from a real number to Scientific Notation:
11. 0.0006633 16. 1,937,000
6.63x10^-4
9.3x10^5
12. 0.0004445 17. 34,570
4.45x10^-4
3.45x10^-1
13. 0.002182 18. 0.00003948
2.18x10^3
3.94x10^5
14. 469.5 19. 894,500
4.69x10^-2
8.94x10^-5
15. 727,400 20. 678.3
7.27x10^-5
6.78x10^-2
19

The Learning Goal for this assignment is:
students will learn what properties are used to describe matter and how matter can change it's form.
Notes Section:
significant numbers you have to multiply or divide. in doing so you have round it to the nearest significant numbers
Question Sig Figs Question Add & Subtract Question Multiple & Divide
1 4 1 55.36 1 20,000
2 4 2 84.2 2 94
3 3 3 115.4 3 300
4 3 4 0.8 4 7
5 4 5 245.53 5 62
6 3 6 34.5 6 0.005
7 3 7 74.0 7 4,000
8 2 8 53.287 8 3,900,000
9 2 9 54.876 9 2
10 2 10 40.19 10 30,000,000
11 3 11 7.7 11 1,200
12 2 12 67.170 12 0.2
13 3 13 81.0 13 0.87
14 4 14 73.290 14 0.049
15 4 15 29.789 15 2,000
16 3 16 39.53 16 0.5
17 4 17 70.58 17 1.9
18 2 18 86.6 18 0.05
19 2 19 64.990 19 230
20 1 20 36.0 20 460,000

Significant Figures Rules
There are three rules on determining how many significant figures are in a
number:
1. Non-zero digits are always significant.
2. Any zeros between two significant digits are significant.
3. A final zero or trailing zeros in the DECIMAL PORTION ONLY are
significant.
Please remember that, in science, all numbers are based upon measurements (except for a very few
that are defined). Since all measurements are uncertain, we must only use those numbers that are
meaningful.
Not all of the digits have meaning (significance) and, therefore, should not be written down. In
science, only the numbers that have significance (derived from measurement) are written.
Rule 1: Non-zero digits are always significant.
If you measure something and the device you use (ruler, thermometer, triple-beam, balance, etc.)
returns a number to you, then you have made a measurement decision and that ACT of measuring
gives significance to that particular numeral (or digit) in the overall value you obtain.
Hence a number like 46.78 would have four significant figures and 3.94 would have three.
Rule 2: Any zeros between two significant digits are significant.
Suppose you had a number like 409. By the first rule, the 4 and the 9 are significant. However, to
make a measurement decision on the 4 (in the hundred's place) and the 9 (in the one's place), you
HAD to have made a decision on the ten's place. The measurement scale for this number would have
hundreds, tens, and ones marked.
Like the following example:
These are sometimes called "captured zeros."
If a number has a decimal at the end (after the one’s place) then all digits (numbers) are significant
and will be counted.
In the following example the zeros are significant digits and highlighted in blue.
960.
70050.
21

Rule 3: A final zero or trailing zeros in the decimal portion ONLY are
significant.
This rule causes the most confusion among students.
In the following example the zeros are significant digits and highlighted in blue.
0.07030
0.00800
Here are two more examples where the significant zeros are highlighted in blue.
When Zeros are Not Significant Digits
4.7 0 x 10−³
6.5 0 0 x 10⁴
Zero Type # 1 : Space holding zeros in numbers less than one.
In the following example the zeros are NOT significant digits and highlighted in red.
0.09060
0.00400
These zeros serve only as space holders. They are there to put the decimal point in its correct
location.
They DO NOT involve measurement decisions.
Zero Type # 2 : Trailing zeros in a whole number.
In the following example the zeros are NOT significant digits and highlighted in red.
200
25000
For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal point)
of the numbers ONLY. Here is what to do:
1) Count the number of significant figures in the decimal portion of each number in the problem. (The
digits to the left of the decimal place are not used to determine the number of decimal places in the
final answer.)
2) Add or subtract in the normal fashion.
3) Round the answer to the LEAST number of places in the decimal portion of any number in the
problem
The following rule applies for multiplication and division:
The LEAST number of significant figures in any number of the problem determines the number of
significant figures in the answer.
This means you MUST know how to recognize significant figures in order to use this rule.

How Many Significant Digits for Each Number?
1) 2359 = ______ 4
2) 2.445 x 10−⁵= ______ 4
3) 2.93 x 10⁴= ______ 3
4) 1.30 x 10−⁷= ______ 3
5) 2604 = ______ 4
6) 9160 = ______ 3
7) 0.0800 = ______ 3
8) 0.84 = ______ 2
9) 0.0080 = ______ 2
10) 0.00040 = ______ 2
11) 0.0520 = ______ 2
12) 0.060 = ______ 2
13) 6.90 x 10−¹= ______ 3
14) 7.200 x 10⁵= ______ 6
15) 5.566 x 10−²= ______ 4
16) 3.88 x 10⁸= ______ 9
17) 3004 = ______ 4
18) 0.021 = ______ 2
19) 240 = ______ 3
20) 500 = ______ 1

For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal point) of the
numbers ONLY. Here is what to do:
1) Count the number of significant figures in the decimal portion of each number in the problem. (The
digits to the left of the decimal place are not used to determine the number of decimal places in the
final answer.)
2) Add or subtract in the normal fashion.
3) Round the answer to the LEAST number of places in the decimal portion of any number in the
problem.
Solve the Problems and Round Accordingly...
1) 43.287 + 5.79 + 6.284 = _______ 53.36
2) 87.54 - 3.3 = _______ 84.24
3) 99.1498 + 6.5397 + 9.7 = _______ 115.4
4) 5.868 - 5.1 = _______ .76
5) 59.9233 + 86.21 + 99.396 = _______ 245.5
6) 7.7 + 26.756 = _______ 34.5
7) 66.8 + 2.3 + 4.8516 = _______ 74
8) 9.7419 + 43.545 = _______ 53.3
9) 4.8976 + 48.4644 + 1.514 = _______ 54.9
10) 4.335 + 35.85 = _______ 40.185
11) 9.448 - 1.7 = _______ 7.74
12) 75.826 - 8.6555 = _______ 67.17
13) 57.2 + 23.814 = _______ 81.01
14) 77.684 - 4.394 = _______ 73.29
15) 26.4496 + 3.339 = _______ 29.8
16) 9.6848 + 29.85 = _______ 39.53
17) 63.11 + 2.5412 + 4.93 = _______ 70.6
18) 11.2471 + 75.4 = _______ 86.64
19) 73.745 - 8.755 = _______ 65
20) 6.5238 + 1.7 + 27.79 = _______ 36.01

The following rule applies for multiplication and division:
The LEAST number of significant figures in any number of the problem determines the number of
significant figures in the answer.
This means you MUST know how to recognize significant figures in order to use this rule.
Solve the Problems and Round Accordingly...
1) 0.6 x 65.0 x 602 = __________
20,000
2) 720 ÷ 7.7 = __________
94 93.50649351
23,478 round it so its one sign number
3) 929 x 0.3 = __________
300
278.7
4) 300 ÷ 44.31 = __________
7
5) 608 ÷ 9.8 = __________ 62
6) 0.06 x 0.079 = __________
0.005
7) 0.008 x 72.91 x 7000 = __________ 4,000
8) 73.94 x 67 x 780 = __________
3,900,000
9) 0.62 x 0.097 x 40 = __________
2
10) 600 x 10 x 5030 = __________
30,000,000
11) 5200 ÷ 4.46 = __________ 1,200
12) 0.0052 x 0.4 x 107 = __________
.2
13) 0.099 x 8.8 = __________
.87
14) 0.0095 x 5.2 = __________
.049
15) 8000 ÷ 4.62 = __________ 2,000
16) 0.6 x 0.8 = __________ 0.5
17) 2.84 x 0.66 = __________
1.9
18) 0.5 x 0.09 = __________
0.05
19) 8100 ÷ 34.84 = __________
230
20) 8.24 x 6.9 x 8100 = __________
460,000
25

Dimensional Analysis
This is a way to convert from one unit of a given substance to
another unit using ratios or conversion units. What this video
www.youtube.com/watch?v=aZ3J60GYo6U
Let’ look at a couple of examples:
1. Convert 2.6 qt to mL.
First we need a ratio or conversion unit so that we can go from quarts to milliliters. 1.00 qt = 946 mL
Next write down what you are starting with
2.6 qt
Then make you conversion tree
2.6 qt
946mL
= 2,459.6mL
1 1
Then fill in the units in your ratio so that you can cancel out the original unit and will be left with the
unit you need for the answer. Cross out units, one at a time that are paired, and one on top one on
the bottom.
2.6 qt mL
qt
Now fill in the values from the ratio.
2.6 qt 946 mL
1.00 qt
Now multiply all numbers on the top and multiply all numbers on the bottom and write them as a
fraction.
2.6 qt 946 mL = 2,459.6 mL
1.00 qt 1.00
Now divide the top number by the bottom number and write that number with the unit that was not
crossed out.

1qt=32 oz 1gal = 4qts 1.00 qt = 946 mL 1L = 1000mL
2. Convert 8135.6 mL to quarts
8,135.6mL 1
1
946mL
=
8,135.6mL
946mL
8.6 qts
3. Convert 115.2 oz to mL
115.2 oz 1qt
1 32 oz
946mL
1 qt
=
108,979.2
32
3,405.6
4. Convert 2.3 g to Liters
2.3g 4 qts 946mL
1 1gal
1qts
=
1L 8,703.2
1,000mL
1,000
8.7L
5. Convert 8.42 L to oz
8.42L
1
1,000mL
1L
1qt
946mL
=
32 oz 269,440
1qt
946
284.820296oz
Go to http://science.widener.edu/svb/tutorial/ chose #7 “Converting Volume” and do 5 more in the
space provided.
1. Convert _________ 946mL to _________ qts
946mL 1qt 946
1 946mL
946
=
1qt
2. Convert _________ 8797.8mLto _________ qts
8797.8 1qt
1
946mL
=
8,797.8
946
9.3qts
3. Convert _________ 316.8oz to _________ L
316.8oz 1qt
946mL
1
32oz
1qt
4. Convert _________ 1.02g to _________ L
1.02g 4qts
946mL
1 1g
1qt
=
1L 299,692.8
32,000
1,000mL
=
1L 3,859.68
1,000mL
1,000
9.3654L
3.85968L
5. Convert _________ 8892.4mLto _________ qts
8892.4mL 1qt 8892.4
1 946mL
946
=
9.4qts

Chapter 4
Unit 2
Atomic Structure
The students will learn what makes up atoms and how are
atoms of one element different from atoms of another element.
Explore the scientific theory of atoms (also known as atomic theory) by
describing changes in the atomic model over time and why those changes
were necessitated by experimental evidence.
Students will be able to draw/identify each atomic model.
Students will be able to compare/contrast the different atomic models.
Students will be able to describe how results of experimental evidence caused
the atomic model to change.
proton
electron
neutron
nucleus
electron cloud
Explore the scientific theory of atoms (also known as atomic theory) by
describing the structure of atoms in terms of protons, neutrons and
electrons, and differentiate among these particles in terms of their mass,
electrical charges and locations within the atom.
Students will compare/contrast the characteristics of subatomic particles.
atomic number
mass number
isotope
atomic mass unit (amu)
atomic mass
28

Chapter 5
Electrons in Atoms
The students will be able to describe the arrangement of
electrons in atoms and predict what will happen when
electrons in atoms absorb or release energy.
Describe the quantization of energy at the atomic level.
Students will participate in activities to view emission spectrums using a
diffraction grating or a spectroscope.
Students will be able to explain how the spectrum lines relate to electron motion.
energy level
atomic orbital
quantum mechanical model
Chapter 6
The Periodic Table
The student will learn what information the periodic table
provides and how periodic trends can be explained.
Relate properties of atoms and their position in the periodic table to the
arrangement of their electrons.
Students will be able to compare and contrast metals, nonmetals, and metalloids.
Students will be able to describe the traits of various families on the periodic
table.
Students will be able to explain periodicity.
Students will write/represent electron configuration of various elements.
Students will be able to use a periodic table to calculate the number of p + , e - , and
n 0 .
Students will be able to calculate the average weight of mass.
periodic law
halogen
metals
noble gas
nonmetals
transition metal
metalloid
atomic radius
alkali metal
ionization energy
alkaline earth metal
electronegativity
29

The Learning Goal for this assignment is:
The students will learn what makes up atoms and how are atoms of one element different from atoms of another
element.
Notes Section
in order for a atom to be a anion atom then it has to have more electrons than protons. and the same aplies for
cations which are more positive ions. For it to be normal then it has to have the equal protons, neutrons, and electrons.
http://www.learner.org/interactives/periodic/basics_interactive.html
30

Atoms Are Building Blocks
Atoms are the basis of chemistry. They are the basis for everything in the Universe. You
should start by remembering that matter is composed of atoms. Atoms and the study of
atoms are a world unto themselves. We're going to cover basics like atomic structure
and bonding between atoms.
Smaller Than Atoms?
Are there pieces of matter that are smaller than atoms?
Sure there are. You'll soon be learning that atoms are
composed of pieces like electrons, protons, and neutrons.
But guess what? There are even smaller particles moving
around in atoms. These super-small particles can be found
inside the protons and neutrons. Scientists have many
names for those pieces, but you may have heard of
nucleons and quarks. Nuclear chemists and physicists
work together at particle accelerators to discover the
presence of these tiny, tiny, tiny pieces of matter.
Even though super-tiny atomic particles exist, you only
need to remember the three basic parts of an atom: electrons, protons, and neutrons.
What are electrons, protons, and neutrons? A picture works best to show off the idea.
You have a basic atom. There are three types of pieces in that atom: electrons, protons,
and neutrons. That's all you have to remember. Three things! As you know, there are
almost 120 known elements in the periodic table. Chemists and physicists haven't
stopped there. They are trying to make new ones in labs every day. The thing that
makes each of those elements different is the number of electrons, protons, and
neutrons. The protons and neutrons are always in the center of the atom. Scientists call
the center region of the atom the nucleus. The nucleus in
a cell is a thing. The nucleus in an atom is a place where
you find protons and neutrons. The electrons are always
found whizzing around the center in areas called shells or
orbitals.
You can also see that each piece has either a "+", "-", or a
"0." That symbol refers to the charge of the particle. Have
you ever heard about getting a shock from a socket, static
electricity, or lightning? Those are all different types of
electric charges. Those charges are also found in tiny particles of matter. The electron
always has a "-", or negative, charge. The proton always has a "+", or positive, charge. If
the charge of an entire atom is "0", or neutral, there are equal numbers of positive and
negative pieces. Neutral means there are equal numbers of electrons and protons. The
third particle is the neutron. It has a neutral charge, also known as a charge of zero. All
atoms have equal numbers of protons and electrons so that they are neutral. If there are
more positive protons or negative electrons in an atom, you have a special atom called
an ion.
31

Looking at Ions
We haven’t talked about ions before, so let’s get down to basics. The
atomic number of an element, also called a proton number, tells you the
number of protons or positive particles in an atom. A normal atom has a
neutral charge with equal numbers of positive and negative particles.
That means an atom with a neutral charge is one where the number of
electrons is equal to the atomic number. Ions are atoms with extra
electrons or missing electrons. When you are missing an electron or
two, you have a positive charge. When you have an extra electron
or two, you have a negative charge.
What do you do if you are a sodium (Na) atom? You have eleven
electrons — one too many to have an entire shell filled. You need to
find another element that will take that electron away from you. When you lose that
electron, you will you’ll have full shells. Whenever an atom has full shells, we say it is
"happy." Let's look at chlorine (Cl). Chlorine has seventeen electrons and only needs
one more to fill its third shell and be "happy." Chlorine will take your extra sodium
electron and leave you with 10 electrons inside of two filled shells. You are now a happy
atom too. You are also an ion and missing one electron. That missing electron gives you
a positive charge. You are still the element sodium, but you are now a sodium ion (Na + ).
You have one less electron than your atomic number.
Ion Characteristics
So now you've become a sodium ion. You have ten electrons.
That's the same number of electrons as neon (Ne). But you
aren't neon. Since you're missing an electron, you aren't really
a complete sodium atom either. As an ion you are now
something completely new. Your whole goal as an atom was
to become a "happy atom" with completely filled electron
shells. Now you have those filled shells. You have a lower
energy. You lost an electron and you are "happy." So what
makes you interesting to other atoms? Now that you have
given up the electron, you are quite electrically attractive.
Other electrically charged atoms (ions) of the opposite charge
(negative) are now looking at you and seeing a good partner to
bond with. That's where the chlorine comes in. It's not only chlorine. Almost any ion with
a negative charge will be interested in bonding with you.
32

Electrovalence
Don't get worried about the big word. Electrovalence is just another word for something
that has given up or taken electrons and become an ion. If you look at the periodic table,
you might notice that elements on the left side usually become positively charged ions
(cations) and elements on the right side get a negative charge (anions). That trend
means that the left side has a positive valence and the right side has a negative
valence. Valence is a measure of how much an atom wants to bond with other atoms. It
is also a measure of how many electrons are excited about bonding with other atoms.
There are two main types of bonding, covalent and electrovalent. You may have heard
of the term "ionic bonds." Ionic bonds are electrovalent bonds. They are just groups of
charged ions held together by electric forces. When in the presence of other ions, the
electrovalent bonds are weaker because of outside electrical forces and attractions.
Sodium and chlorine ions alone have a very strong bond, but as soon as you put those
ions in a solution with H + (Hydrogen ion), OH - (Hydroxide), F - (Fluorine ion) or Mg ++
(Magnesium ion), there are charged distractions that break the Na-Cl bond.
Look at sodium chloride (NaCl) one more time. Salt is a very strong bond when it is
sitting on your table. It would be nearly impossible to break those ionic/electrovalent
bonds. However, if you put that salt into some water (H2O), the bonds break very
quickly. It happens easily because of the electrical attraction of the water. Now you have
sodium (Na + ) and chlorine (Cl - ) ions floating around the solution. You should remember
that ionic bonds are normally strong, but they are very weak in water.
33

Neutron Madness
We have already learned that ions are atoms that are
either missing or have extra electrons. Let's say an atom
is missing a neutron or has an extra neutron. That type of
atom is called an isotope. An atom is still the same
element if it is missing an electron. The same goes for
isotopes. They are still the same element. They are just a
little different from every other atom of the same element.
For example, there are a lot of carbon (C) atoms in the
Universe. The normal ones are carbon-12. Those atoms have 6 neutrons. There are a
few straggler atoms that don't have 6. Those odd ones may have 7 or even 8 neutrons.
As you learn more about chemistry, you will probably hear about carbon-14. Carbon-14
actually has 8 neutrons (2 extra). C-14 is considered an isotope of the element carbon.
Messing with the Mass
If you have looked at a periodic table, you may have noticed that the atomic mass of
an element is rarely an even number. That happens because of the isotopes. If you are
an atom with an extra electron, it's no big deal. Electrons don't have much of a mass
when compared to a neutron or proton.
Atomic masses are calculated by figuring out the
amounts of each type of atom and isotope there are in
the Universe. For carbon, there are a lot of C-12, a
couple of C-13, and a few C-14 atoms. When you
average out all of the masses, you get a number that is a
little bit higher than 12 (the weight of a C-12 atom). The
average atomic mass for the element is actually 12.011.
Since you never really know which carbon atom you are
using in calculations, you should use the average mass
of an atom.
Bromine (Br), at atomic number 35, has a greater variety of isotopes. The atomic mass
of bromine (Br) is 79.90. There are two main isotopes at 79 and 81, which average out
to the 79.90amu value. The 79 has 44 neutrons and the 81 has 46 neutrons. While it
won't change the average atomic mass, scientists have made bromine isotopes with
masses from 68 to 97. It's all about the number of neutrons. As you move to higher
atomic numbers in the periodic table, you will probably find even more isotopes for
each element.
34

Summary
In atoms there are small particles called protons, neutrons, and electrons. With each of these smaller particles they
have postive, negaive, and neutral charges. Protons have positive charges, electrons have negative charges. Protons
will lways have a positiv charge and the elctron will have the same amount of negative charges in order to be a neutral
atom. There is also ions, they have extra electrons or are missing electrons and this is were they start to combine or
share to become a "happy atom". Giving up electrons make an electrically attractive.
37

36

Electron Configuration
Color the sublevel:
s = Red
d = Green
p = Blue
f = Orange
s
red
D
green
p
Blue
F orange
Write in sublevels
1
2
Write period, sublevel and super scripts.
Ctrl Shift =
gives you super scripts
3
4
5
6
7
37

The Learning Goal for this assignment is:
The students will be able to describe the arrangement of electrons in aotms and preidct what will happen when electrons
in atoms absorb or realese energy.
www.youtube.com/watch?v=jtYzEzykFdg
www.youtube.com/watch?
annotation_id=annotation_2076&feature=iv&src_vid=jtYzEzykFdg&v=cOlac8ruD_0
www.youtube.com/watch?
annotation_id=annotation_570977&feature=iv&src_vid=cOlac8ruD_0&v=lR2vqHZWb5A
Notes Section
You must know the the 3 rules aufbau, hund rule, and pauli exclusion principle
knowing all three of these allow you to understand how to do electron configurations in the first place.
for example if you want to find out lithium it is 1s^2 2p^6 3s^2
38

Electron Configuration
In order to write the electron configuration for an atom you must know the 3 rules of
electron configurations.
1. Aufbau
Notation
nO e
where
n is the energy level
O is the orbital type (s, p, d, or f)
e is the number of electrons in that orbital shell
Principle
electrons will first occupy orbitals of the lowest energy level
2. Hund rule
when electrons occupy orbitals of equal energy, one electron enters each orbital until
all the orbitals contain one electron with the same spin.
3. Pauli exclusion principle
an orbital contains a maximum of 2 electrons and
paired electrons will have opposite spin
39

In the space below, write the unabbreviated electron configurations of the following elements:
1) sodium ________________________________________________
1s^2 2s^2 2p^6 3s
2) iron ________________________________________________
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6
d=n-l
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
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 5d^10 6p^6 7s^2 5f^5
In the space below, write the abbreviated electron configurations of the following elements:
6) cobalt ________________________________________________
[Ar] 4s^2 3d^7
7) silver ________________________________________________
[Ag] 5s^2 4d^10
8) tellurium ________________________________________________
[Te] 5s^2 4d^10 5p^4
9) radium ________________________________________________
[Ra] 7s^2
10) lawrencium ________________________________________________
[Lr] 7s^2 5f^14 6d^1
Determine what elements are denoted by the following electron configurations:
11) 1s²s²2p⁶3s²3p⁴ ____________________
sulfur
12) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹ ____________________
rubidium
13) [Kr] 5s²4d¹⁰5p³ ____________________
antimony
14) [Xe] 6s²4f¹⁴5d⁶ ____________________
osmium
15) [Rn] 7s²5f¹¹ ____________________
einsteinium
Identify the element or determine that it is not a valid electron configuration:
16) 1s²2s²2p⁶3s²3p⁶4s²4d¹⁰4p⁵ ____________________
not valid 4d is not right
17) 1s²2s²2p⁶3s³3d⁵ ____________________
not valid 3p is ahead of it
18) [Ra] 7s²5f⁸ ____________________
not valid not a noble gas
19) [Kr] 5s²4d¹⁰5p⁵ ____________________
valid iodine
20) [Xe] ____________________
not valid
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
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
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
7)silver [Kr] 5s 2 4d 9 8)tellurium[Kr] 5s 2 4d 10 5p 4
9)radium [Rn] 7s 2 10)lawrencium[Rn] 7s 2 5f 14 6d 1
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
[Kr] 5s 2 4d 10 5p 3 antimony [Xe] 6s 2 4f 14 5d 6 osmium
[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”)
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)
[Kr] 5s 2 4d 10 5p 5 valid iodine
20)[Xe] not valid (an element can’t be its own electron configuration)
40

ubidium [Rb]
carbon [C]
sulfur [S]
copper [Cu]
xenon
Praseodymium
In the transitional metls there is 10 in the row not 9.
2s does not have 2 positive valence electrons it is neutral.
1s^2
2s^2 2p^1
3s^1
4s^2 3d^10 4p^6
4s^2 3d^4
3s^2 3p^3
2s^2 2p^2
4s^2 3d^7
6s^2 4f^16 5d^8
7s^6 5f^6
2s^2 2p^4
4s
41

Create groups for these Scientist and explain your groupings
(use the information you got from your research)
Areas where the scientists advanced the understanding in such as proton,Neutron, Electron and
provided theories have strengthened our understanding of the atom today. Without the support of
th 17 scientists and more we wouldn't of have made all these discoveries without the traces of their
research well. The information i got from the research shpwed me that each and every scientisit
furthered in a certain category leading in the long run to huge contribution to the atomic theory.
Proton contributors
Eugene goldstein
James Chadwick
Neutron contributors
James Chadwick
Ernest Rutherford
Electron contributors
Robert Millikan
Lothar Myer
J.J. Thomson
Erwin Shrodinger
Niels Bohr
Louis de Barogile
Radioavtivity
Antoine Henri Becquere
Glenn Seaborg
Marie and Pierre Curie
Periodic table/elements
Dmitri Mendeleev
J.W. Dobereiner
Theories
John Dalton
Democritus
Hantaro Nagaoka
42

