Introductory Chemistry Concepts and Critical Thinking, 8E Charles H Corwin Solution Manual by Ace Test Banks

INSTRUCTOR MANUAL AND

TEST ITEM FILE

EIGHTH EDITION

Introductory Chemistry CONCEPTS AND CRITICAL THINKING

CHARLES H. CORWIN

Contents Preface

Learning Objectives and Associated Exercises

vii

Solutions to Even-Numbered Textbook Exercises 1

Introduction to Chemistry

Prerequisite Science Skills

The Metric System

Matter and Energy

Models of the Atom

The Periodic Table

Language of Chemistry

Chemical Reactions

The Mole Concept

Chemical Equation Calculations

Gases

Liquids and Solids

Chemical Bonding

Solutions

Acids and Bases

109

Advanced Problem Solving

119

Chemical Equilibrium

129

Oxidation and Reduction

137

Nuclear Chemistry

145

Organic Chemistry

151

Biochemistry

163

iii

Preface

To the Instructor I offer the following suggestions based on countless correspondences with instructors who teach our subject. I have learned that there are many viable approaches for achieving student success depending on course objectives and classroom facilities. Since I teach in a media classroom, I incorporate PowerPoint® and video presentations, Internet assignments, collaborative learning, chemical demonstrations, as well as traditional techniques. It is my experience that lecture materials have a relatively short lifespan and I must renew each presentation or students sense a lack of freshness. To accompany this Eighth Edition of Introductory Chemistry: Concepts and Critical Thinking there is a plethora of technology including an Instructor Resource Center with pre-built PowerPoint® presentations, animations, interactive activities, Instructor’s Manual lecture outlines, and a Test Bank in Word® format. In terms of a guiding principle, I obtain the best results when I organize an entire course so as to include one or more interesting asides in each lecture. This adds an extra dimension to every presentation; for example, I may provide a media supplement, an Internet assignment, or a chemical demonstration, depending on the topic. A second guiding principle is that the lecture topic should be supported by a corresponding laboratory experiment. For example, when discussing the Bohr atom, I demonstrate emission line spectra in lecture, while students perform the experiment Atomic Fingerprints found in the Pearson Laboratory Manual. ..

Testing For introductory chemistry classes, I have found that frequent testing improves test scores, achieves higher attendance, and more students successfully complete the course. The Test Bank for the Eighth Edition contains over 3000 questions, covering key terms, each learning objective, and general questions that are cumulative in nature. This unparalleled resource has been class-tested on thousands of students and each question and answer have undergone detailed item analysis. Print and Media Resources Introductory Chemistry: Concepts and Critical Thinking, 8e, has numerous ancillaries and the following are available to the instructor upon adoption of the textbook: • Instructor Resource Center • Instructor Manual and Test Bank • MasteringChemistry® Acknowledgments I would like to thank the many reviewers, colleagues, and adoptees who have used the textbook and ancillary materials in previous editions and offered valuable comments that provide accuracy and balanced coverage. I have acknowledged these individuals in the preface of the text, and I welcome your comments for this latest edition and invite you to contact me directly.

Charles H. Corwin Department of Chemistry American River College Sacramento, CA 95841 corwinc@arc.losrios.edu

Learning Objectives and Associated Exercises

Chapter 1 Bloom’s Learning Objectives L1 L2 L2 L2 L5

Name the key term that corresponds to a definition. Describe the early practice of chemistry. Identify the three steps in the scientific method. Describe the modern practice of chemistry. Conclude that chemistry is very relevant in our daily life.

(Sec. 1.1) (Sec. 1.1) (Sec. 1.2) (Sec. 1.3)

Key Terms Exercises 1–4 Exercises 5–12 Exercises 13–16 Exercises 17–18

(PSS.1) (PSS.1) (PSS.2) (PSS.3) (PSS.4) (PSS.5) (PSS.6) (PSS.6) (PSS.7) (PSS.7)

Key Terms Exercises 1–2 Exercises 3–10 Exercises 11–16 Exercises 17–20 Exercises 21–24 Exercises 25–28 Exercises 29–32 Exercises 33–38 Exercises 39–40 Exercises 41–44

(Sec. 2.1) (Sec. 2.1) (Sec. 2.2) (Sec. 2.2) (Sec. 2.3) (Sec. 2.4) (Sec. 2.4) (Sec. 2.5) (Sec. 2.5)

Key Terms Exercises 1–6 Exercises 7–10 Exercises 11–12 Exercises 13–14 Exercises 15–18 Exercises 19–20 Exercises 21–28 Exercises 29–32 Exercises 33–34

Prerequisite Science Skills Bloom’s Learning Objectives L1 L2 L2 L2 L3 L3 L3 L2 L2 L3 L3

Name the key term that corresponds to a definition. Explain why measurements are never exact. Identify instruments for taking measurements. Identify significant digits in a measurement. Round off nonsignificant digits. Apply significant digit rule for adding and subtracting measurements. Apply significant digit rule for multiplying and dividing measurements. Express an ordinary number as a power of 10. Express a power of 10 as an ordinary number. Show an ordinary number in scientific notation. Show scientific notation as an ordinary number.

Chapter 2 Bloom’s Learning Objectives L1 L1 L1 L3 L3 L2 L1 L2 L2 L3

vii

L3 L3 L5 L3 L1 L3 L1 L3 L3 L3 L3

Solve problems that relate length, width, thickness, and volume of a rectangular solid. Express a given volume in milliliters, cubic centimeters, or cubic inches. Estimate the volume of a solid by the displacement of water. Apply the concept of density. Memorize the value for the density of water in grams per milliliter. Solve problems that relate density, mass, and volume. Memorize the freezing point and boiling point of water on the Fahrenheit, Celsius, and Kelvin scales. Express a given temperature in degrees Fahrenheit (°F), degrees Celsius (°C), and Kelvin (K). Apply the concept of specific heat to good and poor conductors of heat. Express heat energy in units of calories, kilocalories, joules, and kilojoules. Solve problems that relate specific heat, mass, and temperature change.

(Sec. 2.6)

Exercises 35–38

(Sec. 2.6) (Sec. 2.7) (Sec. 2.8) (Sec. 2.8) (Sec. 2.8) (Sec. 2.9)

Exercises 39–42 Exercises 43–46 Exercises 47–50 Exercises 51–52 Exercises 53–58 Exercises 59–60

(Sec. 2.9)

Exercises 61–66

(Sec. 2.10) (Sec. 2.10) (Sec. 2.10)

Exercise 67–68 Exercises 69–70 Exercises 71–76

(Sec. 3.1)

Key Terms Exercises 1–6

(Sec. 3.1)

Exercises 7–12

(Sec. 3.2) (Sec. 3.3) (Sec. 3.4) (Sec. 3.4)

Exercises 13–16 Exercises 17–22 Exercises 23–26 Exercises 27–30

(Sec. 3.4)

Exercises 31–34

(Sec. 3.5)

Exercises 35–40

(Sec. 3.5) (Sec. 3.6) (Sec. 3.7) (Sec. 3.8) (Sec. 3.9) (Sec. 3.9) (Sec. 3.9) (Sec. 3.10) (Sec. 3.10)

Exercises 41–42 Exercises 43–50 Exercises 51–58 Exercises 59–62 Exercises 63–64 Exercises 65–66 Exercises 67–68 Exercises 69–74 Exercises 75–78

(Sec. 4.1) (Sec. 4.2) (Sec. 4.2) (Sec. 4.3) (Sec. 4.3)

Key Terms Exercises 1–4 Exercises 5–6 Exercises 7–10 Exercises 11–12 Exercises 13–16

(Sec. 4.4) (Sec. 4.4) (Sec. 4.5) (Sec. 4.5)

Exercises 17–18 Exercises 19–24 Exercises 25–32 Exercises 33–36

(Sec. 4.6)

Exercises 37–44

Chapter 3 Bloom’s Learning Objectives L1 L2 L2 L4 L1 L1 L5 L5 L3 L3 L4 L4 L3 L4 L2 L2 L3 L2

Name the key term that corresponds to a definition. Describe the motion of particles in the solid, liquid, and gaseous states of matter. Describe the effect of temperature on the solid, liquid, and gaseous states of matter. Classify a sample of matter as an element, compound, or mixture. Memorize the names and symbols of 48 common elements. List the properties of metals and nonmetals. Predict whether an element is a metal, nonmetal, or semimetal given its position in the periodic table. Predict whether an element is a solid, liquid, or gas at 25 °C and normal atmospheric pressure. Calculate the number of atoms of each element in a compound given the chemical formula. Apply the law of definite composition to a compound. Classify a property of a substance as physical or chemical. Classify a change in a substance as physical or chemical. Apply the conservation of mass law to chemical changes. Distinguish between potential and kinetic energy. Summarize the relationship of temperature and kinetic energy. Summarize the relationship of kinetic energy and physical state. Apply the conservation of energy law to physical and chemical changes. Identify the following forms of energy: chemical, electrical, mechanical, nuclear, heat, and light.

Chapter 4 Bloom’s Learning Objectives L1 L2 L2 L1 L2 L1 L4 L4 L2 L3 L4 viii

L3 L2 L2 L4 L4 L4 L5 L2 L1

Illustrate the quantum concept applied to matter and energy. Describe the Bohr model of the atom. Explain the relationship between energy levels in an atom and lines in an emission spectrum. Indicate each sublevel within a given energy level. Indicate the number of electrons that can occupy a given sublevel or energy level. Diagram the order of sublevels according to increasing energy. Diagram the predicted electron configurations for selected elements. Describe the relative size, shape, and energy of s and p orbitals. Specify the number of electrons that can occupy a given orbital.

(Sec. 4.7) (Sec. 4.8) (Sec. 4.8)

Exercises 45–50 Exercises 51–52 Exercises 53–66

(Sec. 4.9) (Sec. 4.9)

Exercises 67–70 Exercises 71–72

(Sec. 4.10) (Sec. 4.10) (Sec. 4.11) (Sec. 4.11)

Exercises 73–74 Exercises 75–78 Exercises 79–84 Exercises 85–86

(Sec. 5.1) (Sec. 5.2) (Sec. 5.3)

Key Terms Exercises 1–4 Exercises 5–8 Exercises 9–20

(Sec. 5.3)

Exercises 21–28

(Sec. 5.4)

Chapter 5 Bloom’s Learning Objectives L1 L2 L2 L4

Name the key term that corresponds to a definition. Describe the original periodic law proposed by Mendeleev. Describe the modern periodic law proposed by Moseley. Classify the elements according to their groups (families) and periods (series) in the periodic table. Designate a group of elements in the periodic table using both the American convention (IA–VIIIA) and the IUPAC convention (1–18). Explain the trend in metallic character within a group or period.

Explain the trend in atomic radius within a group or period.

(Sec. 5.4)

Predict a physical property for an element given the values of other elements in the same group. Predict a chemical formula for a compound given the formulas of other compounds containing an element in the same group. Predict the highest energy sublevel for an element given its position in the periodic table. Predict the electron configuration for an element given its position in the periodic table. Predict the number of valence electrons for any representative element. Diagram the electron dot formula for any representative element. Describe the general trends of ionization energy in the periodic table. Predict the group with the highest and the lowest ionization energy. Predict the element in a pair having a higher ionization energy. Predict an ionic charge for a representative element. Diagram the predicted electron configuration for selected ions.

(Sec. 5.5)

Exercises 29–30, 33–34 Exercises 31–32, 35–36 Exercises 37–40

(Sec. 5.5)

Exercises 41–46

(Sec. 5.6)

Exercises 47–54

(Sec. 5.6)

Exercises 55–56

(Sec. 5.7) (Sec. 5.8) (Sec. 5.9) (Sec. 5.9) (Sec. 5.9) (Sec. 5.10) (Sec. 5.10)

Exercises 57–60 Exercises 61–62 Exercises 63–64 Exercises 65–66 Exercises 67–70 Exercises 71–76 Exercises 77–80

(Sec. 6.1)

Key Terms Exercises 1–2

(Sec. 6.1) (Sec. 6.1)

Exercises 1–2 Exercises 3–4

(Sec. 6.2) (Sec. 6.2) (Sec. 6.3) (Sec. 6.4)

Exercises 5–8 Exercises 9–10 Exercises 11–14 Exercises 15–20

(Sec. 6.5) (Sec. 6.5) (Sec. 6.6)

Exercises 21–24 Exercises 25–30 Exercises 31–34

L5 L5 L5 L5 L4 L2 L5 L5 L5 L5

Chapter 6 Bloom’s Learning Objectives L1 L4 L4 L4 L2 L5 L2 L3 L5 L3 L5

Name the key term that corresponds to a definition. Classify a compound as a binary ionic compound, a ternary ionic compound, or a binary molecular compound. Classify an acid as a binary acid or a ternary oxyacid. Classify an ion as a monoatomic cation, a monoatomic anion, a polyatomic cation, or a polyatomic anion. Provide systematic names and formulas for common monoatomic ions. Predict the ionic charge for ions of representative elements. Provide systematic names and formulas for common polyatomic ions. Write the chemical formulas for ionic compounds composed of monoatomic ions and polyatomic ions. Determine the ionic charge on a cation in a binary ionic compound. Write systematic names and formulas for binary ionic compounds. Determine the ionic charge on a cation in a ternary ionic compound. ..

L3 L3 L3 L3 L3

Write systematic names and formulas for ternary ionic compounds. Write names and formulas for ionic compounds using the Latin system. Write systematic names and formulas for binary molecular compounds. Write systematic names and formulas for binary acids. Write systematic names and formulas for ternary oxyacids.

(Sec. 6.6) (Sec. 6.7) (Sec. 6.8) (Sec. 6.9) (Sec. 6.10)

Exercises 31–40 Exercises 41–48 Exercises 49–52 Exercises 53–54 Exercises 55–58

(Sec. 7.1) (Sec. 7.2)

Key Terms Exercises 1–6 Exercises 7–8

(Sec. 7.2) (Sec. 7.3) (Sec. 7.4)

Exercises 7–16 Exercises 17–20 Exercises 21–24

(Sec. 7.5)

Exercises 25–26

(Sec. 7.5)

Exercises 27–28

(Sec. 7.5)

Exercises 29–36

(Sec. 7.6)

Exercises 37, 38, and 41 Exercises 39, 40, and 42 Exercises 43–46

Chapter 7 Bloom’s Learning Objectives L1 L2 L2 L3 L3 L4 L6 L6 L6 L6 L6 L6 L5 L5 L5 L6 L6 L6 L5 L6 L6

Name the key term that corresponds to a definition. Describe four observations that are evidence of a chemical reaction. Identify seven elements that occur naturally as diatomic molecules: H2, N2, O2, F2, Cl2, Br2, I2. Write a chemical equation given the description of a chemical reaction. Convert a chemical reaction into a balanced chemical equation. Classify a chemical reaction as one of the following types: combination, decomposition, single-replacement, double-replacement, or neutralization Propose a balanced equation for the combination reaction of a metal and oxygen gas. Propose a balanced equation for the combination reaction of a nonmetal and oxygen gas. Propose a balanced equation for the combination reaction of a metal and a nonmetal. Propose a balanced equation for the decomposition reaction of a metal hydrogen carbonate. Propose a balanced equation for the decomposition reaction of a metal carbonate. Propose a balanced equation for the decomposition reaction of a compound that releases oxygen gas. Predict whether a metal reacts in a given aqueous salt solution. Predict whether a given metal reacts in an aqueous acid solution. Predict whether a given metal reacts in water at 25 °C. Propose a balanced equation for the single-replacement reaction of a given metal in an aqueous solution. Propose a balanced equation for the single-replacement reaction of a given metal in an aqueous acid. Propose a balanced equation for the single-replacement reaction of an active metal in water. Predict whether an ionic compound dissolves in water given the general rules for solubility. Propose a balanced equation for the double-replacement reaction of two salts in aqueous solution. Propose a balanced equation for the neutralization reaction of an acid and a base.

(Sec. 7.6) (Sec. 7.6) (Sec. 7.7) (Sec. 7.7) (Sec. 7.7) (Sec. 7.8)

(Sec. 7.9)

Exercises 47–48 Exercises 49–50 Exercises 51–52 Exercises 53–56, and 61–62 Exercises 57–58, and 63–64 Exercises 59–60, 65–66 Exercises 67–70

(Sec. 7.10)

Exercises 71–74

(Sec. 7.11)

Exercises 75–78

(Sec. 8.1) (Sec. 8.1)

Key Terms Exercises 1-4 Exercises 1-4

(Sec. 8.2) (Sec. 8.3) (Sec. 8.4) (Sec. 8.5) (Sec. 8.5)

Exercises 5-12 Exercises 13-16 Exercises 17-22 Exercises 23-24 Exercises 25-30

(Sec. 7.8) (Sec. 7.8)

Chapter 8 Bloom’s Learning Objectives L1 L1 L5 L3 L3 L4 L1 L3

Name the key term that corresponds to a definition. Memorize the value of Avogadro’s number: 6.02 x 1023. Determine the mass of Avogadro’s number of atoms for any element by referring to the periodic table. Calculate the moles of a substance given the number of particles. Calculate the molar mass of a substance given its chemical formula. Calculate the mass of a substance given the number of particles. Memorize the value for the molar volume of any gas at STP: 22.4 L/mol. Calculate the density of a gas at STP from its molar mass and molar volume.

L4 L3 L3 L3 L3

Calculate the volume of a gas at STP given its mass, or number of particles. (Sec. 8.6) Calculate the percent composition of a compound given its chemical formula. (Sec. 8.7) Calculate the empirical formula of a compound given its mass composition. (Sec. 8.8) Calculate the empirical formula of a compound given its percent composition. (Sec. 8.8) Calculate the molecular formula of a compound given its empirical formula (Sec. 8.9) and molar mass.

Exercises 31-38 Exercises 39-46 Exercises 47-52 Exercises 53-58 Exercises 59-68

Chapter 9 Bloom’s Learning Objectives L1 L3 L4 L3 L4 L3 L3 L3 L2 L5 L3 L3 L3

Name the key term that corresponds to a definition. Relate the coefficients in a balanced chemical equation to (a) moles of reactants and products, and (b) liters of gaseous reactants and products. Verify the coefficients of a balanced chemical equation using the conservation of mass law. Relate the number of moles of two substances in a balanced equation. Classify the three basic types of stoichiometry problems: mass–mass, mass–volume, and volume–volume. Solve mass–mass stoichiometry problems. Solve mass–volume stoichiometry problems. Solve volume–volume stoichiometry problems. Explain the concept of a limiting reactant. Predict the limiting reactant in a chemical reaction given the number of moles of each reactant. Solve mass–mass stoichiometry problems involving a limiting reactant. Solve volume–volume stoichiometry problems involving a gaseous limiting reactant. Calculate the percent yield for a reaction given the actual yield and theoretical yield.

(Sec. 9.1)

Key Terms Exercises 1–4

(Sec. 9.1)

Exercises 5–6

(Sec. 9.2) (Sec. 9.3)

Exercises 7–12 Exercises 13–18

(Sec. 9.4) (Sec. 9.5) (Sec. 9.6) (Sec. 9.7) (Sec. 9.7)

Exercises 19–28 Exercises 29–36 Exercises 37–48 Exercises 49–50 Exercises 51–58

(Sec. 9.8) (Sec. 9.8)

Exercises 59–66 Exercises 67–74

(Sec. 9.9)

Exercises 75–78

(Sec. 10.1) (Sec. 10.2)

Key Terms Exercises 1-2 Exercises 3-4

(Sec. 10.2) (Sec. 10.3) (Sec. 10.3)

Exercises 5-8 Exercises 9-10 Exercises 11-14

(Sec. 10.4) (Sec. 10.4) (Sec. 10.5) (Sec. 10.5) (Sec. 10.6) (Sec. 10.6) (Sec. 10.7)

Exercises 15-16 Exercises 17-20 Exercises 21-22 Exercises 23-26 Exercises 27-28 Exercises 29-32 Exercises 33-42

Chapter 10 Bloom’s Learning Objectives L1 L1 L2 L3 L2 L4 L3 L3 L3 L3 L3 L3 L3 L2 L3 L1 L5 L3

Name the key term that corresponds to a definition. List five observed properties of a gas. Memorize standard atmospheric pressure in the following units: atm, mm Hg, torr, cm Hg, in. Hg, psi, and kPa. Convert a given gas pressure to a different unit of measurement. Identify the three variables that affect the pressure of a gas. Indicate whether gas pressure increases or decreases for a given change in volume, temperature, or number of moles of gas. Illustrate a graph of the pressure–volume relationship for a gas. Calculate the pressure or volume of a gas after a change in conditions. Illustrate a graph of the volume–temperature relationship for a gas. Calculate the volume or temperature of a gas after a change in conditions. Illustrate a graph of the pressure–temperature relationship for a gas. Calculate the pressure or temperature of a gas after a change in conditions. Calculate the pressure, volume, or temperature of a gas after a change in conditions. Describe the relationship between vapor pressure and temperature. Apply Dalton’s law of partial pressures to a mixture of gases. List five characteristics of an ideal gas according to the kinetic theory of gases. Determine the value of absolute zero from a graph of volume or pressure versus temperature. Calculate the pressure, volume, temperature, or moles of gas from the ideal gas equation. ..

(Sec. 10.8) Exercises 43-46 (Sec. 10.9) Exercises 47-54 (Sec. 10.10) Exercises 55-56 (Sec. 10.10) Exercises 57-62 (Sec. 10.11) Exercises 63-66

Chapter 11 Bloom’s Learning Objectives L1 L2 L2 L2 L2 L5 L2 L2 L2 L3 L3 L2 L3 L3 L5

Name the key term that corresponds to a definition. Identify five observed properties of a liquid. (Sec. 11.1) Explain the concept of an intermolecular bond. (Sec. 11.2) Describe three types of attraction between molecules in a liquid: (Sec. 11.2) temporary dipoles, permanent dipoles, and hydrogen bonds. Describe the relationship between intermolecular attraction in a liquid and (Sec. 11.3) the properties of vapor pressure, boiling point, viscosity, and surface tension. Predict which liquid in a pair has the higher vapor pressure, (Sec. 11.3) boiling point, viscosity, and surface tension. Identify five observed properties of a solid. (Sec. 11.4) Describe three types of crystalline solids: ionic, molecular, and metallic. (Sec. 11.5) Explain the concepts of specific heat, heat of fusion, and heat of vaporization. (Sec. 11.6) Calculate heat changes that involve specific heat, heat of fusion, and (Sec. 11.6) heat of vaporization. Illustrate the bond angle and net dipole in a water molecule. (Sec. 11.7) Explain the unusual physical properties of water. (Sec. 11.8) Write chemical equations for the chemical reactions of water. (Sec. 11.9) Calculate the percentage of water in a hydrate. (Sec. 11.10) Determine the water of hydration for a hydrate. (Sec. 11.10)

Key Terms Exercises 1–4 Exercises 5–6 Exercises 5–6 Exercises 7–12 Exercises 13–22 Exercises 23–26 Exercises 27–32 Exercises 33–34 Exercises 35–44 Exercises 45–50 Exercises 51–60 Exercises 61–70 Exercises 71–76 Exercises 77–78

Chapter 12 Bloom’s Learning Objectives L1 L2 L5 L3 L3 L4 L4 L4 L4 L2 L3 L3 L2 L2 L2 L2 L5 L6

Name the key term that corresponds to a definition. Discuss the role of valence electrons in a chemical bond. Predict whether a bond is ionic or covalent given a formula unit or molecule. Illustrate the formation of an ionic bond between a metal atom and a nonmetal atom. Illustrate the formation of a covalent bond between two nonmetal atoms by sharing valence electrons. Diagram the electron dot formula for a molecule. Diagram the structural formula for a molecule. Diagram the electron dot formula for a polyatomic ion. Diagram the structural formula for a polyatomic ion. Describe the electronegativity trends in the periodic table. Calculate the electronegativity difference in a polar covalent bond. Apply delta notation (δ+ and δ–) to a polar bond. Describe and identify a nonpolar covalent bond. Identify seven elements that occur naturally as diatomic molecules: H2, N2, O2, F2, Cl2, Br2, I2. Describe and identify a coordinate covalent bond. Describe hydrogen bond attraction between two molecules. Determine the shape of a molecule by applying VSEPR theory. Propose how a molecule with polar bonds can be nonpolar.

