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A. C. B. D. Explanation: ‘D’ is the answer because, given the information provided on the chart, it can be seen that the catalyst lowered the required energy for the reaction to activate, and it progressed similarly. The other graphs do not demonstrate the lowering of the activation energy. In fact, ‘B’ demonstrates an increase in the required activation energy, and ‘C’ represents a “multistage” reaction, in which there are intermediate products and where multiple “micro-reactions” occur. This same principle of multiple reactions is demonstrated in ‘A.’ 9. H2O2, hydrogen peroxide, naturally breaks down into H 2 O 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 MnO 2 ? A. a catalyst B. an enhancer C. an inhibitor D. a reactant Explanation: The answer is “a catalyst” because a substance which lowers the activation energy of a reaction and thus increases the rate of reaction is a catalyst, not an enhancer, inhibitor, or reactant. An inhibitor accomplishes the opposite effect, interfering with the action of a catalyst. A reactant is merely a required part of the reaction to form a product, and an “enhancer” has no explicit definition or purpose in chemical reactions. 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. Explanation: When a reaction is at equilibrium, its forward reaction rate and reverse reaction rate have reached an equal state, where the production of products and reactants have become equal. If more reactant is added, this increases the forward reaction rate as the reaction attempts to achieve equilibrium again. This is why ‘B’ is the answer. “The reverse reaction rate decreases” is incorrect, because, in order for the reaction to achieve equilibrium again after more reactants are added, the reaction must speed up the forward reaction to use up the excess reactants. “The forward reaction rate remains the same” is incorrect because of the aforementioned reasoning. “The reverse reaction rate remains the same” is incorrect, because if the forward reaction rate increases to accommodate the excess reactants, equilibrium implies the two rates are equal, so there must be a change in the 145
everse reaction rate as well. 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 Explanation: The answer here is ‘A’ because, the greater the pressure, the greater the ability for the molecules A and B to collide and form the reactant AB. The other answers are incorrect because A + B cannot equal C or D. ‘D’ is wrong because an increase in pressure would favor the reverse reaction, because if you increase pressure, A and C would have a greater chance of colliding with one another, which would consequently increase the rate of reaction. 4HCl (g) + O 2(g) ⇄ 2H2O(l) + 2Cl2(g) + 113 kJ 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 Explanation: The answer here is “increasing the system’s pressure.” This is the answer because heating the mixture will cause the reverse reaction to be favored, since heat is a product of the reaction. Adding water to the system will drive the reaction left, because if more product is added, the reverse reaction increases. Decreasing the oxygen concentration would reduce the concentration of products, but also of reactants. Increasing the pressure increases the rate of reaction between reactants, because of the decreased volume and thus a greater chance of collision (which leads to a higher rate of reaction) 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 Explanation: “Increasing the NO concentration in the flask” is the answer to this question. This is because the larger the volume of NO (a product) in a given space, the greater the chance it has of colliding with CO 2 , which would thus lead to an increased reaction rate. Pumping more CO (reactant) into the flask would shift the reaction right, and raising the pressure would increase the production of product as reactants collide more frequently. Increasing the concentration of reactants would lead to more product; this is simply because the greater the volume of particles within a given space, the greater the chance of collision between said particles. Venting CO 2 would result in a greater forward reaction, again, as the reaction attempts to reach equilibrium, and so NO 2 and CO begin producing more product. 146
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Chemistry Notebook By: Joel J
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Unit 1 Measurement Lab Separation o
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Unit 5 (28 days) Chapter 10 Chemica
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30. Always lubricate glassware (tub
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Chapter 1 Unit 1 Introduction to Ch
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To use the Stair-Step method, find
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Using SI Units Match the terms in C
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Standards of Measurement Fill in th
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SCIENTIFIC NOTATION RULES How to Wr
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Convert each number from Scientific
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Significant Figures Rules There are
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How Many Significant Digits for Eac
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The following rule applies for mult
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1qt=32 oz 1gal = 4qts 1.00 qt = 946
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Chapter 5 Electrons in Atoms The st
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Atoms Are Building Blocks Atoms are
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Electrovalence Don't get worried ab
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Summary Atoms are essentially the b
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Electron Configuration Color the su
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Electron Configuration In order to
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Calcium (Ca) Nickel (Ni) Carbon (C)
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Research the Scientist and summariz
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Using Wikipedia, define the 8 categ
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Ionization Energy Define Ionization
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Ion Size Define Ion Size: Ion size
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Chapter 7 Ionic and Metallic Bondin
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10. Name __________________________
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The Nucleus A typical model of the
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Gamma Radiation After a decay react
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This analysis shows that sodium has
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Metals, with only a few electrons i
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POLAR BONDING results when two diff
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Step 6 Put the atoms in the structu
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Trigonal Planal Molecular Geometry
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Tetrahedral Molecular Geometry Orbi
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Bent Molecular Geometry Orbital Equ
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T-Shaped Molecular Geometry Orbital
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Square Planar Molecular Geometry Or
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sp 3 d 2 AX6 None 90 Octahedral sp
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Chapter 9 Unit 4 Chemical Names and
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Learning Goals: Describe the proper
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Unit 5 Chapter 10 Chemical Quantiti
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www.youtube.com/watch?v=AsqEkF7hcII
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Categories of Reactions All chemica
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List what type the following reacti
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How to Balance Chemical Equations A
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Step 5 Balance the hydrogen atoms n
Page 97 and 98: 6) ___ Mn(NO2)2 + ___ BeCl2 ___ Be
Page 99 and 100: The Learning Goal for this assignme
Page 101 and 102: Stoichiometry and Balanced Equation
Page 103 and 104: Step 1: 200g C3H8 is equal to 4.54
Page 105 and 106: Stoichiometry and balanced equation
Page 107 and 108: %yield = 6.1 tons × 100 = 64% 9.6t
Page 109 and 110: Use the space provided to write out
Page 111 and 112: Unit 6 Chapter 13 States of Matter
Page 115 and 116: uniform internal structure, and amo
Page 117 and 118: Volume Pressure Temperature is Aver
Page 119 and 120: Volume and temperature (in Kelvins)
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Page 125 and 126: Combined Gas Law: This gas law comb
Page 129 and 130: Unit 7 Chapter 16 Solutions The stu
Page 133 and 134: Part 5: Molarity Molarity (M) is th
Page 137 and 138: The Time-Temperature Graph We are g
Page 139 and 140: Step Two: solid ice melts Now, we c
Page 141 and 142: Step Four: liquid water boils Now,
Page 143 and 144: The following table summarizes the
Page 145 and 146: The Learning Goal for this section
Page 147: B. decreasing the concentration of
Page 151 and 152: Unit 8 Chapter 19 Acid and Bases Th
Page 153 and 154: The Learning Goal for this section
Page 155 and 156: This would be a generic base: XOH -
Page 157 and 158: HCl + H2O ⇌ H3O + + Cl¯ HCl - th
Page 159 and 160: Several categories of substances ca
Page 161 and 162: 1. 2.82 2. 8.30 3. 0.00 4. 3.485 5.
Page 165: 2. What is the oxidation number of
Page 170 and 171: Table E Selected Polyatomic Ions Fo
Page 172 and 173: Table H Vapor Pressure of Four Liqu
Page 174 and 175: Table K Common Acids Table N Select
Page 176 and 177: Table R Organic Functional Groups C
Page 178 and 179: Table S Properties of Selected Elem
Page 180 and 181: Table T Important Formulas and Equa
Page 182 and 183: CHEMISTRY EXAM FORMULA AND RESOURCE
Page 184: Activity Series of Metals Name Symb
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