Research the Scientist and summarize their contributions to the Atomic Theory
Antoine Henri Becquerel He was the first one to be able to discover radioactivity. He discovered by accident that
uranium salts are affected by light emiting a penetrating raditation that can be registered on a
photographic plate.
Niels Bohr
Louis de Barogilie
Glenn Seaborg
Hantaro Nagaoka He came up with an alternative planetary model of the atom where the positively charged center is
surrounded by a number of revolving electrons. His model showed that an atom had a very massive
atomic center.
Democritus
Marie and Pierre Curie they started the study of radioactivity, which led to discovery of some elements on the periodic
table of elements like radium and polonium. They thought radioactivity does not depend on how
atoms are arranged into molecules, but in the atom itself.
Eugene Goldstein
Dmitri Mendeleev
J.J. Thomson
James Chadwick
Erwin Shrodinger
John Dalton
Lothar Meyer
proposed a theory that the hydrogen atom based on quantum theory. He proclaimed electrons should move
around the nucleus but only in a directed orbit.
known for research on the quantum theory and predicting the wave nature of electrons. His ideas
were the basis for develpoing the wave mechanics theory. Improved the knowledge of the physcial
nature on the atomic scale.
discovered elements such as plutonium and went on to identify several other and more of the
radioactive "transuranium" elements. discovered over 100 isotopes of elements throughout the
periodic table.
First came up with the idea that matter conisisted of tiny invisible particles called "atomos" He also created
the first atom model but it did not have the essential components of the electrons, protons, and neutrons.
Helped further the topic of atoms.
He is credited with the discovery of the proton aswell. He was a early investiigator of discharge
tubes. Discovered the anode rays.
He arranged 63 elements into the periodic table based on the atomic mass of course. Used it to put
the atomic weight and put them into columns based on physical and chemicl properties. Helps
understand the periodic table of elements more due to his contribution.
discovered electrons and noticed atoms could be divided. He made the plum pudding model These
discoveries helped forward the understanding of the atom and it's generic make up.
James attacked the berryllium atoms with alpha particles. discovered the neutron. Interpreted the
radiation as being composed of particles with neutral electrical charge and the apporximate mass
of a proton.
furthered the research on the Bohr atom model. Used mathematical equations to describe the
chances of finding an electron in a specified position. His model was called the quantum mechanical
model of the atom.
proposed the atomic theory using measurable properties of mass. remains valid in todays modern chemical
thought. the rules are 1) all matter is made of atoms. 2) indivisible and indestructable. 3) compoundsare
formed by combination of two or more different kinds of atoms.
He was one of the pioneers in developing the first periodic table of elements. Took ideas from Mendeleyev
table and formed his own table that showed the atomic weight and the properties of the element.
Robert Millikan
J.W. Dobereiner
Ernest Rutherford
as a result of his oil-drop expiriment he was able to measure the electron's charge. His contribution is
he advanced J.J Thompsons hypothesis. Also specified the size of the electrons charges.
He developed the laws of triad in 1817, he had 3 groups to find out the ralative atomic masses of the
elements.The relative mass of the middle elemen was close to the average of the other two elements
and led to important realization that atomic masses help allign the elemetns when arranging them.
discovered alpha and beta rays, also made the laws of radioactive decay. Identified that alpha
particles were also helium nuclei.
43

The Learning Goal for this Assignment is
Alkali Metals
Any of the elements occupying group iA (1) of the periodic table. They are very reactive, electropositive, monova
metals forming strongly alkaline hydroxides.
Alkali Earth Metals
Six chemical elements in collum (group) 2 of the periodic table.
Transitional Metals
Any of the set of metallic elements occupying a central block ( groups 4b-
Inter Transitional Metals
elements 57-71 (lanthanides) and 89-103 (actinides) it is in two rows below all the other elements.
Metals
solid material that is tupically hard, shiny, malleable, fusible, and ductible, with good electrical and thermal
conductivity.
Metalloids
An intermediate between those of metals and solid nonmetals. They are electrical semiconducters scuh as the elements
garmanium and silicon.
Non Metals
It is a chemical element such as Boron, nitrogen, and carbon. lacks the characteristics of a metal and that is able to form
anions, acidic oxides, acids, and stable compounds with hydrogen.
Noble Gases
Any of the gaseuous elements such as helium, meon, argon, krypton, xenon, and radon. Any of them occupying
0.
44

Using Wikipedia, define the 8 categories of elements on the
left page.
Color your periodic table similar to the one on
pages 168—169 of your book.
alkali metals
alkaline metals
other metals
transitional metals
lanthanoids
metalloids
non metals
halogens
noble gases
unknown elements
actinoids
45

Atomic Size
Define Atomic Size:
The size of the atom.
Explanation:
when the period is going higher and moves left it is decreasing in size because it has less mass and it
affects the amount of electrons it can hold.
46

Define Ionization Energy:
Ionization Energy
The energy required to remove an electron from an atom.
Explanation:
The two factors that determine whether it is has more is the gravity and energy level. the higher the energy level
and the more mass it has and less gravity the closer you get to the left it has a stronger gravtational pull.
47

Define Electronegativity:
Electronegativity
the abiltiy of an atom of an element to attract electrons when the atom is in a compound.
Explanation:
The larger the mass the stronger pull it has to take in more electrons, while having a less powerful pull it has more
electrons. Also can share more.
48

Ion Size
Define Ion Size:
The size of the ion's atom in comparison to the size of the elemental atom.
increase
cation
Anion
Explanation:
This is when atoms gain or lose electrons, the atom becomes an ion. When an atom gains an electron, it
becomes a negatively charged ion that is anion. when it loses an electron it is a positively charged ion that
is a cation.
49

Unit 3
Chapter 25 Nuclear Chemistry
The students will learn what happens when an unstable
nucleus decays and how nuclear chemistry affects their lives.
Explore the theory of electromagnetism by comparing and contrasting the
different parts of the electromagnetic spectrum in terms of wavelength,
frequency, and energy, and relate them to phenomena and applications.
Students will be able to compare and contrast the different parts of the
electromagnetic spectrum.
Students will be able to apply knowledge of the EMS to real world phenomena.
Students will be able to quantitatively compare the relationship between energy,
wavelength, and frequency of the EMS.
amplitude
wavelength
frequency
hertz
electromagnetic radiation
photon
Planck’s constant
Explain and compare nuclear reactions (radioactive decay, fission and
fusion), the energy changes associated with them and their associated
safety issues.
Students will be able to compare and contrast fission and fusion reactions.
Students will be able to complete nuclear decay equations to identify the type of
decay.
Students will participate in activities to calculate half-life.
radioactivity
nuclear radiation
alpha particle
beta particle
gamma ray
positron
½ life
transmutation
fission
fusion
50

Chapter 7
Ionic and Metallic Bonding
The students will learn how ionic compounds form and how
metallic bounding affects the properties of metals.
Compare the magnitude and range of the four fundamental forces
(gravitational, electromagnetic, weak nuclear, strong nuclear).
Students will compare/contrast the characteristics of each fundamental force.
gravity
electromagnetic
strong
weak
Distinguish between bonding forces holding compounds together and other
attractive forces, including hydrogen bonding and van der Waals forces.
Students will be able to compare/contrast traits of ionic and covalent bonds.
Students will be able to compare/contrast basic attractive forces between
molecules.
Students will be able to predict the type of bond or attractive force between
atoms or molecules.
ionic bond
covalent bond
metallic bond
polar covalent bond
hydrogen bond
van der Waals forces
London dispersion forces
Chapter 8
Covalent Bonding
The students will learn how molecular bonding is different
than ionic bonding and electrons affect the shape of a
molecule and its properties.
Interpret formula representations of molecules and compounds in terms of
composition and structure.
Students will be able to interpret chemical formulas in terms of # of atoms.
Students will be able to differentiate between ionic and molecular compounds.
Students will be able to list various VSEPR shapes and identify examples of
each.
Students will be able to predict shapes of various compounds.
Molecule
empirical formula
Atom
Electron
Element
Compound
51

Name ____________________
Juan Roque
Go to the web site www.darvill.clara.net/emag
1. Click on “How the waves fit into the spectrum” and fill in this table:
>: look out for the
RED words on the web site!
Low __________, frequency Long wavelength
High frequency, Short ______________
wavelengths
Radio Waves
microwaves Infra-red Visible Light Ultra-Violet X-rays
Gamma rays
2. Click on “Radio waves”. They are used for _______________________
communication
3. Click on “Microwaves”. They are used for cooking, mobile _________, Wifi _______ speed cameras and _________. radar
4. Click on “Infra-red”. These waves are given off by _____ hot _________. objects They are used for remote controls,
cameras in police ____________ helicopters , and alarm systems.
5. Click on “Visible Light”. This is used in ___ CDplayers and _______ laser printers, and for seeing where we’re going.
6. “UV” stands for “ ________ Ultra ___________”. Violet This can damage the _________ retina in your eyes, and cause
sunburn and even _______ skin cancer. Its uses include detecting forged ______ bank _______. notes
7. X-rays are used to see inside people, and for _________ airport security.
8. Gamma rays are given off by some ________________ radioactive substances. We can use them to kill ________ cancer cells,
which is called R_______________ adiotherapy .
9. My Quiz score is ____%. 94
52

10. Name ________________________________
Go to the web site www.darvill.clara.net/emag
Name How they’re made Uses Dangers
by number of astronomical processesUsed to kill cancer cells
Gamma rays
X-rays
Ultra-violet
Visible light
Infra-red
microwaves
Radio waves
very high-energy electrons are made
When electrons strike a metal target
electrons are liberated from heated
filament and accelerated by high
voltage towards trget.
Given off by the sun in large
quantities
special lamps such as sun beds
white ligth is actually made up of a
whole range of colours, mixed together.
Radiotherapy
Kill microbes and used to sterilise food
To see inside people
airport security
atronomers
Sun tan and detecting forged bank
notes
kill microbes
produces vitamin D
food drug companies sterilise their products
to see things
CD and DVD players
remote controls, video recorders
below the visible red light
heal sprot injuries
anthing else that is given off by hot objects
night sights
extremely high frequency radio waves,
made by various amount of transmitters
Made by various types of transmitters,
depending on the wavelength
mobile phones
wifi
speed cameras
radar
mainly used for communication
Can cause cell damage
cancer
mutations
can cause cell damage
cancer
Large doses can damage
retin in the eyes
can cause sunburn and
skin cancer
can damage the retina
of the eyes
looking at the sun
too much and you can
overheat
ovrexposure can cause
cateracts
large doses can cause
cancer and disorders
_____ Frequency _____ frequency,
Short wavelength ______ Wavelength
53

Learning Goal for this section:
The students will learn what happens when an unstable nucleus decays and how nuclear chemistry affects thier lives.
Notes Section:
Alpha particles He 4 goes through alpha
2 238
92 Uranium >
down by 2
protons>
234
90 Thorium
+
4
2 He
Beta Particle
up by 1
negative
electron
6p
8n
14
6 Carbon
converting nuetron to proton
carbon 14 turns into nitrogen
14
Nitrogen + negative electron
7
positive
electron
positron
5p
3n
8
5 Be > 8
Be + e+
4
4p
4n
Gamma Energy
always associatied with alpha particle or beta
Half-LIfe 100g start 20min
50g 20min 50g of something else
25 40min 75g of something else
it decays and becomes something else
12.5 60min 87.5g of something else
still 100g but of something else
0.75 120min
93.75 of something else
54

The Nucleus
A typical model of the atom is called the Bohr Model, in
honor of Niels Bohr who proposed the structure in 1913. The Bohr atom consists of a central nucleus
composed of neutrons and protons, which is surrounded by electrons which “orbit” around the nucleus.
Protons carry a positive charge of one and have a mass of about 1 atomic mass unit or amu (1 amu =1.7x10-
27 kg, a very, very small number). Neutrons are electrically “neutral” and also have a mass of about 1 amu. In
contrast electron carry a negative charge and have mass of only 0.00055 amu. The number of protons in a
nucleus determines the element of the atom. For example, the number of protons in uranium is 92 and the
number in neon is 10. The proton number is often referred to as Z.
Atoms with different numbers of protons are called elements, and are arranged in the periodic table with
increasing Z.
Atoms in nature are electrically neutral so the number of electrons orbiting the nucleus equals the number of
protons in the nucleus.
Neutrons make up the remaining mass of the nucleus and provide a means to “glue” the protons in place.
Without neutrons, the nucleus would split apart because the positive protons would repel each other. Elements
can have nucleii with different numbers of neutrons in them. For example hydrogen, which normally only has
one proton in the nucleus, can have a neutron added to its nucleus to from deuterium, ir have two neutrons
added to create tritium, which is radioactive. Atoms of the same element which vary in neutron number are
called isotopes. Some elements have many stable isotopes (tin has 10) while others have only one or two. We
express isotopes with the nomenclature Neon-20 or 20 Ne 10, with twenty representing the total number of
neutrons and protons in the atom, often referred to as A, and 10 representing the number of protons (Z).
Alpha Particle
Decay
Alpha decay is a radioactive process in which a
particle with two neutrons and two protons is
ejected from the nucleus of a radioactive atom. The particle is identical to the nucleus of a helium atom.
Alpha decay only occurs in very heavy elements such as uranium, thorium and radium. The nuclei of these
atoms are very “neutron rich” (i.e. have a lot more neutrons in their nucleus than they do protons) which makes
emission of the alpha particle possible.
After an atom ejects an alpha particle, a new parent atom is formed which has two less neutrons and two less
protons. Thus, when uranium-238 (which has a Z of 92) decays by alpha emission, thorium-234 is created
(which has a Z of 90).
Because alpha particles contain two protons, they have a positive charge of two. Further, alpha particles are
very heavy and very energetic compared to other common types of radiation. These characteristics allow alpha
particles to interact readily with materials they encounter, including air, causing many ionizations in a very short
distance. Typical alpha particles will travel no more than a few centimeters in air and are stopped by a sheet of
paper.
55

Beta Particle Decay
Beta decay is a radioactive process in which an electron is emitted from the nucleus of a radioactive
atom Because this electron is from the nucleus of the atom, it is called a beta particle to distinguish it
from the electrons which orbit the atom.
Like alpha decay, beta decay occurs in isotopes which are “neutron rich” (i.e. have a lot more
neutrons in their nucleus than they do protons). Atoms which undergo beta decay are located below
the line of stable elements on the chart of the nuclides, and are typically produced in nuclear reactors.
When a nucleus ejects a beta particle, one of the neutrons in the nucleus is transformed into a proton.
Since the number of protons in the nucleus has changed, a new daughter atom is formed which has
one less neutron but one more proton than the parent. For example, when rhenium-187 decays
(which has a Z of 75) by beta decay, osmium-187 is created (which has a Z of 76). Beta particles
have a single negative charge and weigh only a small fraction of a neutron or proton. As a result, beta
particles interact less readily with material than alpha particles. Depending on the beta particles
energy (which depends on the radioactive atom), beta particles will travel up to several meters in air,
and are stopped by thin layers of metal or plastic.
Positron emission or beta plus decay (β+ decay) is a subtype of radioactive decay called beta decay,
in which a proton inside a radionuclide nucleus is converted into a neutron while releasing a positron
and an electron neutrino (νe). Positron emission is mediated by the weak force.
An example of positron emission (β+ decay) is shown with magnesium-23 decaying into sodium-23:
23 Mg12 → 23 Na11 + e +
Because positron emission decreases proton number relative to neutron number, positron decay
happens typically in large "proton-rich" radionuclides. Positron decay results in nuclear transmutation,
changing an atom of one chemical element into an atom of an element with an atomic number that is
less by one unit.
Positron emission should not be confused with electron emission or beta minus decay (β− decay),
which occurs when a neutron turns into a proton and the nucleus emits an electron and an
antineutrino.
56

Gamma
Radiation
After a decay reaction, the nucleus is often in an
“excited” state. This means that the decay has
resulted in producing a nucleus which still has
excess energy to get rid of. Rather than emitting another beta or alpha particle, this energy is lost by
emitting a pulse of electromagnetic radiation called a gamma ray. The gamma ray is identical in
nature to light or microwaves, but of very high energy.
Like all forms of electromagnetic radiation, the gamma ray has no mass and no charge. Gamma rays
interact with material by colliding with the electrons in the shells of atoms. They lose their energy
slowly in material, being able to travel significant distances before stopping. Depending on their initial
energy, gamma rays can travel from 1 to hundreds of meters in air and can easily go right through
people.
It is important to note that most alpha and beta emitters also emit gamma rays as part of their decay
process. However, their is no such thing as a “pure” gamma emitter. Important gamma emitters
including technetium-99m which is used in nuclear medicine, and cesium-137 which is used for
calibration of nuclear instruments.
Half Life
Half-life is the time required for the quantity of a
radioactive material to be reduced to one-half its
original value.
All radionuclides have a particular half-life, some
of which a very long, while other are extremely
short. For example, uranium-238 has such a
long half life, 4.5x109 years, that only a small fraction has decayed since the earth was formed. In
contrast, carbon-11 has a half-life of only 20 minutes. Since this nuclide has medical applications, it
has to be created where it is being used so that enough will be present to conduct medical studies.
57

The Learning Goal for this assignment is:
Distinguish between bonding forces holding compounds together and other attractive forces, including hydrogen bonding
and Van der Waals forces.
Introduction to Ionic Compounds
Those molecules that consist of charged ions with opposite charges are called IONIC. These ionic
compounds are generally solids with high melting points and conduct electrical current. Ionic
compounds are generally formed from metal and a non-metal elements. See Ionic Bonding below.
Ionic Compound Example
For example, you are familiar with the fairly benign unspectacular behavior of common white
crystalline table salt (NaCl). Salt consists of positive sodium ions (Na + ) & negative chloride ions (Cl - ).
On the other hand the element sodium is a silvery gray metal composed of neutral atoms which react
vigorously with water or air. Chlorine as an element is a neutral greenish-yellow, poisonous, diatomic
gas (Cl2).
The main principle to remember is that ions are completely different in physical and chemical
properties from the neutral atoms of the elements.
The notation of the + and - charges on ions is very important as it conveys a definite meaning.
Whereas elements are neutral in charge, IONS have either a positive or negative charge depending
upon whether there is an excess of protons (positive ion) or excess of electrons (negative ion).
Formation of Positive Ions
Metals usually have 1-4 electrons in the outer energy level. The electron arrangement of a rare gas is
most easily achieved by losing the few electrons in the newly started energy level. The number of
electrons lost must bring the electron number "down to" that of a prior rare gas.
How will sodium complete its octet?
First examine the electron arrangement of the atom. The atomic number is eleven, therefore, there
are eleven electrons and eleven protons on the neutral sodium atom. Here is the Bohr diagram and
Lewis symbol for sodium:
58

This analysis shows that sodium has only one electron in its outer level. The nearest rare gas is neon
with 8 electron in the outer energy level. Therefore, this electron is lost so that there are now eight
electrons in the outer energy level, and the Bohr diagrams and Lewis symbols for sodium ion and
neon are identical. The octet rule is satisfied.
Ion Charge?
What is the charge on sodium ion as a result of losing one electron? A comparison of the atom and
the ion will yield this answer.
Sodium Atom
Sodium Ion
11 p+ to revert to 11 p + Protons are identical in
12 n an octet 12 n
the atom and ion.
Positive charge is
11 e- lose 1 electron 10 e-
caused by lack of
0 charge + 1 charge
electrons.
Formation of Negative Ions
How will fluorine complete its octet?
First examine the electron arrangement of the atom. The atomic number is nine, therefore, there are
nine electrons and nine protons on the neutral fluorine atom. Here is the Bohr diagram and Lewis
symbol for fluorine:
This analysis shows that fluorine already has seven electrons in its outer level. The nearest rare gas
is neon with 8 electron in the outer energy level. Therefore only one additional electron is needed to
complete the octet in the fluorine atom to make the fluoride ion. If the one electron is added, the Bohr
diagrams and Lewis symbols for fluorine and neon are identical. The octet rule is satisfied.
59

Ion Charge?
What is the charge on fluorine as a result of adding one electron? A comparison of the atom and the
ion will yield this answer.
Fluorine Atom Fluoride Ion *
9 p+ to complete 9 p + Protons are identical in
10 n octet 10 n
9 e- add 1 electron 10 e-
0 charge - 1 charge
the atom and ion.
Negative charge is
caused by excess
electrons
* The "ide" ending in the name signifies a simple negative ion.
Summary Principle of Ionic Compounds
An ionic compound is formed by the complete transfer of electrons from a metal to a nonmetal and
the resulting ions have achieved an octet. The protons do not change. Metal atoms in Groups 1-3
lose electrons to non-metal atoms with 5-7 electrons missing in the outer level. Non-metals gain 1-4
electrons to complete an octet.
Octet Rule
Elemental atoms generally lose, gain, or share electrons with other atoms in order to achieve the
same electron structure as the nearest rare gas with eight electrons in the outer level.
The proper application of the Octet Rule provides valuable assistance in predicting and explaining
various aspects of chemical formulas.
Introduction to Ionic Bonding
Ionic bonding is best treated using a simple
electrostatic model. The electrostatic model
is simply an application of the charge
principles that opposite charges attract and
similar charges repel. An ionic compound
results from the interaction of a positive and
negative ion, such as sodium and chloride in
common salt.
The IONIC BOND results as a balance
between the force of attraction between
opposite plus and minus charges of the ions
and the force of repulsion between similar
negative charges in the electron clouds. In
crystalline compounds this net balance of
forces is called the LATTICE ENERGY.
Lattice energy is the energy released in the
formation of an ionic compound.
DEFINITION: The formation of an IONIC
BOND is the result of the transfer of one or
more electrons from a metal onto a nonmetal.
60

Metals, with only a few electrons in the outer energy level, tend to lose electrons most readily. The
energy required to remove an electron from a neutral atom is called the IONIZATION POTENTIAL.
Energy + Metal Atom ---> Metal (+) ion + e-
Non-metals, which lack only one or two electrons in the outer energy level have little tendency to lose
electrons - the ionization potential would be very high. Instead non-metals have a tendency to gain
electrons. The ELECTRON AFFINITY is the energy given off by an atom when it gains electrons.
Non-metal Atom + e- --- Non-metal (-) ion + energy
The energy required to produce positive ions (ionization potential) is roughly balanced by the energy
given off to produce negative ions (electron affinity). The energy released by the net force of
attraction by the ions provides the overall stabilizing energy of the compound.
Notes Section:
61

The Learning Goal for this assignment is:
Students will learn how molecule bonding is different from ionic bonding and electrons affect the shape
of the molecule and its properties.
Introduction to Covalent Bonding:
Bonding between non-metals consists of two electrons shared between two atoms. Using the Wave
Theory, the covalent bond involves an overlap of the electron clouds from each atom. The electrons
are concentrated in the region between the two atoms. In covalent bonding, the two electrons shared
by the atoms are attracted to the nucleus of both atoms. Neither atom completely loses or gains
electrons as in ionic bonding.
There are two types of covalent bonding:
1. Non-polar bonding with an equal sharing of electrons.
2. Polar bonding with an unequal sharing of electrons. The number of shared electrons depends on
the number of electrons needed to complete the octet.
NON-POLAR BONDING results when two identical non-metals equally share electrons between
them. One well known exception to the identical atom rule is the combination of carbon and hydrogen
in all organic compounds.
Hydrogen
The simplest non-polar covalent molecule is hydrogen. Each hydrogen
atom has one electron and needs two to complete its first energy level.
Since both hydrogen atoms are identical, neither atom will be able to
dominate in the control of the electrons. The electrons are therefore
shared equally. The hydrogen covalent bond can be represented in a
variety of ways as shown here:
The "octet" for hydrogen is only 2 electrons since the nearest rare gas is
He. The diatomic molecule is formed because individual hydrogen atoms
containing only a single electron are unstable. Since both atoms are
identical a complete transfer of electrons as in ionic bonding is
impossible.
Instead the two hydrogen atoms SHARE both electrons equally.
Oxygen
Molecules of oxygen, present in about 20% concentration in air are
also covalent molecules. See the graphic on the left of the Lewis Dot
Structure.
There are 6 electrons in the outer shell, therefore, 2 electrons are
needed to complete the octet. The two oxygen atoms share a total of
four electrons in two separate bonds, called double bonds.
The two oxygen atoms equally share the four electrons.
62

POLAR BONDING results when two different non-metals unequally share electrons between them.
One well known exception to the identical atom rule is the combination of carbon and hydrogen in all
organic compounds.
The non-metal closer to fluorine in the Periodic Table has a greater tendency to keep its own electron
and also draw away the other atom's electron. It is NOT completely successful. As a result, only
partial charges are established. One atom becomes partially positive since it has lost control of its
electron some of the time. The other atom becomes partially negative since it gains electron some of
the time.
Hydrogen Chloride
Hydrogen Chloride forms a polar covalent molecule. The graphic
on the left shows that chlorine has 7 electrons in the outer shell.
Hydrogen has one electron in its outer energy shell. Since 8
electrons are needed for an octet, they share the electrons.
However, chlorine gets an unequal share of the two electrons,
although the electrons are still shared (not transferred as in ionic
bonding), the sharing is unequal. The electrons spends more of the
time closer to chlorine. As a result, the chlorine acquires a "partial"
negative charge. At the same time, since hydrogen loses the
electron most - but not all of the time, it acquires a "partial" charge.
The partial charge is denoted with a small Greek symbol for delta.
Water
Water, the most universal compound on all of the earth, has the property of
being a polar molecule. As a result of this property, the physical and
chemical properties of the compound are fairly unique.
Dihydrogen Oxide or water forms a polar covalent molecule. The graphic on
the left shows that oxygen has 6 electrons in the outer shell. Hydrogen has
one electron in its outer energy shell. Since 8 electrons are needed for an
octet, they share the electrons.
Notes Section:
metal and nonmetal make a ionic bond
metal and metal make a covalent bond
left side metal right side nonmetal
C^2H^6O
flourine most electronegative and hydrogen is right above it.
1) C2 x 4 8
H^6 x 1 6
O1 x 6 8 = 25
5)
2) C^2 x 8 16
H^6 x 2 12
O^1 x 8 8 = 36
3) 36-20=16e-
4) 2/ 16= 8 bonds
63