(Sec. 12.1) (Sec. 12.1) (Sec. 12.2)

Key Terms Exercises 1–4 Exercises 5–10 Exercises 11–28

(Sec. 12.3)

Exercises 29–32

(Sec. 12.4) (Sec. 12.4) (Sec. 12.5) (Sec. 12.5) (Sec. 12.6) (Sec. 12.6) (Sec. 12.6) (Sec. 12.7) (Sec. 12.7)

Exercises 33–38 Exercises 33–38 Exercises 39–44 Exercises 39–44 Exercises 45–50 Exercises 51–52 Exercises 53–54 Exercises 55–56 Exercises 57–58

(Sec. 12.8) (Sec. 12.9) (Sec. 12.10) (Sec. 12.10)

Exercises 59–66 Exercises 67–70 Exercises 71–76 Exercises 77–78

(Sec. 13.1)

Key Terms Exercises 1-2

(Sec. 13.1) (Sec. 13.2)

Exercises 3-6 Exercises 7-14

(Sec. 13.3)

Exercises 15-20

Chapter 13 Bloom’s Learning Objectives L1 L2 L2 L5 L5

xii

L3 L2 L4 L4 L4 L3 L3 L3 L3 L3 L3 L3 L3

Illustrate how an ionic compound and a molecular compound dissolve in water. Describe the effect of temperature, stirring, and particle size on the rate of dissolving for a solid compound in water. Interpret a graph that shows temperature versus solubility of a solid compound in water. Interpret a graph of temperature versus solubility and determine whether a solution is saturated, unsaturated, or supersaturated. Distinguish among solutions that are saturated, unsaturated, or supersaturated. Calculate the mass percent concentration of a solution. Write three pairs of unit factors given the mass percent concentration of a solution. Solve problems that involve a mass of solute, mass of solvent, and the mass percent concentration of a solution. Calculate the molar concentration of a solution. Write a pair of unit factors given the molar concentration of a solution. Solve problems that involve a mass of solute, volume of solution, and the molar concentration of a solution. Solve problems that involve a solution undergoing dilution. Solve problems that involve a balanced chemical equation and the molar concentration of a solution.

(Sec. 13.4)

Exercises 21-24

(Sec. 13.5)

Exercises 25-26

(Sec. 13.6)

Exercises 27-36

(Sec. 13.7)

Exercises 37-40

(Sec. 13.7)

Exercises 41-42

(Sec. 13.8) (Sec. 13.8)

Exercises 43-44 Exercises 45-46

(Sec. 13.8)

Exercises 47-52

(Sec. 13.9) (Sec. 13.9) (Sec. 13.9)

Exercises 53-54 Exercises 55-56 Exercises 57-64

(Sec. 13.10) Exercises 65-68 (Sec. 13.11) Exercises 69-70

Chapter 14 Bloom’s Learning Objectives L1 L1 L4 L2 L4 L4 L2 L2 L3 L3 L2 L3 L3 L3 L2 L3 L2 L3 L2 L3

Name the key term that corresponds to a definition. List the general properties of acids and bases. Classify a solution of given pH as one of the following: strongly acidic, weakly acidic, neutral, weakly basic, or strongly basic. Identify an Arrhenius acid and base. Classify a strong or weak acid and base given the degree of ionization. Indicate the acid and base that react to produce a given salt. Identify a Brønsted–Lowry acid and base in a neutralization reaction. Describe the color of a solution with a given pH and a drop of phenolphthalein, methyl red, or bromthymol blue indicator. Solve problems given acid–base titration data. Calculate the mass percent concentration of a solution given the molarity. Describe a standard solution of acid or base. Solve problems that involve standardization of an acid or base. Show the ionization constant equation for water, Kw. Calculate the molar hydroxide ion concentration given the molar hydrogen ion concentration. Describe the relationship between pH and molar hydrogen ion concentration. Calculate pH values and molar hydrogen ion concentrations. Describe the electrical conductivity of strong and weak electrolytes. Show strong and weak electrolytes as ionized or nonionized. Describe the procedure for writing a net ionic equation. Write net ionic equations for given chemical reactions.

(Sec. 14.1) (Sec. 14.1)

Key Terms Exercises 1–2 Exercises 3–4

(Sec. 14.2) (Sec. 14.2) (Sec. 14.2) (Sec. 14.3) (Sec. 14.4)

Exercises 5–8 Exercises 9–10 Exercises 11–14 Exercises 15–20 Exercises 21–26

(Sec. 14.5) (Sec. 14.5) (Sec. 14.6) (Sec. 14.6) (Sec. 14.7) (Sec. 14.7)

Exercises 27–32 Exercises 33–34 Exercises 35–36 Exercises 35–42 Exercises 43–44 Exercises 45–48

(Sec. 14.8) (Sec. 14.9) (Sec. 14.10) (Sec. 14.10) (Sec. 14.11) (Sec. 14.11)

Exercises 49–52 Exercises 53–60 Exercises 61–66 Exercises 67–70 Exercises 71–72 Exercises 73–76

(Sec. 15.1)

Key Terms Exercises 1–8

(Sec. 15.1)

Exercises 9–10

Chapter 15 Bloom’s Learning Objectives L1 L2 L5

xiii

L3 L3

Solve problems that relate moles of substance in the following: (a) number of atoms, molecules, or formula units (b) mass of substance (c) volume of gas Solve problems that relate amounts of substance in the following: (a) mass–mass stoichiometry (b) mass–volume stoichiometry (c) volume–volume stoichiometry (d) solution stoichiometry Solve stoichiometry problems that involve two or more chemical reactions. Solve problems that involve two or more chemical principles.

(Sec. 15.2)

Exercises 11–22

(Sec. 15.3)

Exercises 23–28

(Sec. 15.4) (Sec. 15.5)

Exercises 29–34 Exercises 35–46

(Sec. 16.1)

Key Terms Exercises 1–4

(Sec. 16.1)

Exercises 5–8

(Sec. 16.2)

Exercises 9–10

(Sec. 16.2)

Exercises 11–16

(Sec. 16.3) (Sec. 16.3) (Sec. 16.4) (Sec. 16.4) (Sec. 16.5) (Sec. 16.6) (Sec. 16.6) (Sec. 16.7) (Sec. 16.8)

Exercises 17–18 Exercises 19–20 Exercises 21–26 Exercises 27–28 Exercises 29–34 Exercises 35–36 Exercises 37–40 Exercises 41–44 Exercises 45–46

(Sec. 16.8) (Sec. 16.9)

Exercises 47–52 Exercises 53–56

(Sec. 17.1) (Sec. 17.1)

Key Terms Exercises 1–8 Exercises 1–8

(Sec. 17.2) (Sec. 17.2) (Sec. 17.3)

Exercises 9–14 Exercises 15–16 Exercises 17–22

(Sec. 17.4)

Exercises 23–28

(Sec. 17.5)

Exercises 29–32

Chapter 16 Bloom’s Learning Objectives L1 L2 L2 L3 L3 L2 L2 L3 L3 L3 L3 L3 L3 L3 L3 L3

Name the key term that corresponds to a definition. Explain the effect of collision frequency, collision energy, and orientation of molecules on the rate of a chemical reaction. Explain the effect of concentration, temperature, and a catalyst on the rate of a chemical reaction. Diagram the general energy profile for an endothermic and an exothermic reaction. Label the transition state, energy of activation, and heat of reaction on a given energy profile. Describe the equilibrium concept for a reversible reaction. Express the law of chemical equilibrium as an equation. Write the equilibrium constant expression for a reversible reaction. Calculate an equilibrium constant, Keq, from experimental data. Apply Le Chatelier’s principle to reversible reactions in the gaseous state. Write the equilibrium constant expression for a weak acid or a weak base. Calculate an ionization constant, Ki, from experimental data. Apply Le Chatelier’s principle to solutions of weak acids and weak bases. Write the equilibrium constant expression for a slightly soluble ionic compound. Calculate a solubility product constant, Ksp, from experimental data. Apply Le Chatelier’s principle to a saturated solution of a slightly soluble ionic compound.

Chapter 17 Bloom’s Learning Objectives L1 L2 L3

L2 L2 L3 L3

xiv

L5 L2 L2 L2 L2

Predict whether a redox reaction is spontaneous or nonspontaneous given a list of reduction potentials. Identify the anode and cathode in a given voltaic cell. Identify the oxidation and reduction half-reactions in a given spontaneous electrochemical cell. Identify the anode and cathode in a given electrolytic cell. Identify the oxidation and reduction half-reactions in a given nonspontaneous electrochemical cell.

(Sec. 17.5)

Exercises 33–36

(Sec. 17.6) (Sec. 17.6)

Exercises 37–40 Exercises 37–40

(Sec. 17.7) (Sec. 17.7)

Exercises 41–44 Exercises 41–44

(Sec. 18.1) (Sec. 18.2)

Key Terms Exercises 1–6 Exercises 7–8

(Sec. 18.2) (Sec. 18.3)

Exercises 9–12 Exercises 13–14

(Sec. 18.3)

Exercises 15–16

(Sec. 18.3)

Exercises 17–18

(Sec. 18.4)

Exercises 19–24

(Sec. 18.5) (Sec. 18.6) (Sec. 18.7) (Sec. 18.7) (Sec. 18.8) (Sec. 18.8)

Exercises 25–30 Exercises 31–38

Chapter 18 Bloom’s Learning Objectives L1 L2 L3 L3 L2 L2 L2 L3 L2 L3 L2 L3 L2 L3

Name the key term that corresponds to a definition. Describe the properties of alpha, beta, and gamma radiation. Illustrate the following types of radiation using atomic notation: alpha, beta, gamma, positron, neutron, and proton. Write balanced nuclear equations involving natural radioactivity. Identify products in the decay series for uranium-238 given the radiation emitted. Identify products in the decay series for uranium-235 given the radiation emitted. Identify products in the decay series for thorium-232 and neptunium-237 given the radiation emitted. Calculate the amount of radioactive sample, given the initial amount, elapsed time, and half-life of the radionuclide. Describe several applications of selected radionuclides. Write balanced nuclear equations involving induced radioactivity. Describe the process in a nuclear fission reaction. Write a nuclear equation for a given fission reaction. Describe the process in a nuclear fusion reaction. Write a nuclear equation for a given fusion reaction.

Exercises 39–46 Exercises 47–52

Chapter 19 Bloom’s Learning Objectives L1 L4 L4 L4 L3 L3 L3 L3 L3 L3 L2 L2 L2 L3 L3 L3 L3 L3 L3 L3 L3

Name the key term that corresponds to a definition. Classify a hydrocarbon having single, double, or triple bonds. Classify a hydrocarbon as saturated, unsaturated, or aromatic. Classify a hydrocarbon as an alkane, alkene, alkyne, or arene. Write names and formulas for simple alkanes. Write combustion reactions for alkanes. Write names and formulas for simple alkenes. Write names and formulas for simple alkynes. Write hydrogenation reactions for alkenes and alkynes. Write names and formulas for simple arenes. Describe the concept of an organic functional group. Identify the functional group in each of the following: organic halide, alcohol, phenol, ether, and amine. Identify the functional group in each of the following: aldehyde, ketone, carboxylic acid, ester, and amide. Write names and formulas for simple organic halides Write names and formulas for simple alcohols. Write names and formulas for simple phenols. Write names and formulas for simple ethers. Write names and formulas for simple amines. Write names and formulas for simple aldehydes. Write names and formulas for simple ketones. Write names and formulas for simple carboxylic acids. ..

(Sec. 19.1) (Sec. 19.1) (Sec. 19.1) (Sec. 19.2) (Sec. 19.2) (Sec. 19.3) (Sec. 19.3) (Sec. 19.3) (Sec. 19.4) (Sec. 19.5) (Sec. 19.5)

Key Terms Exercises 1–6 Exercises 1–6 Exercises 1–6 Exercises 7–18 Exercises 19–20 Exercises 21–24 Exercises 25–34 Exercises 35–36 Exercises 37–40 Exercises 41–44 Exercises 41–44

(Sec. 19.5)

Exercises 41–44

(Sec. 19.6) (Sec. 19.7) (Sec. 19.7) (Sec. 19.7) (Sec. 19.8) (Sec. 19.9) (Sec. 19.9) (Sec. 19.10)

Exercises 45–48 Exercises 49–58 Exercises 51–58 Exercises 51–58 Exercises 59–62 Exercises 63–64 Exercises 65–68 Exercises 69–70 xv

L3 L2 L3

Write names and formulas for simple esters. Identify the parent alcohol and acid in an ester. Write names and formulas for simple amides.

(Sec. 19.10) Exercises 71–72 (Sec. 19.10) Exercises 73–74 (Sec. 19.10) Exercises 75–76

Chapter 20 Bloom’s Learning Objectives L1 L2 L2 L2 L2 L2 L2 L2 L2 L4 L4 L2 L2 L2 L2 L2 L2 L2 L4 L2 L2

xvi

Name the key term that corresponds to a definition. Recognize a protein given its structural formula. Recognize a carbohydrate given its structural formula. Recognize a lipid given its structural formula. Recognize a nucleic acid given its structural formula. Describe the primary structure of a protein. Describe the secondary structure of a protein. Describe the tertiary structure of a protein. Identify amino acids and a peptide linkage. Analyze the lock-and-key model for an enzyme. Analyze the action of an enzyme as a catalyst. Recognize the structure of mono-, di-, and polysaccharides. Identify a glycoside linkage between monosaccharides. Recognize the structure of a triglyceride. Recognize the structure of a phospholipid. Recognize the structure of a lipid wax. Recognize the structure of a steroid. Describe the three components of a nucleotide. Distinguish between DNA and RNA nucleotides. Explain the process of replication. Explain the process of transcription.

(Sec. 20.1) (Sec. 20.1) (Sec. 20.1) (Sec. 20.1) (Sec. 20.2) (Sec. 20.2) (Sec. 20.2) (Sec. 20.2) (Sec. 20.3) (Sec. 20.3) (Sec. 20.4) (Sec. 20.4) (Sec. 20.5) (Sec. 20.5) (Sec. 20.5) (Sec. 20.5) (Sec. 20.6) (Sec. 20.6) (Sec. 20.6) (Sec. 20.6)

Key Terms Exercises 1, 3 Exercises 1, 4 Exercises 2, 5 Exercises 2, 6 Exercises 7 Exercises 8 Exercises 9–10 Exercises 11–20 Exercises 21–24 Exercises 25–28 Exercises 29–38 Exercises 39–42 Exercises 43–48 Exercises 49–50 Exercises 51–52 Exercises 53–54 Exercises 55–56 Exercises 57–64 Exercises 65–66 Exercises 67–68

Learning Objectives and Associated Exercises

Chapter 1 Bloom’s Learning Objectives L1 L2 L2 L2 L5

(Sec. 1.1) (Sec. 1.1) (Sec. 1.2) (Sec. 1.3)

Key Terms Exercises 1–4 Exercises 5–12 Exercises 13–16 Exercises 17–18

(PSS.1) (PSS.1) (PSS.2) (PSS.3) (PSS.4) (PSS.5) (PSS.6) (PSS.6) (PSS.7) (PSS.7)

Key Terms Exercises 1–2 Exercises 3–10 Exercises 11–16 Exercises 17–20 Exercises 21–24 Exercises 25–28 Exercises 29–32 Exercises 33–38 Exercises 39–40 Exercises 41–44

Prerequisite Science Skills Bloom’s Learning Objectives L1 L2 L2 L2 L3 L3 L3 L2 L2 L3 L3

vii

Chapter 2 Bloom’s Learning Objectives L1 L1 L1 L3 L3 L2 L1 L2 L2 L3 L3 L3 L5 L3 L1 L3 L1 L3 L3 L3 L3

Name the key term that corresponds to a definition. List the base units and symbols of the metric system. List the prefixes for multiples and fractions of base units. Show the unit equation for a base metric unit and a prefix unit. Show the two unit factors derived from a metric unit equation. Express a given metric measurement using a different metric prefix. Memorize the metric equivalent for inch, pound, quart, and second. Express a given measurement in metric units or English units. Express a given quantity in a sample as a percent-by-mass. Apply percent as a unit factor. Solve problems that relate length, width, thickness, and volume of a rectangular solid. Express a given volume in milliliters, cubic centimeters, or cubic inches. Estimate the volume of a solid by the displacement of water. Apply the concept of density. Memorize the value for the density of water in grams per milliliter. Solve problems that relate density, mass, and volume. Memorize the freezing point and boiling point of water on the Fahrenheit, Celsius, and Kelvin scales. Express a given temperature in degrees Fahrenheit (°F), degrees Celsius (°C), and Kelvin (K). Apply the concept of specific heat to good and poor conductors of heat. Express heat energy in units of calories, kilocalories, joules, and kilojoules. Solve problems that relate specific heat, mass, and temperature change.

(Sec. 2.1) (Sec. 2.1) (Sec. 2.2) (Sec. 2.2) (Sec. 2.3) (Sec. 2.4) (Sec. 2.4) (Sec. 2.5) (Sec. 2.5) (Sec. 2.6)

Key Terms Exercises 1–6 Exercises 7–10 Exercises 11–12 Exercises 13–14 Exercises 15–18 Exercises 19–20 Exercises 21–28 Exercises 29–32 Exercises 33–34 Exercises 35–38

(Sec. 2.6) (Sec. 2.7) (Sec. 2.8) (Sec. 2.8) (Sec. 2.8) (Sec. 2.9)

Exercises 39–42 Exercises 43–46 Exercises 47–50 Exercises 51–52 Exercises 53–58 Exercises 59–60

(Sec. 2.9)

Exercises 61–66

(Sec. 2.10) (Sec. 2.10) (Sec. 2.10)

Exercise 67–68 Exercises 69–70 Exercises 71–76

(Sec. 3.1)

Key Terms Exercises 1–6

(Sec. 3.1)

Exercises 7–12

(Sec. 3.2) (Sec. 3.3) (Sec. 3.4) (Sec. 3.4)

Exercises 13–16 Exercises 17–22 Exercises 23–26 Exercises 27–30

(Sec. 3.4)

Exercises 31–34

(Sec. 3.5)

Exercises 35–40

(Sec. 3.5) (Sec. 3.6) (Sec. 3.7) (Sec. 3.8) (Sec. 3.9) (Sec. 3.9) (Sec. 3.9) (Sec. 3.10) (Sec. 3.10)

Exercises 41–42 Exercises 43–50 Exercises 51–58 Exercises 59–62 Exercises 63–64 Exercises 65–66 Exercises 67–68 Exercises 69–74 Exercises 75–78

Chapter 3 Bloom’s Learning Objectives L1 L2 L2 L4 L1 L1 L5 L5 L3 L3 L4 L4 L3 L4 L2 L2 L3 L2

viii

Chapter 4 Bloom’s Learning Objectives L1 L2 L2 L1 L2 L1 L4 L4 L2 L3 L4 L3 L2 L2 L4 L4 L4 L5 L2 L1

Name the key term that corresponds to a definition. Describe the Dalton model of the atom. Describe the Thomson raisin pudding model of the atom. Specify the relative charge on an electron and a proton. Describe the Rutherford nuclear model of the atom. Specify the relative charge and approximate mass of an electron, proton, and neutron. Analyze atomic notation and indicate protons, neutrons, and electrons. Analyze a given isotope and indicate the number of neutrons. Explain the concept of relative atomic mass. Calculate the atomic mass of an element given the mass and abundance of the naturally occurring isotopes. Analyze the relationship of wavelength, frequency, and energy of light. Illustrate the quantum concept applied to matter and energy. Describe the Bohr model of the atom. Explain the relationship between energy levels in an atom and lines in an emission spectrum. Indicate each sublevel within a given energy level. Indicate the number of electrons that can occupy a given sublevel or energy level. Diagram the order of sublevels according to increasing energy. Diagram the predicted electron configurations for selected elements. Describe the relative size, shape, and energy of s and p orbitals. Specify the number of electrons that can occupy a given orbital.

(Sec. 4.1) (Sec. 4.2) (Sec. 4.2) (Sec. 4.3) (Sec. 4.3)

Key Terms Exercises 1–4 Exercises 5–6 Exercises 7–10 Exercises 11–12 Exercises 13–16

(Sec. 4.4) (Sec. 4.4) (Sec. 4.5) (Sec. 4.5)

Exercises 17–18 Exercises 19–24 Exercises 25–32 Exercises 33–36

(Sec. 4.6) (Sec. 4.7) (Sec. 4.8) (Sec. 4.8)

Exercises 37–44 Exercises 45–50 Exercises 51–52 Exercises 53–66

(Sec. 4.9) (Sec. 4.9)

Exercises 67–70 Exercises 71–72

(Sec. 4.10) (Sec. 4.10) (Sec. 4.11) (Sec. 4.11)

Exercises 73–74 Exercises 75–78 Exercises 79–84 Exercises 85–86

(Sec. 5.1) (Sec. 5.2) (Sec. 5.3)

Key Terms Exercises 1–4 Exercises 5–8 Exercises 9–20

(Sec. 5.3)

Exercises 21–28

(Sec. 5.4)

Chapter 5 Bloom’s Learning Objectives L1 L2 L2 L4

Explain the trend in atomic radius within a group or period.

(Sec. 5.4)

(Sec. 5.5)

Exercises 29–30, 33–34 Exercises 31–32, 35–36 Exercises 37–40

(Sec. 5.5)

Exercises 41–46

(Sec. 5.6)

Exercises 47–54

(Sec. 5.6)

Exercises 55–56

(Sec. 5.7) (Sec. 5.8) (Sec. 5.9) (Sec. 5.9) (Sec. 5.9) (Sec. 5.10) (Sec. 5.10)

Exercises 57–60 Exercises 61–62 Exercises 63–64 Exercises 65–66 Exercises 67–70 Exercises 71–76 Exercises 77–80

L5 L5 L5 L5 L4 L2 L5 L5 L5 L5

Chapter 6 Bloom’s Learning Objectives L1 L4 L4 L4 L2 L5 L2 L3 L5 L3 L5 L3 L3 L3 L3 L3

(Sec. 6.1)

Key Terms Exercises 1–2

(Sec. 6.1) (Sec. 6.1)

Exercises 1–2 Exercises 3–4

(Sec. 6.2) (Sec. 6.2) (Sec. 6.3) (Sec. 6.4)

Exercises 5–8 Exercises 9–10 Exercises 11–14 Exercises 15–20

(Sec. 6.5) (Sec. 6.5) (Sec. 6.6) (Sec. 6.6) (Sec. 6.7) (Sec. 6.8) (Sec. 6.9) (Sec. 6.10)

Exercises 21–24 Exercises 25–30 Exercises 31–34 Exercises 31–40 Exercises 41–48 Exercises 49–52 Exercises 53–54 Exercises 55–58

(Sec. 7.1) (Sec. 7.2)

Key Terms Exercises 1–6 Exercises 7–8

(Sec. 7.2) (Sec. 7.3) (Sec. 7.4)

Exercises 7–16 Exercises 17–20 Exercises 21–24

(Sec. 7.5)

Exercises 25–26

(Sec. 7.5)

Exercises 27–28

(Sec. 7.5)

Exercises 29–36

(Sec. 7.6)

Exercises 37, 38, and 41 Exercises 39, 40, and 42 Exercises 43–46

Chapter 7 Bloom’s Learning Objectives L1 L2 L2 L3 L3 L4 L6 L6 L6 L6 L6 L6 L5 L5 L5 L6 L6 L6 L5

(Sec. 7.6) (Sec. 7.6) (Sec. 7.7) (Sec. 7.7) (Sec. 7.7) (Sec. 7.8) (Sec. 7.8) (Sec. 7.8) (Sec. 7.9)

Exercises 47–48 Exercises 49–50 Exercises 51–52 Exercises 53–56, and 61–62 Exercises 57–58, and 63–64 Exercises 59–60, 65–66 Exercises 67–70

L6 L6

Propose a balanced equation for the double-replacement reaction of two salts in aqueous solution. Propose a balanced equation for the neutralization reaction of an acid and a base.

(Sec. 7.10)

Exercises 71–74

(Sec. 7.11)

Exercises 75–78

Chapter 8 Bloom’s Learning Objectives L1 L1 L5 L3 L3 L4 L1 L3 L4 L3 L3 L3 L3

Name the key term that corresponds to a definition. Memorize the value of Avogadro’s number: 6.02 x 1023. (Sec. 8.1) Determine the mass of Avogadro’s number of atoms for any element by (Sec. 8.1) referring to the periodic table. Calculate the moles of a substance given the number of particles. (Sec. 8.2) Calculate the molar mass of a substance given its chemical formula. (Sec. 8.3) Calculate the mass of a substance given the number of particles. (Sec. 8.4) Memorize the value for the molar volume of any gas at STP: 22.4 L/mol. (Sec. 8.5) Calculate the density of a gas at STP from its molar mass and molar volume. (Sec. 8.5) Calculate the volume of a gas at STP given its mass, or number of particles. (Sec. 8.6) Calculate the percent composition of a compound given its chemical formula. (Sec. 8.7) Calculate the empirical formula of a compound given its mass composition. (Sec. 8.8) Calculate the empirical formula of a compound given its percent composition. (Sec. 8.8) Calculate the molecular formula of a compound given its empirical formula (Sec. 8.9) and molar mass.

