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Ch.1 - Chemical Tools: Experimentation & Measurement

Ch.2 - Atoms, Molecules & Ions

Ch.3 - Mass Relationships in Chemical Reactions

BONUS: Lab Techniques and Procedures

BONUS: Mathematical Operations and Functions

Ch.4 - Reactions in Aqueous Solution

Ch.5 - Periodicity & Electronic Structure of Atoms

Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory

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Ch.8 - Covalent Compounds: Bonding Theories and Molecular Structure

Ch.9 - Thermochemistry: Chemical Energy

Ch.10 - Gases: Their Properties & Behavior

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Ch.12 - Solids and Solid-State Materials

Ch.13 - Solutions & Their Properties

Ch.14 - Chemical Kinetics

Ch.15 - Chemical Equilibrium

Ch.16 - Aqueous Equilibria: Acids & Bases

Ch.17 - Applications of Aqueous Equilibria

Ch.18 - Thermodynamics: Entropy, Free Energy & Equilibrium

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Ch.23 - Organic and Biological Chemistry

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Chapter 15, Problem 154a

Acetic acid tends to form dimers, (CH₃CO₂H₂), because of hydrogen bonding: <IMAGE> The equilibrium constant Kc for this reaction is 1.51⨉10² in benzene solution but only 3.7⨉10^-2 in water solution. (a) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in benzene.

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Chapter 15, Problem 154a

Acetic acid tends to form dimers, (CH3CO2H2), because of hydrogen bonding: <IMAGE> The equilibrium constant Kc for this reaction is 1.51⨉102 in benzene solution but only 3.7⨉10-2 in water solution. (a) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in benzene.

Acetic acid tends to form dimers, (CH₃CO₂H₂), because of hydrogen bonding: <IMAGE> The equilibrium constant Kc for this reaction is 1.51⨉10² in benzene solution but only 3.7⨉10^-2 in water solution. (a) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in benzene.