C 2 H 6 O Ethanol CH 3 CH 2 O
Step 1
Find valence e- for all atoms. Add them together.
C: 4 x 2 = 8
H: 1 x 6 = 6
O: 6
Total = 20
Step 2
Find octet e- for each atom and add them together.
C: 8 x 2 = 16
H: 2 x 6 = 12
O: 8
Total = 36
Step 3
Subtract Step 1 total from Step 2.
Gives you bonding e-.
36 – 20 = 16e-
Step 4
Find number of bonds by diving the number in step 3 by 2
(because each bond is made of 2 e-)
16e- / 2 = 8 bond pairs
These can be single, double or triple bonds.
Step 5
Determine which is the central atom
Find the one that is the least electronegative.
Use the periodic table and find the one farthest
away from Fluorine or
The one that only has 1 atom.
64

Step 6
Put the atoms in the structure that you think it will
have and bond them together.
Put Single bonds between atoms.
Step 7
Find the number of nonbonding (lone pairs) e-.
Subtract step 3 number from step 1.
20 – 16 = 4e- = 2 lone pairs
Step 8
Complete the Octet Rule by adding the lone
pairs.
Add any left over bonds to make double or triple
bonds.
Then, if needed, use any lone pairs to make
double or triple bonds so that all atoms meet
the Octet Rule.
See Step 4 for total number of bonds.
Step 9
Find the formal charges for the atoms in the compound.
Arrange atoms so that all formal charges
are as close to 0 as possible.
Some central atoms do not meet the octet rule.
Boron can sometimes have only 6 electrons and
some elements in Periods 3—7 may exceed the
octet rule.
65

Linear
Molecular Geometry
Orbital Equation Lone Pairs Angle
sp AX 2 none 180
BeCl 2
Berylium
dicloride
Cl
Be
Cl
element bond lone pair
C
66

Trigonal planal
Molecular Geometry
Orbital Equation Lone Pairs Angle
sp 2 AX 3 none 60
BF 3
F
B
Boron triflouride
F
F
element bond lone pair
C
67

Trigonal planar
Molecular Geometry
Orbital Equation Lone Pairs Angle
sp 3 AX 2 E 2 2 104
OF 2
Oxygen diflouride
O
F
F
element bond lone pair
C
68

Tetrahedral
Molecular Geometry
Orbital Equation Lone Pairs Angle
sp 3 AX 4 none 109.5
PO 4
3-
Phosphate
O
O
P
5-5=0
O
6-6=0
O
6-7=1-
element bond lone pair
C
69

Trigonal Pyramidal
Molecular Geometry
Orbital Equation Lone Pairs Angle
sp 3 AX 3 E 2 104.5
PH 3
Phosphorous trihydride
H
P
H
H
element bond lone pair
C
70

linear
Molecular Geometry
Orbital Equation Lone Pairs Angle
sp AX 2 2 180
O 3
trioxide
O
O
O
element bond lone pair
C
71

Trigonal Bi pyramidal
Molecular Geometry
Orbital Equation Lone Pairs Angle
sp 3 d AX 5 none 120/90
PCl 5
Phosphorous
pentacloride
Cl
Cl
Cl
P
Cl
Cl
element bond lone pair
C
C
72

T-shaped
Molecular Geometry
Orbital Equation Lone Pairs Angle
sp 3 d AX 3 E 2 2 90
ClF 3
Chlorine trifloride
F
Cl
F
F
element bond lone pair
C
73

Octahedral
Molecular Geometry
Orbital Equation Lone Pairs Angle
sp 3 d 2 AX 6 none 90
SF 6
Sulfur hexafluoride
F
F
F
S
F
F
F
element bond lone pair
74

Square Planar
Molecular Geometry
Orbital Equation Lone Pairs Angle
sp 3 d 2 AX 4 E 2 2 90
ICl 4
-
Iodine tetrafluoride
ion
Cl
Cl
I
Cl
Cl
element bond lone pair
C
75

Orbitals Equation Lone Pairs
Angle
Name
Sp 3 d AX5 none 120/90 Phosphorous Penta Chloride
Sp 3 d 2 AX6 none 90
Sulfur hexafluoride
Sp 3 d AX3E2 2 90
Chlorine trifluoride
sp AX2 2 180
Trioxide
Sp 3 AX3E 2 104.5
Phosphorous trihydride
Sp 3 AX4 none 109.5
Phosphorous tetroxide
Sp 3 d 2 AX4E2 2 90
Iodine tetrafluoride ion
Sp 3 AX2E2 2 104
Oxygen difluoride
Sp 2 AX3 none 60
Boron trifluoride
sp AX2 none 180
Beryllium dichloride
78

Name Formula Charge
Dichromate Cr₂O₇ 2-
Sulfate SO₄ 2-
Hydrogen Carbonate HCO₃ 1-
Hypochlorite ClO 1-
Phosphate PO₄ 3-
Nitrite NO₂ 1-
Chlorite ClO₂ 1-
Dihydrogen phosphate H₂PO₄ 1-
Chromate CrO₄ 2-
Carbonate CO₃ 2-
Hydroxide OH 1-
Hydrogen phosphate HPO₄ 2-
Ammonium NH₄ 1+
Acetate C₂H₃O₂ 1-
Perchlorate ClO₄ 1-
Permanganate MnO₄ 1-
Chlorate ClO₃ 1-
Hydrogen Sulfate HSO₄ 1-
Phosphite PO₃ 3-
Sulfite SO₃ 2-
Silicate SiO₃ 2-
Nitrate NO₃ 1-
Hydrogen Sulfite HSO₃ 1-
Oxalate C₂O₄ 2-
Cyanide CN 1-
Hydronium H₃O 1+
Thiosulfate S₂O₃ 2-
77

Chapter 9
Unit 4
Chemical Names and Formulas
The students will learn how the periodic table helps them
determine the names and formulas of ions and compounds.
Chapter 22 Hydrocarbon Compounds
The student will learn how Hydrocarbons are named and the
general properties of Hydrocarbons.
Describe how different natural resources are produced and how their rates
of use and renewal limit availability.
Students will explore local, national, and global renewable and nonrenewable
resources.
Students will explain the environmental costs of the use of renewable and
nonrenewable resources.
Students will explain the benefits of renewable and nonrenewable resources.
Nuclear reactors
Natural gas
Petroleum
Refining
Coal
78

Chapter 23 Functional Groups
The student will learn what effects functional groups have on
organic compounds and how chemical reactions are used in
organic compounds.
Describe the properties of the carbon atom that make the diversity of carbon
compounds possible.
Identify selected functional groups and relate how they contribute to
properties of carbon compounds.
Students will identify examples of important carbon based molecules.
Students will create 2D or 3D models of carbon molecules and explain why this
molecule is important to life.
covalent bond
single bond
double bond
triple bond
monomer
polymer
79

http://www.bbc.co.uk/education/guides/zm9hvcw/revision
students will learn what effects functional groups have on
organic compounds and how chemical reactinos are used in
organic compounds.
A homologous series is a family of hydrocarbons with similar
chemical properties who share the same genral formula.
The general formula of the alkanes is CnH2n+2
methane- (natural gases) cooking, heating
propane- used in gas cylinders for BBQ etc
octane- used in petrol for cars.
The longest unbranched chain containing the functional group is th
parent molecule, or simply the longest unbranched chain for alkane
rememeber that, the longest chain can go round a bend.
80

5 with
double bond
2 without
81

Unit 5
Chapter 10 Chemical Quantities
The student will learn why the mole is important and how the
molecular formula of a compound can be determined
experimentally.
Chapter 11 Chemical Reactions
The students will learn how chemical reactions obey the law of
conservation of mass and how they can predict the products
of a chemical reaction.
Characterize types of chemical reactions, for example: redox, acid-base,
synthesis, and single and double replacement reactions.
Students will be able to identify the type of chemical reaction that occurs.
Students will be able to compare/contrast reactants and products of various
types of chemical reactions.
Students will be able to predict the product of various reactants.
Students will be able to write balanced chemical equations for each type of
reaction.
Decomposition
Combustion
Redox
Acid-Base
Synthesis
single-replacement
double-replacement
Differentiate between chemical and nuclear reactions.
Students will compare/contrast chemical and nuclear reactions.
fission
fusion
82

Chapter 12 Stoichiometry
The students will learn how balanced chemical equations are
used in stoichiometric calculations and how to calculate
amounts of reactants and products in a chemical equation.
Apply the mole concept and the law of conservation of mass to calculate
quantities of chemicals participating in reactions.
Students will be able to use a balanced equation to determine mole ratios.
Students will be able to apply law of conservation of mass to chemical equations.
Students will be able to calculate empirical and molecular formulas.
Students will be able to calculate the % composition of a compound.
Students will be able to calculate theoretical yield.
Students will be able to calculate % error.
Students will be able to calculate molar mass.
Students will be able to perform stoichiometric calculations, including limiting
reagents.
mole
Avogadro’ s number
molar mass
gram formula mass
83

www.youtube.com/watch?v=AsqEkF7hcII
www.youtube.com/watch?v=tEn0N4R2dqA
www.youtube.com/watch?v=Pft2CASl0M0
www.youtube.com/watch?v=rwhJklbK8R0
The Mole
1 mol= 6.02x10^23
avogadros number
amu --> molar mass (g)
12.01 + 16.00 + 16.00
Nitruous Oxide
14.01 + 16.00 = 30.01 grams
1 mole of 12 C atoms in 12 g of carbon
1 gram = 1 mole of amu
1 gram = 6.02x10^23 amu
3.5 dozens roses 12 roses
dozen roses
= 42 roses
2.5 mol roses 6.02x10^23
1 mol roses
= 1.505x10^24 roses
84

www.youtube.com/watch?v=BTRm8PwcZ3U
www.youtube.com/watch?v=F9NkYSKJifs
www.youtube.com/watch?v=xPdqEX_WMjo
Molar Mass
525g Ca
1 mol Ca
=
13.1 mol Ca
40g Ca
25 g Au atoms
1 mol Au
6.02x10^23 atoms
197g Au
1 mol
7.65x10^23 atoms Au
6(12g/mol)
72g/mol
12( 1 g/mol)= 12 g/mol
6(16g/mol = 46g/mol
180g/mol
2.5mo C6H12O6
18g
1 mol C6H12O6
450g C6H12O6
85

Categories of Reactions
All chemical reactions can be placed into one of six categories. Here they are, in no
particular order:
1) Synthesis: A synthesis reaction is when two or more simple compounds combine to form a
more complicated one. These reactions come in the general form of:
Reactants--->products H2+O2---->H2O
A+B----> AB
One example of a synthesis reaction is the combination of iron and sulfur to form iron (II) sulfide:
8 Fe + S8 ---> 8 FeS
If two elements or very simple molecules combine with each other, it’s probably a synthesis reaction.
The products will probably be predictable using the octet rule to find charges.
2) Decomposition: A decomposition reaction is the opposite of a synthesis reaction - a
complex molecule breaks down to make simpler ones. These reactions come in the general form:
AB---->A+B
H2O----> H2+O2
One example of a decomposition reaction is the electrolysis of water to make oxygen and hydrogen
gas:
2 H2O ---> 2 H2 + O2
If one compound has an arrow coming off of it, it’s probably a decomposition reaction. The products
will either be a couple of very simple molecules, or some elements, or both.
3) Single displacement: This is when one element trades places with another element in a
compound. These reactions come in the general form of:
A+BC---->B+AC
One example of a single displacement reaction is when magnesium replaces hydrogen in water to
make magnesium hydroxide and hydrogen gas:
Mg + 2 H2O ---> Mg(OH)2 + H2
If a pure element reacts with another compound (usually, but not always, ionic), it’s probably a single
displacement reaction. The products will be the compounds formed when the pure element switches
places with another element in the other compound.
Important note: these reactions will only occur if the pure element on the reactant side of the equation
is higher on the activity series than the element it replaces.
86

4) Double displacement: This is when the anions and cations of two different molecules
switch places, forming two entirely different compounds. These reactions are in the general form:
AB+CD---->AD+CB
One example of a double displacement reaction is the reaction of lead (II) nitrate with potassium
iodide to form lead (II) iodide and potassium nitrate:
Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3
If two ionic compounds combine, it’s probably a double displacement reaction. Switch the cations
and balance out the charges to figure out what will be made.
Important note: These reactions will only occur if both reactants are soluble in water and only one
product is soluble in water.
5) Acid-base: This is a special kind of double displacement reaction that takes place when an
acid and base react with each other. The H + ion in the acid reacts with the OH - ion in the base,
causing the formation of water. Generally, the product of this reaction is some ionic salt and water:
HA+BOH----->AB+H2O
One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium
hydroxide:
HBr + NaOH ---> NaBr + H2O
If an acid and a base combine, it’s an acid-base reaction. The products will be an ionic compound
and water.
6) Combustion: A combustion reaction is when oxygen combines with another compound to
form water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An
example of this kind of reaction is the burning of napthalene:
organic CarbonBase+O2--->H2O+CO
C10H8 + 12 O2 ---> 10 CO2 + 4 H2O
If something that has carbon and hydrogen reacts with oxygen, it’s probably a combustion reaction.
The products will be CO2 and H2O.
Follow this series of questions. When you can answer "yes" to a question, then
stop!
1) Does your reaction have two (or more) chemicals combining to form one chemical? If yes, then it's
a synthesis reaction
2) Does your reaction have one large molecule falling apart to make several small ones? If yes, then
it's a decomposition reaction
3) Does your reaction have any molecules that contain only one element? If yes, then it's a single
displacement reaction
4) Does your reaction have water as one of the products? If yes, then it's an acid-base reaction
5) Does your reaction have oxygen as one of it's reactants and carbon dioxide and water as
products? If yes, then it's a combustion reaction
6) If you haven't answered "yes" to any of the questions above, then you've got a double
displacement reaction.
87

List what type the following reactions are:
1) NaOH + KNO3 --> NaNO3 + KOH
double displacement
2) CH4 + 2 O2 --> CO2 + 2 H2O
Combustion
3) 2 Fe + 6 NaBr --> 2 FeBr3 + 6 Na
single displacement
4) CaSO4 + Mg(OH)2 --> Ca(OH)2 + MgSO4
5) NH4OH + HBr --> H2O + NH4Br
acid-base
double displacement
6) Pb + O2 --> PbO2
synthesis
7) Na2CO3 --> Na2O + CO2
decomposition
88

Determine the Type of Reaction for each equation.
Then predict the products of each of the following chemical reactions. If a reaction will not occur,
explain why not.
Then Balance the equation.
1) __Ag2SO4 + __NaNO3 →
double displacement
2) __NaI 2 + __CaSO4 1 → _1_Na2SO4+_1_CaI2
Na 2
Na2
I 2
I 2
Ca1
Ca 1
S1
S 1
O4
O 4
3) __HNO3 1 + __Ca(OH)2 1 →
H3
N 1
O 5
Ca 1
2+ 2-
4) __CaCO3 →
H 3
N 1
O 5
Ca 1
Ca2+ CO2
_1_Ag2SO4+_1_NaNO3---->_2_AgNO3+_1_Na2SO4
Ag2
Ag1 2
S1
S1
O7 10
O7 10
Na1 2
Na2
N1 2
N1 2
1__H(OH)2+_1_CaNO3
double displacement
decomposition
5) __AlCl3 + __(NH4)PO4 →
__ALPO4+__(NH4)CL3
double displacement
6) __Pb + __Fe(NO3)3 →
Fe+Pb(NO3)3
single displacement
7) __C3H6 + __O2 →
H20+CO2+CH4
combustion
8) __Na + __CaSO4 →
Ca+NaSO4
Single displacement
89

How to Balance Chemical Equations
A chemical equation is a theoretical or written representation of what happens during a chemical
reaction. The law of conservation of mass states that no atoms can be created or destroyed in a
chemical reaction, so the number of atoms that are present in the reactants has to balance the
number of atoms that are present in the products. Follow this guide to learn how to balance chemical
equations.
Step 1
Write down your given equation. For this example, we will use:
C3H8 + O2 --> H2O + CO2
Step 2
Write down the number of atoms that you have on each side of the equation. Look at the subscripts
next to each atom to find the number of atoms in the equation.
Left side: 3 carbon, 8 hydrogen and 2 oxygen
Right side: 1 carbon, 2 hydrogen and 3 oxygen
90

Step 3
Always leave hydrogen and oxygen for last. This means that you will need to balance the carbon
atoms first.
Step 4
Add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon
atoms on the left of the equation.
C3H8 + O2 --> H2O + 3CO2
The coefficient 3 in front of carbon on the right side indicates 3 carbon atoms just as the subscript 3
on the left side indicates 3 carbon atoms.
In a chemical equation, you can change coefficients, but you should never alter the subscripts.
91

Step 5
Balance the hydrogen atoms next. You have 8 on the left side, so you'll need 8 on the right side.
C3H8 + O2 --> 4H2O + 3CO2
On the right side, we added a 4 as the coefficient because the subscript showed that we already
had 2 hydrogen atoms.
When you multiply the coefficient 4 times the subscript 2, you end up with 8.
Step 6
Finish by balancing the oxygen atoms.
Because we've added coefficients to the molecules on the right side of the equation, the number of
oxygen atoms has changed. We now have 4 oxygen atoms in the water molecule and 6 oxygen
atoms in the carbon dioxide molecule. That makes a total of 10 oxygen atoms.
Add a coefficient of 5 to the oxygen molecule on the left side of the equation. You now have 10
oxygen molecules on each side.
C3H8 + 5O2 --> 4H2O + 3CO2.
The carbon, hydrogen and oxygen atoms are balanced. Your equation is complete.
92

1) ___ 2 NaNO3 + ___ 1 PbO ___ 1 Pb(NO3)2 + ___ 1 Na2O
not equal so need 2 moles to balance it out
2) ___ 6 AgI + ___ 1 Fe2(CO3)3 ___ 2 FeI3 + ___ 3 Ag2CO3
6 iodines 2 iron
2 iron
6 iodines
6 Ag
3) ___ 1 C2H4O2 + ___ 2 O2 ___ 2 CO2 + ___ 2 H2O
Now equal because their is the correct amount of hydrogen carbon and oxygen
4) ___ 1 ZnSO4 + ___ 1 Li2CO3 ___ 1 ZnCO3 + ___ 1 Li2SO4
completely balanced
5) ___ 1 V2O5 + ___ 5 CaS ___ 5 CaO + ___ 1 V2S5
calcium is now equal aswell as sulfur
93

6) ___ 1 Mn(NO2)2 + ___ 1 BeCl2 ___ 1 Be(NO2)2 + ___ 1 MnCl2
already balanced
7) ___ 3 AgBr + ___ 1 GaPO4 ___ 1 Ag3PO4 + ___ 1 GaBr3
3 silver on each side now and bromine
8) ___ H2SO4 + ___ B(OH)3 __ B2(SO4)3 + ___ 6 H2O
3 2 1
hydrogens are equal on both sides now
9) ___ 1 S8 + ___ 8 O2 ___ 8 SO2
8 sulfurs and 16 oxygens on both sides
10) 1___ Fe + ___ 2 AgNO3 ___ 1 Fe(NO3)2 + ___
2
Ag
equal silvers on both sides making it balanced
94

1) 2 NaNO3 + PbO Pb(NO3)2 + Na2O
2) 6 AgI + Fe2(CO3)3 2 FeI3 + 3 Ag2CO3
3) C2H4O2 + 2 O2 2 CO2 + 2 H2O
4) ZnSO4 + Li2CO3 ZnCO3 + Li2SO4
5) V2O5 + 5 CaS 5 CaO + V2S5
6) Mn(NO2)2 + BeCl2 Be(NO2)2 + MnCl2
7) 3 AgBr + GaPO4 Ag3PO4 + GaBr3
8) 3 H2SO4 + 2 B(OH)3 B2(SO4)3 + 6 H2O
9) S8 + 8 O2 8 SO2
10) Fe + 2 AgNO3 Fe(NO3)2 + 2 Ag
Additional Notes:
in order to find out the product of the equation, you have to know how to identify what type of reaction it is. There are 6
types of reactions: double displacement, synthesis, decomposition, single displacement, combustion, and acid base.
Use one of the reactions to find out the product of something.
double displacement AB+CD--->AD+CB
single displacement A+BC----->B+AC
synthesis A+B----->AB
decomposition AB----->A+B
combustion needs carbon and hydrogen reacting with oxygen
acid base when a acid and a base combine makes this special type of double displacement
Once the reaction is figured out, the next thing to do is balance the equation so that it will be equal on both sides of the
problem.
95

The Learning Goal for this assignment is:
The students will learn how balanaced chemical equations are used in stoichiometric calculations and how to calculate
amounts of reactants and products in a chemical equation.
Stoichiometry and Balancing Reactions
Stoichiometry is a section of chemistry that involves using relationships between reactants and/or
products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means
element and metron means measure, so stoichiometry literally translated means the measure of
elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to
first understand the relationships that exist between products and reactants and why they exist, which
require understanding how to balanced reactions.
Balancing
In chemistry, chemical reactions are frequently written as an equation, using chemical symbols. The
reactants are displayed on the left side of the equation and the products are shown on the right, with
the separation of either a single or double arrow that signifies the direction of the reaction. The
significance of single and double arrow is important when discussing solubility constants, but we will
not go into detail about it in this module. To balance an equation, it is necessary that there are the
same number of atoms on the left side of the equation as the right. One can do this by raising the
coefficients.
Reactants to Products
A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a
chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as
well as their states, and the proportion for how much of each particle is create relative to one another,
through the stoichiometric coefficient. The following equation demonstrates the typical format of a
chemical equation:
2Na(s) + 2HCl(aq) → 2NaCl(aq) + H2(g)
In the above equation, the elements present in the reaction are represented by their chemical
symbols. Based on the Law of Conservation of Mass, which states that matter is neither created nor
destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and
products, though the elements they are paired up with often change in a reaction. In this reaction,
sodium (Na), hydrogen (H), and chloride (Cl) are the elements present in both reactants, so based on
the law of conservation of mass, they are also present on the product side of the equations.
Displaying each element is important when using the chemical equation to convert between
elements.
Stoichiometric Coefficients
In a balanced reaction, both sides of the equation have the same number of elements. The
stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical
reaction to balance the number of each element on both the reactant and product sides of the
equation. These stoichiometric coefficients are useful since they establish the mole ratio between
reactants and products. In the balanced equation:
2Na(s)+2HCl(aq)→2NaCl(aq)+H2(g)
96

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
mole of H2(g). If we know how many moles of Na we start out with, we can use the ratio of 2 moles
of NaCl to 2 moles of Na to determine how many moles of NaCl were produced or we can use the
ration of 1 mole of H2 to 2 moles of Na to convert to NaCl. This is known as the coefficient factor. The
balanced equation makes it possible to convert information about one reactant or product to
quantitative data about another element. Understanding this is essential to solving stoichiometric
problems.
Example 1
Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.
Solution
___Pb(OH)4 +___H2SO4→___Pb(SO4)2 +___H2O
Start by counting the number of atoms of each element.
Unbalanced
Pb 1 1 Pb
O 8 9 O
H 6 2 H
S 1 2 S
The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does
not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the
equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a
coefficient of 2 should be added in front of H2SO4to have an equal number of sulfur on both sides of
the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4
coefficient should be added in front of H2O where there is a deficiency of oxygen. Count the number
of elements now present on either side of the equation. Since the numbers are the same, the
equation is now balanced.
Pb(OH)4 + 2H2SO4→ Pb(SO4)2 + 4H2O
Balanced
Pb 1 1 Pb
O 8 12 12 9 O
H 6 8 8 2 H
S 1 2 2 2 S
Balancing reactions involves finding least common multiples between numbers of elements present
on both sides of the equation. In general, when applying coefficients, add coefficients to the
molecules or unpaired elements last.
A balanced equation ultimately has to satisfy two conditions.
1. The numbers of each element on the left and right side of the equation must be equal.
2. The charge on both sides of the equation must be equal. It is especially important to pay
attention to charge when balancing redox reactions.
97

Stoichiometry and Balanced Equations
In stoichiometry, balanced equations make it possible to compare different elements through the
stoichiometric factor discussed earlier. This is the mole ratio between two factors in a chemical
reaction found through the ratio of stoichiometric coefficients. Here is a real world example to show
how stoichiometric factors are useful.
Example 2
There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How
many party invitations can be sent?
Solution
The equation for this can be written as
I+2S→IS2
where
I represent invitations,
S represents stamps, and
IS 2 represents the sent party invitations consisting of one invitation and two stamps.
Based on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced equation.
Invitations Stamps Party Invitations Sent
In this example are all the reactants (stamps and invitations) used up? No, and this is normally the
case with chemical reactions. There is often excess of one of the reactants. The limiting reagent, the
one that runs out first, prevents the reaction from continuing and determines the maximum amount of
product that can be formed.
Example 3
What is the limiting reagent in this example?
Solution
Stamps, because there was only enough to send out invitations, whereas there were enough
invitations for 12 complete party invitations. Aside from just looking at the problem, the problem can
be solved using stoichiometric factors.
12 I x 1IS2 = 12 IS2 possible
1I
20 S x 1IS2 = 10 IS2 possible
2S
98

When there is no limiting reagent because the ratio of all the reactants caused them to run out at the
same time, it is known as stoichiometric proportions.
Types of Reactions
There are 6 basic types of reactions.
Combustion: Combustion is the formation of CO2 and H2O from the reaction of a chemical
and O2
Combination (synthesis): Combination is the addition of 2 or more simple reactants to form a
complex product.
Decomposition: Decomposition is when complex reactants are broken down into simpler
products.
Single Displacement: Single displacement is when an element from on reactant switches with
an element of the other to form two new reactants.
Double Displacement: Double displacement is when two elements from on reactants
switched with two elements of the other to form two new reactants.
Acid-Base: Acid- base reactions are when two reactants form salts and water.
Molar Mass
Before applying stoichiometric factors to chemical equations, you need to understand molar mass.
Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual
element as listed in the periodic table established this relationship for atoms or ions. For compounds
or molecules, you have to take the sum of the atomic mass times the number of each atom in order to
determine the molar mass.
Example 4
What is the molar mass of H2O?
Solution
Molar mass = 2 × (1.00g/mol) + 1×(16.0g/mol) = 18.0g/mol
Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of
products or vice versa.
Example 5: Combustion of Propane
Propane (C3H8) burns in this reaction:
C3H8 + 5O2 → 4H2O + 3CO2
If 200 g of propane is burned, how many g of H2Ois produced?
Solution
Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products
you must convert from grams of C3H8 to moles of C3H8 then from moles of C3H8 to moles of H2O.
Then convert from moles of H2O to grams of H2O.
99