Key Terms Exercises 1-4 Exercises 1-4 Exercises 5-12 Exercises 13-16 Exercises 17-22 Exercises 23-24 Exercises 25-30 Exercises 31-38 Exercises 39-46 Exercises 47-52 Exercises 53-58 Exercises 59-68

Chapter 9 Bloom’s Learning Objectives L1 L3 L4 L3 L4 L3 L3 L3 L2 L5 L3 L3 L3

(Sec. 9.1)

Key Terms Exercises 1–4

(Sec. 9.1)

Exercises 5–6

(Sec. 9.2) (Sec. 9.3)

Exercises 7–12 Exercises 13–18

(Sec. 9.4) (Sec. 9.5) (Sec. 9.6) (Sec. 9.7) (Sec. 9.7)

Exercises 19–28 Exercises 29–36 Exercises 37–48 Exercises 49–50 Exercises 51–58

(Sec. 9.8) (Sec. 9.8)

Exercises 59–66 Exercises 67–74

(Sec. 9.9)

Exercises 75–78

(Sec. 10.1) (Sec. 10.2)

Key Terms Exercises 1-2 Exercises 3-4

(Sec. 10.2) (Sec. 10.3) (Sec. 10.3)

Exercises 5-8 Exercises 9-10 Exercises 11-14

(Sec. 10.4)

Exercises 15-16

Chapter 10 Bloom’s Learning Objectives L1 L1 L2 L3 L2 L4 L3

L3 L3 L3 L3 L3 L3 L2 L3 L1 L5 L3

Calculate the pressure or volume of a gas after a change in conditions. Illustrate a graph of the volume–temperature relationship for a gas. Calculate the volume or temperature of a gas after a change in conditions. Illustrate a graph of the pressure–temperature relationship for a gas. Calculate the pressure or temperature of a gas after a change in conditions. Calculate the pressure, volume, or temperature of a gas after a change in conditions. Describe the relationship between vapor pressure and temperature. Apply Dalton’s law of partial pressures to a mixture of gases. List five characteristics of an ideal gas according to the kinetic theory of gases. Determine the value of absolute zero from a graph of volume or pressure versus temperature. Calculate the pressure, volume, temperature, or moles of gas from the ideal gas equation.

(Sec. 10.4) (Sec. 10.5) (Sec. 10.5) (Sec. 10.6) (Sec. 10.6) (Sec. 10.7)

Exercises 17-20 Exercises 21-22 Exercises 23-26 Exercises 27-28 Exercises 29-32 Exercises 33-42

(Sec. 10.8) Exercises 43-46 (Sec. 10.9) Exercises 47-54 (Sec. 10.10) Exercises 55-56 (Sec. 10.10) Exercises 57-62 (Sec. 10.11) Exercises 63-66

Chapter 11 Bloom’s Learning Objectives L1 L2 L2 L2 L2 L5 L2 L2 L2 L3 L3 L2 L3 L3 L5

Chapter 12 Bloom’s Learning Objectives L1 L2 L5 L3 L3 L4 L4 L4 L4 L2 L3 L3 L2

xii

(Sec. 12.1) (Sec. 12.1) (Sec. 12.2)

Key Terms Exercises 1–4 Exercises 5–10 Exercises 11–28

(Sec. 12.3)

Exercises 29–32

(Sec. 12.4) (Sec. 12.4) (Sec. 12.5) (Sec. 12.5) (Sec. 12.6) (Sec. 12.6) (Sec. 12.6) (Sec. 12.7)

Exercises 33–38 Exercises 33–38 Exercises 39–44 Exercises 39–44 Exercises 45–50 Exercises 51–52 Exercises 53–54 Exercises 55–56

L2 L2 L2 L5 L6

Identify seven elements that occur naturally as diatomic molecules: H2, N2, O2, F2, Cl2, Br2, I2. Describe and identify a coordinate covalent bond. Describe hydrogen bond attraction between two molecules. Determine the shape of a molecule by applying VSEPR theory. Propose how a molecule with polar bonds can be nonpolar.

(Sec. 12.7)

Exercises 57–58

(Sec. 12.8) (Sec. 12.9) (Sec. 12.10) (Sec. 12.10)

Exercises 59–66 Exercises 67–70 Exercises 71–76 Exercises 77–78

(Sec. 13.1)

Key Terms Exercises 1-2

(Sec. 13.1) (Sec. 13.2)

Exercises 3-6 Exercises 7-14

(Sec. 13.3)

Exercises 15-20

(Sec. 13.4)

Exercises 21-24

(Sec. 13.5)

Exercises 25-26

(Sec. 13.6)

Exercises 27-36

(Sec. 13.7)

Exercises 37-40

(Sec. 13.7)

Exercises 41-42

(Sec. 13.8) (Sec. 13.8)

Exercises 43-44 Exercises 45-46

(Sec. 13.8)

Exercises 47-52

(Sec. 13.9) (Sec. 13.9) (Sec. 13.9)

Exercises 53-54 Exercises 55-56 Exercises 57-64

Chapter 13 Bloom’s Learning Objectives L1 L2 L2 L5 L5 L3 L2 L4 L4 L4 L3 L3 L3 L3 L3 L3 L3 L3

Name the key term that corresponds to a definition. Describe the effect of temperature and pressure on the solubility of a gas in a liquid. Apply Henry’s law to the solubility of a gas in a liquid. Determine whether a liquid is miscible or immiscible in another liquid by applying the like dissolves like rule. Determine whether a solid is soluble or insoluble in a liquid by applying the like dissolves like rule. Illustrate how an ionic compound and a molecular compound dissolve in water. Describe the effect of temperature, stirring, and particle size on the rate of dissolving for a solid compound in water. Interpret a graph that shows temperature versus solubility of a solid compound in water. Interpret a graph of temperature versus solubility and determine whether a solution is saturated, unsaturated, or supersaturated. Distinguish among solutions that are saturated, unsaturated, or supersaturated. Calculate the mass percent concentration of a solution. Write three pairs of unit factors given the mass percent concentration of a solution. Solve problems that involve a mass of solute, mass of solvent, and the mass percent concentration of a solution. Calculate the molar concentration of a solution. Write a pair of unit factors given the molar concentration of a solution. Solve problems that involve a mass of solute, volume of solution, and the molar concentration of a solution. Solve problems that involve a solution undergoing dilution. Solve problems that involve a balanced chemical equation and the molar concentration of a solution.

(Sec. 13.10) Exercises 65-68 (Sec. 13.11) Exercises 69-70

Chapter 14 Bloom’s Learning Objectives L1 L1 L4 L2 L4 L4 L2 L2 L3 L3 L2

(Sec. 14.1) (Sec. 14.1)

Key Terms Exercises 1–2 Exercises 3–4

(Sec. 14.2) (Sec. 14.2) (Sec. 14.2) (Sec. 14.3) (Sec. 14.4)

Exercises 5–8 Exercises 9–10 Exercises 11–14 Exercises 15–20 Exercises 21–26

(Sec. 14.5) (Sec. 14.5) (Sec. 14.6)

Exercises 27–32 Exercises 33–34 Exercises 35–36 xiii

L3 L3 L3 L2 L3 L2 L3 L2 L3

Solve problems that involve standardization of an acid or base. Show the ionization constant equation for water, Kw. Calculate the molar hydroxide ion concentration given the molar hydrogen ion concentration. Describe the relationship between pH and molar hydrogen ion concentration. Calculate pH values and molar hydrogen ion concentrations. Describe the electrical conductivity of strong and weak electrolytes. Show strong and weak electrolytes as ionized or nonionized. Describe the procedure for writing a net ionic equation. Write net ionic equations for given chemical reactions.

(Sec. 14.6) (Sec. 14.7) (Sec. 14.7)

Exercises 35–42 Exercises 43–44 Exercises 45–48

(Sec. 14.8) Exercises 49–52 (Sec. 14.9) Exercises 53–60 (Sec. 14.10) Exercises 61–66 (Sec. 14.10) Exercises 67–70 (Sec. 14.11) Exercises 71–72 (Sec. 14.11) Exercises 73–76

Chapter 15 Bloom’s Learning Objectives L1 L2 L5 L3

L3 L3

Name the key term that corresponds to a definition. Describe several techniques for solving chemical problems, including: unit analysis, algebraic analysis, concept maps, and visualization. Estimate a “ballpark” answer before using a calculator to obtain a precise answer. Solve problems that relate moles of substance in the following: (a) number of atoms, molecules, or formula units (b) mass of substance (c) volume of gas Solve problems that relate amounts of substance in the following: (a) mass–mass stoichiometry (b) mass–volume stoichiometry (c) volume–volume stoichiometry (d) solution stoichiometry Solve stoichiometry problems that involve two or more chemical reactions. Solve problems that involve two or more chemical principles.

(Sec. 15.1)

Key Terms Exercises 1–8

(Sec. 15.1)

Exercises 9–10

(Sec. 15.2)

Exercises 11–22

(Sec. 15.3)

Exercises 23–28

(Sec. 15.4) (Sec. 15.5)

Exercises 29–34 Exercises 35–46

(Sec. 16.1)

Key Terms Exercises 1–4

(Sec. 16.1)

Exercises 5–8

(Sec. 16.2)

Exercises 9–10

(Sec. 16.2)

Exercises 11–16

(Sec. 16.3) (Sec. 16.3) (Sec. 16.4) (Sec. 16.4) (Sec. 16.5) (Sec. 16.6) (Sec. 16.6) (Sec. 16.7) (Sec. 16.8)

Exercises 17–18 Exercises 19–20 Exercises 21–26 Exercises 27–28 Exercises 29–34 Exercises 35–36 Exercises 37–40 Exercises 41–44 Exercises 45–46

(Sec. 16.8) (Sec. 16.9)

Exercises 47–52 Exercises 53–56

Chapter 16 Bloom’s Learning Objectives L1 L2 L2 L3 L3 L2 L2 L3 L3 L3 L3 L3 L3 L3 L3 L3

xiv

Chapter 17 Bloom’s Learning Objectives L1 L2 L3

L2 L2 L3 L3

L5 L5 L2 L2 L2 L2

Name the key term that corresponds to a definition. Explain the concept of an oxidation number. Solve for the oxidation number of an element in each of the following: (a) metals and nonmetals (b) monoatomic and polyatomic ions (c) ionic and molecular compounds. Identify the oxidized and reduced substances in a given redox reaction. Identify the oxidizing and reducing agents in a given redox reaction. Write a balanced chemical equation for a redox reaction using the oxidation number method. Write a balanced chemical equation for a redox reaction using the half-reaction method: (a) in an acidic solution (b) in a basic solution. Predict the stronger oxidizing agent and reducing agent given a list of reduction potentials. Predict whether a redox reaction is spontaneous or nonspontaneous given a list of reduction potentials. Identify the anode and cathode in a given voltaic cell. Identify the oxidation and reduction half-reactions in a given spontaneous electrochemical cell. Identify the anode and cathode in a given electrolytic cell. Identify the oxidation and reduction half-reactions in a given nonspontaneous electrochemical cell.

(Sec. 17.1) (Sec. 17.1)

Key Terms Exercises 1–8 Exercises 1–8

(Sec. 17.2) (Sec. 17.2) (Sec. 17.3)

Exercises 9–14 Exercises 15–16 Exercises 17–22

(Sec. 17.4)

Exercises 23–28

(Sec. 17.5)

Exercises 29–32

(Sec. 17.5)

Exercises 33–36

(Sec. 17.6) (Sec. 17.6)

Exercises 37–40 Exercises 37–40

(Sec. 17.7) (Sec. 17.7)

Exercises 41–44 Exercises 41–44

(Sec. 18.1) (Sec. 18.2)

Key Terms Exercises 1–6 Exercises 7–8

(Sec. 18.2) (Sec. 18.3)

Exercises 9–12 Exercises 13–14

(Sec. 18.3)

Exercises 15–16

(Sec. 18.3)

Exercises 17–18

(Sec. 18.4)

Exercises 19–24

(Sec. 18.5) (Sec. 18.6) (Sec. 18.7) (Sec. 18.7) (Sec. 18.8) (Sec. 18.8)

Exercises 25–30 Exercises 31–38

Chapter 18 Bloom’s Learning Objectives L1 L2 L3 L3 L2 L2 L2 L3 L2 L3 L2 L3 L2 L3

Exercises 39–46 Exercises 47–52

Chapter 19 Bloom’s Learning Objectives L1 L4 L4 L4 L3 L3

(Sec. 19.1) (Sec. 19.1) (Sec. 19.1) (Sec. 19.2) (Sec. 19.2)

Key Terms Exercises 1–6 Exercises 1–6 Exercises 1–6 Exercises 7–18 Exercises 19–20 xv

L3 L3 L3 L3 L2 L2 L2 L3 L3 L3 L3 L3 L3 L3 L3 L3 L2 L3

Write names and formulas for simple alkenes. Write names and formulas for simple alkynes. Write hydrogenation reactions for alkenes and alkynes. Write names and formulas for simple arenes. Describe the concept of an organic functional group. Identify the functional group in each of the following: organic halide, alcohol, phenol, ether, and amine. Identify the functional group in each of the following: aldehyde, ketone, carboxylic acid, ester, and amide. Write names and formulas for simple organic halides Write names and formulas for simple alcohols. Write names and formulas for simple phenols. Write names and formulas for simple ethers. Write names and formulas for simple amines. Write names and formulas for simple aldehydes. Write names and formulas for simple ketones. Write names and formulas for simple carboxylic acids. Write names and formulas for simple esters. Identify the parent alcohol and acid in an ester. Write names and formulas for simple amides.

(Sec. 19.3) (Sec. 19.3) (Sec. 19.3) (Sec. 19.4) (Sec. 19.5) (Sec. 19.5)

Exercises 21–24 Exercises 25–34 Exercises 35–36 Exercises 37–40 Exercises 41–44 Exercises 41–44

(Sec. 19.5)

Exercises 41–44

(Sec. 19.6) (Sec. 19.7) (Sec. 19.7) (Sec. 19.7) (Sec. 19.8) (Sec. 19.9) (Sec. 19.9) (Sec. 19.10) (Sec. 19.10) (Sec. 19.10) (Sec. 19.10)

Exercises 45–48 Exercises 49–58 Exercises 51–58 Exercises 51–58 Exercises 59–62 Exercises 63–64 Exercises 65–68 Exercises 69–70 Exercises 71–72 Exercises 73–74 Exercises 75–76

Chapter 20 Bloom’s Learning Objectives L1 L2 L2 L2 L2 L2 L2 L2 L2 L4 L4 L2 L2 L2 L2 L2 L2 L2 L4 L2 L2

xvi

CHAPTER

Introduction to Chemistry

Section 1.1 Evolution of Chemistry 2.

Thales stated water was the single element that composed earth, air, and space.

Aristotle stated air, earth, fire, water, and ether were five basic elements responsible for everything in nature.

Step 2 of the scientific method is to analyze the data and propose a tentative hypothesis to explain the experimental observations.

Robert Boyle is generally considered the founder of the scientific method.

10. A hypothesis is an initial proposal that is tentative, whereas a theory is a proposal that has been extensively tested and verified. 12. (a) is a natural law because the relationship is stated as an equation. (b) is a scientific theory because the number of molecules cannot be counted. (c) is a natural law because gas temperature and pressure can be measured. (d) is a scientific theory because it is a model that explains the atom.

Introduction to Chemistry

Section 1.2 Modern Chemistry 14. Antoine Lavoisier is considered the founder of modern chemistry. 16. Agriculture and medicine are industries in which chemistry plays a crucial role. Other industries in which chemistry plays an important role include the pharmaceutical, electronics, paper, construction, and petrochemical industries. Section 1.3 Learning Chemistry 18. A solution to the nine-dot problem using three straight lines is shown below; the unconscious assumption regards the angle of the lines and the size of the dots. •

•

Challenge Exercises 20. By staring at the point where the blocks intersect, we can “flip” the image and view the blocks as stacking upward, or stacking downward.

Chapter 1

CHAPTER

Prerequisite Science Skills

PSS

Section PSS.1 Measurements 2. 4.

The diameter of a 5¢ nickel coin is approximately 2 cm. (a) (c)

Unit meter liter

Quantity length volume

Unit (b) kilogram (d) millisecond

Quantity mass time

(b)

2.05 cm and (c) 2.00 cm each have an uncertainty of ± 0.05 cm.

(d) 25.000 g has a mass uncertainty of ± 0.001 g; thus, an electronic balance.

10.

(b)

25.0 mL and (c) 25.5 mL each have an uncertainty of ± 0.5 mL.

Section PSS.2 Significant Digits 12.

Measurement (a) 500 cm (b) 5000 g (c) 0.05 mL (d) 0.005 s

Significant Digits 1 significant digit 1 significant digit 1 significant digit 1 significant digit

14.

Measurement (a) 5.0 cm (b) 5.05 g (c) 5.02 × 10–1 mL (d) 1.0 × 10–2 s

Significant Digits 2 significant digits 3 significant digits 3 significant digits 2 significant digits

16.

Measurement (a) 0.5 cm (b) 0.50 g (c) 1.00 × 101 mL (d) 1.000 × 103 s

Significant Digits 1 significant digit 2 significant digits 3 significant digits 4 significant digits

Prerequisite Science Skills

Section PSS.3 Rounding Off Nonsignificant Digits 18.

20.

(a) (b) (c) (d)

Example 20.155 0.204 500 2055 0.2065

Rounded Off 20.2 0.205 2060 (2.06 × 103) 0.207

Example (a) 1.454 × 101 (b) 1.455 × 102 (c) 1.508 × 10–3 (d) 1.503 × 10–4

Rounded Off 1.45 × 101 1.46 × 102 1.51 × 10–3 1.50 × 10–4

Section PSS.4 Adding and Subtracting Measurements 22.

(a) (b)

0.4 g + 0.44 g + 0.444 g = 1.284 g 15.5 g + 7.50 g + 0.050 g = 23.050 g

rounds to 1.3 g rounds to 23.1 g

24.

(a) (b)

242.197 g − 175 g = 67.197 g 27.55 g − 14.545 g = 13.005 g

rounds to 67 g rounds to 13.01 g

Section PSS.5 Multiplying and Dividing Measurements 26.

(a) (b) (c) (d)

28.

(a)

3.65 cm × 2.10 cm = 7.665 cm2 8.75 cm × 1.15 cm = 10.0625 cm2 16.5 cm × 1.7 cm = 28.05 cm2 21.1 cm × 20 cm = 422 cm2

rounds to 7.67 cm2 rounds to 10.1 cm2 rounds to 28 cm2 rounds to 400 cm2

26.0 cm2 10.1 cm = 2.5743 cm 9.95 cm3 (b) 0.15 cm2 = 66.333 cm 131.78 cm3 2 (c) 19.25 cm = 6.8457 cm 131.78 cm3 2 (d) 19.2 cm = 6.8635 cm

rounds to 2.57 cm rounds to 66 cm rounds to 6.846 cm2 rounds to 6.86 cm2

Section PSS.6 Exponential Numbers 30.

(a)

10 × 10 × 10 × 10 = 104

(b)

1 10

Chapter PSS

× 10 × 10 × 10

1 4

= ( 10 )

= 10–4

32.

(a)

3 × 3 × 3 × 3 = 34

(b)

1 3

(a) (b)

Number 1 × 1012 1 × 10–22

Scientific Notation 1,000,000,000,000 0.000 000 000 000 000 000 000 1

(a) (b)

Number 100,000,000,000,000,000 0.000 000 000 000 001

Scientific Notation 1 × 1017 1 × 10–15

(a) (b)

Scientific Notation 1 × 100 1 × 10–10

Number 1 0.000 000 000 1

34.

36.

38.

× 3 × 3 × 3

1 4

= ( 3 ) = 3–4

Section PSS.7 Scientific Notation 40.

Ordinary Number (a) 1,010,000,000,000,000 (b) 0.000 000 000 000 456 (c) 94,500,000,000,000,000 (d) 0.000 000 000 000 000 019 50

42.

2.69 × 1019 helium atoms

44.

6.64 x 10–24 g/helium atom

Scientific Notation 1.01 × 1015 4.56 × 10–13 9.45 × 1016 1.950 × 10–17

General Exercises 46.

10.00 mL (± 0.01 mL)

48.

3.00 × 108 meters per second The velocity must be expressed in scientific notation because the rounded value, 300,000,000 meters per second, has only one significant digit.

50.

126.457 g + 131.60 g = 258.057 g

52.

Exponential Number 0.170 × 102 0.00350 × 10–1

(a) (b)

rounds to 258.06 g Scientific Notation 1.70 × 101 3.50 × 10–4

.. Prerequisite Science Skills 5

Challenge Exercises 54.

Mass of a neutron = 1.6749 × 10–24 g Mass of a proton = 1.6726 × 10–24 g Mass difference: (1.6749 × 10–24 g) – (1.6726 × 10–24 g) = 2.3 × 10–27 g

56.

1 metric ton = 2200 lb = 2.200 × 103 lb 1 English ton = 2000 lb = 2.000 × 103 lb Mass difference: 2.200 × 103 lb – 2.000 × 103 lb = 0.200 × 103 lb = 2.00 × 102 lb

Chapter PSS

CHAPTER

The Metric System

Section 2.1 Basic Units and Symbols 2. (a) (c) 4. (a) (c) 6. (a) (c) 8. (a) (c) 10. (a) (c)

Basic Unit

Quantity

Basic Unit

meter liter

length volume

(b) gram (d) second

mass time

Unit

Symbol

Unit

Symbol

petameter femtoliter

Pm fL

(b) gigagram (d) nanosecond

Gg ns

Symbol

Unit

Symbol

Unit

cm dL

centimeter deciliter

Instrument

Quantity

micrometer buret

length volume

Quantity

Unit/Symbol

1 x 1012 m 1 x 10–6 L

terameter microliter

(b) mg (d) µs Instrument

Quantity

milligram microsecond Quantity

(b) electronic balance mass (d) atomic clock time Quantity (b) 1 x 106 g (d) 1 x 10–12 s

Unit/Symbol megagram picosecond

The Metric System

Section 2.2 Metric Conversion Factors 12.

Unit Equation (a) (c)

14.

Unit Equation

1 Gm = 1 x 109 m 1 L = 1 x 101 dL

(b) 1 Tg = 1 x 1012 g (d) 1 s = 1 x 102 cs

Unit Factors

1 Gm (a) 1 x 109 m

and

1 x 109 m 1 Gm

(b)

1 Tg 1 x 1012 g

and

1 x 1012 g 1 Tg

1L 1 x 101 dL

and

1 x 101 dL 1L

1s (d) 1 x 102 cs

and

1 x 102 cs 1s

(c)

Section 2.3 Metric–Metric Conversions 16.

18.

5.00 Mm

(b)

5.00 mg

1g x 1 x 103 mg = 5.00 x 10–3 g

(c)

5.00 mL

1L x 1000 mL = 5.00 x 10–3 L

(d) 5.00 ds

1s x 10 ds

(a)

7.50 km

1000 m 1 km

(b)

750 Mg

1 x 106 g 1 Mg

(c)

0.750 pL

1L x 1 x 1012 pL

(d) 0.000 750 ms

1 x 106 m = 5.00 x 106 m 1 Mm

(a)

Chapter 2

= 5.00 x 10–1 s (0.500 s)

1 Gm x 1 x 109 m = 7.50 x 10–6 Gm 1 Tg 1 x 1012 g

x x

= 7.5 x 10–4 Tg

1 x 106 µL = 7.50 x 10–7 µL 1L

1s 1 x 103 ms

1 x 109 ns 1s

= 7.50 x 102 ns

Section 2.4 Metric–English Conversions 20.

22.

24.

(a)

100 cm 1m

1 in. 2.54 cm

(b)

1 kg

1000 g 1 kg

1 lb 454 g

= 2.20 lb

(c)

1000 mL 1L

(d)

1 sec 1.00 s

= 1.00 sec

(a)

65 in.

2.54 cm 1 in.

170 cm

(b)

65 lb

454 g 1 lb

3.0 x 104 g

(c)

65.0 mL

1 qt 946 mL

= 0.0687 qt

(d)

6.50 x 10–3 s

(a)

800.0 m

100 cm 1m

1 in. 2.54 cm

(b)

0.375 kg

1000 g 1 kg

1 lb 454 g

= 0.826 lb

(c)

1250 mL

1 qt 946 mL

1 gal 4 qt

= 0.330 gal

(d)

1.52 x 103 ds

1s 10 ds

1 min 60 s

= 2.53 min

2.54 cm 1 in.

26.

2722 ft

12 in. 1 ft

28.

21 km 1L

0.621 mi 1 km

1 qt 946 mL

1.00 sec 1s

1 yd 36 in.

= 1.09 yd

= 1.06 qt

= 6.50 x 10–3 sec

1L 0.264 gal

1m 100 cm

1 yd 36 in. = 875 yd

= 829.7 m

= 49 mi/gal

The Metric System

Section 2.5 The Percent Concept 30.

55.5 g stainless steel

32.

2.40 gal ethanol 20.0 gal gasohol

34.

10.5 g chromium 100 g stainless steel = 5.83 g chromium

x 100% = 12.0% ethanol in gasohol

mass of coin = 3.051 g copper and zinc mass of zinc = 0.153 g zinc mass of copper = 3.051 g – 0.153 g = 2.898 g copper 2.898 g Percent of copper: 3.051 g

x 100% = 94.98% copper

Section 2.6 Volume by Calculation 36.