Step 1: 200g C3H8 is equal to 4.54 mol C3H8.
Step 2: Since there is a ratio of 4:1 H2O to C3H8, for every 4.54 mol C3H8 there are 18.18 mol
H2O.
Step 3: Convert 18.18 mol H2O to g H2O 18.18 mol H2O is equal to 327.27 g H2O.
Variation in Stoichiometric Equations
Almost every quantitative relationship can be converted into a ratio that can be useful in data
analysis.
Density
Density (ρ) is calculated as mass/volume. This ratio can be useful in determining the volume of a
solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse
relationship would be used.
Volume x (Mass/Volume) = Mass
Mass x (Volume/Mass) = Volume
Percent Mass
Percent establishes a relationship as well. A percent mass states how many grams of a mixture are of
a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams
are of the stated element or compound. This is useful in determining mass of a desired substance in
a molecule.
Example 6
A substance is 5% carbon by mass. If the total mass of the substance is 10.0 grams, what is the
mass of carbon in the sample? How many moles of carbon are there?
Solution
10 g sample x 5 g carbon = 0.5 g carbon
100 g sample
0.5g carbon x 1 mol carbon = 0.0416 mol carbon
12.0g carbon
Molarity
Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it
is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical
equations and dilutions.
100

Example 7
How much 5M stock solution is needed to prepare 100 mL of 2M solution?
Solution
100 mL of dilute solution (1 L/1000 mL) (2 mol/1L solution) (1 L stock solution/5 mol solution) (1000
ml stock solution/1L stock solution) = 40 mL stock solution.
These ratios of molarity, density, and mass percent are useful in complex examples ahead.
Determining Empirical Formulas
An empirical formula can be determined through chemical stoichiometry by determining which
elements are present in the molecule and in what ratio. The ratio of elements is determined by
comparing the number of moles of each element present.
Example 8
1. Find the molar mass of the empirical formula CH2O.
12.0g C + (1.00g H) * (2H) + 16.0g O = 30.0 g/mol CH2O
2. Determine the molecular mass experimentally. For our compound, it is 120.0 g/mol.
3. Divide the experimentally determined molecular mass by the mass of the empirical formula.
(120.0 g/mol) / (30.0 g/mol) = 3.9984
4. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a
slight error in the experimentally determined molecular mass. If the answer is not close to a whole
number, there was either an error in the calculation of the empirical formula or a large error in the
determination of the molecular mass.
5. Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular
formula.
CH2O * 4 =?
C: 1 * 4 = 4
H: 2 * 4 = 8
O 1 * 4 = 4
CH2O * 4 = C4H8O4
6. Check your result by calculating the molar mass of the molecular formula and comparing it to the
experimentally determined mass.
molar mass of C4H8O4= 120.104 g/mol
experimentally determined mass = 120.056 g/mol
% error = | theoretical - experimental | / theoretical * 100%
% error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100%
% error = 0.040 %
101

Stoichiometry and balanced equations make it possible to use one piece of information to calculate
another. There are countless ways stoichiometry can be used in chemistry and everyday life. Try and
see if you can use what you learned to solve the following problems.
Problem 1
Why are the following equations not considered balanced?
a. H2O(l)→H2(g)+O2(g)
b. Zn(s)+Au + (aq) →Zn 2+ b. because Au is not present in the product.
(aq) +Ag(s)
Problem 2
Hydrochloric acid reacts with a solid chunk of aluminum to produce hydrogen gas and aluminum ions.
Write the balanced chemical equation for this reaction.
Problem 3
Given a 10.1M stock solution, how many mL must be added to water to produce 200 mL of 5M
solution?
Problem 4
If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide and water, what is the
limiting reagent and how many moles of water are produced? The unbalanced equation is provided
below.
CH4(g)+O2(g)→CO2(g)+H2O(l)
C 1
H 4
O 2 4
C 1
H 2 4
O 3 4
0.502gmethane
102

Theoretical and Actual Yields
Key Terms
(Excess reagent, limiting reagent)
Theoretical and actual yields
Percentage or actual yield
Skills to Develop
Use stoichiometric calculation to determine excess and limiting reagents in a chemical reaction
and explain why.
Calculate theoretical yields of products formed in reactions that involve limiting reagents.
Evaluate percentage or actual yields from known amounts of reactants
Theoretical and Actual Yields
Reactants not completely used up are called excess reagents, and the reactant that completely
reacts is called the limiting reagent. This concept has been illustrated for the reaction:
2Na+Cl2 →2NaCl
Amounts of products calculated from the complete reaction of the limiting reagent are called
theoretical yields, whereas the amount actually produced of a product is the actual yield. The ratio of
actual yield to theoretical yield expressed in percentage is called the percentage yield.
percent yield = actual yield / theoretical yield ×100
Chemical reaction equations give the ideal stoichiometric relationship among reactants and products.
Thus, the theoretical yield can be calculated from reaction stoichiometry. For many chemical
reactions, the actual yield is usually less than the theoretical yield, understandably due to loss in the
process or inefficiency of the chemical reaction.
Example 1
Methyl alcohol can be produced in a high-pressure reaction
CO(g) + 2H2(g) →CH3OH(l)
If 6.1 metric tons of methyl alcohol is obtained from 1.2 metric tons of hydrogen reacting with excess
amount of CO, estimate the theoretical and the percentage yield?
Hint:
To calculate the theoretical yield, consider the reaction
CO(g) + 2H2(g) → CH3OH(l)
28.0 + 4.0 = 32.0 (stoichiometric masses in, g, kg, or tons)
1.2 tons H2 × 32.0 CH3OH = 9.6 tons CH3OH
4.0 H2
Thus, the theoretical yield from 1.2 metric tons (1.2x10 6 g) of hydrogen gas is 9.6 tons. The actual
yield is stated in the problem, 6.1 metric tons. Thus, the percentage yield is
103

%yield = 6.1 tons × 100 = 64%
9.6tons
Discussion
Due to chemical equilibrium or the mass action law, the limiting reagent may not be completely
consumed. Thus, a lower yield is expected in some cases. Losses during the recovery process of the
product will cause an even lower actual yield.
Example 2
A solution containing silver ion, Ag + , has been treated with excess of chloride ions Cl − . When dried,
0.1234 g of AgCl was recovered. Assuming the percentage yield to be 98.7%, how many grams of
silver ions were present in the solution?
Hint:
The reaction and relative masses of reagents and product are:
The calculation,
Ag + (aq) + Cl − (aq) → AgCl(s)
107.868 + 35.453 = 143.321
0.1234 g AgCl ×107.868 g Ag + =0.09287 g Ag +
143.321g AgCl
shows that 0.1234 g dry AgCl comes from 0.09287g Ag + ions. Since the actual yield is only 98.7%,
the actual amount of Ag + ions present is therefore
0.09287 g Ag + = 0.09409 g Ag +
0.987
Discussion
One can also calculate the theoretical yield of AgCl from the percentage yield of 98.7% to be
0.1234 g AgCl =0.1250 g AgCl
0.987
From 0.1250 g AgCl, the amount of Ag + present is also 0.09409 g.
Stoichiometry - A Review
Skills Taught
evaluate molecular weight for a given formula
evaluate weight (mass) percentages of elements for a given formula
evaluate amounts (in mass and mole units) produced in a chemical reaction from given
conditions
classify reactions by types: combination, combustion, displacement, formation, etc
determine the chemical formula when weight percentages are given and then evaluate the
mole percentages of elements in the formula
determine the chemical formula when weight percentages are given and molecular weight is
known
determine the amount produced, the actual yield, and other stoichiometry quantities for a given
reaction
104

Review Purposes
To get an overall view of stoichiometry.
Apply skills learned to perform quantitative chemical analysis.
Apply theories and rules of chemistry to solve problems.
Assess areas of strength and weakness for review purposes.
Improve problem solving strategy and learning efficiency.
Stoichiometry
STOICHIOMETRY is the quantitative relationship of reactants and products. This unit has been
divided into the following objects. A brief review is given here so that you can get a birds'-eye or
overall view of stoichiometry.
1.Amounts of substances
Express amounts of substance in mass units of g, kg, tons, and convert them to moles,
kilomoles, or millimoles.
2.Chemical formulas
Represent a substance with a formula that reflects its chemical composition, structure, and
bonding; evaluate weight and mole percentages of elements in a substance; and determine
chemical formula by elemental analysis.
3.Reaction features
Define some common features of chemical reactions; classify chemical reactions by common
features such as combination, combustion, decomposition, displacement, and redox reactions.
4.Reaction equations
Express quantitative relationships using chemical reaction equations; evaluate quantities of
reactants and products in a chemical reaction; and solve reaction stoichiometry problems.
5.Excess and limiting reagents
Define excess and limiting reagents; determine excess and limiting reagents in a reaction
mixture; and determine quantities produced in a chemical reaction.
6.Yields
Define theoretical and actual yields due to limiting reagent; apply the concept of limiting
reagent to evaluate theoretical yield; convert actual yield to percentage yields.
105

Use the space provided to write out the steps you take to solve different types of problems.
Use any additional space for notes. These 2 pages should be full when you turn in your notebook.
Steps to solve problems
Mole to Mole:
In order to balance an equation you need to know the amount of the element on
both sides.
for example: 4_KNO3 ----> 4_KNO2 +2_O2
K1 4
N1 4
O3 12
K1 4
N1 4
O4 12
Since both sides at first weren't equal you had to find a balance for both sides,
but oxygen wasn't equal so you had to multiply each side by 4 to be able to
equalize the moles. You also had to alter the amount of K and Nitrogen to be
able to change Oxygen on the left side of the equation.
Mole to mole: in order to be able to change the mole of something you have to
know the given mole first to convert it to the moles you actually have in the
problem, then take that and convert it to the different elements amount of moles
In this problems it thier is 6 moles of KNO3 it goes to 4 moles then 1 to 4 moles
of KNO2 then to 2 moles of O2
106

You need the problem to be balanced in order for yuo to be albe to even sovle
so you have to get the grams or mass of the element so you can figure out the
total mount of a substance you got or moles depending on the type of reaction
and problem.
Solving a problem where it is in a word problem you have to identify the element or know the reactant already. Once
you figure out the compound then you have to find out the charges by what group it is in. group one elements have a
charge of positive one while elements in group 17 have a negative one charge. Note: flourine is the least electronega
element.
107

Unit 6
Chapter 13 States of Matter
The students will learn what are the factors that determine and
characteristics that distinguish gases liquids and solids and
how substances change from one state to another.
Differentiate among the four states of matter.
Students will measure the physical characteristics of matter such as temperature
and density.
Students will compare and contrast the physical characteristics of the 4 states of
matter.
solid
liquid
gas
plasma
Relate temperature to the average molecular kinetic energy.
Students will be able to compare and contrast the motion of particles of a sample
at various temperatures.
Kinetic energy
Kinetic theory
Temperature
Describe phase transitions in terms of kinetic molecular theory.
Students will be able to identify and describe phase changes.
Students will be able to compare and contrast the change in particle motion
for phase changes.
Students will be able to interpret heating/cooling curves and phase diagrams.
melting point
freezing point
boiling point
condensation
sublimation
phase diagram
kinetic molecular theory
108

Chapter 14 The Behavior of Gases
The students will learn how gases respond to changes in
pressure, volume, and temperature and why the ideal gas law
is useful even though ideal gases do not exist.
Interpret the behavior of ideal gases in terms of kinetic molecular theory.
Students will be able to describe the behavior of an ideal gas.
Students will participate in activities to apply the Ideal Gas Law and its
component laws to predict gas behavior.
Students will be able to perform temperature/pressure conversions.
Compressibility
Boyle’s Law
Charles’s Law
Gay-Lussac’s Law
Combine Gas Law
Ideal Gas Law
Partial pressure
Dalton’s Law of partial pressure
Diffusion
Effusion
Graham’s Law of effusion
Chapter 15 Water and Aqueous Systems
The students will learn how the interactions between water
molecules account for the unique properties of water and how
aqueous solutions form.
Discuss the special properties of water that contribute to Earth's suitability
as an environment for life: cohesive behavior, ability to moderate
temperature, expansion upon freezing, and versatility as a solvent.
Students will be able to prepare a solution of known molarity
Students will participate in activities to calculate molarity
Surface tension
Surfactant
Aqueous solutionSolvent
Solute
109

The Learning Goal for this assignment is:
Difference among the four states of matter.
Take note over the following chapter. Use the Headings provided to organize your notes. Define and number all highlighted vocabulary (total 23 ) as well
as summarize the sections. You may add pictures where needed. The pictures should be an appropriate size. Use Arial 12 for all text. This document
should be 3 pages and should be saved as a pdf before you submit it into Angel.
Chapter 13 States of Matter
Pages 420 - 439
13.1 The Nature of Gases
Kinetic Theory and a Model for Gases
1.Kinetic energy- energy of an object because it has motion.
2.Kinetic theory-all matter consists of tiny particles that are in constant motion.
Q: What are the three assumptions of the kinetic theory as it applies to gases?
The particles in a gas are considered to be small, hard spheres with an insignificant volume.
The motion of the particles in a gas is rapid, constant, and random.
All collisions between particles in a gas are perfectly elastic.
Kinetic theory is the constant motion of matter that keeps on moving and the kinetic theory is applied
to gases with particles that are far apart in distance and empty space in it.
Gas Pressure
3. Gas pressure- results from the force exerted by gas per unit surface area of an object.
4. Vacuum- empty space with no particles and no pressure.
5. Atmospheric Pressure- collisions of atoms and molecules in air with objects.
6. Barometer- Device that is used to measure atmospheric pressure.
7. Pascal- SI unit of pressure.
8. Standard atmosphere- the pressure required to support 760mm of mercury barometer at 25-degree
C.
How does kinetic theory explain gas pressure?
Gas pressure is the result of billions of rapidly moving particles in a gas simultaneously colliding with
an object.
Gas pressure is caused by the mixture of gases. Atmospheric pressure is decreased as you make an
incline farther up and Earth’s atmosphere decreases the higher you elevate. The Barometer is used
to measure the pressure and depends on the pressure of particles colliding in the air to determine the
atmosphere pressure.
Kinetic Energy and Temperature
As a substance is heated, its particles absorb energy, some of which is stored within a particle.
What is the relationship between the temperature is kelvins and the average kinetic energy of
particles?
110

The Kelvin temperature of a substance is directly proportional to the average kinetic energy of the
particles of the substance.
a particle in motion of liquids and solids determines the temperature.
13.2 The Nature of Liquids
A Model of Liquids
What factors determine the physical properties of a liquid?
The interplay between the disruptive motions of a particle in a liquid and
the attractions among the particles determines the physical properties of liquids.
There is no attraction between particles in a gas while liquid particles are attracted to each other.
Pressure of liquid barely has an effect on the volume. Liquids and solids are condensed states of
matters. Gases are far apart and not clustered like solids.
Evaporation
9. Vaporization- the conversion of a liquid to a gas or vapor.
10. Evaporation- occurs at the surface of a liquid that is not boiling.
What is the relationship between evaporation and kinetic energy?
During evaporation, only those molecules with a certain minimum kinetic energy can escape from the
surface of the liquid.
Most molecules in liquids don’t have kinetic energy to overcome attractive forces to escape into a
gaseous state.
Vapor Pressure
11. Vapor Pressure- measure of force exerted by gas above a liquid.
An increase in temperature of a contained liquid increases the vapor pressure.
Boiling Point
12. Boiling point- the temperature at which vapor pressure of the liquid is just equal to the external
pressure on the liquid.
13. Normal boiling point- boiling point of a liquid at a pressure of 101.3kPa.
Rate of evaporation of a liquid from an open container increases as the liquid is heated up. It allows
for more particles in the liquids surface to overcome the attractive forces that keep them in the liquid
state and from changing into a gas.
13.3 The Nature of Solids
A Model of Solids
14. Melting point- the temperature at which a solids changes into a liquid.
15. Freezing point- the temperature at which a liquid changes into a solid.
How are the structures and properties of solids related?
The general properties of solids reflect the orderly arrangement of their particles and the fixed
locations of their particle.
When a solid is heated, the particles vibrate more rapidly as their kinetic energy increases.
Temperature is strong enough to get passed the attractions that can hold them in a fixed position. At
a freezing point, the liquid and solid phases are in equilibrium.
Crystal Structure and Unit Cell
16. Crystal- the particles are arranged in an orderly, repeating, three-dimensional pattern called and
crystal lattice.
17. Unit Cell- Smallest group of particles within a crystal that retains the geometric shape of the
crystal.
18. Allotropes- two or more different molecular forms of the same element in the same physical state.
19. Amorphous solid- lacks and ordered internal surface.
20. Glass- transparent fusion product of inorganic substances that have cooled to a rigid state without
crystallizing.
2111

Allotropes are consisted of atoms of the same element they have different properties because their
structure is different.
13.4 Changes of State
Sublimation
21. Sublimation- The change of a substance from a solid to a vapor without passing through the liquid
state.
This occurs when solids, like liquids, have a vapor pressure. Dry ice hace a low
Temperature of -78 degrees celsius.
Example: organic chemist use sublimation to separate mixtures and to purify
compounds.
Phase Diagrams
22. Phase diagram- the conditions of temperature and pressure at which a substance exists as a
solid, liquid, or gas (vapor).
23. Triple Point- describes the only set of conditions at which all three phases can exist in equilibrium
with one another.
A diagram in which it shows the phases of elements. Such as water in shows the boiling point the
melting point and what it would take to make it a gas or ice. It gives the number at which it would
change. Triple point shows the coordinates to pinpoint the degree of temperature to make it freeze or
melt etc.
Example: this diagram shows that depending on the pressure and the
Other factor of temperature makes a direct correlation with what the
State of matter the substance will be.
112
3

Name: Juan Roque
Name: Jayde Pasco
Grade: 10
States of Matter Project
You and you lab partner are going to create a study aid in the form of a game for the
information in Chapter 13 States of Matter.
First, each of you, independently from each other, will summarize the chapter on 3
pages of a pdf which will be submitted in Angel by the end of class on Wednesday Feb
22.
Second, you and you lab partner will be given a game platform which you will use for
your questions and answers, either Jeopardy or Kahoot.
Third, you will fill in the information at the bottom of this page with your username,
passwords and/or websites so that you do not forget this and I have a copy in case
anything gets misplaced. This page will be submitted into Angel as a Word Document
on Wednesday February 22 during class.
Fourth, you will use your notes to generate the questions and answers.
Finally, you will give me access to your game by putting the website or Game Number
on this page adding this page to your 3 pages of notes and resubmitting it in Angel as
a pdf by the end of the class on Friday Feb 24.
This page is due by the end of class on Wednesday February 22.
This Project is due by the end of class on Friday February 24.
Jeopardy (https://jeopardylabs.com)
Password: alrightfam
Edit Link: jeopardylabs.com/edit/states-of-matter697
Play Link:
Kahoot (https://getkahoot.com)
Username:
Email:
Password:
Game PIN:
jeopardylabs.com/play/states-of-matter697
4
113

temperature
average kinetic energy
pressure
volume
if one constant we can predict how it can change the other thing
keep volume constant increase in temp we can tell the increase in pressure.
114

up
down
Temperature stays constant and if we change the volume how will it change the pressure
as I change the pressure how will it increase the temperature.
sqeezing the balloon the preasure will increase volume decreases.
this is called inverse relationships- when one thing goes up then one thing has to go down.
3x2 = 2x3
multiply and you have a value of 6.
decrease volume the pressure increases vice versa
pressure decrease increase in volume.
115

keep the pressure constant
heat up the particles and they move around more and the pressure increases.
increase the temp volume is increase
decrease the volume decrease the temperature
2 8
2= 8
change the temperature you can tell the volume.
116

as we increase the temperature increase the pressure
volume remains constant.
heat up and they hit the side of the balloons more if the container is the same.
3 = 9
3 9
directly proportional to kelvin temperature.
117

direct relationship
so divide
temperature constant
decrease volume
increase pressure
inverse relationship
pressure constant
r=ratio
n= number of moles
Ideal Gas Law
PxV=nRT
R=0.08205 L atom / mol x K
= 8.3145 L kPa / mol x K
= 8.3145 J / mol x K
1.987 cal / mol x K
= 62.364 L torr/ mol x K
R = .0821 atm
R= 8.31 kPa
R= 62.4 torrs
Notes: 1 atm= 101.32 atm
1 J= 1 L kPa
1 cal= 4.182 J
1 atm= 760 torr
What pressure will be exerted by .450 mol of gas at 25C if it is contained
in a .650L vessel?
Px.650L= .450mol(.0821)(25 298K)
Px.650L=11.00961
=16.92307692
=17.0 atm
118

P1V1=P2V2
atm
1 atmosphere is equal to 760 torr
2.605atm X 7730mL= 2.95 mL
1980torr X 7730mL
axb=c
20136.65= 2.95x
2.95 2.95
6830mL
1980torr
1 atm 2.605263 atm
760torr
119

kelvin equals everything stops
K=
to find Kelvin to Celsius
+273
2920mL
Vl = 3550mL
85.4C 294.0K
V1 = 3550
358.4K 294.0K
294.0K x X= 358.4K X 3550mL
294.0K x X = 1272320K x mL
294.0K 294.0K
4.33L 4.337619048mL
2920mL = 6660mL
x 853.0K
6660mL x X= 2920mL x 8530K
6660mL x = 2490760mLxK
6660mL 6660mL
x= 373.987988K 374K
120

108kPa= X
41C 22C
108kPa=X
314K 314K
41+273= 314K 22+273= 295
314K= 31860
314K 314K
101.464968
101kPa
101kPa
198kPa= 225kPa
27C x
27
273
300kPa
198kPa= 225kPa
300kPa x 340.9090909
341K
PxV
P
T
V
T
PV
Combined Gas Law
155kPa= 1.00L= 605kPaX
25C 125C
V= .342L
155kPa 1.00L= 605kPa
298K
398K
Ideal gas law
PxV=nRT
121

The Learning Goal for this assignment is:
The students will learn how the interaction between water molecules account for the
unique properties of water and how aqueous solutions form
Page 124
Take note over the following chapter. Use the Headings provided to organize your notes. Define and number all highlighted vocabulary (total 22 ) as well
as summarize and take notes over the sections. You may add pictures where needed. The pictures should be an appropriate size. Use Arial 12 for all
text. This document should be 2 pages and should be saved as a pdf before you submit it into Angel.
Chapter 15 Water and Aqueous Systems
Pages 488 - 507
15.1 Water and Its Properties
Water in the Liquid State
Many unique and important properties of water – including its high surface tension, low vapor pressure,
and high boiling point – result from hydrogen bonding
Surface Tension - The inward force or pull that tends to minimize the surface area of a liquid. Surfactant –
Any substance that interferes with the hydrogen bonding between water molecules and thereby reduces
surface tension
Water is a simple molecule requiring three atoms; the oxygen atom forms a covalent bond with each of
the hydrogen atoms. In a water molecule, the bond polarities are equal, but the two poles do not cancel
each other because a water molecule is bent. The molecules as a whole is polar; the negative end
attracts the positive end. Hydrogen bonding also occurs – the oxygen atom has a partial negative charge.
The surface of water acts like a skin – the ability of water molecules to form hydrogen bonds with other
molecules that surround them on all sides. Water forms nearly spherical drops on a leaf. Water molecules
at the surface of the water drop cannot for hydrogen bonds with molecules in the air, so they are drawn
into the body of the liquid. Water drops bead up on some surfaces, is added to water, the drop spreads
out. So basically a surfactant causes the beads of water to collapse and spread out. An extensive network
of hydrogen bonds holds the molecules in liquid water to one another. These hydrogen bonds must be
broken before water changes states so the tendency of these molecules to escape is low and evaporation
is slow. Molecular compounds of low molar mass are usually gases or liquids with low boiling points at
normal atmospheric pressure.
Water in the Solid State
The structure of ice is a regular open framework of water molecules in a hexagonal arrangement
As a typical liquid cools, it begins to contract and its density increases gradually; it increases because the
molecules of the liquid move closer together so that a given volume of the liquid contains more molecules
and thus more mass. A typical solid sinks in its own liquid because the density of the solid is greater than
that of the corresponding liquid. As water begins to cool, it behaves initially like a typical liquid, it contracts
slightly and its density gradually increases. Below 4 ° C, water no longer behaves like a liquid and at 0° it
forms as ice. The unique properties of ice are a result of hydrogen bonding. Extensive hydrogen bonding
in ice holds the water molecules farther apart in a more ordered arrangement than in liquid water. The
hexagonal symmetry of a snowflakes reflects the structure of the ice crystal. A layer of ice acts like an
insulator acts.
15.2 Homogeneous Aqueous Systems
Solutions
Substances that dissolve most readily in water include ionic compounds and polar covalent compounds.
Aqueous solution: water that contains dissolved substances Solvent: In a solution, the dissolving medium
Solute: The dissolved particles in a solution. Salvation: The process by which the positive and negative
ions of an ionic solid become surrounded by solvent molecules
Water dissolves so many of the substances that in comes in contact with making it an aqueous solution. A
solvent dissolves the solute, and the solute becomes dispersed in the solvent. They can be any state of
matter. Solutions are homogeneous mixtures and stable mixtures. Nonpolar covalent compounds, such as

methane and compounds found in oil, grease and gasoline don’t dissolve in water but in gasoline. When
an ionic solid dissolves, the ions become solvated or surrounded by solvent molecules. Water molecules
are in continuous motion because they have kinetic energy. Some ionic compounds, the attractions
among the ions in the crystals are stronger than the attractions exerted by water. These compounds
cannot be solvated to any significant extent and are therefore nearly insoluble. Oil and gasoline do not mix
with water because they are composed of nonpolar molecules. The attractive forces that hold two
molecules in oil together are similar in magnitude to the forces that hold two molecules in gasoline
together. Molecules in oil can easily separate and replace molecules in gasoline to form a solution. Polar
solvents dissolve ionic compounds and nonpolar compounds - like dissolves like.
Electrolytes and Nonelectrolytes
All ionic compounds are electrolytes because they dissociate into ions.
Electrolyte: A compound that conducts an electric current when it is in an aqueous solution or in the molten
state. Nonelectrolyte: A compound that conduct an electric current in either an aqueous solution or the molten
state. Strong electrolyte: . Weak electrolyte: Conducts an electric current poorly because only a fraction of the
solute in the solution exists as ions.
Conduction of an electric current requires ions that are mobile and thus able to carry changes through a liquid.
The solution conducts an electric current if it contains ions. Some polar molecular compounds are
nonelectrolytes in the pure state but become electrolytes when they dissolve in water. This change occurs
because such compounds ionize in solutions. Not all electrolytes conduct an electric current to the same
degree. In a solution that contains a strong electrolyte all or nearly all of the solute exists as ions. The ions
move in solution and conduct an electric current. Inorganic bases and acids are strong electrolytes. Organic
acid and bases are weak electrolytes. Electrolytes are essential to metabolic processes as it carries electrical
impulses internally and to other cells. However, an electrolyte imbalance can occur if you become dehydrated.
Hydrates
The forces holding the water molecules in hydrates are not very strong, so the water is easily lost and
regained.
Water of hydration: Water contained in a crystal. Hydrate: A compound that contains water of hydration.
Effloresce: . Anhydrous: A substance that does not contain water. Hygroscopic: Hydrates or other
compounds that remove moisture from air. Desiccant: Substance sued to absorb moisture from air and
create dry atmosphere. Deliquescent: They remove sufficient water from the air to dissolve completely
and form solutions.
Water can be driven from a hydrate from heating it. Water molecules are an integral part of the crystal
structures. In writing the formula of a hydrate, use a dot to connect the formula of the compound and the
number of water molecules per formula unit. The water molecules in hydrates are held by weak forces so
hydrates often have vapor pressure. As the crystals lose water of hydration, they effloresce and become
coated with a white powder of anhydrous sodium carbonate. Hydrated ionic compounds that have a low
vapor pressure remove water from moist air to form higher hydrates. A desiccator may contain calcium
chloride. Substance that must be kept dry are stored inside.
15.3 Heterogeneous Aqueous Systems
Suspensions
A suspension differs from a solution because the particles of a suspension are much larger and do not
stay suspended indefinitely.
Suspension: A mixture from which particles settle out upon standing
Suspension are heterogeneous because at least two substances can be identified.
Colloids
Colloids have particles smaller than those in suspensions and larger than those in solutions.
Colloids: A heterogeneous mixture containing particles that range in size from 1 n to 1000 nm. Typical
effect: the scattering of visible light by colloidal particles. Brownian motion: Chaotic movement of colloidal
particles. Emulsion: A colloidal dispersion of a liquid in a liquid.
Particles are spread or dispersed. They’re cloudy so they can be distinguished by Tyndall effect and by
the observation of Brownian motion. They are also subject to coagulation or clumping together and they
can be emulsified or made stable. Coagulation are when they absorb ions from the from the dispersing
medium onto the surface.
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Unit 7
Chapter 16 Solutions
The students will learn what properties are used to describe
the nature of solutions and how to quantify the concentration
of a solution.
Chapter 17 Thermochemistry
The student will learn how energy is converted in a chemical
or physical process and how to determine the amount of
energy is absorbed or released in that process.
Differentiate among the various forms of energy and recognize that they can
be transformed from one form to others.
Students will participate in activities to investigate and describe the
transformation of energy from one form to another (i.e. batteries, food, fuels,
etc.)
Explore the Law of Conservation of Energy by differentiating among open,
closed, and isolated systems and explain that the total energy in an isolated
system is a conserved quantity.
Students will be able to calculate various energy changes:
o q = mc∆t
o ∆Hfus
o ∆Hmelt
Thermochemistry
Heat
System
Surrounding
Law of conservation of energy
Bond Making is exothermic
Bond Breaking is endothermic
Heat capacity
Specific heat
Calorimetry
Enthalpy
Thermochemical equation
Molar heat of (fusion, solidification,
vaporization, condensation, solution)
Distinguish between endothermic and exothermic chemical processes.
Students will be able to recognize exothermic and endothermic reactions through
experimentation.
Students will participate in activities (Pasco) to create exothermic and
endothermic graphs.
Endothermic
Exothermic
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Create and interpret potential energy diagrams, for example: chemical
reactions, orbits around a central body, motion of a pendulum
Students will participate in activities (Pasco) to create exothermic and
endothermic graphs.
Students will be able to interpret exothermic and endothermic reaction graphs.
Potential energy diagram
Thermochemical equations
Chapter 18 Reaction Rates and Equilibrium
The student will learn how the rate of a chemical reaction can
be controlled, what the role of energy is and why some
reactions occur naturally and others do not.
Explain how various factors, such as concentration, temperature, and
presence of a catalyst affect the rate of a chemical reaction.
Students will be able to describe how each factor may affect the rate of a
chemical reaction.
Students will be able to compare the relative effect of each factor on the rate of a
chemical reaction.
Rate
Collision theory
Activation energy
Catalyst
Activated complex
Inhibitor
Explain the concept of dynamic equilibrium in terms of reversible processes
occurring at the same rates.
Students will be able to describe a system in dynamic equilibrium.
Students will be able to describe how factors may affect the equilibrium of a
reaction.
Reversible reaction
Chemical equilibrium
Le Chatelier principle
Explain entropy’s role in determining the efficiency of processes that convert
energy to work.
Students will be able to describe the change in entropy of a reaction.
Students will be able to determine if a reaction is spontaneous
Entropy
Law of disorder
Spontaneous/nonspontaneous reaction
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The Learning Goal for this assignment is:
The students will learn what properties are used to describe the nature of solutions and how to quantify
the concentration of a solution.
Defining Concentration
Measures of Concentration
Concentration is defined as the amount of dissolved solute in a given amount of solvent or
solution. There are several terms that describe concentration. Some of these terms are relative;
that is, they can be used only to compare the concentration of one solution to another. Dilute and
concentrated are two such terms. A dilute solution contains less dissolved solute than a
concentrated solution (in equal volumes of solution).
The terms saturated, unsaturated, and supersaturated are terms that describe concentration more
precisely.
Saturated: The maximum amount of solute is dissolved in a given amount of solvent at a
particular temperature. Such solutions are stable.
Unsaturated: Less than the maximum amount of solute is dissolved in a given amount of
solvent at a particular temperature. Such solutions are stable.
Supersaturated: More than the maximum amount of solute is dissolved in a given amount of
solvent at a particular temperature. Such solutions are unstable.
Look at the solubility curve shown below:
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The solubility of NaNO3 is 86.0 g/100 mL H2O (at 20 °C). If you
prepare a solution of 50.0 g NaNO3 dissolved in 100 mL H2O, an unsaturated solution results
(Point A on the graph).
continue adding NaNO to the solution until 86.0 g are dissolved, a saturated solution results
(Point B).
heat the solution to 50 °C, 113 grams NaNO3 can be dissolved (Point C). When the solution
cools back down to 20 °C, it will be supersaturated (Point D).
Quantifying Concentration
To describe the concentration of a solution even more precisely, various measures of concentration
can be used. Some of the ways concentration can be quantified include calculating the
Mass of solute per solution mass (expressed as a percent or parts per million)
Moles of solute per kilogram solvent (molality)
Mass of solute per liter of solution (grams/liter)
Moles of solute per liter of solution (molarity)

Part 1: Mass Percent
Mass percent (also called percent by mass, weight percent, or percent by weight) compares the
mass of the solute to the entire mass of the solution.
Notes:
Mass percent is when the mass of solute is divided by the mass of the entire solution x
100.
Part 2: Parts per Million
Parts per million (ppm) is another measure of concentration. It is similar to mass percent. But
mass percent indicates the number of grams of solute per 100 g solution. Parts per million
indicates the number of grams of solute per 1,000,000 g solution. This measure of concentration is
often used to express the concentrations of very dilute solutions.
Notes:
Parts per million is the mass of solute (whats getting dissolved) by the mass of the solution x 10 to
the 6th.
This Part measure 3: Molality of concentration is used for very dilute solutions.
Molality (m) is the ratio of the moles of solute to the kilograms of solvent. Note: this is the first
measure of concentration that is concerned with the mass of the solvent, not the mass of the
solution as a whole.
Notes:
Molality is moles of solute divided by the kg of solvent, make sure to convert units of measurements
before actually doing it. Kg is mass of solvent.
Part 4: Grams per Liter
To express the concentration of a solution in grams per liter, you must know the mass of the solute
and the volume of the solution, not just the volume of the solvent.
Notes:
Grams per Liter is the mass of solute divided by the volume of the solution. Remember to have the
right units before doing anything math. So convert with the right units.
Suppose you wanted to know what the concentration would be before making the solution. Could that
be done? In order to relate the volume or mass of solvent to the volume of solution, you would have
to know the density of the solution. You will see how solution density can be used to calculate
molarity in the next section.
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Part 5: Molarity
Molarity (M) is the most common measure of concentration. The concentration of most solutions
you see in the lab are expressed in terms of molarity. Just like g/L, molarity calculations require
that you know either the volume or density of the resulting solution.
Notes:
Molarity is not Molality, it is moles of solute divded by the liters of solution. The units of measurements
are moles/liter.
Using Density to Calculate Molarity
If the volume of the resulting solution is not known, molarity is calculated as follows:
Convert the volume of solvent to grams. (The simulation does this step for you.)
Determine the total mass of the solution (mass of solute + mass of solvent).
Convert the solution mass to volume in milliliters, using its density (volume = mass / density).
Convert the solution volume to liters (divide by 1000).
Convert solute grams to moles.
Calculate the molarity (moles solute / L solution).
Reinforcing What You've Learned
1. 35.0 g potassium dichromate are dissolved in 354.0 g distilled water. What is the concentration
of the resulting solution, expressed as percent by mass? 9.00%
2. A chemist discovers that 2587.0 g distilled water are contaminated with 13.0 g NaNO3. What is
the concentration of NaNO3, expressed in ppm? 5.00 X 10 3
3. 175.00 g H2SO4 are added to 61.49 g distilled H2O. What is the molality of the resulting solution?
29.04m
4. 225 g NaBr are dissolved in 525 g distilled water. If the volume of the resulting solution is
584 mL, what is its concentration when expressed in g/L? 385g/L
5. 470.0 g Na2CO3 are dissolved in 4230.0 g distilled water. What is the molarity of the resulting
solution (D = 1.1029 g/mL)? 1.1561 M
Applying What You've Learned
6. Think about the equations for mass percent and parts/million. Consider how you might convert
between these two measures of concentration.
Mass percent = ppm ÷ 10,000 ppm = mass percent X .0001
7. Think about the equations for grams/liter and molarity. Consider how you might convert
between these two measures of concentration.
grams / L = molarity X molar mass
molarity = grams / L ÷ molar mass
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Glossary
Concentration: Amount of dissolved solute in a given amount of solvent or solution.
Density: Amount of matter per unit volume. Density is calculated by dividing an object's mass by
its volume.
Mass: Amount of matter an object contains or, more scientifically, the measure of an object's
resistance to changes in motion. The SI (Systéme International) unit for mass is the kilogram. In
the lab, mass is often measured in grams.
Mass percent: Also referred to as percent by mass and, occasionally, weight percent or percent
by weight. Mathematically, mass % = (mass of solute / mass of entire solution) x 100.
Molar mass: Mass of a compound, calculated by adding up the individual masses for its component
atoms, which are obtained from the periodic table of the elements. Molar mass is expressed in
grams/mole and is sometimes referred to as molecular mass (for molecular compounds) or formula
mass (for ionic compounds).
Molality (m): Moles of solute per kilogram of solvent. Mathematically, m = moles of solute / kilogram of
solvent.
Molarity (M): Moles of solute per liter of solution. Mathematically, M = moles of solute / liter of
solution.
Mole (mol): Counting unit used to express the large numbers of particles, such as atoms or
molecules, that are involved in chemical processes. One mole of particles contains 6.02 x 10
particles. The mass of one mole of an element, in grams, is equivalent to the atomic mass for that
element, as indicated on the periodic table.
Parts per million (ppm): Measure of concentration often used for dilute solutions. Mathematically,
ppm = (mass of solute / mass of solution) x 10.
Solubility: Measure of the maximum amount of solute that can be dissolved in a given amount of
solvent at a given temperature, forming a stable solution.
Solute: Dissolved substance in a solution. The solute is generally the solution component present
in the lesser amount.
Solution: Homogeneous mixture in which one substance has been dissolved in another.
Solvent: Substance in which a solute is dissolved to form a solution. The solvent is generally the
solution component present in the greater amount.
Volume: Amount of space an object occupies. The SI (Systéme International) unit for volume is
the cubic meter. In the lab, volume is often measured in cubic centimeters, milliliters, or liters.
SAS Curriculum Pathways VLab #866
Page 131

The Learning Goal for this assignment is:
The student will learn how energy is converted in a chemical or physical process and how to
determine the amount of energy is absorbed or released in that process.
The System and the Surroundings in Chemistry
Thermochemistry
The system is the part of the universe we wish to focus our attention on. In the world of chemistry, the
system is the chemical reaction. For example:
2H2 + O2 ---> 2H2O
The system consists of those molecules which are reacting.
The surroundings are everything else; the rest of the universe. For example, say the above reaction is
happening in gas phase; then the walls of the container are part of the surroundings.
There are two important issues:
1. a great majority of our studies will focus on the change in the amount of energy, not the
absolute amount of energy in the system or the surroundings.
2. regarding the direction of energy flow, we have a "sign convention."
Two possibilities exist concerning the flow of energy between system and surroundings:
1. The system can have energy added to it, which increases its amount and lessens the energy
amount in the surroundings.
2. The system can have energy removed from it, thereby lowering its amount and increasing the
amount in the surroundings.
We will signify an increase in energy with a positive sign and a loss of energy with a negative sign.
Also, we will take the point-of-view from the system. Consequently:
1. When energy (heat or work) flow out of the system, the system decreases in its amount. This
is assigned a negative sign and is called exothermic.
2. When energy (heat or work) flows into the system, the system increases its energy amount.
This is assigned a positive sign and is called endothermic.
We do not discuss chemical reactions from the surrounding's point-of-view. Only from the system's.
Notes:
Page 132
Thermochemistry is about the amount of energy in a substance not in the system or surroundings
The energy flow is always going to have a sign convention
When it comes the system and surrounding, they can have energy added and removed. When
added it will increase amount and lessen energy in the surroundings; when removed its the
opposite. Exothermic is when energy flows out the system (decreasing) and endothermic is when
it flows into the system.

Specific Heat
Here is the definition of specific heat:
the amount of heat necessary for 1.00 gram of a substance to change 1.00 °C
Note the two important factors:
1. It's 1.00 gram of a substance
2. and it changes 1.00 °C
Keep in mind the fact that this is a very specific value. It is only for one gram going one degree. The
specific heat is an important part of energy calculations since it tells you how much energy is needed
to move each gram of the substance one degree.
Every substance has its own specific heat and each phase has its own distinct value. In fact, the
specific heat value of a substance changes from degree to degree, but we will ignore that.
The units are often Joules per gram-degree Celsius (J/g*°C). Sometimes the unit J/kg K is also used.
This last unit is technically the most correct unit to use, but since the first one is quite common, you
will need to know both.
I will ignore calorie-based units almost entirely.
Here are the specific heat values for water:
Phase J g¯1 °C¯1 J kg¯1
K¯1
Gas 2.02 2.02 x 10 3
Liquid 4.184 4.184 x 10 3
Solid 2.06 2.06 x 10 3
Notice that one set of values is simply 1000 times bigger than the other. That's to offset the influence
of going from grams to kilograms in the denominator of the unit.
Notice that the change from Celsius to Kelvin does not affect the value. That is because the specific
heat is measured on the basis of one degree. In both scales (Celsius and Kelvin) the jump from one
degree to the next are the same "distance." Sometimes a student will think that 273 must be involved
somewhere. Not in this case.
Specific heat values can be looked up in reference books. Typically, in the classroom, you will not be
asked to memorize any specific heat values. However, you may be asked to memorize the values for
the three phases of water.
As you go about the Internet, you will find different values cited for specific heats of a given
substance. For example, I have seen 4.186 and 4.187 used in place of 4.184 for liquid water. None of
the values are wrong, it's just that specific heat values literally change from degree to degree. What
happens is that an author will settle on one particular value and use it. Often, the one particular value
used is what the author used as a student.
Hence, 4.184.
Page 133

The Time-Temperature Graph
We are going to heat a container that has 72.0 grams of ice (no liquid water yet!) in it. To make the
illustration simple, please consider that 100% of the heat applied goes into the water. There is no loss
of heat into heating the container and no heat is lost to the air.
Let us suppose the ice starts at -10.0 °C and that the pressure is always one atmosphere. We will
end the example with steam at 120.0 °C.
There are five major steps to discuss in turn before this problem is completely solved. Here they are:
1. the ice rises in temperature from -10.0 to 0.00 °C.
2. the ice melts at 0.00 °C.
3. the liquid water then rises in temperature from zero to 100.0 °C.
4. the liquid water then boils at 100.0 °C.
5. the steam then rises in temperature from 100.0 to 120.0 °C
Each one of these steps will have a calculation associated with it. WARNING: many homework and
test questions can be written which use less than the five steps. For example, suppose the water in
the problem above started at 10.0 °C. Then, only steps 3, 4, and 5 would be required for solution.
To the right is the type of graph which is typically used to
show this process over time.
You can figure out that the five numbered sections on the
graph relate to the five numbered parts of the list just above
the graph.
Also, note that numbers 2 and 4 are phases changes: solid
to liquid in #2 and liquid to gas in #4.
Q=mcΔT
where ΔT is (Tf – Ti)
Here are some symbols that will be used, A LOT!!
Δt = the change in temperature from start to finish in degrees Celsius (°C)
m = mass of substance in grams
c = the specific heat. Its unit is Joules per gram X degree Celsius (J / g °C is one way to write
the unit; J g¯1 °C¯1 is another)
q = the amount of heat involved, measured in Joules or kilojoules (symbols = J and kJ)
mol = moles of substance.
ΔH is the symbol for the molar heat of fusion and ΔH is the symbol for the molar heat of
vaporization.
We will also require the molar mass of the substance. In this example it is water, so the molar mass is
18.0 g/mol.
Notes:
Page 134
Specific heat is basically the amount of heat for a substance to change heat. It will always be 1
gram of a substance and it will change to 1 Celsius.
Time-Temperature Graph shows the change from ice to water to vapor over time.
It will always be -10 if you're going to start with ice and ends at 0 because ice melts.
For water, it can be watever (my pun) degree, but once it is 100 it starts to boil becoming vapor.
Steam rises in temperature from 100 to 120.
Memorize the key notes above and Q=mc(DELTA)T and (Delta)(T-T)
Q=mC(DELTA)T is mass of substance times specific heat times the change in temperature
change in temperature is inital minus the final (THERE CAN BE NO NEGATIVE)

Step One: solid ice rises in temperature
As we apply heat, the ice will rise in temperature until it
arrives at its normal melting point of zero Celsius.
Once it arrives at zero, the Δt equals 10.0 °C.
Here is an important point: THE ICE HAS NOT MELTED
YET.
At the end of this step we have SOLID ice at zero
degrees. It has not melted yet. That's an important point.
Each gram of water requires a constant amount of energy
to go up each degree Celsius. This amount of energy is
called specific heat and has the symbol c.
72.0 grams of ice (no liquid water yet!) has changed 10.0 °C. We need to calculate the energy
needed to do this.
This summarizes the information needed:
Δt = 10 °C
The mass = 72.0 g
c = 2.06 Joules per gram-degree Celsius
The calculation needed, using words & symbols is:
q = (mass) (Δt) (c)
Why is this equation the way it is?
Think about one gram going one degree. The ice needs 2.06 J for that. Now go the second degree.
Another 2.06 J. Go the third degree and use another 2.06 J. So one gram going 10 degrees needs
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
on until 72 grams.
With the numbers in place, we have:
q = (72.0 g) (10 °C) (2.06 J/g °C)
So we calculate and get 1483.2 J. We won't bother to round off right now since there are four more
calculations to go. Maybe you can see that we will have to do five calculations and then sum them all
up.
One warning before going on: three of the calculations will yield J as the unit on the answer and two
will give kJ. When you add the five values together, you MUST have them all be the same unit.
In the context of this problem, kJ is the preferred unit. You might want to think about what 1483.2 J is
in kJ.
Notes:
Ice has not melted when it reached 0 degrees, it is solid ice at zero.
The "c" in mcΔt is the specific heat and in this stage it is 2.06 Joules per gram-degree Celsius
Make sure to have the right units of measurement when you're done calculating the math so you
won't make any mistakes
Page 135

Step Two: solid ice melts
Now, we continue to add energy and the ice begins to
melt.
However, the temperature DOES NOT CHANGE. It
remains at zero during the time the ice melts.
Each mole of water will require a constant amount of
energy to melt. That amount is named the molar heat of
fusion and its symbol is ΔHf. The molar heat of fusion is
the energy required to melt one mole of a substance at its
normal melting point. One mole of solid water, one mole
of solid benzene, one mole of solid lead. It does not
matter. Each substance has its own value.
During this time, the energy is being used to overcome water molecules' attraction for each other,
destroying the three-dimensional structure of the ice.
The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion
between calories and Joules is 4.184 J = 1.000 cal.
Sometimes you also see this number expressed "per gram" rather than "per mole." For example,
water's molar heat of fusion is 6.02 kJ/mol. Expressed per gram, it is 334.16 J/g.
Typically, the term "heat of fusion" is used with the "per gram" value.
72.0 grams of solid water is 0.0 °C. It is going to melt AND stay at zero degrees. This is an important
point. While the ice melts, its temperature will remain the same. We need to calculate the energy
needed to do this.
This summarizes the information needed:
ΔHf = 6.02 kJ/mol
The mass = 72.0 g
The molar mass of H2O = 18.0 gram/mol
The calculation needed, using words & symbols is:
q = (moles of water) (ΔHf)
We can rewrite the moles of water portion and make the equation like this:
q = (grams water / molar mass of water) (ΔHf)
Why is this equation the way it is?
Think about one mole of ice. That amount of ice (one mole or 18.0 grams) needs 6.02 kilojoules of
energy to melt. Each mole of ice needs 6.02 kilojoules. So the (grams water / molar mass of water) in
the above equation calculates the amount of moles.
With the numbers in place, we have:
q = (72.0 g / 18.0 g mol¯1 ) (6.02 kJ / mol)
So we calculate and get 24.08 kJ. We won't bother to round off right now since there are three more
calculations to go. We're doing the second step now. When all five are done, we'll sum them all up.
Page 136

Step Three: liquid water rises in temperature
Once the ice is totally melted, the temperature can now
begin to rise again.
It continues to go up until it reaches its normal boiling
point of 100.0 °C.
Since the temperature went from zero to 100, the Δt is
100.
Here is an important point: THE LIQUID HAS NOT
BOILED YET.
At the end of this step we have liquid water at 100 degrees. It has not turned to steam yet.
Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount
of energy is called specific heat and has the symbol c. There will be a different value needed,
depending on the substance being in the solid, liquid or gas phase.
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
water WILL NOT BOIL. We need to calculate the energy needed to do this.
This summarizes the information needed:
Δt = 100.0 °C (100.0 °C – 0.0 °C)
The mass = 72.0 g
c = 4.184 Joules per gram-degree Celsius
The calculation needed, using words & symbols is:
q = (mass) (Δt) (c)
Why is this equation the way it is?
Think about one gram going one degree. The liquid water needs 4.184 J for that. Now go the second
degree. Another 4.184 J. Go the third degree and use another 4.184 J. So one gram going 100
degrees needs 4.184 x 100 = 418.4 J. Now we have 72 grams, so gram #2 also needs 418.4, gram
#3 needs 418.4 and so on until 72 grams.
With the numbers in place, we have:
q = (72.0 g) (100.0 °C) (4.184 J/g °C)
So we calculate and get 30124.8 J. We won't bother to round off right now since there are two more
calculations to go. We will have to do five calculations and then sum them all up.
Notes:
The liquid rises in temperature once the ice has melted.
Remember that the liquid has not boiled yet, similar to when the ice had not melted
yet.
At the end of this step, water will be at 100 degrees not yet steam.
To calculate the information, use Q=mcΔt and use the right units
Page 137