38.

(a)

(b)

1 in.3

x x

1000 mL 1L

1 cm3 1 mL

= 1000 cm3 (exactly)

2.54 cm 1 in.

2.50 mm

1m x 1000 mm

5.00 cm x

100 cm 1m

2.54 cm 1 in.

= 0.250 cm

0.250 cm = 6.25 cm3

40.

(2.75 cm3) (2.75 cm x cm x cm) (4.95 cm) (2.45 cm) = (4.95 cm) (2.45 cm)

42.

0.500 L

1000 mL 1L

1 mL x 1 cm3

1 cm3 1 mL

= 0.227 cm

⎛⎜ 1 in. ⎞⎟3 ⎝2.54 cm⎠

= 45.8 in.3

Section 2.7 Volume by Displacement 44.

Volume of gemstone: 6.5 mL – 5.0 mL = 1.5 mL

46.

Volume of hydrogen gas: 75.5 mL – 43.0 mL = 32.5 mL

Chapter 2

= 16.4 mL

Section 2.8 The Density Concept 48. (a) 50.

Substance

Sink or Float

redwood

sink

Gas (a)

(b)

Rise or Drop

laughing gas

Substance

Sink or Float

bamboo

float

Gas

drop

Rise or Drop

(b)

ammonia

1 cm3 1 mL

rise

52.

The density of water is 1.00 kg/L.

54.

(a)

25.5 g 4.5 mL = 5.7 g/mL

(b)

95.5 g (3.55 cm) (2.50 cm) (1.75 cm)

(a)

5.00 cm3

2.18 g 1 cm3

= 10.9 g

(b)

2.50 cm3

1.59 g 1 cm3

= 3.98 g

(a)

1.00 kg

1000 g 1 kg

1 cm3 2.33 g

1 mL 1 cm3

= 429 mL

(b)

1.00 kg

1000 g 1 kg

1 cm3 4.51 g

1 mL 1 cm3

= 222 mL

56.

58.

= 6.15 g/mL

Section 2.9 Temperature 60.

62.

Temperature Scale

Boiling Point of Water

(a) (b) (c)

Fahrenheit Celsius Kelvin

212 °F 100 °C 373 K

(a)

⎛⎜19 °C x 180 °F ⎞⎟ + 32 °F = 66 °F 100 °C⎠ ⎝

(b)

⎛⎜–175 °C x 180 °F ⎞⎟ + 32 °F = –283 °F 100 °C⎠ ⎝

The Metric System

64.

66.

(a)

273 K – 273 =

0 °C

(b)

100 K – 273 = –173 °C

329 K − 273 = 56 °C 180 °F ⎞ 100 °C⎟⎠ + 32 °F = 133 °F

⎛⎜56 °C x ⎝

Section 2.10 The Heat Concept 68.

In Table 2.5, water is listed as the liquid that is the worst conductor of heat.

70.

120 kJ

1000 J 1 kJ

1 cal x 4.184 J

72.

35.5 g

0.0920 cal 1 g x °C

(50.0 – 25.0) °C

74.

35.7 cal 75.0 g x (43.5 – 29.0) °C

0.0328 cal / g x °C

76.

52.5 cal

1 g x °C 0.0382 cal

1 (35.5 – 25.0) °C

= 131 g lead

1 x 109 Bytes 1 GB

1 MB 1 x 106 Bytes

= 8.5 x 105 MB

1 kcal 1000 cal

29 kcal

= 81.7 cal

General Exercises 78.

850 GB

80.

2.0 TB x

1 x 1012 Bytes 1 GB 1 flash drive x x = 125 flash drives 9 1 TB 16 GB 1 x 10 Bytes

82.

1.00 ds

84.

3.26 yr x

1s 10 ds

30 pulses 1s

= 3 pulses

365 d 24 hr 60 min 60 s 3.00 x 108 m x x x x 1 yr 1d 1 hr 1 min 1s

1 km 1000 m

= 3.08 x 1013 km

Chapter 2

86.

100.0 yd

0.914 m 1 yd

= 91.4 m

160.0 ft

1 yd 3 ft

0.914 m 1 yd

= 48.7 m

Area of a football field: 91.4 m x 48.7 m = 4,450 m2

88.

1 km 1500 m x 1000 m

1 mile 1.61 km

= 0.932 mile

Note: A 1500-meter race (0.932 mile) is less than a mile.

90.

1 drop water

1 mL water 1.00 g water x 20 drops water x 1 mL water

3.34 x 1022 molecules 1 g water = 2 x 1021 molecules

92.

Specific gravity of jet fuel:

0.775 g /mL 1.00 g /mL = 0.775

Challenge Exercises 94.

1.00 g 1 mL

1 lb 1 mL x 454 g 1 cm3

96.

13.6 g 1 mL

1 kg x 1000 g

⎛⎜2.54 cm⎞⎟ 3 x ⎛⎜12 in.⎞⎟ 3 = 62.4 lb/ft3 ⎝ 1 in. ⎠ ⎝ 1 ft ⎠

1 mL ⎛100 cm⎞ 3 4 3 1 cm3 x ⎜⎝ 1 m ⎟⎠ = 1.36 x 10 kg/m

The Metric System

CHAPTER

Matter and Energy

Section 3.1 Physical States of Matter 2.

The gaseous state has a variable shape and variable volume.

The liquid state has a variable shape and does not compress significantly.

The liquid state has closely packed particles, which are free to move around.

10.

12.

(a) (b)

Change of State solid to liquid liquid to gas

Term melting vaporizing

(a) (b)

Change of State solid to liquid liquid to gas

Heat/Cool heating heating

The process of sublimation (solid to gas) requires heating, and the solid absorbs heat in the process.

Section 3.2 Elements, Compounds, and Mixtures 14.

16.

Example (a) iron metal (b) oxygen gas (c) iron oxide (d) iron ore (a) (b) (c) (d)

Classification element element compound mixture

Example silver bar silver ore silver oxide silver alloy

Classification element heterogeneous mixture compound homogeneous mixture

Matter and Energy

Section 3.3 Names and Symbols of the Elements 18.

20.

(a) (c) (e) (g)

Element strontium oxygen fluorine copper

Symbol Sr O F Cu

(b) (d) (f) (h)

Element xenon zinc sodium nickel

Symbol Xe Zn Na Ni

(a) (c) (e) (g)

Symbol Cl Cd Co Cr

Element chlorine cadmium cobalt chromium

(b) (d) (f) (h)

Symbol Ne Ge Ra Te

Element neon germanium radium tellurium

(a) (c) (e) (g)

Element helium potassium selenium lead

Atomic Number 2 19 34 82

(b) (d) (f) (h)

Element carbon copper silver radon

Atomic Number 6 29 47 86

22.

Section 3.4 Metals, Nonmetals, and Semimetals 24.

Property (a) dense solid (b) low boiling point (c) ductile (d) reacts with metals

Classification metal nonmetal metal nonmetal

26.

Property (a) shiny solid (b) brittle solid (c) low density (d) forms alloys

Classification metal nonmetal nonmetal metal

(a) (b) (c) (d)

Element C Ca Cl Cu

Classification nonmetal metal nonmetal metal

(a) (b) (c) (d)

Element argon antimony arsenic astatine

Classification nonmetal semimetal semimetal semimetal

28.

30.

Chapter 3

32. (a) (b) (c) (d)

Element S Se Si Sn

Physical State solid solid solid solid

(a) (b) (c) (d)

Element helium manganese iodine mercury

Physical State gas solid solid liquid

34.

Section 3.5 Compounds and Chemical Formulas 36.

38.

40.

42.

(a)

Chemical Formula C12H17N4OS

Chemical Composition 12 carbon atoms 17 hydrogen atoms 4 nitrogen atoms 1 oxygen atom 1 sulfur atom

(b)

C17H20N4O6

17 carbon atoms 20 hydrogen atoms 4 nitrogen atoms 6 oxygen atoms

(a) (b)

Chemical Composition 20 carbon, 30 hydrogen, 1 oxygen 31 carbon, 46 hydrogen, 2 oxygen

Chemical Formula vitamin A: C20H30O vitamin K: C31H46O2

(a) (b)

Chemical Formula CH3(CH2)16COOH C3H5(C17H33COO)3

Total Atoms stearic acid: 56 atoms triolein: 167 atoms

From the law of constant composition, we can predict the mass ratio of the elements in natural and synthetic vitamin C is 9:1:12.

Section 3.6 Physical and Chemical Properties 44. (a) (c)

Property physical physical

Property (b) physical (d) chemical

(a) (c)

Property physical chemical

Property (b) chemical (d) physical

46.

Matter and Energy

48.

50.

(a) (c)

Property chemical physical

Property (b) physical (d) physical

(a) (c)

Property chemical physical

Property (b) physical (d) chemical

Section 3.7 Physical and Chemical Changes 52. (a) (c)

Change chemical physical

Change (b) physical (d) chemical

(a) (c)

Change physical physical

Change (b) chemical (d) chemical

(a) (c)

Change physical physical

Change (b) chemical (d) chemical

(a) (c)

Change chemical chemical

Change (b) physical (d) physical

54.

56.

58.

Section 3.8 Conservation of Mass 60.

5.00 g iron + 2.77 g phosphorus = 7.77 g iron phosphide

62.

0.750 g mercury oxide – 0.695 g mercury = 0.055 g oxygen

Section 3.9 Potential and Kinetic Energy 64.

Particles in the gaseous state have high kinetic energy and the most movement.

66.

If the temperature increases, the motion of particles will increase.

68.

If the hydrogen gas in the steel cylinder cools from 50 °C to 0 °C, there is less kinetic energy and the motion of hydrogen molecules will decrease.

Chapter 3

Section 3.10 Conservation of Energy 70.

From the law of conservation of energy, 1500 kcal of heat energy is required to vaporize the water to steam at 100 °C.

72.

1230 cal + 1590 cal = 2820 cal is required to heat the solid iron from 1535 °C to 2000 °C.

74.

From the law of conservation of energy, 94.8 kcal of heat is released when hydrogen and oxygen gases react to produce 25.0 g of water.

76.

Energy Conversion (a) uranium converts water to steam (b) steam drives a turbine (c) a turbine turns a generator (d) a generator makes electricity

Forms of Energy nuclear → heat heat → mechanical mechanical → mechanical mechanical → electrical

78.

Energy Conversion (a) lead–acid battery making electricity (b) electricity turning a starter motor (c) starter motor turning a flywheel (d) gasoline exploding into hot gases

Forms of Energy chemical → electrical electrical → mechanical mechanical → mechanical chemical → mechanical

General Exercises 80.

A post-1982 penny is an example of a heterogeneous mixture because the zinc core of the penny and the copper coating have different physical properties.

82.

A dry-cleaning solvent contains different chlorofluorocarbons, but it is a homogenous mixture because the properties are constant for a given sample.

84.

The element antimony is below arsenic in Group V/15; thus, antimony has similar chemical properties and can be substituted for arsenic.

86.

The three most abundant elements in the human body are carbon, hydrogen, and oxygen.

88. (a) (b) (c) (d)

Element roentgenium, Rf copernicium, Cn flerovium, Fl livermorium, Lv

Atomic Number 111 112 114 116

Matter and Energy

90.

From the conservation of energy law, we know that the energy released when elements react and combine is equal to the energy to decompose the compound. Thus, 27,200 joules of energy is released when hydrogen and nitrogen gases react and combine to produce 10.0 g of ammonia.

92.

From the law of conservation of mass and energy, the product should weigh slightly more than the reactants. In theory, a small amount of heat energy is converted into mass; in practice, the mass gain is undetectable.

94.

The total mass and energy for all substances before a chemical reaction is exactly equal to the total mass and energy following the chemical reaction.

Challenge Exercises 96.

A change of chemical formula, but not physical state, is an example of a chemical change. For example, H2O in the form of gaseous steam can be decomposed to give hydrogen and oxygen gases.

Chapter 3

CHAPTER

Models of the Atom

Section 4.1 Dalton Model of the Atom 2.

Aristotle argued that matter is continuous in 350 B.C.

Dalton used the law of conservation of mass, and the law of definite proportion (i.e., law of definite composition) to support the atomic theory.

(a) (b)

“Atoms are indestructible” proved to be invalid with the discovery of nuclear fission; that is, the splitting of an atomic nucleus. “Atoms of the same element are identical” proved to be invalid with the discovery of isotopes.

Section 4.2 Thomson Model of the Atom 8.

The pudding represents a homogeneous atom with a positive charge.

10.

The proton (p+) is the simplest positive particle.

12.

The relative charge on a proton (p+) is plus one (+1).

Section 4.3 Rutherford Model of the Atom 14.

Electrons are located outside the nucleus.

16.

The approximate size of a nucleus is 1 × 10–13 cm.

18.

The relative charge of an electron and proton is –1 and +1, respectively.

Models of the Atom

Section 4.4

Atomic Notation

20.

Isotope

Neutrons

(a)

2 1H

2–1=1n

(c)

18 8O

18 – 8 = 10 n

Isotope

Neutrons

(a)

hydrogen-2

2–1=1n

(c)

cobalt-60

60 – 27 = 33 n

22.

Isotope

Neutrons

(b)

3 2 He

3–2=1n

(d)

22 10Ne

22 – 10 = 12 n

(b)

Isotope

Neutrons

carbon-14

14 – 6 = 8 n

(d) iodine-131

131 – 53 = 78 n

24. Atomic Notation

26.

Atomic Number

Mass Number

Number of Protons

Number of Neutrons

Number of Electrons

19 9

31 15

59 28

210 82

210

128

15 e– e–

(b)

18 n 17 p+

17 e–

18 e–

(d)

78 n 53 p+

53 e–

(a)

16 n 15 p+

(c)

22 n 18 p+

Chapter 4

Section 4.5

Atomic Mass

28.

The atomic mass is the weighted average of the individual isotopic masses that occur naturally for an element.

30.

Carbon-12 is the current reference isotope for the atomic mass scale.

32.

27Al has only one naturally occurring isotope with a mass of 26.98 amu.

34. 36.

31 P has only one naturally occurring isotope with a mass of 30.97 amu.

38.

Mg-24: Mg-25: Mg-26:

23.985 amu 24.986 amu 25.983 amu

Zn-64: Zn-66: Zn-67: Zn-68: Zn-70:

63.929 amu × 65.926 amu × 66.927 amu × 67.925 amu × 69.925 amu × Atomic Mass

Section 4.6

× 0.7870 × 0.1013 × 0.1117 Atomic Mass 0.4889 0.2781 0.0411 0.1857 0.0062

= = = =

18.88 amu 2.531 amu 2.902 amu 24.31 amu

= = = = = =

31.25 amu 18.33 amu 2.75 amu 12.61 amu 0.43 amu 65.37 amu

The Wave Nature of Light

40.

Blue light has a shorter wavelength than green light.

42.

Green light has a lower frequency than blue light.

44.

Green light has lower energy than blue light.

46.

A wavelength of 1250 nm has lower energy than 1100 nm.

Section 4.7 48.

An electron represents the quantum nature of electrical energy.

50.

52.

The Quantum Concept

(a) (b)

Example spiral staircase elevated ramp

Characteristic quantized continuous

(a) (b)

Example 10-mL volumetric pipet 10-mL graduated cylinder

Characteristic quantized continuous

Models of the Atom

Section 4.8

Bohr Model of the Atom

54.

The transition from energy level 3 to 2 is least energetic because the electron drops from a lower energy level than 5 to 2, or 4 to 2.

56.

The transition from energy level 2 to 1 is least energetic because the electron drops from a lower energy level than 4 to 1, or 3 to 1.

58.

Energy Level Change 4 to 2

Spectral Line Color blue-green line

60.

Energy Level Change 4 to 3

Type of Emission infrared energy

62.

The violet line in the emission spectrum has the shortest wavelength.

64.

66.

(a) (b)

Energy Level Change 3 to 2 4 to 2

Number of Photons 100 100

(a) (b) (c)

Energy Level Change 5 to 1 5 to 2 5 to 4

Spectral Line Color ultraviolet line violet line infrared line

Section 4.9 68.

Energy Levels and Sublevels

Fine lines in the emission spectra of the elements suggest sublevels within main energy levels.

70. (a) (b) (c) (d)

Energy Level 1st level 2nd level 3rd level 4th level

Sublevels 1s 2s 2p 3s 3p 3d 4s 4p 4d

72.

Sublevel (a) an s sublevel (b) a p sublevel (c) a d sublevel (d) an f sublevel

74.

The maximum number of electrons in the fourth energy level is equal to the sum of the maximum number of electrons in the 4s, 4p, 4d, and 4f sublevels, that is, 2 e– + 6 e– + 10 e– + 14 e– = 32 e–.

Chapter 4

Max. Electrons 2 e– 6 e– 10 e– 14 e–

Section 4.10 76.

Electron Configuration

The 4d sublevel follows the 5s sublevel.

78. (a) (b) (c) (d)

Element B Ar Mn Ni

(a) (b) (c) (d)

Electron Configuration 1s2 2s22p5 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d5 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5

80.

Electron Configuration 1s2 2s2 2p1 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 4s2 3d5 1s2 2s2 2p6 3s2 3p6 4s2 3d8

Element F Ar Tc I

Models of the Atom

Section 4.11 82.

(a)

1s, 2s, 2px orbital set

84. (a) (c)

86.

(b)

Orbitals

Larger Orbital

2s or 3s 2px or 2py

3s same size

3px, 3py, 3pz orbital set

Orbitals (b) 2px or 3px (d) 4py or 4pz

Description (a) (b)

88.

Quantum Mechanical Model of the Atom

Chapter 4

3px same size

Orbital

spherical orbital in the 6th energy level dumbbell-shaped orbital in the 3rd energy level

Sublevel (a) 1s (b) 2p (c) 3d (d) 4f

Larger Orbital

Max. # of Electrons 2 e– 6 e– 10 e– 14 e–

6s 3p

General Exercises 90.

Let X = the mass of Ga-71 isotope Abundance of Ga-71: 100% – 60.10% Ga-71 isotope: Ga-69 isotope: Both isotopes:

X × 0.3990 68.92 amu × 0.6010 atomic mass 0.3990 X 0.3990 X

39.90% = = =

0.3990 X 41.42 amu 69.72 amu

+ 41.42 amu = 69.72 amu = 69.72 amu – 41.42 amu 0.3990 X = 28.30 amu 28.30 amu X = 0.3990

mass of Ga-71 isotope

70.93 amu

92.

The periodic table lists a mass number for Pm (147), and not an atomic mass. Therefore, we conclude that promethium has no stable isotopes.

94.

Wavelength (a) 280 nm (b) 950 nm

Region (see Figure 4.9) ultraviolet infrared

Challenge Exercises 96.

If the average mass of bromine is approximately 80 amu and one isotope is Br-79, given an equal abundance the other isotope must be Br-81.

98.

An atom is more stable with a completely filled d sublevel than a partially filled sublevel. If one of the copper 4s electrons drops into the 3d sublevel, the 3d sublevel is completely filled, and thus the copper atom is more stable.

Models of the Atom

CHAPTER

The Periodic Table

Section 5.1 Classification of Elements 2.

Argon was not discovered until 1894 (see Figure 3.7).

Einsteinium was not synthesized until 1952 (see Figure 3.7).

Section 5.2 The Periodic Law Concept 6.

After Moseley’s discovery in 1913, the periodic law states that physical and chemical properties tend to repeat periodically when elements are arranged according to increasing atomic number.

Note that Te precedes I in the periodic table, even though the atomic mass of Te (127.60) is greater than I (126.90).

Section 5.3 Groups and Periods of Elements 10.

Horizontal rows in the periodic table are referred to as periods or series.

12.

The term for the elements that belong to Groups 3–12 is the transition elements.

14.

The elements in the series that follow element 89 are called the actinides.

16.

The term for the elements that include Th–Lr is the actinides.

18.

The elements on the right side of the periodic table are nonmetals.

20.

The elements Po and At are radioactive semimetals.

The Periodic Table

22.

24.

26.

28.

(a) (b) (c) (d)

Family boron group oxygen group nickel group copper group

Group Number IIIA/13 VIA/16 VIII/10 IB/11

(a) (c) (e) (g)

IUPAC Group 2 Group 6 Group 11 Group 16

(a) (c)

Element Si Br

Element (b) As (d) Rn

(a) (c)

Element K Sc

Element (b) Ca (d) Pm

American IIA VIB IB VIA

(b) (d) (f) (h)

IUPAC Group 4 Group 8 Group 12 Group 18

American IVB VIII IIB VIIIA

Section 5.4 Periodic Trends 30.

Proceeding from left to right in the periodic table, the metallic character of an element generally decreases.

32.

Proceeding from left to right in the periodic table, the atomic radius generally decreases.

34.

Metallic Character

(a) B < Al (b) Na < K (c) Mg < Ba (d) H < Fe (Note: The element with the greater metallic character is in bold.) 36.

Atomic Radius Atomic Radius (a) Rb > Sr (b) As > Se (c) Pb > Bi (d) I > Xe (Note: The element with the larger atomic radius is in bold.)

Chapter 5

Section 5.5 Properties of Elements 38.

40.

42.

44.

46.

Predicted atomic radius for Ra:

217 pm + (217 – 215) pm = 219 pm

Predicted density for Ra:

3.65 g/mL + (3.65 – 2.63) g/mL = 4.67 g/mL

Predicted melting point for Ra:

725 °C – (769 – 725) °C = 681 °C

Predicted atomic radius of Rn:

210 pm + (210 – 202) pm = 218 pm

Predicted density of Rn:

5.86 g/L + (5.86 – 3.74) g/L = 7.98 g/L

Predicted boiling point of Rn:

–107 °C – (–152 – 107) °C = –62 °C

(a) (c)

Compound lithium oxide gallium oxide

(a) (c)

Compound cadmium oxide cadmium selenide Compound arsenic sulfide and,

(a)

Formula Li2O Ga2O3

Compound (b) barium oxide (d) tin oxide

Formula BaO SnO2

Formula CdO CdSe

Compound (b) cadmium sulfide (d) cadmium telluride

Formula CdS CdTe

Formulas As2S3 As2S5

Compound antimony sulfide and,

Formulas Sb2S3 Sb2S5

(b)

Section 5.6 Blocks of Elements 48.

The elements in Groups IIIA/3 through VIIIA/18 are filling p sublevels.

50.

The inner transition elements are filling f sublevels.

52.

The elements in the actinide series are filling a 5f sublevel.

54. (a) (c) (e) (g)

Element He U Be Si

Highest Sublevel 1s 5f 2s 3p

(b) (d) (f) (h)

Element K Pd Co Pt

Highest Sublevel 4s 4d 3d 5d

The Periodic Table

56. (a) (b) (c) (d) (e) (f)

Element B Ti Na O Ge Ba

(g) (h)

Pd Kr

Electron Configuration 1s2 2s2 2p1 1s2 2s2 2p6 3s2 3p6 4s2 3d 2 1s2 2s2 2p6 3s1 1s2 2s2 2p4 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d8 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

Core Notation [He] 2s2 2p1 [Ar] 4s2 3d 2 [Ne] 3s1 [He] 2s2 2p4 [Ar] 4s2 3d10 4p2 [Xe] 6s2 [Kr] 5s2 4d8 [Ar] 4s2 3d10 4p6

Section 5.7 Valence Electrons 58.

60.

(a) (c)

Group IIA/2 VIA/16

Valence Electrons 2 6

Group (b) IVA/14 (d) VIIIA/18

Valence Electrons 4 8

(a) (c) (e) (g)

Element He Se Cs Sb

Valence Electrons 2 6 1 5

(b) (d) (f) (h)

Element Pb Ne Ga Br

Valence Electrons 4 8 3 7

Element

Electron Dot Formula

Section 5.8 Electron Dot Formulas 62.

Element

Electron Dot Formula

(a)

(b)

(c)

(d) Ne

(e)

(f)

(g)

(h)

Chapter 5

Ga Br

Section 5.9 Ionization Energy 64.

Proceeding from left to right in the periodic table, the ionization energy increases.

66.

The Group VIIIA/18 elements have the highest ionization energy.

68.

Ionization Energy Ionization Energy (a) Ga < Ge (b) Si < P (c) Br < Cl (d) As > Sb (Note: The element with the higher ionization energy is in bold.)

70.

Ionization Energy Ionization Energy (a) Mg < Si (b) Pb < Bi (c) Ca < Ga (d) P < Cl (Note: The element with the lower ionization energy is in bold.)

Section 5.10 Ionic Charges 72.