Step Four: liquid water boils
Now, we continue to add energy and the water begins to
boil.
However, the temperature DOES NOT CHANGE. It
remains at 100 during the time the water boils.
Each mole of water will require a constant amount of
energy to boil. That amount is named the molar heat of
vaporization and its symbol is ΔH. The molar heat of
vaporization is the energy required to boil one mole of a
substance at its normal boiling point. One mole of liquid water, one mole of liquid benzene, one mole
of liquid lead. It does not matter. Each substance has its own value.
During this time, the energy is being used to overcome water molecules' attraction for each other,
allowing them to move from close together (liquid) to quite far apart (the gas state).
The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion
between calories and Joules is 4.184 J = 1.000 cal.
Typically, the term "heat of vaporization" is used with the "per gram" value.
72.0 grams of liquid water is at 100.0 °C. It is going to boil AND stay at 100 degrees. This is an
important point. While the water boils, its temperature will remain the same. We need to calculate the
energy needed to do this.
This summarizes the information needed:
ΔH = 40.7 kJ/mol
The mass = 72.0 g
The molar mass of H2O = 18.0 gram/mol
The calculation needed, using words & symbols is:
q = (moles of water) (ΔH)
We can rewrite the moles of water portion and make the equation like this:
q = (grams water / molar mass of water) (ΔH)
Why is this equation the way it is?
Think about one mole of liquid water. That amount of water (one mole or 18.0 grams) needs 40.7
kilojoules of energy to boil. Each mole of liquid water needs 40.7 kilojoules to boil. So the (grams
water / molar mass of water) in the above equation calculates the amount of moles.
With the numbers in place, we have:
q = (72.0 g / 18.0 g mol¯1 ) (40.7 kJ / mol)
So we calculate and get 162.8 kJ. We won't bother to round off right now since there is one more
calculation to go. We're doing the fourth step now. When all five are done, we'll sum them all up.
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Step Five: steam rises in temperature
Once the water is completely changed to steam, the
temperature can now begin to rise again.
It continues to go up until we stop adding energy. In this
case, let the temperature rise to 120 °C.
Since the temperature went from 100 °C to 120°C, the Δt
is 20°C.
Each gram of water requires a constant amount of energy
to go up each degree Celsius. This amount of energy is called specific heat and has the symbol c.
There will be a different value needed, depending on the substance being in the solid, liquid or gas
phase.
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
needed to do this.
This summarizes the information needed:
Δt = 20 °C
The mass = 72.0 g
c = 2.02 Joules per gram-degree Celsius
The calculation needed, using words & symbols is:
q = (mass) (Δt) (c)
Why is this equation the way it is?
Think about one gram going one degree. The liquid water needs 2.02 J for that. Now go the second
degree. Another 2.02 J. Go the third degree and use another 2.02 J. So one gram going 20 degress
needs 2.02 x 20 = 44 J. Now we have 72 grams, so gram #2 also needs 44, gram #3 needs 44 and
so on until 72 grams.
I hope that helped.
With the numbers in place, we have:
q = (72.0 g) (20 °C) (2.02 J/g °C)
So we calculate and get 2908.8 J. We won't bother to round off right now since we still need to sum
up all five values.
Notes:
The water has completely changed to steam, so the temperature can rise all over
again.
Remember that the Δt has to be positive, so if you have to reverse the numbers you
can
The value of "c" during this stage will be 2.02 Joules per gram degree Celsius
Use Q=mcΔt
Use the correct measurements and don't round for anything
Page 139

The following table summarizes the five steps and their results. Each step number is a link back to
the explanation of the calculation.
Converting to kJ gives us this:
1.4832 kJ
24.08 kJ
30.1248 kJ
162.8 kJ
2.9088 kJ
Step q 72.0 g of H2O
1 1483.2 J Δt = 10 (solid)
2 24.08 kJ melting
3 30124.8 J Δt = 100 (liquid)
4 162.8 kJ boiling
5 2908.8 J Δt = 20 (gas)
Summing up gives 221.3968 kJ and proper significant digits gives us 221.4 kJ for the answer.
Notice how all units were converted to kJ before continuing on. Joules is a perfectly fine unit; it's just
that 221,396.8 J is an awkward number to work with. Usually Joules is used for values under 1000,
otherwise kJ is used.
By the way, on other sites you may see kj used for kilojoules. I've also seen Kj used. Both of these
are wrong symbols. kJ is the only correct symbol.
Enthalpy
When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal
to the change in enthalpy. Enthalpy (H) is the sum of the internal energy (U) and the product of
pressure and volume (PV) given by the equation:
H=U+PV
When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal
to the change in enthalpy. Enthalpy is a state function which depends entirely on the state
functions T, P and U. Enthalpy is usually expressed as the change in enthalpy (ΔH) for a process
between initial and final states:
ΔH=ΔU+ΔPVΔ
If temperature and pressure remain constant through the process and the work is limited to pressurevolume
work, then the enthalpy change is given by the equation:
ΔH=ΔU+PΔV
Also at constant pressure the heat flow (q) for the process is equal to the change in enthalpy defined
by the equation:
ΔH=q
By looking at whether q is exothermic or endothermic we can determine a relationship between ΔH
and q. If the reaction absorbs heat it is endothermic meaning the reaction consumes heat from the
surroundings so q>0 (positive). Therefore, at constant temperature and pressure, by the equation
above, if q is positive then ΔH is also positive. And the same goes for if the reaction releases heat,
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then it is exothermic, meaning the system gives off heat to its surroundings, so q (the 4..18 came from the stage it where it was) -19340 =
2832.789 (T1-179173.7145) --> 159833.7145 = 2832.786T --> divide to find t so 56.42279879. Use the
Problem least amount 3 of significant figures, so 4 which is 56.42 degrees Celsius is your final temperature.
Calculate the enthalpy (ΔH) for the process in which 45.0 grams of water is converted from liquid at
10° C to vapor at 25° C.
To calculate entha
Solution 1
amount of heat= (mass of substance) x (specific heat capacity) x (change of temperature)
amount of heat=(1gram) (4.18 joules)/ (grams °C) (10°C)
amount of heat=41.8 joules
Solution 2
Q=mcΔT
-19340 J = (677.7g) (4.18J) (T f – 63.25°C)
-19340 J = (2832.789) (T f – 63.25°C)
-19340 J = 2832.789 T f – 179173.7145
159833.7145 = 2832.786T f
56.42279879 = T f
T f = 56.42 °C
Solution 3
Part 1: Heating water from 10.0 to 25.0 °C
ΔkJ = 45.0g H 20 x (4.184J/gH 2O °C) x (25.0 - 10.0) °C x 1kJ/1000J = 2.82 kJ
Part 2: Vaporizing water at 25.0 °C
ΔkJ = 45.0 g H 2O x 1 mol H 2O/18.02 g H 2O x 44.0 kJ/1 mol H 2O = 110 kJ
Part 3: Total Enthalpy Change
ΔH = 2.82 kJ + 110kJ
ΔH = 112.82 kJ
Page 141

The Learning Goal for this section is:
Explain how various factors, such as concentration, temperature, and percent of a catalyst affect the
rate of a chemical reaction.
You are going to answer these 15 questions first in the order they are given to you. This will be a quiz grade. You are then going to explain the
chemical concepts that are used in each question. You can do this in the order given or group them by concepts. You can use your book or information
found on the internet but all information must be written in your own word. The font needs to be Arial 12. This is due on Monday April 24 by midnight in
the drop box. This document should be a 6-page pdf.
1. B 6. B 11. A
2. C 7. B 12. D
3. C 8. D 13. C
4. D 9. A 14. A
5. A 10. B 15. D
1 This graph represents the change in energy for two laboratory trials of the same reaction.
Which factor could explain the energy difference between the trials?
A Heat was added to trial #2.
B A catalyst was added to trial #2.
C Trial #1 was stirred.
D Trial #1 was cooled.
Answer: B
- The answer is B because the trial one had to use way more energy than the trial two.
- This is because a catalyst was added to trial 2 and that is why it didn’t have to use so much
energy to get a reaction.
- Catalysts speed up reactions, meaning the substance doesn’t need to use so much energy to
carry out the reaction it is making.
- The reaction in trial didn’t need so much energy as trial one to react, and that means that there
was a catalyst added so the reaction could occur faster.
2 Consider this balanced chemical equation:
Which will increase the rate of the reaction?
A increasing pressure on the reaction
B decreasing concentration of the reactants
C adding a catalyst to the reaction
D decreasing the temperature of the reaction
2H2O2 (aq) → 2H2O(l) + O2 (g)
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- Answer: C
- The answer is C because to increase the rate of a reaction, you would have to use a catalyst
to move the rate up
- The definition of a catalyst is “A substance that increases the rate of a chemical reactions
without itself undergoing any permanent chemical change.”
- In most equations, by simply increasing the pressure or temperature, the reaction rate will be
increased.
- -Here in this equation, by increasing the pressure of 2H2O2, that will make the reaction occur
not as fast than it normally would.
3 For the reaction
A + (aq) + B — (aq) → AB (s)
increasing the temperature increases the rate of the reaction. Which is the best explanation for this
happening?
A The pressure increases, which in turn increases the production of products.
B The concentration of reactants increases with an increase in temperature.
C The average kinetic energy increases, so the likelihood of more effective collisions between ions
increases.
D Systems are more stable at high temperatures.
- The Collison Theory us when atoms, ions and molecules react in order to form products by
colliding with one another’s. It explains why some reactants are extremely slow at room
temperature.
- If having enough kinetic energy those who do not have enough energy will bounce apart,
unchanged
- Activation Energy – minimum energy that particles who collide must have to react
- -The answer would be letter D because, when the temperature is increased, that makes the
pressure increase too, and when the pressure is increased, the rate of the reaction would also
be increasing, thus explaining how the rate of reaction is increasing.
- -Both Temperature and rate of reaction have a relationship with the pressure of the equation.
- This problem is similar to question number 2, when the pressure of temperature is increased,
the rate will happen faster.
- Here when the temperature gets increased, that will let A and B come together in a quicker
way than expected because of the heat.
4 Which statement explains why the speed of some reactions is increased when the surface area of
one or all the reactants is increased?
A increasing surface area changes the electronegativity of the reactant particles
B increasing surface area changes the concentration of the reactant particles
C increasing surface area changes the conductivity of reactant particles
D increasing surface area enables more reactant particles to collide
- The answer is letter D because when the surface area is increased the concentration does get
changed because it will decrease.
- If the concentration of the reactant particles were to decrease, the substance would have a
lower electronegativity.
- When I was figuring out the problem, i thought of how the concentration of a substance gets
watered down with ice, because there is more substance in the container. this would be the
same case when you increase the surface area there will be more room for the particles to be
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moving from one side to another, and when particles are moving more, they are way more
prone to collide into one another because they are moving more.
- -The surface area does have a relation to how much the particles in the substance collide, and
the concentration of the substance.
- -When the surface area is increased, the particles have more space to be moving around in,
and therefore do have to collide into one another so much.
-
catalyst
C6H6 + Br2 → C6H5Br + HBr
5 Which of the following changes will cause an increase in the rate of the above reaction?
A increasing the concentration of Br2
B decreasing the concentration of C6H6
C increasing the concentration of HBr
D decreasing the temperature
- Answer A
- Increasing the concentration of a reactant will increase the rate of the reaction.
- Increasing the concentration of a product will not increase the rate of the reaction.
- Br2 and C6H6 are reactants, and HBr is a product.
- If the reaction is exothermic, heat energy is released as the reaction occurs. When heat
energy is released, decreasing the temperature will increase the rate of the reaction
2CO + O2 → 2CO2
6 If the above reaction takes place inside a sealed reaction chamber, then which of these procedures
will cause a decrease in the rate of reaction?
A raising the temperature of the reaction chamber
B increasing the volume inside the reaction chamber
C removing the CO2 as it is formed
D adding more CO to the reaction chamber
- Answer: B
- Increasing the volume of the chamber will make the particles more farther apart and spread
out, this will decrease the rate of the reaction
- Raising the temperature of the reaction temperature is not the correct answer because
temperature is a one of the key elements that affect the reaction rate
- The answer would be letter B because when the surface area or volume is decreased, the
reaction will occur much slower.
- - When water is stored in a container, the smaller or bigger it is makes it either take longer or
less time to become ice.
- - This is kind of what the enclosed container does to this reaction, there is less room for the
reaction to occur.
- -Also, adding or removing parts of the equation, are not going to do anything, just change what
is reacting.
7 A catalyst can speed up the rate of a given chemical reaction by
A increasing the equilibrium constant in favor of products.
B lowering the activation energy required for the reaction to occur.
C raising the temperature at which the reaction occurs.
D increasing the pressure of reactants, thus favoring products.
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- Answer: B
- -The answer would be letter B because as mentioned previously, what changes the rate of a
reaction is either a change, either rise or decrease in temperature or also an increase or
decrease in the pressure.
- - Letters C and D talk about increasing temperature or an increase in pressure, but because
this question is asking about what a catalyst does to a chemical reaction, these are not the
answers, the answer is B because that is what a catalyst does.
- -The definition of a catalyst is literally “a substance that increases the rate or a chemical
reaction without itself undergoing any permanent chemical change” and because of this letter
B would be the correct answer choice.
- Activation energy is a barrier that reactants must go through before products form
- Lowering it will allow the catalyst it take charge to begin the speeding process at a much
farther rate if the activation energy was risen
-
8 Which reaction diagram shows the effect of using the appropriate catalyst in a chemical reaction?
A
C
B
D
- Answer D
- The answer is Graph D because the catalyst added to one of the lines is lower, and this is
because the catalyst lowered the rate necessary for the reaction to happen.
- The other graphs have some other types of changes but none of them show what a catalyst
would do to the slope of a graph.
- Letters A and C don’t make sense because they fluctuate more than once as if the energy
lowered then peaked, then lowered again, yet raised back up then down, this would only
happen if there were two reactions happening.
9 H2O2, hydrogen peroxide, naturally breaks down into H2O and O2 over time. MnO2, manganese
dioxide, can be used to lower the energy of activation needed for this reaction to take place and, thus,
increase the rate of reaction. What type of substance is MnO2?
A a catalyst
B an enhancer
C an inhibitor
D a reactant
- Answer: A
- The answer would be letter a because MnO2 is speeding up the reaction, and a catalyst is
what does that.
Page 145

- An enhancer is “a DNA sequence that increases the level of transcription of a gene that is
located nearby on the same chromosome” and that is not what MnO2 is doing in this equation
that we are taking into account.
- It could be a reactant, but it makes more sense that is would be a catalyst because a reactant
simply reacts, and it doesn’t do anything else to the equation, as in making its reaction rate go
faster or slower.
10 When a reaction is at equilibrium and more reactant is added, which of the following changes is
the immediate result?
A The reverse reaction rate remains the same.
B The forward reaction rate increases.
C The reverse reaction rate decreases.
D The forward reaction rate remains the same.
- Answer B
- The answer would be letter B because by adding more reactant, the forward reaction would be
the one to decrease.
- It would not have anything to do with a reverse reaction because there is nothing being taken
out of the reaction.
- A reverse reaction is when “reactants form products that, in turn, react together to give the
reactants back”, and that is clearly not what is going on in this problem.
11 In which of the following reactions involving gases would the forward reaction be favored by an
increase in pressure?
A A + B ⇄AB
B A + B ⇄ C + D
C 2A + B ⇄ C + 2D
D AC ⇄ A + C
- Answer A
- The answer is letter A because an increase in pressure would mean that the equation is just
the simplest form for reacting.
- Letter C does not even have an equivalent equation.
- Letter D would actually be a reverse reaction because the product is given before the reactants
are given.
12 Which action will drive the reaction to the right?
A heating the equilibrium mixture
B adding water to the system
C decreasing the oxygen concentration
D increasing the system’s pressure
4HCl(g) + O2(g) ⇄ 2H2O(l) + 2Cl2(g) + 113 kJ
- Answer D
- The answer is letter D because by increasing the system’s pressure, there is a change being
made to the equation, which would move it from one side of the equation to the other, such as
the right side which is where this is wanting to be moved to.
- By heating the mixture, you would be increasing the temperature but that would not mean that
the reactants would form the product they do.
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- Decreasing the oxygen, as letter C says does not do anything other than simply decrease the
oxygen in the equation.
- Another thing that wouldn’t change a thing would be adding water to the system’s pressure,
that would just add more factors to the pressure but not necessarily mean it would make the
reactants produce the product.
NO2(g) + CO(g) ⇄ NO(g) + CO2(g)
13 The reaction shown above occurs inside a closed flask. What action will shift the reaction to the
left?
A pumping CO gas into the closed flask
B raising the total pressure inside the flask
C increasing the NO concentration in the flask
D venting some CO2 gas from the flask
- Answer C
- The answer is letter C because by increasing the concentration in the flask, that would make
the the reaction happen on the left side of the equation.
- It would not be letter B because raising the pressure would make the reaction shift to the right,
not to the left where we want it to be.
- it would also not be letter D because venting CO2 because that would not shift the reaction like
we want it to shift.
- It would also not make sense to pump carbon dioxide gas into the closed flask, because that
would do something else, but not shift the reaction from the right to the left as the question is
asking to happen.
NH4Cl(s) + heat ⇄ NH3(g) + HCl(g)
14 What kind of change will shift the reaction above to the right to form more products?
A a decrease in total pressure
B an increase in the concentration of HCl
C an increase in the pressure of NH3
D a decrease in temperature
Answer: A
- The answer to this question is letter A because the pressure and the temperature are the two
things to have the biggest effect on an equation like this.
- By increasing the pressure or concentration, the reaction would be moved to the left, so if it
wants to be moved to the right you would have to decrease the pressure, as letter A says.
15 In a sealed bottle that is half full of water, equilibrium will be attained when water molecules
A cease to evaporate.
B begin to condense.
C are equal in number for both the liquid and the gas phase.
D evaporate and condense at equal rates.
- Answer: D
- The answer would be letter D because when water molecules are evaporating but they’re are
being condensed at the same rate, it’s as if nothing is happening to the water molecules.
Page 147

Unit 8
Chapter 19 Acid and Bases
The student will learn what are the different ways chemists
define aids and bases, what the pH of a solution means and
how chemist use acid-base reactions.
Relate acidity and basicity to hydronium and hydroxyl ion concentration and
pH.
Students will be able to use a pH scale to identify substances as acids or bases.
Students will be able to use various equipment (probeware, universal pH, etc.) to
identify the pH of substances.
Students will be able to calculate H3O+ and OH- concentration of various
substances.
pH scale
Hydronium ion
Arrhenius acid/base
Lewis acid/base
Bronsted-Lowry acid/base
Strong acid/base
Weak acid/base
Neutralization reaction
Titration
Chapter 20 Oxidation-Reduction Reactions
The student will learn what happens during oxidation and
reduction and how to balance redox equations.
Describe oxidation-reduction reactions in living and non-living systems.
Students will be able to compare and contrast redox reactions.
Students will be able to assign oxidation numbers to redox reactions.
Students will be able to write half reactions
Oxidation
148

Reduction
Oxidation reduction reaction
Oxidation number
Half reaction
Electrochemical process
Battery
Cathode
Anode
Electrolysis
149

The Learning Goal for this section is:
The students will learn what are the different ways chemists define acids and bases, what the pH of a solution
means and how chemist use acid-base reactions
Acids and Bases
The Observable Properties of Acids and Bases
The words acid and alkaline (an older word for base) are derived from direct sensory experience.
Acid Property #1:
The word acid comes from the Latin word acere, which means "sour." All acids taste sour. Well The student will lea
known from ancient times were vinegar, sour milk and lemon juice. Aspirin (scientific name:
acetylsalicylic acid) tastes sour if you don't swallow it fast enough. Other languages derive their word
for acid from the meaning of sour. So, in France, we have acide. In Germany, we have säure from
saure and in Russia, kislota from kisly.
Base Property #1:
The word "base" has a more complex history (see below) and its name is not related to taste. All
bases taste bitter. For example, mustard is a base. It tastes bitter. Many medicines, because they are
bases, taste bitter. This is the reason cough syrups are advertised as having a "great grape taste."
The taste is added in order to cover the bitterness of the active ingredient in cough syrup.
Acid Property #2:
Acids make a blue vegetable dye called litmus turn red.
Base Property #2:
Bases are substances which will restore the original blue color of litmus after having been reddened
by an acid.
Acid Property #3:
Acids destroy the chemical properties of bases.
Base Property #3:
Bases destroy the chemical properties of acids.
Neutralization is the name for this type of reaction.
Acid Property #4:
Acids conduct an electric current.
Base Property #4:
Bases conduct an electric current.
This is a common property shared with salts. Acids, bases and salts are grouped together into a
category called electrolytes, meaning that a water solution of the given substance will conduct an
electric current.
Non-electrolyte solutions cannot conduct a current. The most common example of this is sugar
dissolved in water.
150

So far, the properties have an obvious relationship: taste, color change, mutual destruction, and
response to electric current. This last property is related, but in a less obvious way. The property
below identifies a unique chemical reaction that acids and bases engage in.
Acid Property #5:
Upon chemically reacting with an active metal, acids will evolve hydrogen gas (H2). The key word, of
course, is active. Some metals, like gold, silver or platinum, are rather unreactive and it takes rather
extreme conditions to get these "unreactive" metals to react. Not so with the metals in this property.
They include the alkali metals (Group I, Li to Rb), the alkaline earth metals (Group II, Be to Ra), as
well as zinc and aluminum. Just bring the acid and the metal together at anything close to room
temperature and you get a reaction. Here's a sample reaction:
Zn + 2 HCl(aq) ---> ZnCl2 + H2
Another common acid reaction some sources mention is that acids react with carbonates (and
bicarbonates) to give carbon dioxide gas:
HCl + NaCO3 ---> CO2 + H2O + NaCl
Base Property #5:
Bases feel slippery, sometimes people say soapy. This is because they dissolve the fatty acids and
oils from your skin and this cuts down on the friction between your fingers as you rub them together.
In essence, the base is making soap out of you. Yes, bases are involved in the production of soap! In
the early years of soap making, the soaps were very harsh on the skin and clothes due to the high
base content. Even today, people with very sensitive skin must sometimes use a non-soap-based
product for bathing.
It was not until more modern times that the chemical nature (as opposed to observable properties) of
acids and bases began to be explored. That leads to this property that is not directly observable by
the senses.
Acid Property #6:
Acids produce hydrogen ion (H + ) in solution. A more correct formula for what is produced is that of the
hydronium ion, H3O + . Both formulas are used interchangeably.
Acid base theories: Svante Arrhenius
I. Introduction
The basic idea is that certain substances remain ionized in solution all the time. Today, everyone
accepts this without question, but it was the subject of much dissention and disagreement in 1884,
when a twenty-five-year-old Arrhenius presented and defended his dissertation.
II. The Acid Base Theory
Acid - any substance which delivers hydrogen ion (H + ) to the solution.
Base - any substance which delivers hydroxide ion (OH¯) to the solution.
Here is a generic acid dissociating, according to Arrhenius:
HA ---> H + + A¯
151

This would be a generic base:
XOH ---> X + + OH¯
When acids and bases react according to this theory, they neutralize each other, forming water and a
salt:
HA + XOH ---> H2O + XA
Keeping in mind that the acid, the base and the salt all ionize, we can write this:
Finally, we can drop all spectator ions, to get this:
H + + A¯ + X + + OH¯ ---> H2O + X + + A¯
H + + OH¯ ---> H2O
These ideas covered all of the known acids at the time (the usual suspects like hydrochloric acid,
acetic acid, and so on) and most of the bases (sodium hydroxide, potassium hydroxide, calcium
hydroxide and so on). HOWEVER, and it is a big however, the theory did not explain why ammonia
(NH3) was a base. There are other problems with the theory also.
III. Problems with Arrhenius' Theory
1. The solvent has no role to play in Arrhenius' theory. An acid is expected to be an acid in any
solvent. This was found to not be the case. For example, HCl is an acid in water, behaving in
the manner Arrhenius expected. However, if HCl is dissolved in benzene, there is no
dissociation, the HCl remaining as un-dissociated molecules. The nature of the solvent plays a
critical role in acid-base properties of substances.
2. All salts in Arrhenius' theory should produce solutions that are neither acidic or basic. This is
not the case. If equal amounts of HCl and ammonia react, the solution is slightly acidic. If equal
amounts of acetic acid and sodium hydroxide are reacted, the resulting solution is basic.
Arrhenius had no explanation for this.
3. The need for hydroxide as the base led Arrhenius to propose the formula NH4OH as the
formula for ammonia in water. This led to the misconception that NH4OH is the actual base,
not NH3.
In fact, by 1896, several years before Arrhenius announced his theory, it had been recognized that
characteristic base properties where just as evident in such solvents as aniline, where no hydroxide
ions were possible.
4. H + , a bare proton, does not exist for very long in water. The proton affinity of H2O is about 799
kJ/mol. Consequently, this reaction:
H2O + H + ---> H3O +
happens to a very great degree. The "concentration" of free protons in water has been estimated to
be 10¯130 M. A rather preposterous value, indeed.
The Arrhenius theory of acids and bases will be fully supplanted by the theory proposed
independently by Johannes and Thomas Lowry in 1923.
152

The acid base theory of Brønsted and Lowry
I. Introduction
In 1923, within several months of each other, Johannes Nicolaus Brønsted (Denmark) and Thomas
Martin Lowry (England) published essentially the same theory about how acids and bases behave.
Since they came to their conclusions independently of each other, both names have been used for
the theory name.
II. The Acid Base Theory
Using the words of Brønsted:
". . . acids and bases are substances that are capable of splitting off or taking up hydrogen ions,
respectively."
Or an acid-base reaction consists of the transfer of a proton from an acid to a base. KEEP THIS
THOUGHT IN MIND!!
Here is a more recent way to say the same thing:
An acid is a substance from which a proton can be removed.
A base is a substance that can remove a proton from an acid.
Remember: proton, hydrogen ion and H + all mean the same thing
Very common in the chemistry world is this definition set:
An acid is a "proton donor."
A base is a "proton acceptor."
In an acid, the hydrogen ion is bonded to the rest of the molecule. It takes energy (sometimes a little,
sometimes a lot) to break that bond. So the acid molecule does not "give" or "donate" the proton, it
has it taken away. In the same sense, you do not donate your wallet to the pickpocket, you have it
removed from you.
The base is a molecule with a built-in "drive" to collect protons. As soon as the base approaches the
acid, it will (if it is strong enough) rip the proton off the acid molecule and add it to itself.
Now this is where all the fun stuff comes in that you get to learn. You see, some bases are stronger
than others, meaning some have a large "desire" for protons, while other bases have a weaker drive.
It's the same way with acids, some have very weak bonds and the proton is easy to pick off, while
other acids have stronger bonds, making it harder to "get the proton."
One important contribution coming from Lowry has to do with the state of the hydrogen ion in solution.
In Brønsted's announcement of the theory, he used H + . Lowry, in his paper (actually a long letter to
the editor) used the H3O + that is commonly used today.
III. Sample Equations written in the Brønsted-Lowry Style
A. Reactions that proceed to a large extent:
153

HCl + H2O ⇌ H3O + + Cl¯
HCl - this is an acid, because it has a proton available to be transferred.
H2O - this is a base, since it gets the proton that the acid lost.
Now, here comes an interesting idea:
H3O + - this is an acid, because it can give a proton.
Cl¯ - this is a base, since it has the capacity to receive a proton.
Notice that each pair (HCl and Cl¯ as well as H2O and H3O + differ by one proton (symbol = H + ). These
pairs are called conjugate pairs.
HNO3 + H2O ⇌ H3O + + NO3¯
The acids are HNO3 and H3O + and the bases are H2O and NO3¯.
Remember that an acid-base reaction is a competition between two bases (think about it!) for a
proton. If the stronger of the two acids and the stronger of the two bases are reactants (appear on the
left side of the equation), the reaction is said to proceed to a large extent.
Here are some more conjugate acid-base pairs to look for:
H2O and OH¯
HCO3¯ and CO3 2¯
H2PO4¯ and HPO4 2¯
HSO4¯ and SO4 2¯
NH4 + and NH3
CH3NH3 + and CH3NH2
HC2H3O2 and C2H3O2¯
B. Reactions that proceed to a small extent:
If the weaker of the two acids and the weaker of the two bases are reactants (appear on the left side
of the equation), the reaction is said to proceed to only a small extent:
HC2H3O2 + H2O ⇌ H3O + + C2H3O2¯
NH3 + H2O ⇌ NH4 + + OH¯
Identify the conjugate acid base pairs in each reaction.
HC 2H 3O 2 and C 2H 3O 2¯
is one conjugate pair.
H 2O and H 3O + is the other.
NH 3 and NH 4
+
is one pair.
H 2O and OH¯ is the other.
Notice that H 2O in the first equation is acting as a base and in the second equation is acting as an acid.
154

IV. Problems with the Theory
This theory works very nicely in all protic solvents (water, ammonia, acetic acid, etc.), but fails to
explain acid base behavior in aprotic solvents such as benzene and dioxane. That job will be left for a
more general theory, such as the Lewis Theory of Acids and Bases.
The Lewis theory of acids and bases
I. Introduction
Lewis gives his definition of an acid and a base:
"We are inclined to think of substances as possessing acid or basic properties, without having a
particular solvent in mind. It seems to me that with complete generality we may say that a basic
substance is one which has a lone pair of electrons which may be used to complete the stable group
of another atom, and that an acid is one which can employ a lone pair from another molecule in
completing the stable group of one of its own atoms."
"In other words, the basic substance furnishes a pair of electrons for a chemical bond, the acid
substance accepts such a pair."
It is important to make two points here:
1. NO hydrogen ion need be involved.
2. NO solvent need be involved.
The Lewis theory of acids and bases is more general than the "one sided" nature of the Bronsted-
Lowry theory. Keep in mind that Bronsted-Lowry, which defines an acid as a proton donor and a base
as a proton acceptor, REQUIRES the presence of a solvent, specifically a protic solvent, of which
water is the usual example. Since almost all chemistry is done in water, the fact that this limits the
Bronsted-Lowry definition is of little practical consequence.
The Lewis definitions of acid and base do not have the constraints that the Bronsted-Lowry theory
does and, as we shall see, many more reactions were seen to be acid base in nature using the Lewis
definition than when using the Bronsted-Lowry definitions.
II. The Acid Base Theory
The modern way to define a Lewis acid and base is a bit more concise than above:
Acid: an electron acceptor.
Base: an electron donor.
A "Lewis acid" is any atom, ion, or molecule which can accept electrons and a "Lewis base" is any
atom, ion, or molecule capable of donating electrons. However, a warning: many textbooks will say
"electron pair" where I have only written "electron." The truth is that it sometimes is an electron pair
and sometimes it is not.
It turns out that it may be more accurate to say that "Lewis acids" are substances which are electrondeficient
(or low electron density) and "Lewis bases" are substances which are electron-rich (or high
electron density).
155

Several categories of substances can be considered Lewis acids:
1. positive ions
2. having less than a full octet in the valence shell
3. polar double bonds (one end)
4. expandable valence shells
Several categories of substances can be considered Lewis bases:
1. negative ions
2. one of more unshared pairs in the valence shell
3. polar double bonds (the other end)
4. the presence of a double bond
Sören Sörenson and the pH scale
I. Short Historical Introduction
In the late 1880's, Svante Arrhenius proposed that acids were substances that delivered hydrogen ion
to the solution. He has also pointed out that the law of mass action could be applied to ionic
reactions, such as an acid dissociating into hydrogen ion and a negatively charged anion.
This idea was followed up by Wilhelm Ostwald, who calculated the dissociation constants (the
modern symbol is Ka) of many weak acids. Ostwald also showed that the size of the constant is a
measure of an acid's strength.
By 1894, the dissociation constant of water (today called Kw) was measured to the modern value of
1 x 10¯14 .
In 1904, H. Friedenthal recommended that the hydrogen ion concentration be used to characterize
solutions. He also pointed out that alkaline (modern word = basic) solutions could also be
characterized this way since the hydroxyl concentration was always 1 x 10¯14 ÷ the hydrogen ion
concentration. Many consider this to be the real introduction of the pH scale.
III. The Introduction of pH
Sörenson defined pH as the negative logarithm of the hydrogen ion concentration.
pH = - log [H + ]
Remember that sometimes H3O + is written, so
pH = - log [H3O + ]
means the same thing.
So let's try a simple problem: The [H + ] in a solution is measured to be 0.010 M. What is the pH?
The solution is pretty straightforward. Plug the [H + ] into the pH definition:
pH = - log 0.010
An alternate way to write this is:
pH = - log 10¯2
Since the log of 10¯2 is -2, we have:
pH = - (- 2)
Which, of course, is 2.
156

Let's discuss significant figures and pH.
Another sample problem: Calculate the pH of a solution in which the [H3O + ] is 1.20 x 10¯3 M.
For the solution, we have:
pH = - log 1.20 x 10¯3
This problem can be done very easily using your calculator. However, be warned about putting
numbers into the calculator.
So you enter (-), log, 1.20, X10 n , (-), 3, enter.
The answer, to the proper number of significant digits is: 2.921.
III. Significant Figures in pH
Here is the example problem: Calculate the pH of a solution where the [H + ] is 0.00100 M. (This could
also be a pOH problem. The point being made is the same.)
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
is 3.
You would be graded wrong!! Why? Because the pH is not written to reflect the number of significant
figures in the concentration.
Notice that there are three sig figs in 0.00100. (Hopefully you remember significant figures, since you
probably studied them months ago before getting to acid base stuff. THEY ARE STILL IMPORTANT!)
So, our pH value should also reflect three significant figures.
However, there is a special rule to remember with pH (and pOH) values. The whole number portion
DOES NOT COUNT when figuring out how many digits to write down.
Let's phrase that another way: in a pH (and a pOH), the only place where significant figures are
contained is in the decimal portion.
So, the correct answer to the above problem is 3.000. Three sig figs and they are all in the decimal
portion, NOT (I repeat NOT) in the whole number portion.
Practice Problems
Convert each hydrogen ion concentration into a pH. Identify each as an acidic pH or a basic pH.
1. 0.0015 2.82
2. 5.0 x 10¯9
8.30
3. 1.0 0.00
4. 3.27 x 10¯4
3.845
5. 1.00 x 10¯12
12.00
6. 0.00010
4.00
acid
base
acid
base
acid
157

1. 2.82
2. 8.30
3. 0.00
4. 3.485
5. 12.000
6. 4.00
Sörenson also just mentions the reverse direction. That is, suppose you know the pH and you want to
get to the hydrogen ion concentration ([H + ])?
Here is the equation for that:
[H + ] = 10¯pH
That's right, ten to the minus pH gets you back to the [H + ] (called the hydrogen ion concentration).
This is actually pretty easy to do with the calculator. Here's the sample problem: calculate the [H + ]
from a pH of 2.45.
This problem can be done very easily using your calculator. However, be warned about putting
numbers into the calculator.
So you enter 2nd, 10 x , (-), 2.45, enter.
The answer, to the proper number of significant digits is: .00355.
The pH of an acidic pond is 5. What is the hydrogen ion concentration (moles per liter)?
The answer is:
pH = -log (hydrogen ion concentration)
The answer was .00001. Thus, 5 = -log (.00001).
We'll take the formula that you started with (pH = -log([H+])) and work to the answer (solve for [H+]).
pH = - log ([H+]) Given.
pH = log ([H+] (-1) ) Since logarithms are like exponents, when you multiply a log by
something, you can just move it to the inside of log as an exponent.
10 pH = 10 log ([H+] (-1)) Take each side to tenth power.
10 pH = [H+] (-1) Since "log" is just another notation for "log base 10", when you
raise a log to the tenth power, the log cancels out.
[H+] = 10 (-pH)
Take the reciprocal of both sides.
That is the general form. To answer the specific question,
5 = - log ([H+])
5 = log ([H+] (-1) )
10 5 = [H+] (-1)
10 (-5) = [H+]
[H+]
= .00001 mol/L
158

On your calculator you would input 10, ^, (-), 5 and you would get 0.00001.
This is also the way to find the amount of OH + that are present in a base.
To find the pH: -log(concentration)
To find the concentration: 10 -pH
Define these terms:
pH scale
IIt stands for "potential of hydrogen" it measure the acidity or alkalinity of water soluble substances.
pH value goes from 1 to 14, below 7 indicates acidity, 1 most acidic.
Hydronium ion
The ion H30+, consisting of protonated water molecule and present in all aqueous acids.
Arrhenius acid/base
it is a substance that dissociates in water to form hydrogen ions H+. An Arrhenius base is a substance
that dissociates in water to form hydroxide OH-.
Lewis acid/base
A compound or ionic species that can accept an electron pair from a donor compound. A compound or ionic
species that can donate an electron pair to an acceptor compound.
Bronsted-Lowry acid/base
An acid is a proton (hydrogen ion) donor. A base is a proton (hydrogen ion) acceptor.
Strong acid/base
strong acid is one that completely ionizes (dissociates) in a solution (provided there is suffiecient sulvent).
a strong base breaks apart 100% in a solution.
Weak acid/base
a weak base does not ionize fully in an aqueous solution. An acid that dissociates incompletely, releasing only
some of its hydrogen atoms into the solution.
Neutralization reaction
it is a chemical reaction in which an acid and a base react quantitatively with each other.
Titration
A technique where a solution of konwn concentration is used to determine the concentration of an unknown
solution.
159

The Learning Goal for this assignment is:
The student will learn what happens during oxidation and reduction and how to balance redox equations.
1. Oxidation Numbers
Redox Reactions
Oxidation and reduction
Every atom, ion or polyatomic ion has a formal oxidation number associated with it. This value
compares the number of protons in an atom (positive charge) and the number of electrons assigned
to that atom (negative charge).
In many cases, the oxidation number reflects the actual charge on the atom, but there are many
cases where it does not. Think of oxidation numbers as a bookkeeping exercise simply to keep track
of where electrons go.
2. Reduction
Reduction means what it says: the oxidation number is reduced in reduction.
This is accomplished by adding electrons. The electrons, being negative, reduce the overall oxidation
number of the atom receiving the electrons.
3. Oxidation
Oxidation is the reverse process: the oxidation number of an atom is increased during oxidation.
This is done by removing electrons. The electrons, being negative, make the atom that lost them
more positive.
I use this mnemonic to help me remember which is which: LEO the lion says GER.
LEO = Loss of Electrons is Oxidation
GER = Gain of Electrons is Reduction
Another well-known mnemonic is this: OIL RIG
OIL = Oxidation Is Loss (of Electrons)
RIG = Reduction Is Gain (of Electrons)
Another way is to simply remember that reduction is to reduce the oxidation number. Therefore,
oxidation must increase the value.
4. Reduction-Oxidation Reactions
There are many chemical reactions in which one substance gets reduced in oxidation number
(reduction) while another participating substance gets increased in oxidation number (oxidation).
Such a reaction is called called a REDOX reaction. The RED, of course, comes from REDuction and
OX from OXidation. However, it is pronounced re-dox and not red-ox.
Here is a simple example of a redox reaction:
160

Ag+ + Cu ---> Ag + Cu2+
I have deliberately not balanced it and I have also written it in net ionic form. I have found that kids
studying redox get confused by net ionic form and how to change a full equation into net ionic form.
Redox equations need to be balanced but, except for the simplest ones, it cannot be done by
inspection (also called trial and error). I take that back, complex ones can be done by trial and error. It
typically takes quite a bit of work, especially when compared to how long it takes when the proper
technique is used.
There is a technique used to balance redox reactions. It is called "balancing by half-reactions." The
basic plan will be to split the full equation into two simpler parts (called half-reactions), balance them
following several standard steps, then recombine the balanced half-reactions into the final answer.
This is another technique called the "ion-electron method." I plan to ignore it.
Notes:
OIL - OXIDATION IS LOSS (OF ELECTRONS) oxidation number is reduced in reduction.
RIG - REDUCTION IS GAIN (OF ELECTRONS) oxidation number is increased during oxidation.
5. Some Definitions
Oxidizing Agent - that substance which oxidizes somebody else. It is reduced in the process.
Reducing Agent - that substance which reduces somebody else. It is oxidized in the process.
It helps me to remember these definitions by the opposite nature of what happens. By that, I mean
the oxidizing agent gets reduced and the reducing agent gets oxidized.
6. Rules for Assigning Oxidation Numbers
The Oxidation Number of an element corresponds to the number of electrons, e - , that an atom loses,
gains, or appears to use when joining with other atoms in compounds. In determining the Oxidation
Number of an atom, there are seven guidelines to follow:
1. The Oxidation Number of an individual atom is 0. This includes diatomic elements such as
O2 or others like P4 and S8
2. The total Oxidation Number of all atoms in: a neutral species is 0 and in an ion is equal to the
ion charge.
3. Group 1 metals have an Oxidation Number of +1 and Group 2 an Oxidation Number of +2
4. The Oxidation Number of fluorine is -1 in compounds
5. Hydrogen generally has an Oxidation Number of +1 in compounds, except hydrides
6. Oxygen generally has an Oxidation Number of -2 in compounds, except peroxides
7. In binary metal compounds, Group 17 elements have an Oxidation Number of -1, Group 16 of -
2, and Group 15 of -3.
Note: The sum of the Oxidation Number s is equal to zero for neutral compounds and equal to the
charge for polyatomic ion species.
Now, some examples:
1. What is the oxidation number of Cl in HCl?
Since H = +1, the Cl must be -1 (minus one).
161

2. What is the oxidation number of Na in Na2O?
Since O = -2, the two Na must each be +1.
3. What is the oxidation number of Cl in ClO¯?
The O is -2, but since a -1 must be left over, then the Cl is +1.
4. What is the oxidation number for each element in KMnO4?
K = +1 because KCl exists. We know the Cl = -1 because HCl exists.
O = -2 by definition
Mn = +7. There are 4 oxygens for a total of -8, K is +1, so Mn must be the rest.
5. What is the oxidation number of S in SO4 2¯
O = -2. There are four oxygens for -8 total. Since -2 must be left over, the S must = +6.
Please note that, if there is no charge indicated on a formula, the total charge is taken to be zero.
Practice Problems
Find oxidation numbers
1. N in NO3¯
2. C in CO3 2¯
3. Cr in CrO4 2¯
N is 5+ because the charge is -1 so -6+5=-1
C is 4+ because the charge is -2 so -6+4= -2
Cr is +6 because the charge is -2 so -8+6=-2
4. Cr in Cr2O7 2¯
5. Fe in Fe2O3
6. Pb in PbOH +
7. V in VO2 +
8. V in VO 2+
9. Mn in MnO4¯
10. Mn in MnO4 2¯
Cr is -8 because the charge is -2 so -16-14=-2
Fe is 3+ because the charge is 0 so 6-6=0
2+
5+
4+
7+
6+
Notes:
Depending on the charge + or - number of the problem, it can vary what the oxidation number of an element can be.
Oxygen is normally -2 it all depends on the ion charge. The oxidation number of something like Magnesium can change
completely to something its not in charge wise so that it can complete the problem.
162

7. Half Reactions
A half-reaction is simply one which shows either reduction OR oxidation, but not both. Here is the
example redox reaction used in a different file:
Ag + + Cu → Ag + Cu 2+
It has BOTH a reduction and an oxidation in it. That is why we call it a redox reaction, from REDuction
and OXidation.
What you must be able to do is look at a redox reaction and separate out the two half-reactions in it.
To do that, identify the atoms which get reduced and get oxidized. Here are the two half-reactions
from the above example:
Ag+ → Ag
Cu → Cu 2+
The silver is being reduced, its oxidation number going from +1 to zero. The copper's oxidation
number went from zero to +2, so it was oxidized in the reaction. In order to figure out the halfreactions,
you MUST be able to calculate the oxidation number of an atom.
Keep in mind that a half-reaction shows only one of the two behaviors we are studying. A single halfreaction
will show ONLY reduction or ONLY oxidation, never both in the same equation.
Also, notice that the reaction is read from left to right to determine if it is reduction or oxidation. If you
read the reaction in the opposite direction (from right to left) it then becomes the other of our two
choices (reduction or oxidation). For example, the silver half-reaction above is a reduction, but in the
reverse direction it is an oxidation, going from zero on the right to +1 on the left.
There will be times when you want to switch a half-reaction from one of the two types to the other. In
that case, rewrite the entire equation and swap sides for everything involved. If I needed the silver
half-reaction to be oxidation, I'd write Ag → Ag+ rather than just doing it mentally.
The next step is that both half-reactions must be balanced. However, there is a twist. When you
learned about balancing equation, you made equal the number of atoms of each element on each
side of the arrow. That still applies, but there is one more thing: the total amount of charge on each
side of the half-reaction MUST be the same.
When you look at the two half-reactions above, you will see they are already balanced for atoms with
one Ag on each side and one Cu on each side. So, all we need to do is balance the charge. To do
this you add electrons to the more positive side. You add enough to make the total charge on each
side become EQUAL.
To the silver half-reaction, we add one electron:
To the copper half-reaction, we add two electrons:
Ag+ + e¯ ---> Ag
Cu ---> Cu 2+ + 2e¯
163

One point of concern: notice that each half-reaction wound up with a total charge of zero on each
side. This is not always the case. You need to strive to get the total charge on each side EQUAL, not
zero.
One more point to make before wrapping this up. A half-reaction is a "fake" chemical reaction. It's just
a bookkeeping exercise. Half-reactions NEVER occur alone. If a reduction half-reaction is actually
happening (say in a beaker in front of you), then an oxidation reaction is also occurring. The two halfreactions
can be in separate containers, but they do have to have some type of "chemical
connection" between them.
Half-Reaction Practice Problems
Balance each half-reaction for atoms and charge:
4) I3¯ → I¯
+2e-
1) Cl2 → 2Cl¯
2) Sn → Sn 2+ 2e-
3) Fe 2+ → Fe 3+ e-
+2e-
+2cl
5) ICl2¯ → I¯ (I'm being mean on this one. Hint: the iodine is the only thing reduced or oxidized.)
Separate each of these redox reactions into their two half-reactions (but do not balance):
6) Sn + NO3¯ →SnO2 + NO2
+NO3
7) HClO + Co →Cl2 + Co 2+
---->Cl^2 Co^2+
8) NO2 →NO3¯ + NO 2
--->NO^3- ---->No2
Here are the two half-reactions to be combined:
Here is the rule to follow:
Ag+ + e¯ → Ag
Cu → Cu 2+ + 2e¯
the total electrons MUST cancel when the two half-reactions are added.
Another way to say it:
the number of electrons in each half-reaction MUST be equal when the two half-reactions are
added.
What that means is that one (or both) equations must be multiplied through by a factor. The value of
the factor is selected so as to make the number of electrons equal.
In our example problem, the top reaction (the one with silver) must be multiplied by two, like this:
2Ag+ + 2e¯ → 2Ag
164

Notice that each separate substance is increased by the factor amount. Occasionally, a student will
multiply ONLY the electrons by the factor. That is incorrect.
When the two half-reactions are added, we get:
2Ag+ + 2e¯ + Cu → 2Ag + Cu 2+ + 2e¯
With two electrons on each side, they may be canceled, resulting in:
2Ag+ + Cu → 2Ag + Cu 2+
This is the correct answer. Notice that there are two silvers on each side and one copper. Notice also
that the total charge on each side is +2. It is balanced for both atoms and charge. Sometimes, I am
asked if the order matters, if the Cu could be first on the left-hand side. The answer is that the order
does not matter. There happen to be certain styles about where particular substances are put in the
final answer, but these are only styles. They do not affect the chemical correctness of the answer.
Notes:
Each separate substances increased by a factor amount. Such as the problem above it is factored by 2 times.
Practice Problems
Separate into half-reactions, balance them and then recombine.
1) Sn + Cl2 ---> Sn 2+ + Cl¯
2) Fe 2+ + I3¯ ---> Fe 3+ + I¯
1. N in NO 3¯
The O is -2 and three of them makes -6. Since -1 is left over, the N must be +5
2. C in CO 3
2¯
The O is -2 and three of them makes -6. Since -2 is left over, the C must be +4
3. Cr in CrO 4
2¯
The O is -2 and four of them makes -8. Since -2 is left over, the Cr must be +6
4. Cr in Cr 2 O 7
2¯
The O is -2 and seven of them makes -14. Since -2 is left over, the two Cr must be +12
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.
5. Fe in Fe 2 O 3
The O is -2 and three of them makes -6. Each Fe must then be +3
6. +2 9. +7
7. +5 10. +6
8. +4
1) Cl 2 + 2e¯ →2Cl¯ 5) ICl 2¯
+ 2e¯ →I¯ + 2Cl¯
2) Sn →Sn 2+ + 2e¯ 6) Sn →SnO 2 and NO 3¯
→NO 2
3) Fe 2+ →Fe 3+ + e¯ 7) HClO →Cl 2 and Co ---> Co 2+
4) I 3¯ → 3I¯ + 2e¯ 8) NO 2 →NO 3¯
and NO 2 →NO
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
to +5 (oxidation) and in the second, the N goes from +4 to +2 (reduction).
By the way, we could flip the reaction so that NO 3¯
and NO are reacting together to produce only one product, the NO 2 . In that case, the two halfreactions
would be reversed.
165

Reference Tables for Physical Setting/CHEMISTRY
Table A
Standard Temperature and Pressure
Name Value Unit
Standard Pressure 101.3 kPa kilopascal
1 atm atmosphere
Standard Temperature 273 K kelvin
0°C degree Celsius
Table D
Selected Units
Symbol Name Quantity
m meter length
g gram mass
Pa pascal pressure
K kelvin temperature
Table B
Physical Constants for Water
mol
J
mole
joule
amount of
substance
energy, work,
quantity of heat
Heat of Fusion
Heat of Vaporization
Specific Heat Capacity of H 2
O()
Specific Heat Capacity of H 2
O(s)
Specific Heat Capacity of H 2
O(g)
Table C
Selected Prefixes
Factor Prefix Symbol
10 3 kilo- k
334 J/g
2260 J/g
4.18 J/g•K
2.10 J/g•K
2.01 J/g•K
s second time
min minute time
h hour time
d day time
y year time
L liter volume
ppm parts per million concentration
M
molarity
solution
concentration
u atomic mass unit atomic mass
10 –1 deci- d
10 –2 centi- c
10 –3 milli- m
10 –6 micro- μ
10 –9 nano- n
10 –12 pico- p
R1

Table E
Selected Polyatomic Ions
Formula Name Formula Name
H 3
O +
hydronium
CrO 4
2–
chromate
Hg 2
2+
mercury(I)
Cr 2
O 7
2–
dichromate
NH 4
+
C 2
H 3
O
–
2 –}
CH 3
COO
CN –
CO 3
2–
HCO
–
3
C 2
O
2–
4
ClO –
ammonium
acetate
cyanide
carbonate
hydrogen
carbonate
oxalate
hypochlorite
MnO 4
–
NO
–
2
NO
–
3
O
2–
2
OH –
PO 4
3–
SCN –
SO 3
2–
permanganate
nitrite
nitrate
peroxide
hydroxide
phosphate
thiocyanate
sulfite
ClO 2
–
chlorite
SO 4
2–
sulfate
ClO 3
–
chlorate
HSO 4
–
hydrogen sulfate
ClO 4
–
perchlorate
S 2
O 3
2–
thiosulfate
Table F
Solubility Guidelines for Aqueous Solutions
Ions That Form
Soluble Compounds
Group 1 ions
(Li + , Na + , etc.)
ammonium (NH 4 + )
nitrate (NO 3 – )
acetate (C 2
H 3
O 2 – or
CH 3
COO – )
hydrogen carbonate
(HCO 3 – )
chlorate (ClO 3 – )
halides (Cl – , Br – , I – )
Exceptions
when combined with
Ag + , Pb 2+ , or Hg 2
2+
sulfates (SO 4 2– ) when combined with Ag + ,
Ca 2+ , Sr 2+ , Ba 2+ , or Pb 2+
Ions That Form
Insoluble Compounds*
Exceptions
carbonate (CO 3 2– ) when combined with Group 1
ions or ammonium (NH 4 + )
chromate (CrO 4 2– ) when combined with Group 1
ions, Ca 2+ , Mg 2+ , or
ammonium (NH 4 + )
phosphate (PO 4 3– ) when combined with Group 1
ions or ammonium (NH 4 + )
sulfide (S 2– ) when combined with Group 1
ions or ammonium (NH 4 + )
hydroxide (OH – ) when combined with Group 1
ions, Ca 2+ , Ba 2+ , Sr 2+ , or
ammonium (NH 4 + )
*compounds having very low solubility in H 2 O
R2