Group IVA/14 VIA/16

Ionic Charge 4– 2–

Group (b) VA/15 (d) VIIA/17

Ionic Charge 3– 1–

Ion

Ionic Charge

Ion

Ionic Charge

Cs ion O ion

1+ 2–

Ion

Isoelectronic

(a) (c)

Al3+ S2–

Ne Ar

(a) (b) (c) (d)

Ion Ti2+ Zn2+ Y3+ Cs+

Electron Configuration – Core Notation [Ar] 3d2 [Ar] 3d10 [Kr] [Xe]

(a) (b) (c) (d)

Ion Br– Te2– As3– O2–

Electron Configuration – Core Notation [Ar] 4s2 3d10 4p6, or [Kr] [Kr] 5s2 4d10 5p6, or [Xe] [Ar] 4s2 3d10 4p6, or [Kr] [He] 2s2 2p6, or [Ne]

(a) (c) 74. (a) (c) 76.

78.

80.

(b) Ga ion (d) I ion Ion (b) Ca2+ (d) N3–

3+ 1– Isoelectronic Ar Ne

The Periodic Table

General Exercises 82.

84.

(a) (c)

European Group IIA Group IVA

IUPAC Group 2 Group 14

Predicted atomic radius of Ru: Predicted density of Os:

(a) (b) (c) (d) 88.

Element W Bi Ra Ac

IUPAC Group 12 Group 4

135 pm + 126 pm = 131 pm 2

12.45 g/mL + (12.45 – 7.87) g/mL = 17.03 g/mL

Predicted melting point of Fe: 86.

European (b) Group IIB (d) Group IVB

2334 °C – (3033 – 2334) °C = 1635 °C Electron Configuration [Xe] 6s2 4f14 5d4 [Xe] 6s2 4f14 5d10 6p3 [Rn] 7s2 [Rn] 7s2 6d1

Ionic Charges of Hydrogen Ions H+ Hydrogen loses an electron similar to the Group IA/1 elements. H– Hydrogen gains an electron similar to the Group VIIA/17 elements.

Challenge Exercises 90.

Mendeleev predicted the element gallium (Ga), which was not discovered until 1875.

92.

Magnesium has two valence electrons in a filled s sublevel. Aluminum has one electron in an unfilled p sublevel. The ionization energy for aluminum is less than that of magnesium because of the single electron in a p sublevel.

Chapter 5

CHAPTER

Language of Chemistry

Section 6.1 Classification of Compounds 2.

(a) (c)

Compound NH3 NaClO

Classification binary molecular ternary ionic

(a) (c)

Ion H3O+ S2–

Classification Ion polyatomic cation (b) Sr2+ monoatomic anion (d) SO42–

Compound (b) KI (d) H2SO4(aq)

Classification binary ionic ternary acid Classification monoatomic cation polyatomic anion

Section 6.2 Monoatomic Ions 6. (a) (b) (c) (d) 8.

Monoatomic Cation Ba2+ Zn2+ Co3+ Cu+

Stock System Name barium ion zinc ion cobalt(III) ion copper(I) ion

Monoatomic Anion (a) O2– (b) S2– (c) N3– (d) P3–

Systematic Name oxide ion sulfide ion nitride ion phosphide ion

Monoatomic Cation lithium ion silver ion iron(III) ion tin(IV) ion

Chemical Formula Li+ Ag+ Fe3+ Sn4+

10. (a) (b) (c) (d)

Language of Chemistry

Section 6.3 Polyatomic Ions 12. (a) (b) (c) (d) 14.

Oxyanion CO32– HCO3– SO42– HSO4–

Systematic Name carbonate ion hydrogen carbonate ion sulfate ion hydrogen sulfate ion

Polyatomic Ion (a) chlorate ion (b) perchlorate ion (c) permanganate ion (d) acetate ion

Chemical Formula ClO3– ClO4– MnO4– C2H3O2–

Section 6.4 Writing Chemical Formulas 16. (a) (b) (c) (d)

Constituent Ions Li+ + Cl– Ag+ + O2– Cr3+ + I– Sn2+ + N3–

Chemical Formula LiCl Ag2O CrI3 Sn3N2

(a) (b) (c) (d)

Constituent Ions Na+ + C2H3O2– Al3+ + SO32– Hg2+ + CN– Cr3+ + ClO––

Chemical Formula NaC2H3O2 Al2(SO3)3 Hg(CN)2 Cr(ClO)3

(a) (b) (c) (d)

Constituent Ions Pb4+ + SO42– Sn2+ + ClO2– Co2+ + OH– Hg22+ + PO43–

Chemical Formula Pb(SO4)2 Sn(ClO2)2 Co(OH)2 (Hg2)3(PO4)2

18.

20.

Section 6.5 Binary Ionic Compounds 22.

24.

Binary Ionic Compound (a) LiBr (b) SrI2 (c) Na3N (d) AlF3 (a) (b) (c) (d) Chapter 6

Stock System Name lithium bromide strontium iodide sodium nitride aluminum fluoride

Binary Ionic Compound MnCl2 NiF2 SnBr2 CrP

Stock System Name manganese(II) chloride nickel(II) fluoride tin(II) bromide chromium(III) phosphide ..

26.

28.

30.

(a) (b) (c) (d)

Binary Ionic Compound copper(II) sulfide iron(III) phosphide mercury(I) iodide lead(II) oxide

Chemical Formula CuS FeP Hg2I2 PbO

(a) (b)

Binary Ionic Compound beryllium oxide calcium selenide

Chemical Formula BeO CaSe

(a) (b)

Binary Ionic Compound zirconium oxide titanium sulfide

Chemical Formula ZrO2 TiS2

Section 6.6 Ternary Ionic Compounds 32.

34.

36.

38.

40.

(a) (b) (c) (d)

Ternary Ionic Compound KClO3 Mg(NO3)2 Ag2SO4 Al2(CrO4)3

Stock System Name potassium chlorate magnesium nitrate silver sulfate aluminum chromate

(a) (b) (c) (d)

Ternary Ionic Compound Cu2SO4 Fe2(CrO4)3 Hg2(NO2)2 Pb(C2H3O2)4

Latin System Name cuprous sulfate ferric chromate mercurous nitrite plumbic acetate

(a) (b) (c) (d)

Ternary Ionic Compound chromium(III) chlorate lead(II) sulfite tin(IV) carbonate iron(II) chromate

Chemical Formula Cr(ClO3)3 PbSO3 Sn(CO3)2 FeCrO4

(a) (b)

Ternary Ionic Compound lanthanum nitrate scandium nitrite

Chemical Formula La(NO3)3 Sc(NO2)3

(a) (b)

Ternary Ionic Compound cobalt(III) sulfate iron(III) selenate

Chemical Formula Co2(SO4)3 Fe2(SeO4)3

Language of Chemistry

Section 6.7 Latin System of Nomenclature 42. (a) (b) (c) (d)

Monoatomic Cation Fe2+ Fe3+ Pb2+ Pb4+

Latin System Name ferrous ion ferric ion plumbous ion plumbic ion

(a) (b) (c) (d)

Monoatomic Cation mercurous ion mercuric ion stannous ion stannic ion

Chemical Formula Hg22+ Hg2+ Sn2+ Sn4+

(a) (b) (c) (d)

Ionic Compound Cu2S Fe2O3 Hg2(NO2)2 Pb(C2H3O2)4

Latin System Name cuprous sulfide ferric oxide mercurous nitrite plumbic acetate

Ionic Compound (a) cuprous iodide (b) ferric nitride (c) cobaltous phosphate (d) stannic hypochlorite

Chemical Formula CuI FeN Co3(PO4)2 Sn(ClO)4

44.

46.

48.

Section 6.7 Binary Molecular Compounds 50.

52.

(a) (b) (c) (d)

Binary Molecular Compound chlorine dioxide sulfur tetrafluoride iodine monochloride nitrogen monoxide

Chemical Formula ClO2 SF4 ICl NO

(a) (b) (c) (d)

Binary Molecular Compound BrCl SF6 I4O9 Cl2O7

Systematic Name bromine monochloride sulfur hexafluoride tetraiodine nonaoxide dichlorine heptaoxide

Chapter 6

Section 6.8 Binary Acids 54.

(a) (b)

Binary Acid HF(aq) HI(aq)

Systematic Name hydrofluoric acid hydroiodic acid

Section 6.9 Ternary Oxyacids 56.

58.

(a) (b)

Systematic Name carbonic acid nitrous acid

Ternary Oxyacid H2CO3(aq) HNO2(aq)

(a) (b)

Ternary Oxyacid HClO4(aq) H2SO3(aq)

Systematic Name perchloric acid sulfurous acid

General Exercises 60. (a) (b) (c) (d)

Substance chlorine gas molecules chloride ions hypochlorite ions chlorine compounds

Ionic Charge 0 1– 1– 0

(Note: The total ionic charge on a compound is zero.) 62. (a) (b)

Polyatomic Anion SCN?– S2O3?–

Valence Electrons 6 + 4 + 5 = 15 e– 2(6) + 3(6) = 30 e–

(thus, 1–) (thus, 2–)

(Note: The polyatomic anion with a charge of 1– is SCN– because the total of the valence electrons is an odd number (15). 64. IONS

fluoride ion

copper(II)

CuF2

oxide ion

nitride ion

CuO

Cu3N2

ion

copper(II) fluoride

copper(II) oxide

cobalt(II)

CoF2

CoO

Co3N2

ion

cobalt(II) fluoride

cobalt(II) oxide

cobalt(II) nitride

lead(II)

PbF2

PbO

Pb3N2

ion

lead(II) fluoride

lead(II) oxide

lead(II) nitride

copper(II) nitride

Language of Chemistry

66. IONS

68. 70. 72. 74.

cyanide ion

sulfite ion

permanganate ion

mercuric

Hg(CN)2

HgSO3

Hg(MnO4)2

ion

mercuric cyanide

mercuric sulfite

mercuric permanganate

ferric

Fe(CN)3

Fe2(SO3)3

Fe(MnO4)3

ion

ferric cyanide

ferric sulfite

ferric permanganate

stannic

Sn(CN)4

Sn(SO3)2

Sn(MnO4)4

ion

stannic cyanide

stannic sulfite

stannic permanganate

(a)

Compound NaI

Suffix Ending –ide

(b)

Binary Acid HI(aq)

(a)

Compound NaIO2

Suffix Ending –ite

(b)

Ternary Oxyacid Suffix Ending HIO2(aq) –ous acid

(a)

Compound NaIO3

Suffix Ending –ate

(b)

Ternary Oxyacid Suffix Ending HIO3(aq) –ic acid

(a) (b) (c) (d)

Chemical Name acetic acid aqueous nitrogen trihydride aqueous magnesium hydroxide aqueous sodium hydrogen sulfate

Suffix Ending –ic acid

Chemical Formula HC2H3O2(aq) NH3(aq) Mg(OH)2(aq) NaHSO4(aq)

76.

Binary Compound (a) boron tribromide (b) trisilicon tetranitride (c) diarsenic trioxide (d) diantimony pentaoxide

Chemical Formula BBr3 Si3N4 As2O3 Sb2O5

78.

Given Formula TiSiO4

Predicted Formula ZrSiO4

Challenge Exercises 80.

Given Formula LuCl3

Chapter 6

Predicted Formula LrCl3

CHAPTER

Chemical Reactions

Section 7.1 Evidence of Chemical Reactions 2.

(a) (b)

The yellow flame is evidence of a chemical reaction. The gas bubbles are evidence of a chemical reaction.

(a) (b)

The water vapor is evidence of a chemical reaction. The flame-extinguishing gas is evidence of a chemical reaction.

(a) (b)

The precipitate is evidence of a chemical reaction. The blue solution is evidence of a chemical reaction.

Section 7.2 Writing Chemical Equations 8.

Fe(s) + Cl2(g)

10.

Cd(HCO3)2(s) →

12.

Mn(s) + Ni(NO3)2(aq) →

14.

K2CrO4(aq) + CaSO4(aq) →

16.

HNO3(aq) + LiOH(aq) → LiNO3(aq) + HOH(l)

→

FeCl2(s) CdCO3(s) + CO2(g) + H2O(g) Mn(NO3)2(aq) + Ni(s) CaCrO4(s) + K2SO4(aq)

Section 7.3 Balancing Chemical Equations 18.

(a) (b) (c) (d) (e)

4 Co(s) + 3 O2(g) → 2 Co2O3(s) 2 LiClO3(s) → 2 LiCl(s) + 3 O2(g) Cu(s) + 2 AgC2H3O2(aq) → Cu(C2H3O2)2(aq) + 2 Ag(s) Pb(NO3)2(aq) + 2 LiCl(aq) → PbCl2(s) + 2 LiNO3(aq) 3 H2SO4(aq) + 2 Al(OH)3(aq) → Al2(SO4)3(aq) + 6 HOH(l)

Chemical Reactions

20.

(a) (b) (c) (d) (e)

3 Sn(s) + 2 P(s) → Sn3P2(g) Fe2(CO3)3(s) → Fe2O3(s) + 3 CO2(g) 2 Fe(s) + 3 Cd(NO3)2(aq) → 2 Fe(NO3)3(aq) + 3 Cd(s) Co(NO3)2(aq) + H2S(g) → CoS(s) + 2 HNO3(aq) 2 HClO4(aq) + Ba(OH)2(aq) → Ba(ClO4)2(aq) + 2 HOH(l)

Section 7.4 Classifying Chemical Reactions 22.

Refer to the chemical reactions in Exercise 18. (a) combination reaction (b) decomposition reaction (c) single-replacement reaction (d) double-replacement reaction (e) neutralization reaction

24.

Refer to the chemical reactions in Exercise 20. (a) combination reaction (b) decomposition reaction (c) single-replacement reaction (d) double-replacement reaction (e) neutralization reaction

Section 7.5 Combination Reactions General Form: A + Z → AZ 26.

metal + oxygen gas → metal oxide (a) Sn(s) + O2(g) → SnO2(s) (b) 2 Pb(s) + O2(g) → 2 PbO(s)

28.

nonmetal + oxygen gas → nonmetal oxide (a) 2 N2(g) + 5 O2(g) → 2 N2O5(g) (b) 2 Cl2(g) + 3 O2(g) → 2 Cl2O3(g)

30.

metal + nonmetal → ionic compound (a) 2 Fe(s) + 3 F2(g) → 2 FeF3(s) (b) 3 Pb(s) + 4 P(s) → Pb3P4(s)

32.

(a) (b)

S(s) + O2(g) → SO2(g) 2 S(s) + 3 O2(g) → 2 SO3(g)

34.

(a) (b)

2 Sr + O2 → 2 SrO 4 Al + 3 O2 → 2 Al2O3

36.

(a) 3 Zn + 2 P → Zn3P2 (b) 2 Al + 3 S → Al2S3

Chapter 7

Section 7.6 Decomposition Reactions General Form: AZ →

A+Z

38.

metal hydrogen carbonate → metal carbonate + water + carbon dioxide (a) 2 AgHCO3(s) → Ag2CO3(s) + H2O(g) + CO2(g) (b) Ni(HCO3)2(s) → NiCO3(s) + H2O(g) + CO2(g)

40.

metal carbonate → metal oxide + carbon dioxide (a) Cr2(CO3)3(s) → Cr2O3(s) + 3 CO2(g) (b) Pb(CO3)2(s) → PbO2(s) + 2 CO2(g)

42.

metal carbonate → metal oxide + carbon dioxide (a) Li2CO3(s) → Li2O(s) + CO2(g) (b) CdCO3(s) → CdO(s) + CO2(g)

44.

oxygen–containing compounds → oxygen gas (a) Sn(ClO3)2(s) → SnCl2(s) + 3 O2(g) (b) 2 PbO2(s) → 2 PbO(s) + O2(g)

46.

oxygen–containing compounds → oxygen gas (a) 2 AlPO4(s) → 2 AlPO3(s) + O2(g) (b) 2 SnSO4(s) → 2 SnSO3(s) + O2(g)

Section 7.7 The Activity Series Concept 48.

Element (a) Hg (b) Pb (c) Ni (d) Cr

Solution Cd(NO3)2(aq) Cd(NO3)2(aq) Cd(NO3)2(aq) Cd(NO3)2(aq)

Observation no reaction: Cd > Hg no reaction: Cd > Pb no reaction: Cd > Ni reaction: Cr > Cd

50.

Element (a) Hg (b) Pb (c) Ni (d) Cr

Solution HCl(aq) HCl(aq) HCl(aq) HCl(aq)

Observation no reaction: (H) > Hg reaction: Pb > (H) reaction: Ni > (H) reaction: Cr > (H)

Water H2O(l) H2O(l) H2O(l) H2O(l)

Observation reaction: Ba is an active metal no reaction: Mn is not an active metal no reaction: Sn is not an active metal reaction: K is an active metal

52.

(a) (b) (c) (d)

Element Ba Mn Sn K

Chemical Reactions

Section 7.8 Single-Replacement Reactions General Form: A + BZ → AZ + B or: A + BZ → NR 54.

metal1 + aqueous solution1 → metal2 + aqueous solution2 (a) Cd(s) + 2 AgNO3(aq) → 2 Ag(s) + Cd(NO3)2(aq) (b) Ag(s) + Cd(NO3)2(aq) → NR

56.

metal1 + aqueous solution1 → metal2 + aqueous solution2 (a) Fe(s) + HgSO4(aq) → Hg(l) + FeSO4(aq) (b) Hg(l) + FeSO4(aq) → NR

58.

metal + aqueous acid → aqueous solution + hydrogen gas (a) Zn(s) + H2CO3(aq) → ZnCO3 (aq) + H2(g) (b) Cd(s) + 2 HC2H3O2(aq) → Cd(C2H3O2)2(s) + H2(g)

60.

metal + water → metal hydroxide + hydrogen gas (a) 2 Cs(s) + 2 H2O(l) → 2 CsOH(aq) + H2(g) (b) Ra(s) + 2 H2O(l) → Ra(OH)2(aq) + H2(g)

62.

metal1 + aqueous solution1 → metal2 + aqueous solution2 (a) Mg(s) + NiSO4(aq) → MgSO4(aq) + Ni(s) (b) 2 Al(s) + 3 SnSO4(aq) → Al2(SO4)3(aq) + 3 Sn(s)

64.

metal + aqueous acid → aqueous solution + hydrogen gas (a) Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g) (b) 2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)

66.

metal + water → metal hydroxide + hydrogen gas (a) Mg(s) + H2O(l) → NR (b) Sr(s) + 2 H2O(l) → Sr(OH)2(aq) + H2(g)

Section 7.9 Solubility Rules 68.

(a) (c)

Al(NO3)3 Co(OH)2

soluble insoluble

(b) Na2SO4 (d) FePO4

soluble insoluble

70.

(a) (c)

SrCO3 NiS

insoluble insoluble

(b) Ca(OH)2 (d) HgBr2

soluble soluble

Chapter 7

Section 7.10 Double-Replacement Reactions General Form: AX + BZ → AZ + BX or: AX + BZ → NR 72.

aqueous solution1 + aqueous solution2 → precipitate + aqueous solution3 (a) CrI3(aq) + 3 NaOH(aq) → Cr(OH)3(s) + 3 NaI(aq) (b) NiSO4(aq) + Hg2(NO3)2(aq) → Hg2SO4(s) + Ni(NO3)2(aq)

74.

aqueous solution1 + aqueous solution2 → precipitate + aqueous solution3 (a) 2 AgC2H3O2(aq) + SrI2(aq) → 2 AgI(s) + Sr(C2H3O2)2(aq) (b) 2 AlBr3(aq) + 3 Na2CO3(aq) → Al2(CO3)3(s) + 6 NaBr(aq)

Section 7.11 Neutralization Reactions General Form: HX + BOH → BX + HOH 76.

aqueous acid + aqueous base → aqueous salt + water (a) H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 HOH(l) (b) 2 HC2H3O2(aq) + Sr(OH)2(aq) → Sr(C2H3O2)2(aq) + 2 HOH(l)

78.

aqueous acid + aqueous base → aqueous salt + water (b) (b)

H2SO4(aq) + 2 LiOH(aq) H2SO3(aq) + 2 LiOH(aq)

→ →

Li2SO4(aq) + 2 HOH(l) Li2SO3(aq) + 2 HOH(l)

General Exercises 80.

Yes. The sum of the subscripts must be the same for conservation of mass.

82.

(a) (b)

3 FeO(l) + 2 Al(l) → Al2O3(l) + 3 Fe(l) 3 MnO2(l) + 4 Al(l) → 2 Al2O3(l) + 3 Mn(l)

84.

(a) (b)

PCl5(s) + 4 H2O(l) → H3PO4(aq) + 5 HCl(aq) TiCl4(s) + 2 H2O(g) → TiO2(s) + 4 HCl(g)

86.

(a) (b)

2 CH4O(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g) 2 C3H8O(l) + 9 O2(g) → 6 CO2(g) + 8 H2O(g)

Challenge Exercises 88.

Industrial Preparation of Chlorine: 4 HCl(g) + O2(g)

→

2 Cl2(g) + 2 H2O(g)

Chemical Reactions

CHAPTER

The Mole Concept

Section 8.1 Avogadro’s Number 2. (a) (c)

Element Be B

Average Mass 9.01 amu 10.81 amu

Element (b) Ba (d) Br

Average Mass 137.33 amu 79.90 amu

(a) (c)

Element Be B

Mass 9.01 g 10.81 g

Element (b) Ba (d) Br

Mass 137.33 g 79.90 g

Section 8.2 Mole Calculations I 6.

(a) (b)

1 mol Si = 6.02 × 1023 atoms 1 mol SiH4 = 6.02 × 1023 molecules

(a) (b)

6.02 × 1023 atoms Se = 1 mol Se 6.02 × 1023 molecules SeO2 = 1 mol SeO2

10.

(a)

0.250 mol Ca

6.02 × 1023 atoms Ca 1 mol Ca

(b)

0.500 mol F2

6.02 × 1023 molecules F2 1 mol F2

(c)

0.750 mol CaF2 ×

= 1.51 × 1023 atoms Ca

= 3.01 × 1023 molecules F2

6.02 × 1023 formula units CaF2 1 mol CaF2 = 4.52 × 1023 formula units CaF2

The Mole Concept

12.

1 mol Co 6.02 × 1023 atoms Co

(a)

1.00 × 1022 atoms Co ×

(b)

2.00 × 1023 molecules CO2 ×

(c)

3.00 × 1024 formula units CoCO3 ×

= 0.0166 mol Co

1 mol CO2 = 0.332 mol CO2 6.02 × 1023 molecules CO2 1 mol CoCO3 23 6.02 × 10 formula units CoCO

= 4.98 mol CoCO3 Section 8.3 Molar Mass 14.

16.

(a) (b) (c) (d)

Element Ga Ge Sb Se

Molar Mass 69.72 g/mol 72.63 g/mol 121.75 g/mol 78.97 g/mol

(a)

Compound Fe(C2H3O2)2

(b)

Fe3(PO4)2

Molar Mass 55.85 g + 2(12.01 g + 12.01 g + 1.01 g + 1.01 g + 1.01 g + 16.00 g + 16.00 g) = 173.95 g/mol 3(55.85 g) + 2(30.97 g + 16.00 g + 16.00 g + 16.00 g + 16.00 g) = 357.49 g/mol 3(79.90 g Br) + 8(16.00 g O) = 367.70 g/mol 3(12.01 g) + 5(1.01 g) + 3(16.00 g) + 3(14.01 g + 16.00 g + 16.00 g) = 227.11 g/mol

Section 8.4 Mole Calculations II 18.

(a)

MM of Kr = 83.80 g/mol 1.21 × 1024 atoms Kr ×

(b)

1 mol Kr 83.80 g Kr × 1 mol Kr = 168 g Kr 23 6.02 × 10 atoms Kr

MM of N2O = 2(14.01 g N) + 16.00 g O = 44.02 g/mol 6.33 × 1022 molecules N2O ×

1 mol N2O 44.02 g N2O × 1 mol N O 23 2 6.02 × 10 molecules N2O = 4.63 g N2O

Chapter 8

(c)

MM of Mg(ClO4)2

= 24.31 g Mg + 2(35.45 g Cl) + 8(16.00 g O) = 223.21 g/mol

4.17 × 1021 formula units Mg(ClO4)2 ×

× 20.

(a)

1 mol Mg(ClO4)2 6.02 × 1023 formula units Mg(ClO4)2

223.21 g Mg(ClO4)2 1 mol Mg(ClO4)2

= 1.55 g Mg(ClO4)2

MM of Pt = 195.08 g/mol 1 mol Pt 6.02 × 1023 atoms Pt 7.57 g Pt × 195.08 g Pt × = 2.34 × 1022 atoms Pt 1 mol Pt

(b)

MM of C2H6 = 2(12.01 g C) + 6(1.01 g H) = 30.08 g/mol 1 mol C2H6 6.02 × 1023 molecules C2H6 3.88 g C2H6 × 30.08 g C H × 1 mol C2H6 2 6 = 7.77 × 1022 molecules C2H6

(c)

MM of AlCl3 = 26.98 g Al + 3(35.45 g Cl) = 133.33 g/mol 0.152 g AlCl3

1 mol AlCl3 6.02 × 1023 formula units AlCl3 × 133.33 g AlCl × 1 mol AlCl3 3 = 6.86 × 1020 formula units AlCl3

22.

(a)

22.99 g 1 mol Na –23 1 mol Na × 6.02 × 1023 atoms = 3.82 × 10 g/atom

(b)

72.63 g 1 mol Ge 1 mol Ge × 6.02 × 1023 atoms

(c)

MM of SO3 = 32.07 g S + 3(16.00 g O) = 80.07 g/mol

= 1.21 × 10–22 g/atom

1 mol SO3 80.07 g × 1 mol SO3 6.02 × 1023 molecules

= 1.33 × 10–22 g/molecule

(d) MM of NO = 14.01 g N + 16.00 g O = 30.01 g/mol 30.01 g 1 mol NO ×

1 mol NO = 4.99 × 10–23 g/molecule 6.02 × 1023 molecules

The Mole Concept

Section 8.5 Molar Volume 24.

STP conditions: 273 K and 1 atm

26.

(a)

MM of Xe = 131.29 g/mol Density of Xe (at STP):

(b)

1 mol × 22.4 L = 5.86 g/L

MM of Cl2 = 2(35.45 g Cl) = 70.90 g/mol Density of Cl2 (at STP):

(c)

131.29 g 1 mol

70.90 g 1 mol

1 mol 22.4 L = 3.17 g/L

MM of C2H6 = 2(12.01 g C) + 6(1.01 g H) = 30.08 g/mol Density of C2H6 (at STP):

30.08 g 1 mol

1 mol × 22.4 L

= 1.34 g/L

(d) MM of C4H10 = 4(12.01 g C) + 10(1.01 g H) = 58.14 g/mol Density of C4H10 (at STP):

28.

2.14 g × L

58.14 g 1 mol 1 mol × 22.4 L = 2.60 g/L 22.4 L mol = 47.9 g/mol

(a)

MM of ozone:

(b)

MM of silane:

1.43 g × L

22.4 L mol = 32.0 g/mol

(c)

MM of nitric oxide:

1.34 g × L

22.4 L mol = 30.0 g/mol

3.86 g × L

22.4 L mol = 86.5 g/mol

Molecules

Mass

Volume at STP

nitrogen, N2

6.02 × 1023

28.02 g

22.4 L

hydrogen, H2

6.02 × 1023

2.02 g

22.4 L

ammonia, NH3

6.02 × 1023

17.04 g

22.4 L

(d) MM of Freon–22:

30. Gas

Chapter 8

Section 8.6 Mole Calculations III 32.

(a)

MM of NO = 14.01 g N + 16.00 g O = 30.01 g/mol 2.75 g NO

1 mol NO 30.01 g NO

6.02 × 1023 molecules NO 1 mol NO

= 5.52 × 1022 molecules NO (b)

1 L NH3 1 mol NH3 6.02 × 1023 molecules NH3 70.5 mL NH3 × 1000 mL NH × 22.4 L NH × 1 mol NH3 3 3 = 1.89 × 1021 molecules NH3

34.

(a)

MM of N2O3 = 2(14.01 g N) + 3(16.00 g O) = 76.02 g/mol 5.33 L N2O3

(b)

1 mol N2O3 × 22.4 L N O 2 3

76.02 g N2O3 1 mol N2O3

= 18.1 g N2O3

MM of C4H10 = 4(12.01 g C) + 10(1.01 g H) = 58.14 g/mol

1.82 × 1023 molecules C4H10 ×

1 mol C4H10 × 6.02 × 1023 molecules C4H10

58.14 g C4H10 1 mol C4H10

= 17.6 g C4H10 36.

(b)

MM of N2 = 2(14.01 g N) = 28.02 g/mol 1 mol N2 22.4 L N2 5.05 g N2 × 28.02 g N × 1 mol N = 4.04 L N2 2 2

(b)

4.18 × 1024 molecules C2H6 ×

1 mol C2H6 22.4 L C2H6 × 1 mol C H 23 2 6 6.02 × 10 molecules C2H6 = 156 L C2H6

38. Gas

Molecules

Mass

Volume at STP

ozone, O3

2.50 × 1022

1.99 g

0.930 L

carbon dioxide, CO2

2.50 × 1022

1.83 g

0.931 L

carbon monoxide, CO

2.50 × 1022

1.16 g

0.930 L

The Mole Concept

Section 8.7 Percent Composition 40.

The law of definite composition states that a compound always contains the same elements in the same proportion by mass. Thus, the percent chlorine in a pillar of salt 60.66%.

42.

MM of C14H10O4

= 14(12.01 g C) + 10(1.01 g H) + 4(16.00 g O) = 168.14 g C + 10.10 g H + 64.00 g O = 242.24 g/mol 168.14 g C 242.24 g C7H6O3 × 100% = 69.41% C 10.10 g H 242.24 g C7H6O3 × 100% = 4.17% H 64.00 g O 242.24 g C7H6O3 × 100% = 26.42% O

44.

MM NaC5H8NO4

= 22.99 g Na + 5(12.01 g C) + 8(1.01 g H) + 14.01 g N + 4(16.00 g O) = 22.99 g Na + 60.05 g C + 8.08 g H + 14.01 g N + 64.00 g O = 169.13 g/mol 22.99 g Na 169.13 g NaC5H8NO4 × 100% = 13.59% Na 60.05 g C 169.13 g NaC5H8NO4 × 100% = 35.51% C 8.08 g H 169.13 g NaC5H8NO4 × 100% = 4.78% H 14.01 g N 169.13 g NaC5H8NO4 × 100% = 8.284% N 64.00g O 169.13 g NaC5H8NO4 × 100% = 37.84% O

Chapter 8

46.

MM of C55H70MgN4O6 = 55(12.01 g C) + 70(1.01 g H) + 24.31 g Mg + 4(14.01 g N) + 6(16.00 g O) = 660.55 g C + 70.7 g H + 24.31 g Mg + 56.04 g N + 96.00 g O = 907.6 g/mol 660.55 g C 907.6 g C55H70MgN4O6 × 100% = 72.78% C 70.7 g H 907.6 g C55H70MgN4O6 × 100% = 7.79% H 24.31 g Mg 907.6 g C55H70MgN4O6 × 100% = 2.678% Mg 56.04 g N 907.6 g C55H70MgN4O6 × 100% = 6.175% N 96.00 g O 907.6 g C55H70MgN4O6 × 100% = 10.58% O

Section 8.8 Empirical Formula 48.

1 mol Co 1.115 g Co × 58.93 g Co = 0.0189 mol Co 2.025 g CoxSy – 1.115 g Co = 0.910 g S 1 mol S 0.910 g S × 32.07 g S = 0.0284 mol S Co 0.0189 S 0.0284 0.0189

0.0189

= Co1.00S1.50

The empirical formula is Co2S3.

The Mole Concept

50.

1.435 g Hg ×

1 mol Hg 200.59 g Hg = 0.007154 mol Hg

1.550 g HgxOy – 1.435 g Hg = 0.115 g O

52.

0.115 g O ×

1 mol O 16.00 g O = 0.00719 mol O

Hg 0.007154

O 0.00719

0.007154

= Hg1.000O1.01

The empirical formula is HgO.

1 mol Ni 0.500 g Ni × 58.69 g Ni = 0.00852 mol Ni 0.704 g NixOy – 0.500 g Ni = 0.204 g O 1 mol O 0.204 g O × 16.00 g O = 0.0128 mol O

54.

Ni 0.00852

O 0.0128

0.00852

32.0 g O

1 mol O × 16.00 g O

1.33

= 2.00 mol O

= V1.00O1.50

The empirical formula is V2O3.

1 mol Bi 89.7 g Bi × 208.98 g Bi = 0.429 mol Bi 10.3 g O ×

1 mol O 16.00 g O = 0.644 mol O

Bi 0.429 O 0.644 0.429

The empirical formula is Ni2O3.

1 mol V 50.94 g V = 1.33 mol V

68.0 g V

V 1.33 O 2.00

56.

= Ni1.00O1.50

Chapter 8

0.429

= Bi1.00O1.50

The empirical formula is Bi2O3.

58.

30.7 g C ×

1 mol C 12.01 g C = 2.56 mol C

7.74 g H ×

1 mol H 1.01 g H = 7.66 mol H

20.5 g O ×

1 mol O 16.01 g O = 1.28 mol O

1 mol S 41.0 g S × 32.07 g S = 1.28 mol S C 2.56 H 7.66 O 1.28 S 1.28 1.28

1.28

= C2.00H5.98O1.00S1.00 The empirical formula is C2H6OS.

Section 8.9 Molecular Formula 60.

MM of C10H12NO

= 10(12.01 g C) + 12(1.01 g H) + 14.01 g N + 16.00 g O = 120.10 g C + 12.12 g H + 14.01 g N + 16.00 g O = 162.23 g/mol (C10H12NO)n 325 g/mol = C10H12NO 162.23 g/mol

Quinine:

n≈2

The molecular formula of quinine is (C10H12NO)2 or C20H24N2O2. 62.

MM of C3H8N

= 3(12.01 g C) + 8(1.01 g H) + 14.01g N = 36.03 g C + 8.08 g H + 14.01 g N = 58.12 g/mol

Hexamethylene diamine:

(C3H8N)n 115 g/mol C3H8N = 58.12 g/mol

n≈2

The molecular formula of the compound is (C3H8N)2 or C6H16N2. 64.

Empirical Formula 54.5 g C ×

1 mol C 12.01 g C = 4.54 mol C

9.15 g H

1 mol H × 1.01 g H

= 9.06 mol H

36.3 g O

1 mol O × 16.00 g O

= 2.27 mol O

C 4.54 H 9.06 O 2.27 2.27

2.27

= C2.00H3.99O1.00

2.27

The Mole Concept

The empirical formula is C2H4O.

Molecular Formula MM of C2H4O = 2(12.01 g C) + 4(1.01 g H) + 16.00 g O = 44.06 g/mol (C2H4O)n 88 g/mol = C2H4O 44.06 g/mol

Dioxane:

n≈2

The molecular formula of dioxane is (C2H4O)2 or C4H8O2. 66.

Empirical Formula 85.0 g Hg

1 mol Hg × 200.59 g Hg = 0.424 mol Hg

15.0 g Cl

1 mol Cl × 35.45 g Cl

Hg 0.424 Cl 0.423 0.423

0.423

= 0.423 mol Cl

= Hg1.00Cl1.00

The empirical formula is HgCl.

Molecular Formula MM of HgCl = 200.59 g Hg + 35.45 g Cl = 236.04 g/mol HgCl Mercurous chloride:

(HgCl)n 470 g/mol HgCl = 236.04 g/mol

n≈2

The molecular formula of mercurous chloride is (HgCl)2 or Hg2Cl2.

68.

Empirical Formula 40.0 g C ×

1 mol C 12.01 g C = 3.33 mol C

6.72 g H ×

1 mol H 1.01 g H = 6.65 mol H

53.3 g O ×

1 mol O 16.00 g O = 3.33 mol O

C 3.33 H 6.65 O 3.33 3.33

Chapter 8

3.33

= C1.00H2.00O1.00

The empirical formula is CH2O.

Molecular Formula MM of CH2O = 12.01 g C + 2(1.01 g H) + 16.00 g O = 30.03 g/mol (CH2O)n 180 g/mol CH2O = 30.03 g/mol n ≈ 6

Allose:

The molecular formula of allose is (CH2O)6 or C6H12O6.

General Exercises 70.

1 mol Ga 0.500 g Ga × 69.72 g Ga = 0.00717 mol Ga 0.672 g GaxOy – 0.500 g Ga = 0.172 g O 0.172 g O

1 mol O × 16.00 g O

Ga 0.00717 O 0.0108 0.00717

0.00717

= 0.0108 mol O

= Ga1.00O1.51

The empirical formula is Ga2O3.

1 mol Cu 6.02 × 1023 atoms Cu × = 9 × 1017 atoms Cu 63.55 g Cu 1 mol Cu

72.

0.0001 g Cu ×

74.

1 mol furry moles ×

6.02 × 1023 furry moles 17 cm 1m × × 1 mol furry moles 1 furry mole 100 cm × 1 km 1000 m

= 1.0 × 1020 km

9.5 × 1012 km (1 light year) < < < 1.0 × 1020 km (1 mol of furry moles). Thus, a mole of furry moles is many times longer than the distance light travels in a year.

The Mole Concept

76.

MM of C6H12O6 = 6(12.01 g C) + 12(1.01 g H) + 6(16.00 g O) = 180.18 g/mol 1 mol C6H12O6 6 mol C 6.02 × 1023 atoms C 1.00 g C6H12O6 × 180.18 g C H O × 1 mol C H O × 1 mol C 6 12 6 6 12 6 = 2.00 × 1022 atoms C

78.

1 mol Fe2O3 159.70 g Fe2O3 1 mol Fe 10.0 g Fe × 55.85 g Fe × × 2 mol Fe 1 mol Fe2O3 = 14.3 g Fe2O3

Challenge Exercises 80.

MM of C2H5OH = 2(12.01 g C) + 6(1.01 g H) + 16.00 g O = 46.08 g/mol 1 mol C2H5OH 1 molecule C2H5OH × 6.02 × 1023 molecules C H OH 2 5 1 mL C2H5OH × 0.789 g C H OH 2 5

82.

1 atom Al (0.255 nm)3

⎛1 × 109 nm⎞3 ⎜ 1m ⎟ ⎝ ⎠

Chapter 8

46.08 g C2H5OH 1 mol C2H5OH

= 9.70 × 10–23 mL C2H5OH

⎛⎜ 1 m ⎞⎟3 × ⎝100 cm⎠ =

1 cm3 2.70 g Al ×

26.98 g Al 1 mol Al

6.03 × 1023 atoms Al mol Al

CHAPTER

Chemical Equation Calculations

Section 9.1 Interpreting a Chemical Equation 2.

General Equation: 2 A + 3 B → (a)

2 mol A ×

(b)

1LD

2 mol C 2 mol A 3LB 1LD

2C + D

2 mol C

3LB

General Equation: 3 A + B → 2 C (a) conservation of mass law: 1.50 g A + 1.65 g B = 3.15 g C (b) conservation of mass law: 9.45 g C – 4.50 g A = 4.95 g B

(a)

P4(s) MM of P4 MM of O2 MM of P2O3 123.88 g P4 123.88 g P4

(b) MM of P4 MM of O2 MM of P2O5 123.88 g P4 123.88 g P4

+ +

+ 3 O2(g) → 2 P2O3(s) = 4(30.97 g P) = 2(16.00 g O) = 2(30.97 g P) + 3(16.00 g O)

3(32.00 g O2) 96.00 g O2 219.88 g

P4(s)

+ +

→ →

2(109.94 g P2O3) 219.88 g P2O3 219.88 g

+ 5 O2(g) → 2 P2O5(s) = 4(30.97 g P) = 2(16.00 g O) = 2(30.97 g P) + 5(16.00 g O)

5(32.00 g O2) 160.00 g O2 283.88 g

→

→ → →

= 123.88 g/mol = 32.00 g/mol = 109.94 g/mol

= 123.88 g/mol = 32.00 g/mol = 141.94 g/mol

2(141.94 g P2O5) 283.88 g P2O5 283.88 g

Chemical Equation Calculation 57

Section 9.2 Mole–Mole Relationships 8.

2 H2O2(l)

→

2 H2O(l) + O2(g)

Moles of O2 produced: 2 mol H2O 2 mol H2O2

5.00 mol H2O2 × 10.

2 LiClO3(s)

→

5.00 mol H2O

2 LiCl(s) + O2(g)

Moles of O2 produced: 5.00 mol LiClO3 12.

2 mol LiCl 2 mol LiClO3

= 5.00 mol LiCl

3 Ba(s) + N2(g) → Ba3N2(s) Moles of Ba(s) that react: 0.100 mol Ba3N2

3 mol Ba 1 mol Ba3N2 = 0.300 mol Ba

1 mol N2 1 mol Ba3N2 = 0.100 mol N2

Moles of N2(g) that react: 0.100 mol Ba3N2 14.

2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g) Moles of C4H10(g) that react: 2 mol C4H10 2.50 mol H2O × 10 mol H O = 0.500 mol C4H10 2 Moles of O2(g) that react: 13 mol O2 2.50 mol H2O × 10 mol H O = 3.25 mol O2 2

Section 9.3 Types of Stoichiometry Problems 16.

Given mass

Unknown mass

Type of Stoichiometry mass–mass problem

18.

Given volume

Unknown mass

Type of Stoichiometry mass–volume problem

20.

Given volume

Unknown volume

Type of Stoichiometry volume–volume problem

Chapter 9

Section 9.4 Mass–Mass Problems 22.

2 Zn(s) + O2(g)

2 ZnO(s)

→

MM of Zn = 65.39 g/mol MM of ZnO = 81.39 g/mol 1 mol ZnO 1.50 g ZnO × 81.39 g ZnO 24.

1 mol O2 2 mol ZnO ×

32.00 g O2 1 mol O2

= 0.295 g O2

2 Bi(s) + 3 Cl2(g) → 2 BiCl3(s) MM of Cl2 = 70.90 g/mol MM of BiCl3 = 315.33 g/mol 1 mol BiCl3 3 mol Cl2 70.90 g Cl2 3.52 g BiCl3 × 315.33 g BiCl × 2 mol BiCl × 1 mol Cl = 1.19 g Cl2 3 3 2

26.

Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s) MM of Ag = 107.87 g/mol MM of Cu = 63.55 g/mol 1 mol Ag 1 mol Cu 63.55 g Cu 1.50 g Ag × 107.87 g Ag × 2 mol Ag × 1 mol Cu = 0.442 g Cu

28.

2 Co(s) + 3 HgCl2(aq) → 2 CoCl3(aq) + 3 Hg(l) MM of HgCl2 = 271.49 g/mol MM of Hg = 200.59 g/mol 3 mol HgCl2 271.49 g HgCl2 1 mol Hg 5.11 g Hg × 200.59 g Hg × 3 mol Hg × 1 mol HgCl = 6.92 g HgCl2 2

30.

2 Na3PO4(aq) + 3 Ca(OH)2(aq) → Ca3(PO4)2(s) + 6 NaOH(aq) MM of Ca(OH)2 = 74.10 g/mol MM of Ca3(PO4)2 = 310.18 g/mol 1 mol Ca3(PO4)2 3 mol Ca(OH)2 74.10 g Ca(OH)2 2.39 g Ca3(PO4)2 × 310.18 g Ca (PO ) × 1 mol Ca (PO ) × 1 mol Ca(OH) 3 4 2 3 4 2 2 = 1.71 g Ca(OH)2

Chemical Equation Calculation 59

Section 9.5 Mass–Volume Problems 32.

2 NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g) MM of NaHCO3 = 84.01 g/mol 1 L CO2 1 mol CO2 × 1000 mL CO2 22.4 L CO2 ×

25.0 mL CO2 ×

34.

Fe2(CO3)3(s)

→

291.73 g/mol

1 L CO2 1 mol CO2 × 1000 mL CO2 22.4 L CO2 × ×

36.

= 0.188 g NaHCO3

Fe2O3(s) + 3 CO2(g)

MM of Fe2(CO3)3 = 50.0 mL CO2 ×

84.01 g NaHCO3 1 mol NaHCO3

2 mol NaHCO3 1 mol CO2

291.73 g Fe2(CO3)3 1 mol Fe2(CO3)3

1 mol Fe2(CO3)3 3 mol CO2

= 0.217 g Fe2(CO3)3

2 HgO(s) → 2 Hg(l) + O2(g) MM of HgO =

216.59 g/mol

50.0 mL O2 ×

1 L O2 1 mol O2 1000 mL O2 × 22.4 L O2

2 mol HgO × 1 mol O2

216.59 g HgO 1 mol HgO

= 0.967 g HgO 38.

2 H2O2(l) → 2 H2O(l) + O2(g) MM of H2O2 = 34.02 g/mol 1 L O2 1 mol O2 50.0 mL O2 × 1000 mL O × 22.4 L O 2 2

2 mol H2O2 34.02 g H2O2 × 1 mol O2 1 mol H2O2 = 0.152 g H2O2

Chapter 9

Section 9.6 Volume–Volume Problems 40.

3 H2(g) + N2(g) → 2 NH3(g) 50.0 L H2 ×

42.

2 L NH3 3 L H2 = 33.3 L NH3

2 CO(g) + O2(g) → 2 CO2(g) 2 mL CO2 2 mL CO = 10.0 mL CO2

10.0 mL CO ×

44.

2 Cl2(g) + 3 O2(g) → 2 Cl2O3(g) 1.75 L Cl2O3

46.

1000 mL O2 1 L O2

= 2630 mL O2

2 SO2(g) + O2(g) → 2 SO3(g) 25.0 L O2

48.

3 L O2 × 2 L Cl O 2 3

2 L SO2 1 L O2 = 50.0 L SO2

2 N2(g) + 5 O2(g) → 2 N2O5(g) 500.0 cm3 N2O5

5 cm3 O2 3 2 cm3 N2O5 = 1250 cm O2

Section 9.7 Limiting Reactant Concept 50.

N2(g) + O2(g) → 2 NO(g) 0.125 mol N2

2 mol NO 1 mol N2

= 0.250 mol NO

0.125 mol O2

2 mol NO 1 mol O2

= 0.250 mol NO

N2 or O2 are each a limiting reactant because each produces the same amount of product. This reaction produces 0.250 mol NO.

Chemical Equation Calculation 61

52.

2 NO(g) + O2(g) → 2 NO2(g) 1.50 mol NO

2 mol NO2 2 mol NO

= 1.50 mol NO2

2.50 mol O2

2 mol NO2 1 mol O2

= 5.00 mol NO2

The limiting reactant is NO because it produces less product. This reaction produces 1.50 mol NO2.

54.

2 H2(g) + O2(g) → 2 H2O(l) 5.00 mol H2

2 mol H2O 2 mol H2

= 5.00 mol H2O

1.50 mol O2

2 mol H2O 1 mol O2

= 3.00 mol H2O

The limiting reactant is O2 because it produces less water. This reaction produces 3.00 mol H2O.

56.

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) 1.00 mol C2H6 ×

6 mol H2O 2 mol C2H6

= 3.00 mol H2O

3.00 mol O2

6 mol H2O 7 mol O2

= 2.57 mol H2O

The limiting reactant is O2 because it produces less water. This reaction produces 2.57 mol H2O.

Chapter 9

58.

2 Co(s) + 3 S(s) → (a)

Co2S3(s)

1.00 mol Co

1 mol Co2S3 2 mol Co

= 0.500 mol Co2S3

1.00 mol S

1 mol Co2S3 3 mol S

= 0.333 mol Co2S3

The limiting reactant is S because it produces less product. This reaction produces 0.333 mol Co2S3. After reaction: mol of Co = 1.00 – 2(0.333) mol of S = 1.00 – 3(0.333) mol of Co2S3 = 0.00 + 0.333

(b)

= 0.333 mol = 0.000 mol = 0.333 mol

2.00 mol Co ×

1 mol Co2S3 2 mol Co

= 1.00 mol Co2S3

3.00 mol S

1 mol Co2S3 3 mol S

= 1.00 mol Co2S3

Neither Co or S is a limiting reactant because each produces the same amount of product. This reaction produces 1.00 mol CoS. After reaction: mol of Co = mol of S = mol of Co2S3 =

2.00 – 2.00 3.00 – 3.00 0.00 + 1.00

= 0.00 mol = 0.00 mol = 1.00 mol

Section 9.8 Limiting Reactant Problems 60.

FeO(l) + Mg(l) → Fe(l) + MgO(s) 1 mol FeO 1 mol Fe 50.0 g FeO × 71.85 g FeO × 1 mol FeO 1 mol Mg 1 mol Fe 20.0 g Mg × 24.31 g Mg × 1 mol Mg

× ×

55.85 g Fe 1 mol Fe = 38.9 g Fe 55.85 g Fe 1 mol Fe = 45.9 g Fe

Therefore, FeO is the limiting reactant because it will be consumed before Mg. This reaction produces 38.9 g Fe.

Chemical Equation Calculation 63

62.

Fe2O3(l) + 2 Al(l) → 2 Fe(l) + Al2O3(s) 1 mol Fe2O3 2 mol Fe 55.85 g Fe 50.0 g Fe2O3 × 159.70 g Fe O × 1 mol Fe O × 1 mol Fe = 35.0 g Fe 2 3 2 3 25.0 g Al ×

1 mol Al 2 mol Fe 55.85 g Fe 26.98 g Al × 2 mol Al × 1 mol Fe = 51.8 g Fe

Therefore, Fe2O3 is the limiting reactant because it will be consumed before Al. This reaction produces 35.0 g Fe. 64.

Mg(OH)2(s) + H2SO4(l) → MgSO4(s) + 2 H2O(l) 1 mol Mg(OH)2 0.605 g Mg(OH)2 × 58.33 g Mg(OH) 2

1 mol MgSO4 × 1 mol Mg(OH) 2

120.38 g MgSO4 1 mol MgSO4

= 1.25 g MgSO4 1.00 g H2SO4

1 mol H2SO4 98.09 g H2SO4

1 mol MgSO4 1 mol H2SO4

= 1.23 g MgSO4 Therefore, H2SO4 is the limiting reactant because it will be consumed before Mg(OH)2. This reaction produces 1.23 g MgSO4. 66.