150.
140.
Table G
Solubility Curves at Standard Pressure
KI
NaNO 3
130.
120.
KNO 3
110.
100.
Solubility (g solute/100. g H 2
O)
90.
80.
70.
60.
HCl
NH 4
Cl
KCl
50.
40.
30.
NaCl
KClO 3
NH 3
20.
10.
SO 2
0
0 10. 20. 30. 40. 50. 60. 70. 80. 90. 100.
Temperature (°C)
R3

Table H
Vapor Pressure of Four Liquids
200.
propanone
ethanol
150.
water
Vapor Pressure (kPa)
100.
101.3 kPa
ethanoic
acid
50.
0
0 25 50. 75 100. 125
R4

Table I
Heats of Reaction at 101.3 kPa and 298 K
Reaction
ΔH (kJ)*
CH 4
(g) + 2O 2
(g) CO 2
(g) + 2H 2
O() –890.4
C 3
H 8
(g) + 5O 2
(g) 3CO 2
(g) + 4H 2
O() –2219.2
2C 8
H 18
() + 25O 2
(g) 16CO 2
(g) + 18H 2
O() –10943
2CH 3
OH() + 3O 2
(g) 2CO 2
(g) + 4H 2
O() –1452
C 2
H 5
OH() + 3O 2
(g) 2CO 2
(g) + 3H 2
O() –1367
C 6
H 12
O 6
(s) + 6O 2
(g) 6CO 2
(g) + 6H 2
O() –2804
2CO(g) + O 2
(g) 2CO 2
(g) –566.0
C(s) + O 2
(g) CO 2
(g) –393.5
4Al(s) + 3O 2
(g) 2Al 2
O 3
(s) –3351
N 2
(g) + O 2
(g) 2NO(g) +182.6
N 2
(g) + 2O 2
(g) 2NO 2
(g) +66.4
2H 2
(g) + O 2
(g) 2H 2
O(g) –483.6
2H 2
(g) + O 2
(g) 2H 2
O() –571.6
N 2
(g) + 3H 2
(g) 2NH 3
(g) –91.8
2C(s) + 3H 2
(g) C 2
H 6
(g) –84.0
2C(s) + 2H 2
(g) C 2
H 4
(g) +52.4
2C(s) + H 2
(g) C 2
H 2
(g) +227.4
H 2
(g) + I 2
(g) 2HI(g) +53.0
KNO 3
(s) H 2 O K + (aq) + NO 3 – (aq) +34.89
NaOH(s) H 2 O Na + (aq) + OH – (aq) –44.51
NH 4
Cl(s) H 2 O NH 4 + (aq) + Cl – (aq) +14.78
NH 4
NO 3
(s) H 2 O NH 4 + (aq) + NO 3 – (aq) +25.69
NaCl(s) H 2 O Na + (aq) + Cl – (aq) +3.88
LiBr(s) H 2 O Li + (aq) + Br – (aq) –48.83
H + (aq) + OH – (aq) H 2
O() –55.8
*The ΔH values are based on molar quantities represented in the equations.
A minus sign indicates an exothermic reaction.
Most
Active
Least
Active
Table J
Activity Series**
Metals Nonmetals Most
Active
Li
Rb
K
Cs
Ba
Sr
Ca
Na
Mg
Al
Ti
Mn
Zn
Cr
Fe
Co
Ni
Sn
Pb
H 2
Cu
Ag
Au
F 2
Cl 2
Br 2
I 2
**Activity Series is based on the hydrogen
standard. H 2 is not a metal.
Least
Active
R5

Table K
Common Acids
Table N
Selected Radioisotopes
HCl(aq)
Formula
HNO 2
(aq)
HNO 3
(aq)
H 2
SO 3
(aq)
H 2
SO 4
(aq)
H 3
PO 4
(aq)
H 2
CO 3
(aq)
or
CO 2
(aq)
CH 3
COOH(aq)
or
HC 2
H 3
O 2
(aq)
Name
hydrochloric acid
nitrous acid
nitric acid
sulfurous acid
sulfuric acid
phosphoric acid
carbonic acid
ethanoic acid
(acetic acid)
Nuclide Half-Life Decay
Mode
Nuclide
Name
198 Au 2.695 d β – gold-198
14 C 5715 y β – carbon-14
37 Ca 182 ms β + calcium-37
60 Co 5.271 y β – cobalt-60
137 Cs 30.2 y β – cesium-137
53 Fe 8.51 min β + iron-53
220 Fr 27.4 s α francium-220
3 H 12.31 y β – hydrogen-3
131 I 8.021 d β – iodine-131
37 K 1.23 s β + potassium-37
42 K 12.36 h β – potassium-42
Table L
Common Bases
85 Kr 10.73 y β – krypton-85
16 N 7.13 s β – nitrogen-16
Formula
NaOH(aq)
KOH(aq)
Ca(OH) 2
(aq)
NH 3
(aq)
Name
sodium hydroxide
potassium hydroxide
calcium hydroxide
aqueous ammonia
19 Ne 17.22 s β + neon-19
32 P 14.28 d β – phosphorus-32
239 Pu 2.410 × 10 4 y α plutonium-239
226 Ra 1599 y α radium-226
222 Rn 3.823 d α radon-222
90 Sr 29.1 y β – strontium-90
Table M
Common Acid–Base Indicators
Approximate
Indicator pH Range Color
for Color Change
Change
methyl orange 3.1–4.4 red to yellow
bromthymol blue 6.0–7.6 yellow to blue
phenolphthalein 8–9 colorless to pink
litmus 4.5–8.3 red to blue
bromcresol green 3.8–5.4 yellow to blue
thymol blue 8.0–9.6 yellow to blue
99 Tc 2.13 × 10 5 y β – technetium-99
232 Th 1.40 × 10 10 y α thorium-232
233 U 1.592 × 10 5 y α uranium-233
235 U 7.04 × 10 8 y α uranium-235
238 U 4.47 × 10 9 y α uranium-238
Source: CRC Handbook of Chemistry and Physics, 91 st ed., 2010–2011,
CRC Press
Source: The Merck Index, 14 th ed., 2006, Merck Publishing Group
R6

Table O
Symbols Used in Nuclear Chemistry
Name Notation Symbol
alpha particle
4
2
He or 4 2 α α
beta particle
0
–1
e or 0
–1 β β–
gamma radiation
0
0
γ γ
neutron
1
0
n n
proton
1
1
H or 1 1 p p
positron
0
+1
e or 0
+1 β β+
Table P
Organic Prefixes
Prefix
meth- 1
eth- 2
prop- 3
but- 4
pent- 5
hex- 6
hept- 7
oct- 8
non- 9
dec- 10
Number of
Carbon Atoms
Table Q
Homologous Series of Hydrocarbons
Name General Examples
Formula Name Structural Formula
R7
alkanes C n
H 2n+2
ethane
alkenes C n
H 2n
ethene
alkynes C n
H 2n–2
ethyne
Note: n = number of carbon atoms
H H
H C C H
H H
H
H
C C
H
H
H C C H

Table R
Organic Functional Groups
Class of
Compound
Functional
Group
General
Formula
Example
halide
(halocarbon)
F (fluoro-)
Cl (chloro-)
Br (bromo-)
I (iodo-)
R X
(X represents
any halogen)
CH 3
CHClCH 3
2-chloropropane
alcohol
OH
R
OH
CH 3
CH 2
CH 2
OH
1-propanol
ether
O
R O R′
CH 3
OCH 2
CH 3
methyl ethyl ether
aldehyde
O
C H
R
O
C H
O
CH 3
CH 2
C H
propanal
ketone
O
C
O
R C R′
O
CH 3
CCH 2
CH 2
CH 3
2-pentanone
organic acid
O
C OH
R
O
C OH
O
CH 3
CH 2
C OH
propanoic acid
ester
O
C O
O
R C O R′
O
CH 3
CH 2
COCH 3
methyl propanoate
amine
N
R
R′
N R′′
CH 3
CH 2
CH 2
NH 2
1-propanamine
amide
O
C NH
R
O R′
C NH
O
CH 3
CH 2
C NH 2
propanamide
Note: R represents a bonded atom or group of atoms.
R8

0
6.941
+1
Li
3
2-1
Na
39.0983
K +1
19
2-8-8-1
85.4678 +1
Rb
Cs
(223)
Fr
87
-18-32-18-8-1
+1
Ra
88
-18-32-18-8-2
39
138.9055
La
57
2-8-18-18-9-2
+2 (227)
Ac
89
-18-32-18-9-2
47.867
Ti
22
2-8-10-2
91.224
Zr
40
2-8-18-10-2
+3 178.49
Hf
72
*18-32-10-2
+3 (261)
Rf
104
+2
+3
+4
+4
+4
50.9415
V
23
2-8-11-2
+2
+3
+4
+5
51.996
Cr
24
2-8-13-1
95.94
Mo
42
2-8-18-13-1
183.84
W
74
-18-32-12-2
+2
+3
+6
+6
+6
54.9380
Mn
25
2-8-13-2
+2
+3
+4
+7
55.845
Fe
26
2-8-14-2
+2
+3 58.9332
Co
27
2-8-15-2
+2
+3
58.693
Ni
28
2-8-16-2
+2
+3 63.546 Cu
2-8-18-1
107.868
Ag
47
2-8-18-18-1
79
+1
+2
+1
65.409
Zn
30
2-8-18-2
10.81
+3 12.011
B
5
2-3
26.98154
Al
13
2-8-3
+2 69.723
Ga
31
2-8-18-3
+3
+3
–4
+2
+4
C
6
2-4
28.0855
Si
14
2-8-4
72.64
Ge
32
2-8-18-4
Pb
–4
+2
+4
+2
+4
74.9216
As
33
2-8-18-5
Sb
–3
+3
15.9994 O
–2 18.9984
8
2-6 2-7
78.96
Se
34
2-8-18-6
127.60
Te
52
2-8-18-18-6
(209)
Po
84
-18-32-18-6
–2
+4
+6
–2
+4
+6
+2
+4
F
79.904
Br
35
2-8-18-7
126.904
l
53
2-8-18-18-7
(210)
At
85
-18-32-18-7
( ? )
Uus
117
4.00260 0
He
2
2
–1 20.180
Ne
10
2-8
0
22.98977
11
2-8-1
1
1.00794 +1
–1
H
1
1
1
37
2-8-18-8-1
–1
+1
+5
–1
+1
+5
+7
83.798
Kr
36
2-8-18-8
131.29
Xe
54
2-8-18-18-8
(222)
Rn
86
-18-32-18-8
0
+2
0
+2
+4
+6
0
132.905
55
2-8-18-18-8-1
Symbol
Relative atomic masses are based
Group on 12 C = 12 (exact)
Group
2
13 14 15 16 17 18
Atomic Number
+1
+1
9.01218 +2
Be
4
2-2
24.305
Mg
12
2-8-2
40.08
Ca
20
2-8-8-2
87.62
Sr
38
2-8-18-8-2
137.33
Ba
56
2-8-18-18-8-2
(226)
+2
+2
+2
+2
3
44.9559
Sc
21
2-8-9-2
88.9059
Y
2-8-18-9-2
+3
+3
4
KEY
92.9064
Nb +3
+5
41
2-8-18-12-1
180.948
Ta
73
-18-32-11-2
(262)
105
5
Periodic Table of the Elements
Atomic Mass
Electron Configuration
+4
Db
+5
6
(266)
Sg
106
12.011 2-4
–4
6
C
+2
+4
(98)
Tc
43
2-8-18-13-2
186.207
Re
75
-18-32-13-2
(272)
Bh
107
7
Group
+4
+6
+7
+4
+6
+7
8
101.07
Ru
44
2-8-18-15-1
190.23
Os
76
-18-32-14-2
(277)
Hs
108
+3
+3
+4
Selected Oxidation States
Note: Numbers in parentheses
are mass numbers of the most
stable or common isotope.
9
102.906
Rh
45
2-8-18-16-1
192.217
Ir
77
-18-32-15-2
(276)
Mt
109
+3
+3
+4
106.42
Pd
46
2-8-18-18
195.08
Pt
78
-18-32-17-1
+2
+4
+2
+4
196.967
Au
-18-32-18-1
(281)
Ds (280) Rg
110
10
29
111
11 12
+1
+3
112.41
Cd
48
2-8-18-18-2
200.59
Hg
80
-18-32-18-2
(285)
Cn
112
+2 114.818
In
+1
+2
49
2-8-18-18-3
204.383
Tl
81
-18-32-18-3
(284)
Uut
113**
+3
+1
+3
118.71
Sn
50
2-8-18-18-4
207.2
82
-18-32-18-4
(289)
Uuq
114
+2
+4
+2
+4
14.0067 –3
–2
N
–1
7
2-5
30.97376
P
15
2-8-5
121.760
51
2-8-18-18-5
208.980
Bi
83
-18-32-18-5
(288)
Uup
115
+1
+2
+3
+4
+5
–3
+3
+5
+5
–3
+3
+5
+3
+5
32.065
S
16
2-8-6
(292)
Uuh
116
–2
+4
+6
35.453
Cl
17
2-8-7
–1
+1
+5
+7
39.948
Ar
18
2-8-8
18
(294)
Uuo
118
140.116
Ce
58
232.038
Th
90
+3
+4
140.908
Pr +3
59
144.24
Nd
60
+4 231.036
Pa +4 238.029 +5
U +3
+4
+5
+6
91
92
+3
(145)
Pm
61
+3
150.36
Sm
62
+2
+3
151.964
Eu
63
+2
+3
157.25
Gd
64
+3
158.925
(237)Np (244) Pu (243) Am (247) Cm +3 (247) Bk +3
+3
+4
+5
+6
93 94
+3
+4
+5
+6
65
+3
+4
+5
+6
95 96 97
Tb
+3
+4
162.500
Dy
66
(251)
+3
164.930
Ho
67
+3
167.259
Er
68
Cf +3 (252) Es (257) Fm
100
98 99
+3
+3
+3
168.934
Tm +3
69
(258)
Md
101
+2
+3
173.04
Yb
70
(259)
No
102
+2
+3
+2
+3
174.9668
Lu
71
(262)
Lr
103
+3
+3
*denotes the presence of (2-8-) for elements 72 and above
**The systematic names and symbols for elements of atomic numbers 113 and above
will be used until the approval of trivial names by IUPAC.
Source: CRC Handbook of Chemistry and Physics, 91 st ed., 2010–2011, CRC Press
9
R9
Period
1
2
3
4
5
6
7

Table S
Properties of Selected Elements
First
Atomic Symbol Name Ionization
Electro- Melting Boiling* Density** Atomic
Number Energy negativity Point Point (g/cm 3 ) Radius
(kJ/mol) (K) (K) (pm)
1 H hydrogen 1312 2.2 14 20. 0.000082 32
2 He helium 2372 — — 4 0.000164 37
3 Li lithium 520. 1.0 454 1615 0.534 130.
4 Be beryllium 900. 1.6 1560. 2744 1.85 99
5 B boron 801 2.0 2348 4273 2.34 84
6 C carbon 1086 2.6 — — .— 75
7 N nitrogen 1402 3.0 63 77 0.001145 71
8 O oxygen 1314 3.4 54 90. 0.001308 64
9 F fluorine 1681 4.0 53 85 0.001553 60.
10 Ne neon 2081 — 24 27 0.000825 62
11 Na sodium 496 0.9 371 1156 0.97 160.
12 Mg magnesium 738 1.3 923 1363 1.74 140.
13 Al aluminum 578 1.6 933 2792 2.70 124
14 Si silicon 787 1.9 1687 3538 2.3296 114
15 P phosphorus (white) 1012 2.2 317 554 1.823 109
16 S sulfur (monoclinic) 1000. 2.6 388 718 2.00 104
17 Cl chlorine 1251 3.2 172 239 0.002898 100.
18 Ar argon 1521 — 84 87 0.001633 101
19 K potassium 419 0.8 337 1032 0.89 200.
20 Ca calcium 590. 1.0 1115 1757 1.54 174
21 Sc scandium 633 1.4 1814 3109 2.99 159
22 Ti titanium 659 1.5 1941 3560. 4.506 148
23 V vanadium 651 1.6 2183 3680. 6.0 144
24 Cr chromium 653 1.7 2180. 2944 7.15 130.
25 Mn manganese 717 1.6 1519 2334 7.3 129
26 Fe iron 762 1.8 1811 3134 7.87 124
27 Co cobalt 760. 1.9 1768 3200. 8.86 118
28 Ni nickel 737 1.9 1728 3186 8.90 117
29 Cu copper 745 1.9 1358 2835 8.96 122
30 Zn zinc 906 1.7 693 1180. 7.134 120.
31 Ga gallium 579 1.8 303 2477 5.91 123
32 Ge germanium 762 2.0 1211 3106 5.3234 120.
33 As arsenic (gray) 944 2.2 1090. — 5.75 120.
34 Se selenium (gray) 941 2.6 494 958 4.809 118
35 Br bromine 1140. 3.0 266 332 3.1028 117
36 Kr krypton 1351 — 116 120. 0.003425 116
37 Rb rubidium 403 0.8 312 961 1.53 215
38 Sr strontium 549 1.0 1050. 1655 2.64 190.
39 Y yttrium 600. 1.2 1795 3618 4.47 176
40 Zr zirconium 640. 1.3 2128 4682 6.52 164
R10

First
Atomic Symbol Name Ionization
Electro- Melting Boiling* Density** Atomic
Number Energy negativity Point Point (g/cm 3 ) Radius
(kJ/mol) (K) (K) (pm)
41 Nb niobium 652 1.6 2750. 5017 8.57 156
42 Mo molybdenum 684 2.2 2896 4912 10.2 146
43 Tc technetium 702 2.1 2430. 4538 11 138
44 Ru ruthenium 710. 2.2 2606 4423 12.1 136
45 Rh rhodium 720. 2.3 2237 3968 12.4 134
46 Pd palladium 804 2.2 1828 3236 12.0 130.
47 Ag silver 731 1.9 1235 2435 10.5 136
48 Cd cadmium 868 1.7 594 1040. 8.69 140.
49 In indium 558 1.8 430. 2345 7.31 142
50 Sn tin (white) 709 2.0 505 2875 7.287 140.
51 Sb antimony (gray) 831 2.1 904 1860. 6.68 140.
52 Te tellurium 869 2.1 723 1261 6.232 137
53 I iodine 1008 2.7 387 457 4.933 136
54 Xe xenon 1170. 2.6 161 165 0.005366 136
55 Cs cesium 376 0.8 302 944 1.873 238
56 Ba barium 503 0.9 1000. 2170. 3.62 206
57 La lanthanum 538 1.1 1193 3737 6.15 194
Elements 58–71 have been omitted.
72 Hf hafnium 659 1.3 2506 4876 13.3 164
73 Ta tantalum 728 1.5 3290. 5731 16.4 158
74 W tungsten 759 1.7 3695 5828 19.3 150.
75 Re rhenium 756 1.9 3458 5869 20.8 141
76 Os osmium 814 2.2 3306 5285 22.587 136
77 Ir iridium 865 2.2 2719 4701 22.562 132
78 Pt platinum 864 2.2 2041 4098 21.5 130.
79 Au gold 890. 2.4 1337 3129 19.3 130.
80 Hg mercury 1007 1.9 234 630. 13.5336 132
81 Tl thallium 589 1.8 577 1746 11.8 144
82 Pb lead 716 1.8 600. 2022 11.3 145
83 Bi bismuth 703 1.9 544 1837 9.79 150.
84 Po polonium 812 2.0 527 1235 9.20 142
85 At astatine — 2.2 575 — — 148
86 Rn radon 1037 — 202 211 0.009074 146
87 Fr francium 393 0.7 300. — — 242
88 Ra radium 509 0.9 969 — 5 211
89 Ac actinium 499 1.1 1323 3471 10. 201
Elements 90 and above have been omitted.
*boiling point at standard pressure
**density of solids and liquids at room temperature and density of gases at 298 K and 101.3 kPa
— no data available
Source: CRC Handbook for Chemistry and Physics, 91 st ed., 2010–2011, CRC Press
R11

Table T
Important Formulas and Equations
d = density
m
Density d = m = mass
V
V = volume
Mole Calculations number of moles =
given mass
gram-formula mass
measured value – accepted value
Percent Error % error = × 100
accepted value
mass of part
Percent Composition % composition by mass = × 100
mass of whole
mass of solute
parts per million = × 1000000
mass of solution
Concentration
molarity =
moles of solute
liter of solution
Combined Gas Law
P
P = pressure
1
V 1
P
= 2
V 2
V = volume
T 1
T 2 T = temperature
M A
= molarity of H + M B
= molarity of OH –
Titration M A
V A
= M B
V B
V A
= volume of acid V B
= volume of base
q = mCΔT q = heat H f
= heat of fusion
Heat q = mH f
m = mass H v
= heat of vaporization
q = mH v
C=specific heat capacity
ΔT = change in temperature
Temperature
K = °C + 273
K = kelvin
°C = degree Celsius
R12

Collier County CHEMISTRY EXAM FORMULA AND RESOURCE PACKET
GENERAL
D m V
[ ExperimentalValue AcceptedVa lue]
% error
x100
AcceptedVa lue
% yield
ExperimentalYield
TheoreticalYield
x100
CONCENTRATIONS
moles of solute
M = Molarity
liters of solution
KEY
P = pressure
V = volume
T = temperature
n = number of moles
m = mass
M = molar mass (grams/mole)
D = density
KE = kinetic energy
Avogadro’s Number = 6.02 x 10 23
GASES, LIQUIDS, SOLUTIONS
m = Molality
M1V1 M2V2
S1
P1
S 2
P 2
ACID/BASE
pH = - log[H + ]
[H + ]=10 -pH
moles of solute
kilograms of solvent
Gas constant
R 8.314 L kPa L atm L mmHg
0.0821 62.4
K mol K mol K mol
1 atm = 760 mmHg = 760 torr = 101.3 kPa
K = o C + 273
o C = K - 273
STP = Standard Temperature and Pressure = 0 o C
and 1 atm
P V
1 1
P2V
2
pOH = - log [OH - ]
[OH - ]= 10 -pOH
pH + pOH = 14
Kw = [H + ] x [OH - ] = 1.0 x 10 -14 M 2
V
T
1
1
P
T
1
1
V
T
P
T
2
2
2
2
Or V1T2 = V2T1
Or P1T2 = P2T1
THERMOCHEMISTRY
ΔH= mCΔT, where ΔT = T f - T
P1V
1
T
1
P2V
2
T
2
Or
P1V1T2=P2V2T1
q = mCΔT
PV
nRT
Specific Heat of Water = 4.18 J/g*˚C or 1.0 cal/g*˚C
Specific Heat of Ice = 2.1 J/g*˚C or 0.5 cal/g*˚C
Specific Heat of Steam = 2.0 J/g*˚C or 0.4 cal/g*˚C
P
Total
P
1
P
2
Rate A
Rate B
...
Molar MassB
Molar MassA
R13

CHEMISTRY EXAM FORMULA AND RESOURCE PACKET
Solubility of Compounds at 25 o C and 1 atm
acetate
bromide
carbonate
chlorate
chloride
hydroxide
iodide
nitrate
oxide
perchlorate
phosphate
sulfate
sulfide
aluminum S S - S S I S S I S I S d
ammonium S S S S S S S S - S S S S
barium S S I S S S S S sS S I I d
calcium S S I S S S S S sS S I sS I
copper(II) S S - S S I S S I S I S I
iron(II) S S I S S I S S I S I S I
iron(III) S S - S S I S S I S I sS d
lithium S S sS S S S S S S S sS S S
magnesium S S I S S I S S I S I S d
potassium S S S S S S S S S S S S S
silver sS I I S I - I S I S I sS I
sodium S S S S S S S S S S S S S
strontium S S I S S S S S S S I I I
zinc S S I S S I S S I S I S I
S=soluble
sS = slightly soluble in water
I = insoluble in water
d = decomposes in water
- = no such compound
R14

CHEMISTRY EXAM FORMULA AND RESOURCE PACKET
Common Polyatomic Ions
1- Charge 2- Charge 3- Charge
Formula Name Formula Name Formula Name
Dihydrogen
Phosphate
Hydrogen
phosphate
Phosphite
Acetate Oxalate Phosphate
Hydrogen
sulfite
Sulfite
Hydrogen
sulfate
Sulfate
Hydrogen
carbonate
Carbonate
Nitrite Chromate 1+ Charge
Nitrate Dichromate Formula Name
Cyanide Silicate Ammonium
Hydroxide
Permanganate
Hypochlorite
Chlorite
Chlorate
Perchlorate
R15

Activity Series of Metals
Name
Symbol
D
Lithium
Li
e
Potassium
K
c
r
Barium
Ba
e
Calcium
Ca
a
Sodium
Na
s
i
Magnesium
Mg
n
Aluminum
Al
g
Zinc
Zn
Iron
Fe
A
c
Nickel
Ni
t
Tin
Sn
i
v
Lead
Pb
i
(Hydrogen)
(H)*
t
Copper
Cu
y
Mercury
Hg
Silver
Ag
Gold
Au
*Metals from Li to Na will replace H from acids and water; from Mg to
Pb they will replace H from acids only.
Decreasing
Activity
Activity Series of Nonmetal (Halogens)
Name
Symbol
Fluorine F 2
Chlorine Cl 2
Bromine Br 2
Iodine I 2
R16