2 Al(OH)3(s) + 3 H2SO4(l) → Al2(SO4)3(aq) + 6 H2O(l) 3.00 g Al(OH)3

1 mol Al(OH)3 78.01 g Al(OH)3

6 mol H2O 2 mol Al(OH)3

18.02 g H2O 1 mol H2O

= 2.08 g H2O 1.00 g H2SO4

1 mol H2SO4 98.09 g H2SO4

6 mol H2O 3 mol H2SO4

= 0.367 g H2O Therefore, H2SO4 is the limiting reactant because it will be consumed before Al(OH)3. This reaction produces 0.367 g H2O.

Chapter 9

68.

N2(g) + 2 O2(g) → 2 NO2(g) 95.0 mL N2

2 mL NO2 1 mL N2

190 mL NO2

45.0 mL O2

2 mL NO2 2 mL O2

45.0 mL NO2

(1.90 × 102 mL NO2)

Therefore, O2 is the limiting reactant because it will be consumed before N2. This reaction produces 45.0 mL NO2. 70.

2 N2(g) + 3 O2(g) → 2 N2O3(g) 45.0 mL N2

2 mL N2O3 2 mL N2

45.0 mL N2O3

70.0 mL O2

2 mL N2O3 3 mL O2

46.7 mL N2O3

Therefore, N2 is the limiting reactant because it will be consumed before O2. This reaction produces 45.0 mL N2O3. 72.

2 SO2(g) + O2(g) → 2 SO3(g) 1.25 L SO2

2 L SO3 2 L SO2

1.25 L SO3

3.00 L O2

2 L SO3 1 L O2

6.00 L SO3

Therefore, SO2 is the limiting reactant because it will be consumed before O2. This reaction produces 1.25 L SO3. 74.

4 HCl(g) + O2(g) → 2 Cl2(g) + 2 H2O(g) 10.0 L HCl

2 L Cl2 4 L HCl

5.00 L Cl2

50.0 L O2

2 L Cl2 1 L O2

100 L Cl2

(1.00 × 102 mL Cl2)

Therefore, HCl is the limiting reactant because it will be consumed before O2. This reaction produces 5.00 L Cl2.

Chemical Equation Calculation 65

Section 9.9 Percent Yield 76.

Actual yield: 12.5 g PbI2; theoretical yield: 13.9 g PbI2 12.5 g 13.9 g

Percent yield: 78.

× 100%

= 89.9%

Actual yield: 0.725 g K2CO3; theoretical yield: 0.690 g K2CO3 0.725 g 0.690 g

Percent yield:

× 100%

= 105%

General Exercises 80.

The units of molar volume are liters per mole (L/mol).

82.

(NH4)2Cr2O7(s) → Cr2O3(s) + 4 H2O(l) + N2(g) MM of (NH4)2Cr2O7 =

252.10 g/mol

1.54 g (NH4)2Cr2O7

1 mol (NH4)2Cr2O7 252.10 g (NH4)2Cr2O7

× 84.

22.4 L N2 1 mol N2

1 mol N2 1 mol (NH4)2Cr2O7

0.137 L N2 (137 mL N2)

3 MnO2(s) + 4 Al(s) → 3 Mn(s) + 2 Al2O3(s) MM of Mn = 54.94 g/mol MM of MnO2 = 86.94 g/mol

1.00 kg Mn ×

1000 g Mn 1 kg Mn ×

3 mol MnO2 1 mol Mn 54.94 g Mn × 3 mol Mn × =

86.

86.94 g MnO2 1 mol MnO2

1580 g MnO2

Sb2S3(s) + 6 HCl(aq) → 2 SbCl3(aq) + 3 H2S(g) MM of Sb2S3 = 339.71 g/mol MM of SbCl3 = 228.10 g/mol 1 mol Sb2S3 2 mol SbCl3 228.10 g SbCl3 3.00 g Sb2S3 × 339.71 g Sb S × 1 mol Sb S × 1 mol SbCl = 4.03 g SbCl3 2 3 2 3 3

Chapter 9

Challenge Exercises 88.

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) MM of C3H8 =

44.11 g/mol

1 mol C3H8 3 mol CO2 22.4 L CO2 10.0 g C3H8 × 44.11 g C H × 1 mol C H × 1 mol CO = 15.2 L CO2 at STP 3 8 3 8 2

90.

2 H2O(l) → 2 H2(g) + O2(g) MM of H2O = 18.02 g/mol 100.0 mL H2O ×

1.00 g H2O 1 mL H2O

1 mol H2O 1 mol O2 × 18.02 g H2O 2 mol H2O × = 62.2 L O2 at STP

Chemical Equation Calculation 67

22.4 L O2 1 mol O2

CHAPTER

Gases

Section 10.1 Properties of Gases 2.

The following are observed properties of gases: (a) Gases have a variable volume. (b) Gases expand infinitely. (c) Gases compress uniformly. (d) Gases have low densities.

Section 10.2 Atmospheric Pressure 4.

(a) (b)

Units centimeters of mercury inches of mercury

Standard Pressure 76 mm Hg 29.9 in. Hg

(a)

5.00 atm

76 cm Hg 1 atm

3.80 x 102 cm Hg

(b)

5.00 atm

29.9 in. Hg 1 atm

1.50 x 102 in. Hg

(a)

104 kPa

76 cm Hg 101 kPa

78.3 cm Hg

(b)

104 kPa

14.7 psi 101 kPa

15.1 psi

Gases

Section 10.3 Variables Affecting Gas Pressure 10.

(a)

Change volume decreases

Observation pressure increases

Explanation molecules are closer together and collide more frequently

(b)

temperature decreases

pressure decreases

molecules are moving slower and collide with lower frequency and less energy

(c)

moles of gas decrease

pressure decreases

fewer molecules have fewer collisions

(a) (b) (c)

Change volume decreases temperature decreases moles of gas decreases

Observation pressure increases pressure decreases pressure decreases

12.

14.

Increasing the temperature of a gas causes gas molecules to move faster, and they collide with the container more frequently and with more energy. As a result, the pressure of the gas increases.

Section 10.4 Boyle’s Law: Pressure–Volume Relationships 16.

Pressure vs. Reciprocal of Volume P

0 18.

20.

P1 × Vfactor

= P2

15.0 psi

555 mL 275 mL

V1 × Pfactor

= V2

125 mL ×

Chapter 10

705 mm Hg 385 mm Hg

30.3 psi

229 mL

1/V

Section 10.5 Charles’s Law: Volume–Temperature Relationships 22.

Volume vs. Celsius Temperature V

0 24.

t, (°C)

V1 × Tfactor = V2 100 °C + 273 = 373 K 20 °C + 273 = 293 K 293 K 2.50 L × 373 K = 1.96 L

26.

× Vfactor = T2

25 °C + 273 = 298 K 298 K ×

175 mL 125 mL

= 417 K

417 K – 273

= 144 °C

Section 10.6 Gay–Lussac’s Law: Pressure–Temperature Relationships 28.

Pressure vs. Celsius Temperature

0 30.

t, (°C)

× Tfactor = P2

100 °C + 273 = 373 K 10 °C + 273 = 283 K 283 K 455 cm Hg × 373 K

= 345 cm Hg

Gases

32.

× Pfactor = T2

25 °C + 273 = 298 K 298 K ×

0.250 atm 0.500 atm = 149 K

149 K – 273 = –124 °C

Section 10.7 Combined Gas Law 34. initial final

P 772 mm Hg 760 mm Hg

V 100.0 mL V2 Pfactor

T 21 °C + 273 = 294 K 273 K

Tfactor

772 mm Hg 273 K 100.0 mL × 760 mm Hg × 294 K

= 94.3 mL

P 650 mm Hg 350 mm Hg

T –25 °C + 273 = 248 K 25 °C + 273 = 298 K

36. initial final

V 25.0 mL V2 Pfactor

Tfactor

650 mm Hg 350 mm Hg

298 K × 248 K

25.0 mL ×

= 55.8 mL

38. initial final

P 760 mm Hg P2 P1

V 1250 mL 255 mL ×

760 mm Hg ×

Chapter 10

Vfactor

T 273 K 300 °C + 273 = 573 K Tfactor

1250 mL 573 K × 255 mL 273 K

= 7820 mm Hg

40. initial final

P 760 mm Hg 350 mm Hg

V 50.0 mL 350.0 mL

T 273 K T2

Pfactor

Vfactor

273 K

350 mm Hg × 760 mm Hg

350.0 mL 50.0 mL

= 880 K

880 K – 273 = 607 °C 42. initial final

P 760 mm Hg 655 mm Hg

V 500.0 mL 250.0 mL

T 273 K T2

Pfactor

Vfactor

273 K

655 mm Hg × 760 mm Hg

250.0 mL 500.0 mL

= 118 K

118 K – 273 = –155°C Section 10.8 The Vapor Pressure Concept 44.

The vapor pressure of alcohol is much greater than the vapor pressure of mercury at 50 °C. Thus. Thus, the vapor pressure of alcohol can be measured by placing a drop of alcohol on a column of mercury in a barometer.

46. (a) (b)

Temperature 75 °C 100 °C

Vapor Pressure 289.1 mm Hg = 0.3804 atm 760.0 mm Hg = 1.000 atm

Section 10.9 Dalton’s Law of Partial Pressures 48.

Pnitrogen = 587 mm Hg Poxygen = 158 mm Hg Pargon = 7 mm Hg Pnitrogen + Poxygen + Pargon = Patmosphere 587 mm Hg + 158 mm Hg + 7 mm Hg = 752 mm Hg

Gases

50.

760 in. Hg = 3800 mm Hg 1 atm Pnitrogen = 1850 mm Hg Phydrogen = 1150 mm Hg Ptotal = 5.00 atm ×

Pnitrogen + Phydrogen + Pammonia = Ptotal Pammonia = Ptotal – Pnitrogen – Phydrogen Pammonia = 3800 mm Hg – 1850 mm Hg – 1150 mm Hg = 800 mm Hg 52.

The term “wet” gas refers to a gas collected over water. A “wet” gas contains water vapor in the gaseous sample; a “dry” gas does not contain water vapor. The pressure exerted by a “dry” gas is equal to the pressure of the “wet” gas minus the vapor pressure of water at the given temperature.

54.

Ptotal = 775 mm Hg Pwater vapor = 31.8 mm Hg at 30 °C Pwater vapor + Pxenon = Ptotal Pxenon = Ptotal – Pwater vapor Pxenon = 775 mm Hg – 31.8 mm Hg = 743 mm Hg

Section 10.10 Ideal Gas Behavior 56.

An ideal gas at absolute zero (0 K) occupies zero volume (0 mL).

58.

A real gas behaves most like an ideal gas under the conditions of high temperature and low pressure.

60.

An ideal gas obeys the gas laws under all conditions. A real gas deviates from ideal gas behavior, and does not strictly obey the gas laws under all conditions.

62. (a) (b) (c) (d)

Description highest kinetic energy lowest kinetic energy fastest molecules slowest molecules

Gas all gases are equal all gases are equal H2 molecules (lightest molecules) O2 molecules (heaviest molecules)

Note: Since all three gases are at the same temperature, each has the same kinetic energy.

Chapter 10

Section 10.11 Ideal Gas Law 64.

PV n = RT

V = 10.0 L T = 100 K P = 3.50 atm

n =

66.

3.50 atm × 10.0 L × 100 K

PV T = nR

1 mol · K 0.0821 atm · L = 4.26 mol He

P = 2.25 atm V = 5.00 L n = 1.25 mol

T =

2.25 atm × 5.00 L × 1.25 mol

1 mol · K 0.0821 atm · L

= 110 K

110 K – 273 = –163 °C

General Exercises 68.

76.0 cm Hg

70.

Ptotal

= 51.0 psi

17.2 cm alcohol 1 cm Hg + 13.5 psi

Ptotal = 65.0 psi × 72.

= 1310 cm alcohol

+ 0.5 psi

= 65.0 psi

1 atm 14.7 psi = 4.42 atm

Phydrogen + Pwater vapor = Ptotal Phydrogen = Ptotal – Pwater vapor Phydrogen = 758 mm Hg – 17.5 mm Hg = 741 mm Hg

initial final

P 741 mm Hg 760 mm Hg 95.0 mL ×

V 95.0 mL V2 741 mm Hg 760 mm Hg

T 20 °C + 273 = 293 K 273 K 273 K 293 K

= 86.3 mL

Gases

74.

P =

nRT V

n = 3.38 × 1022 molecules

1 mol 6.02 × 1023 molecules = 0.0561 mol

1 atm P = 255 mm Hg × 760 mm Hg = 0.336 atm

T = 100 °C + 273 = 373 K V =

76.

78.

0.0561 mol × 373 K 0.336 atm

0.0821 atm · L = 5.12 L 1 mol · K

Propane, C3H8, has a lower molecular mass than butane, C4H10; therefore, C3H8 molecules have the fastest velocity. PV n = RT

1 atm P = 710 mm Hg × 760 mm Hg = 0.934 atm 1L V = 855 mL × 1000 mL = 0.855 L T = 155 °C + 273 = 428 K

n =

0.934 atm × 0.855 L 428 K

1 mol · K 0.0821 atm · L

0.0227 mol

0.0227 mol F2

38.00 g F2 1 mol F2

0.863 g F2

Challenge Exercises 80.

165 cm2

82.

gRT MM = P V

⎛⎜ 1 in. ⎞⎟2 ⎝2.54 cm⎠

14.7 lb 1 in.2

376 lb

g = 5.40 g V = 1.00 L P = 1 atm T = 0 °C + 273 = 273 K

5.40 g × 273 K MM = 1 atm × 1.00 L

0.0821 atm · L 1 mol · K

= 121 g/mol

The molar mass of CF2Cl2 is closest to the calculated value of 121 g/mol.

Chapter 10

84.

MM =

gRT PV

g = 2.85 g 1L V = 750 mL × 1000 mL = 0.750 L P = 760 mm Hg = 1 atm T = 100 °C + 273 = 373 K

2.85 g × 373 K MM = 1 atm × 0.750 L

0.0821 atm · L = 116 g/mol 1 mol · K

Gases

CHAPTER

Liquids and Solids

Section 11.1 Properties of Liquids 2.

General Observed Properties of Liquids True, liquids have a variable shape. True, most liquids flow readily. False, liquids do not expand significantly with changes in temperature or pressure; however, they may undergo a change in physical state. (d) False, liquids are about 1000 times more dense than gases. (a) (b) (c)

(a) (b) (c) (d)

Substance Ne Ne Ar Ar

Temperature –225 °C –255 °C –175 °C –200 °C

Physical State gas solid gas solid

Section 11.2 The Intermolecular Bond Concept 6.

A hydrogen bond is a specific example of permanent dipole attraction between molecules. A permanent dipole is caused by electrons being drawn closer to one atom than another within a molecule. If a molecule has an O–H bond such as H2O, or an N–H bond such as NH3, a hydrogen bond results between the H atom and the O or N atom on another molecule.

Section 11.3 Vapor Pressure, Boiling Point, Viscosity, and Surface Tension 8.

The boiling point of water occurs when the total pressure exerted by the water molecules in the gaseous state above the liquid equals the pressure exerted by the atmosphere.

10.

A water molecule at the surface of the liquid is attracted to adjacent molecules. This attraction, called surface tension, causes a drop of water to have the smallest possible surface area, and thus assumes a spherical shape.

Liquids and Solids

12.

14.

16.

18. 20.

22.

If the molecules in a liquid are weakly attracted: (a) the vapor pressure is high. (b) the boiling point is low. (c) the viscosity is low. (d) the surface tension is low. (a) (b)

Liquid CH3COOH or C2H5Cl C2H5OH or CH3OCH3

Higher Boiling Point CH3COOH (stronger attraction) C2H5OH (stronger attraction)

(a) (b)

Liquid CH3COOH or C2H5Cl C2H5OH or CH3OCH3

Higher Surface Tension CH3COOH (stronger attraction) C2H5OH (stronger attraction)

When the vapor pressure of the liquid is equal to the atmospheric pressure, the liquid begins to boil. (a) (b)

Temperature 15 °C 30 °C

Vapor Pressure ~70 mmHg ~300 mmHg

The temperature at which the vapor pressure of a liquid is equal to the atmospheric pressure is the boiling point. Thus, the boiling point of methanol is 65 °C because its vapor pressure at this temperature is 1 atm.

Section 11.4 Properties of Solids 24.

26.

(a) (b) (c) (d)

General Observed Properties of Solids True, solids have a fixed shape. True, solids can have a crystalline structure. True, solids do not compress significantly. False, solids are more dense than liquids (H2O and NH3 are exceptions).

(a) (b)

Substance Hg Hg

Temperature ice water, 0 °C boiling water, 100 °C

Physical State liquid liquid

Section 11.5 Crystalline Solids 28. 30.

Most solids are noncrystalline heterogeneous mixtures. For example, rocks, wood, concrete, and most naturally occurring minerals are noncrystalline. (a) (b) (c)

Properties malleable, ductile, conductor high melting point, conductor low melting point, nonconductor

Chapter 11

Type of Solid metallic solid ionic solid molecular solid

32. (a) (b) (c) (d)

Crystalline Solid Zn ZnO P4 IBr

Classification metallic solid ionic solid molecular solid molecular solid

Section 11.6 Changes of Physical State 34.

Cooling Curve for Acetone

—Gas

Bp (°C) Hvapor —Liquid

Mp (°C) Hfusion

—Solid

Heat Lost (cal) 36.

Energy required to melt 25.0 g ice to water at 0 °C: 25.0 g

38.

80.0 cal 1g

= 2.00 × 103 cal (2.00 kcal)

Energy released when water cools: 65.5 g

1.00 cal × (55.5 – 0.0)°C = 3640 cal (3.64 kcal) 1 g × °C

Energy released when water solidifies: 65.5 g

80.0 cal = 5240 cal (5.24 kcal) 1g

Total heat energy released: 3640 cal + 5240 cal = 8880 cal (8.88 kcal)

Liquids and Solids

40.

Energy released when steam condenses: 155 g

540 cal = 83,700 cal (83.7 kcal) 1g

Energy released when water cools: 155 g

1.00 cal × (100.0 – 0.0)°C = 15,500 cal (15.5 kcal) 1 g × °C

Energy released when water solidifies: 155 g

80.0 cal = 12,400 cal (12.4 kcal) 1g

Total heat energy released: 83,700 cal + 15,500 cal + 12,400 cal = 111,600 cal (111.6 kcal)

42.

Energy released when steam cools: 90.5 g

0.48 cal × (110.0 – 100.0)°C = 434 cal (0.434 kcal) 1 g × °C

Energy released when steam condenses: 90.5 g

540 cal = 48,900 cal (48.9 kcal) 1g

Energy released when water cools: 90.5 g

1.00 cal × (100.0 – 0.0)°C = 9050 cal (9.05 kcal) 1 g × °C

Energy released when water solidifies: 90.5 g

80.0 cal = 7240 cal (7.24 kcal) 1g

Total heat energy released: 434 cal + 48,900 cal + 9050 cal + 7240 cal = 65,600 cal (65.6 kcal)

Chapter 11

Section 11.7 Structure of Water 44.

Water has an unusually high boiling point.

46.

Water has an unusually high surface tension.

48.

The intermolecular bond existing between water molecules is a hydrogen bond.

50.

Net Dipole for a Water Molecule Shown with Standard Notation O H

Section 11.8 Physical Properties of Water 52.

Liquid ammonia has the highest heat of fusion; water has the highest heat of vaporization.

54.

Since a solid cube of ammonia floats in liquid ammonia, the liquid is more dense than the solid.

56.

58.

60.

(a) (b)

Liquid H2O or H2Se H2S or H2Te

Higher Boiling Point H2O (hydrogen bond attraction) H2Te (larger size, more attraction)

(a) (b)

Liquid H2O or H2Se H2S or H2Te

Higher Heat of Vaporization H2O (hydrogen bond attraction) H2Te (larger size, more attraction)

Predicted Values for Radioactive H2Po The physical properties if H2Po can be estimated by comparison to H2Te. That is, we can estimate an increase in the values from H2Te to H2Po by comparison to the increase in the values from H2Se to H2Te. Thus, Mp: Bp: Hfusion: Hvapor:

–48.9 + [–48.9 – (–60.4)] = ~ –37.4 °C –2.2 + [–2.2 – (–41.5)] = ~ 37.1 °C 1670 + (1670 – 899) = ~ 2441 cal/mol 5570 + (5570 – 4620) = ~ 6520 cal/mol

Liquids and Solids

Section 11.9 Chemical Properties of Water electrolysis →

62.

2 H2O(l)

64.

(a) (b) (c)

2 Rb(s) + 2 H2O(l) → 2 RbOH(aq) + H2(g) Cs2O(s) + H2O(l) → 2 CsOH(aq) P2O5(s) + 3 H2O(l) → 2 H3PO4(aq)

66.

(a) (b) (c)

Ba(s) + 2 H2O(l) → Ba(OH)2(aq) + H2(g) CaO(s) + H2O(l) → Ca(OH)2(aq) N2O3(g) + H2O(l) → 2 HNO2(aq)

68.

(a)

2 C4H10(g) + 13 O2(g)

spark →

(b)

Co(C2H3O2)2 • 4H2O(s)

Δ →

(c)

2 HNO3(aq) + Ba(OH)2(aq) →

(a)

C4H9OH(l) + 6 O2(g)

spark →

(b)

KAl(SO4)2 • 12H2O(s)

Δ →

KAl(SO4)2(s) + 12 H2O(g)

(c)

H3PO4(aq) + 3 NaOH(aq)

→

Na3PO4(aq) + 3 HOH(l)

70.

2 H2(g) + O2(g)

8 CO2(g) + 10 H2O(g) Co(C2H3O2)2(s) + 4 H2O(g) Ba(NO3)2(aq) + 2 HOH(l) 4 CO2(g) + 5 H2O(g)

Section 11.10 Hydrates 72.

74.

(a) (c) (b) (d)

Chemical Formula MgSO4 • 7H2O MnSO4 • H2O Co(C2H3O2)2 • 4H2O Cr(NO3)3 • 3H2O

Systematic Name magnesium sulfate heptahydrate manganese(II) sulfate monohydrate cobalt(II) acetate tetrahydrate chromium(III) nitrate trihydrate

Systematic Name (a) sodium carbonate decahydrate (b) nickel(II) nitrate hexahydrate (c) cobalt(III) iodide octahydrate (d) chromium(III) acetate monohydrate

Chapter 11

Chemical Formula Na2CO3 • 10H2O Ni(NO3)2 • 6H2O CoI3 • 8H2O Cr(C2H3O2)3 • H2O

76.

(a)

Percentage of water in SrCl2 • 6H2O MM of SrCl2 = 87.62 g + 2(35.45 g) = 158.52 g 6(18.02 g) Percentage of water: 158.52 g + 6(18.02 g)

(b)

× 100% = 40.55% H2O

Percentage of water in K2Cr2O7 • 2H2O MM of K2Cr2O7 = 2(39.10 g) + 2(52.00 g) + 7(16.00 g) = 294.20 g 2(18.02 g) Percentage of water: 294.20 g + 2(18.02 g) × 100% = 10.91% H2O

(c)

Percentage of water in Co(CN)3 • 3H2O MM of Co(CN)3 = 58.93 g + 3(12.01 g) + 3(14.01 g) = 136.99 g 3(18.02 g) Percentage of water: 136.99 g + 3(18.02 g) × 100% = 28.30% H2O

(d) Percentage of water in Na2CrO4 • 4H2O MM of Na2CrO4 = 2(22.99 g) + 52.00 g + 4(16.00 g) = 161.98 g 4(18.02 g) Percentage of water: 161.98 g + 4(18.02 g) × 100% = 30.80% H2O 78.

(a)

Δ →

NiCl2 • XH2O(s) 21.7 g H2O

NiCl2(s) + XH2O(g)

1 mol H2O 18.02 g H2O = 1.20 mol H2O

1 mol NiCl2 78.3 g NiCl2 × 129.59 g NiCl = 0.604 mol NiCl2 2 1.20 NiCl2 • 0.604 H2O

1.20 0.604 = 1.99 ≈ 2 Chemical Formula: NiCl2 • 2H2O

Liquids and Solids

(b)

Δ →

Ni(NO3)2 • XH2O(s) 37.2 g H2O

Ni(NO3)2(s) + X H2O(g)

1 mol H2O × 18.02 g H O = 2.06 mol H2O 2

1 mol Ni(NO3)2 62.8 g Ni(NO3)2 × 182.71 g Ni(NO ) = 0.344 mol Ni(NO3)2 3 2 2.06 Ni(NO3)2 • 0.344 H2O

2.06 0.344 = 5.99 ≈ 6

Chemical Formula: Ni(NO3)2 • 6H2O

(c)

CoSO4 • XH2O(s) 10.4 g H2O

Δ →

CoSO4(s) + X H2O(g)

1 mol H2O × 18.02 g H O = 0.577 mol H2O 2

1 mol CoSO4 89.6 g CoSO4 × 155.00 g CoSO = 0.578 mol CoSO4 4 0.578 CoSO4 • 0.577 H2O

0.578 0.577

= 1.00 ≈ 1

Chemical Formula: CoSO4 • H2O

(d) CrI3 • XH2O(s)

Δ →

CrI3(s) + XH2O(g)

1 mol H2O 27.2 g H2O × 18.02 g H O 2 72.8 g CrI3

= 1.51 mol H2O

1 mol CrI3 × 432.70 g CrI 3

1.51 CrI3 • 0.168 H2O

= 0.168 mol CrI3

1.51 0.168 = 8.99 ≈ 9 Chemical Formula: CrI3 • 9H2O

Chapter 11

General Exercises 80.

Water covers about three-fourths (~75%) of Earth’s surface.

82.

Liquid (a) C6H14 (b) CH3CH2CH2–F (c) CH3–COOH (d) CH3CH2–O–CH2CH3

84.

The boiling point of ethanol at 0.500 atm (380 mm Hg) is approximately 65 °C.

86.

Ammonia, NH3, has N–H bonds, so molecules can hydrogen bond and have a strong intermolecular attraction. Thus, ammonia has high surface tension, minimal surface area, and ammonia “raindrops” on Saturn will be spherical.

Intermolecular Attraction dispersion force dipole force hydrogen bond dipole force

Challenge Exercises 88.

Mass of ethylene glycol: 5.00 L

1000 mL 1L

1.11 g 1 mL

= 5,550 g

Energy required to heat ethylene glycol: 5,550 g

0.561 cal × 1 g ∞ °C

× (197.6 – 25.0)°C = 537,000 cal (537 kcal)

Energy required to vaporize ethylene glycol: 5,550 g

201 cal 1g

= 1,120,000 cal (1,120 kcal)

Total heat energy required: 537,000 cal + 1,120,000 cal = 1,660,000 cal (1,660 kcal)

Liquids and Solids

CHAPTER

Chemical Bonding

Section 12.1 The Chemical Bond Concept 2.

Atom H Cl

Valence Electrons 1 valence e– 7 valence e–

Atom H Cl

Covalent Bond 2 valence e– 8 valence e–

10.

(a) (c)

Compound ZnBr2 IF7

Bond ionic covalent

Compound (b) NH3 (d) PbSO4

Bond covalent ionic

(a) (c)

Substance CH3CH2OH NiO

Particle molecule formula unit

Substance (b) NaClO3 (d) H2SO4

Particle formula unit molecule

(a) (c)

Substance C3 H 6 O Fe3O4

Particle molecule formula unit

Substance (b) Co (d) P4

Particle atom molecule

Chemical Bonding

Section 12.2 Ionic Bonds 12.

(a) (c)

Ion Na ion Ga ion

Ionic Charge 1+ (Group 1) 3+ (Group 13)

Ion (b) Ba ion (d) Pb ion

Ionic Charge 2+ (Group 2) 4+ (Group 14)

(a) (c)

Ion Cl ion Se ion

Ionic Charge 1– (Group 17) 2– (Group 16)

Ion (b) I ion (d) P ion

Ionic Charge 1– (Group 17) 3– (Group 15)

(b) (d) (a) (c)

Ion K+ Ca2+ Sc3+ Ti4+

Electron Configuration 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6

Ion (a) Cl– (b) S2– (c) P3– (d) Si4–

Electron Configuration 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6

Ion Se2– Br– Rb+ Sr2+

Isoelectronic Noble Gas Kr Kr Kr Kr

22.

Ion (a) Sc3+ (b) K+ (c) Ti4+ (d) Ba2+

Isoelectronic Noble Gas Ar Ar Ar Xe

24.

Ion (a) Br– (b) O2– (c) Se2– (d) N3–

Isoelectronic Noble Gas Kr Ne Kr Ne

14.

16.

18.

20. (a) (b) (c) (d)

90 Chapter 12

26.

(a) (b) (c) (d)

Al atom radius > Al ion radius Pb atom radius > Pb ion radius Se atom radius < Se ion radius N atom radius < N ion radius

28.

The true statements are (a) and (d). The corrected statements are: (b) Cobalt atoms lose electrons and sulfur atoms gain electrons. (c) The ionic radius of a cobalt ion is less than its atomic radius.

Section 12.3 Covalent Bonds 30.

(a) (b)

The bond length in H–Br is less than the sum of the atomic radii. The bond length in S=O is less than the sum of the atomic radii.

32.

The true statements are (b) and (c). The corrected statements are: (a) Valence electrons are shared between carbon and oxygen atoms. (d) Energy is required to break a C—O covalent bond.

Section 12.4 Electron Dot Formulas of Molecules 34.

Molecule

Valence Electrons

(a)

Cl2

7 + 7 = 14 e–

(b)

6 + 6 = 12 e–

(c)

1 + 7 = 8 e–

(d) PH3

5 + 3(1) = 8 e–

Electron Dot

Structural Formula

Cl Cl

H I

H H P H

H H

Chemical Bonding

36.

38.

Molecule

Valence Electrons

(a)

HOCl

1 + 6 + 7 = 14 e–

(b)

SO2

6 + 2(6) = 18 e–

(c)

CS2

4 + 2(6) = 16 e–

Electron Dot

Structural Formula

H O Cl

O S

(d) C2H2

2(4) + 2(1) = 10 e–

H C

C H

Molecule

Valence Electrons

Electron Dot

H C

Structural Formula

(a)

(b)

CCl4

HONO2

4 + 4(7) = 32 e–

1 + 3(6) + 5 = 24 e–

CH3OH

4 + 4(1) + 6 = 14 e–

(d) HOCN

1 + 6 + 4 + 5 = 16 e–

(c)

92 Chapter 12

Cl C Cl Cl

H O N O

H O N

O H

H H C O H H

H O C

C H

Section 12.5 Electron Dot Formulas of Polyatomic Ions 40.

Polyatomic Ion Valence Electrons (a)

BrO–

7 + 6 + 1 = 14 e–

(b)

BrO2–

7 + 2(6) + 1 = 20 e–

(c)

BrO3–

7 + 3(6) + 1 = 26 e–

Electron Dot

O –

42.

7 + 4(6) + 1 = 32 e–

Polyatomic Ion Valence Electrons

O –

O ]–

O Br O

O –

O ]–

O O

–

PO43–

5 + 4(6) + 3 = 32 e–

HPO42–

1 + 5 + 4(6) + 2 = 32 e–

PO33–

(d) HPO32–

5 + 3(6) + 3 = 26 e–

1 + 5 + 3(6) + 2 = 26 e–

O ]–

Electron Dot

Structural Formula

O [ O

O ] 3–

P O

H O P O O

O P O (c)

Br O

(b)

[ O

O 3– O P O (a)

O ]–

[ Br

O Br

O O Br (d) BrO4–

Structural Formula

2–

O [H–O

O ] 2–

3–

[ O

H O P O O

O ] 3–

P O

2–

[H–O

O ] 2–

Chemical Bonding

44.

Polyatomic Ion Valence Electrons

Electron Dot H

(a)

(b)

PH4+

SeO32–

5 + 4(1) – 1 = 8 e–

6 + 3(6) + 2 = 26 e–

H P H H

Structural Formula H [ H – P – H ]+

[ O – Se – O ] 2–

O Se O 2– O

O [ O – C – O ] 2–

O C O 2– (c)

CO32–

(d) BO33–

4 + 3(6) + 2 = 24 e–

5 + 3(6) + 3 = 26 e–

[ O – B – O ] 3–

O B O 3– O

Section 12.6 Polar Covalent Bonds 46.

Within a group of elements, the electronegativity increases from bottom to top in the periodic table.

48.

Nonmetals are more electronegative than semimetals.

50.

More Electronegative More Electronegative (a) Se < Br (b) C > B (c) Te < S (d) Ba < Be (Note: The more electronegative element is in bold.)

52. (a) (c)

Bond H—Cl N—O

(a) (c)

Polar Bonds Using Delta Notation δ+ H—S δ– (b) δ– O—S δ+ (d) δ+ S—Cl δ– δ+ N—F δ–

54.

Polarity 3.0 – 2.1 = 0.9 3.5 – 3.0 = 0.5

Bond (b) H—Br (d) C—O

Polarity 2.8 – 2.1 = 0.7 3.5 – 2.5 = 1.0

(Note: δ– indicates the more electronegative atom and δ+ indicates the more electropositive atom.)

94 Chapter 12

Section 12.7 Nonpolar Covalent Bonds 56.

Bond Polarity (a) I—C 2.5 – 2.5 = 0 (b) C—S 2.5 – 2.5 = 0 (c) S—H 2.5 – 2.1 = 0.4 (d) H—Br 2.8 – 2.1 = 0.7 Thus, (a) and (b) are nonpolar.

Classification nonpolar nonpolar polar polar

58.

O2, F2, and I2 occur naturally as diatomic molecules.

Section 12.8 Coordinate Covalent Bonds (Note: Coordinate covalent bonds are indicated by a dash, —.) 60.

62.

Molecule

Valence Electrons

HIO

1 + 7 + 6 = 14 e–

Molecule

Valence Electrons

Electron Dot H

64.

Polyatomic Ion Valence Electrons

PH4+ 66.

1 + 7 + 2(6) = 20 e–

5 + 4(1) – 1 = 8 e–

Polyatomic Ion Valence Electrons

Electron Dot

HIO2

5 + 4(6) + 3 = 32 e–

Coord. Cov. Bond H

Electron Dot

Coord. Cov. Bond

H P H H

Electron Dot

O 3– O P O PO43–

Coord. Cov. Bond

H P H

Coord. Cov. Bond

O O P O

3–

Chemical Bonding

Section 12.9 Hydrogen Bonds H 68.

H—N: – – – H— N: H

Hydrogen bond

70.

A hydrogen bond is ~50% longer than a polar covalent bond.

Section 12.10 Shapes of Molecules 72.

Formula CBr4

Electron Pair tetrahedral

Molecular Shape tetrahedral

Bond Angle 109.5°

74.

Formula PH3

Electron Pair tetrahedral

Molecular Shape trigonal pyramidal

Bond Angle 107°

76.

Formula Cl2O

Electron Pair tetrahedral

Molecular Shape bent

Bond Angle 104.5°

78.

In SiF4, each of the four Si–F bonds is polar, but each fluorine atom pulls equally in opposite directions to give a nonpolar molecule. All four F atoms are at the corners of a tetrahedron and the molecular shape is described as tetrahedral.

General Exercises 80. (a) (b) (c) (d)

Substance Pu O2 Pu2O3 H2O2

Particle atom molecule formula unit molecule

(a) (b) (c) (d)

Ions 1 Sc3+ and 1 N3– 1 Ti4+ and 2 O2– 2 NH4+ and 1 CO32– 3 Hg22+ and 2 PO43–

Chemical Formula ScN TiO2 (NH4)2CO3 (Hg2)3(PO4)2

82.

96 Chapter 12

84.

The radius of a chloride ion is about twice that of a chlorine atom because the chloride ion has one more electron than proton. The extra electron repels other valence electrons, and the ion becomes larger.

86.

Bond Sb—Cl

Polarity 3.0 – 1.9 = 1.1 (polar)

88.

Bond S—I

Polarity 2.5 – 2.5 = 0 (nonpolar)

90.

Polar Bond and Delta Notation

δ+ As—Cl δ–

92.

Molecule

Electron Dot

SbH3

94.

Valence Electrons

5 + 3(1) = 8 e–

Polyatomic Ion Valence Electrons

H Sb H

Structural Formula H Sb H

Electron Dot

Structural Formula [ O Si O ] 2–

O Si O 2–

SiO32–

4 + 3(6) + 2 = 24 e–

96.

Formula SiO2

Electron Pair linear

Molecular Shape linear

Bond Angle 180°

98.

Formula

Electron Pair

Molecular Shape

Bond Angle

CO32–

trigonal planar

120°

Chemical Bonding

Challenge Exercises 100.

Molecule

Valence Electrons

Electron Dot

XeO3

8 + 3(6) = 26 e–

O Xe O O

98 Chapter 12

Structural Formula O Xe O

CHAPTER

Solutions

Section 13.1 Gases in Solution 2.

(a) (b)

Change temperature of soda increases partial pressure of CO2 increases

solubility × pressure factor = new solubility 1.90 mL N2 1 dL blood

Result solubility of CO2 gas decreases solubility of CO2 gas increases

1.00 atm

× ×

4.50 atm 1.00 atm

8.55 mL N2 1 dL blood

0.55 g N2O 0.12 g N2O

4.6 atm

Section 13.2 Liquids in Solution 8. 10.

12.

14.

(a) (b)

nonpolar solute + polar solvent = nonpolar solute + nonpolar solvent =

immiscible miscible

(a) (b) (c) (d)

Solvent C3H7OH C5H12 C6H4(CH3)2 C2H3Cl3

Classification polar solvent nonpolar solvent nonpolar solvent nonpolar solvent

(a) (b) (c) (d)

Solvent C7 H 8 C2H5OH CH3COOH C2HCl3

Miscible or Immiscible immiscible with H2O miscible with H2O miscible with H2O immiscible with H2O

Salad oil is a nonpolar liquid, whereas water is a polar liquid. Because oil and water differ in polarity, the two liquids are immiscible and the less dense oil layer floats on the water layer.

Solutions

Section 13.3 Solids in Solution 16.

18.

20.

(a) (b) (c)

polar solute + polar solvent nonpolar solute + polar solvent ionic solute + polar solvent

= soluble = insoluble = soluble

(a) (b) (c) (d)

Compound trichloroethylene, C2HCl3 iron(III) nitrate, Fe(NO3)3 tartaric acid, H2C4H4O6 dodecane, C12H26

Soluble or Insoluble soluble in C6H14 insoluble in C6H14 insoluble in C6H14 soluble in C6H14

(a) (b) (c) (d)

Vitamin vitamin B3, C6H6N2O vitamin C, C6H8O6 vitamin D, C27H44O vitamin K, C31H46O2

Soluble or Insoluble water soluble water soluble fat soluble fat soluble

Section 13.4 The Dissolving Process 22.

Sucrose crystal, C12H22O11, dissolving in water: water

water

crystal of sucrose water

Water molecules attack the exposed edges and corners of the sugar crystal. 24.

(a)

Cobalt(II) sulfate, CoSO4, dissolved in water: Co2+

(b)

100

SO 42–

Nickel(II) nitrate, Ni(NO3)2, dissolved in water:

Chapter 1413

Section 13.5 Rate of Dissolving 26.

(a) (b) (c) (d)

Factor using refrigerator water shaking the solution using powdered sugar using tap water

Rate of Dissolving decreases increases increases no effect

Section 13.6 Solubility and Temperature 28. 30. 32. 34. 36.

(a) (a) (a) (a) (a) (b) (c)

~85 g LiCl/100 g H2O ~100 g LiCl/100 g H2O ~35 °C ~60 °C supersaturated at 40 °C saturated at 50 °C unsaturated at 60 °C

(b) (b) (b) (b)

~110 g C12H22O11/100 g H2O ~130 g C12H22O11/100 g H2O ~40 °C ~55 °C

Section 13.7 Unsaturated, Saturated, and Supersaturated Solutions 38.

The solubility of KCl is ~35 g/100 g water at 20 °C (Figure 13.5). (a) ~35 g KCl remains in solution (b) ~15 g KCl (50 g – 35 g) crystallizes from solution

40.

(a) (b) (c)

42.

25 g H2O

supersaturated at 25 °C saturated at 55 °C unsaturated at 75 °C

110 g sugar = 28 g sugar can dissolve 100 g H2O The solution contains 25 g of sugar; thus, the solution is unsaturated at 30 °C. ×

Section 13.8 Mass/Mass Percent Concentration 44.

mass of solute mass of solution

× 100% = m/m %

(a)

20.0 g KI 20.0 g KI + 100.0 g water × 100% = 16.7%

(b)

2.50 g AgC2H3O2 2.50 g AgC2H3O2 + 95.0 g water × 100% = 2.56%

(c)

5.57 g SrCl2 5.57 g SrCl2 + 225.0 g water × 100% = 2.42%

50.0 g C12H22O11 (d) 50.0 g C H O + 200.0 g water × 100% = 20.0% 12 22 11

Solutions

101

46. (a)

(b)

(c)

Solution

Unit Factors

3.35% MgCl2

3.35 g MgCl2 100 g solution

and

100 g solution 3.35 g MgCl2

96.65 g H2O 100 g solution

and

100 g solution 96.65 g H2O

3.35 g MgCl2 96.65 g H2O

and

96.65 g H2O 3.35 g MgCl2

5.25 g Cd(NO3)2 100 g solution

and

100 g solution 5.25 g Cd(NO3)2

94.75 g H2O 100 g solution

and

100 g solution 94.75 g H2O

5.25 g Cd(NO3)2 94.75 g H2O

and

94.75 g H2O 5.25 g Cd(NO3)2

6.50 g Na2CrO4 100 g solution

and

100 g solution 6.50 g Na2CrO4

93.50 g H2O 100 g solution

and

100 g solution 93.50 g H2O

6.50 g Na2CrO4 93.50 g H2O

and

93.50 g H2O 6.50 g Na2CrO4

7.25 g ZnSO4 100 g solution

and

100 g solution 7.25 g ZnSO4

92.75 g H2O 100 g solution

and

100 g solution 92.75 g H2O

7.25 g ZnSO4 92.75 g H2O

and

92.75 g H2O 7.25 g ZnSO4

5.25% Cd(NO3)2

6.50% Na2CrO4

(d) 7.25% ZnSO4

48.

102

(a)

35.0 g HCl

(b)

10.5 g HC2H3O2

Chapter 1413

100 g solution = 5.00 g HCl

× ×

100 g solution 4.50 g HC2H3O2

7.00 × 102 g solution =

233 g solution

50.

52.

(a)

10.0 g solution

2.50 g K2CO3 100 g solution

0.250 g K2CO3

(b)

50.0 g solution

5.00 g Li2SO4 100 g solution

2.50 g Li2SO4

(a)

250.0 g solution

99.10 g H2O 100 g solution

247.8 g H2O

(b)

100.0 g solution

95.00 g H2O 100 g solution

95.00 g H2O

Section 13.9 Molar Concentration 54.

(a)

MM of KCl = 39.10 g/mol + 35.45 g/mol = 74.55 g/mol

1.00 g KCl 75.0 mL solution (b)

1 mol KCl 74.55 g KCl

1000 mL solution = 0.179 M KCl 1 L solution

MM of Na2CrO4 = 2(22.99 g/mol) + 52.00 g/mol + 4(16.00 g/mol) = 161.98 g/mol

1.00 g Na2CrO4 1 mol Na2CrO4 1000 mL solution = 0.0823 M Na2CrO4 75.0 mL solution × 161.98 g Na2CrO4 × 1 L solution (c)

MM of MgBr2 = 24.31 g/mol + 2(79.90 g/mol) = 184.11 g/mol

20.0 g MgBr2 1 mol MgBr2 250.0 mL solution × 184.11 g MgBr2

1000 mL solution = 0.435 M MgBr2 1 L solution

(d) MM of Li2CO3 = 2(6.94 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 73.89 g/mol 20.0 g Li2CO3 1 mol Li2CO3 1000 mL solution × × = 1.08 M Li2CO3 250.0 mL solution 73.89 g Li2CO3 1 L solution

Solutions

103

56.

Solution (a)

Unit Factors 0.150 mol KBr 1 L solution 0.150 mol KBr 1000 mL solution

0.150 M KBr

58.

0.150 M Ca(NO3)2

(c)

0.333 M Sr(C2H3O2)2

(a)

MM of NaF

10.0 g NaF (b)

and

1 L solution 0.150 mol KBr 1000 mL solution 0.150 mol KBr

0.150 mol Ca(NO3)2 1 L solution and 1 L solution 0.150 mol Ca(NO3)2 0.150 mol Ca(NO3)2 1000 mL solution and 0.150 mol Ca(NO ) 1000 mL solution 3 2

(b)

(d) 0.333 M NH4Cl

and

0.333 mol Sr(C2H3O2)2 1 L solution and 1 L solution 0.333 mol Sr(C2H3O2)2 0.333 mol Sr(C2H3O2)2 1000 mL solution and 0.333 mol Sr(C H O ) 1000 mL solution 2 3 2 2 0.333 mol NH4Cl 1 L solution 0.333 mol NH4Cl 1000 mL solution

and and

1 L solution 0.333 mol NH4Cl 1000 mL solution 0.333 mol NH4Cl

= 22.99 g/mol + 19.00 g/mol = 41.99 g/mol 1 mol NaF 41.99 g NaF

1 L solution 0.275 mol NaF

= 0.866 L solution

MM of CdCl2 = 112.41 g/mol + 2(35.45 g/mol) = 183.31 g/mol

1 mol CdCl2 1 L solution 10.0 g CdCl2 × 183.31 g CdCl × 0.275 mol CdCl = 0.198 L solution 2 2 (c)

MM of K2CO3 = 2(39.10 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 138.21 g/mol

1 mol K2CO3 1 L solution 10.0 g K2CO3 × 138.21 g K CO × 0.408 mol K CO = 0.177 L solution 2 3 2 3 (d) MM of Fe(ClO3)3 = 55.85 g/mol + 3(35.45 g/mol) + 9(16.00 g/mol) = 306.20 g/mol 1 mol Fe(ClO3)3 10.0 g Fe(ClO3)3 × 306.20 g Fe(ClO ) × 3 3

104

Chapter 1413

1 L solution 0.408 mol Fe(ClO3)3 = 0.0800 L solution

60.

(a)

MM of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol 1.00 L solution ×

(b)

= 4.00 g NaOH

MM of LiHCO3 = 6.94 g/mol + 1.01 g/mol + 12.01 g/mol + 3(16.00 g/mol) = 67.96 g/mol

1.00 L solution ×

(c)

0.100 mol NaOH 40.00 g NaOH × 1 mol NaOH 1 L solution

0.100 mol LiHCO3 67.96 g LiHCO3 × 1 L solution 1 mol LiHCO3

= 6.80 g LiHCO3

MM of CuCl2 = 63.55 g/mol + 2(35.45 g/mol) = 134.45 g/mol

0.500 mol CuCl2 134.45 g CuCl2 25.0 mL solution × 1000 mL solution × 1 mol CuCl2 = 1.68 g CuCl2 (d) MM of KMnO4 = 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol) = 158.04 g/mol 0.500 mol KMnO4 25.0 mL solution × 1000 mL solution × 62.

MM of CaSO4

158.04 g KMnO4 1 mol KMnO4 = 1.98 g KMnO4

= 40.08 g/mol + 32.07 g/mol + 4(16.00 g/mol) = 136.15 g/mol

0.209 g CaSO4 1 mol CaSO4 1000 mL solution × × = 0.0154 M CaSO4 100.0 mL solution 136.15 g CaSO4 1 L solution

64.

(a)

0.453 g NaCl 50.320 g solution × 100% = 0.900%

(b)

MM of NaCl

0.453 g NaCl 50.0 mL solution

= 22.99 g/mol + 35.45 g/mol = 58.44 g/mol ×

1 mol NaCl 58.44 g NaCl

1000 mL solution 1 L solution

= 0.155 M NaCl

Solutions

105

Section 13.10 Dilution of a Solution 66.

50.0 mL solution

0.85 mol HC2H3O2 500.0 mL solution

17 moL HC2H3O2 1000 mL solution

1000 mL solution 1 L solution

Algebraic Solution: M1 ×

17 M × 50.0 mL 500.0 mL

68.

0.85 mol HC2H3O2

1.7 mol HCl 1 L solution

1.7 M HC2H3O2

5.00 L solution

0.10 mol HNO3 1 L solution

0.50 mol HNO3

1 L solution 16 mol HNO3

0.031 L HNO3

Algebraic Solution: M1 ×

0.10 M × 5.00 L 16 M

0.031 L HNO3 (31 mL HNO3)

Section 13.11 Solution Stoichiometry 70.

Hg(NO3)2(aq) (a)

2 KI(aq)

HgI2(s)

→

2 KNO3(aq)

1 mol Hg(NO3)2 0.170 mol KI 25.0 mL KI × 1000 mL KI × 2 mol KI

(b) 25.0 mL KI ×

0.170 mol KI 1000 mL KI ×

1 mol HgI2 2 mol KI

1000 mL Hg(NO3)2 0.209 mol Hg(NO3)2

10.2 mL Hg(NO3)2

454.39 g HgI2 1 mol HgI2 =

106

Chapter 1413

0.966 g HgI2

General Exercises 72.

Smoke demonstrates the Tyndall effect when it contains small particles ranging in size from 1–100 nm.

74.

0.0019 g N2 100 g blood

76.

0.63 g Cl2 500 mL solution × 100 mL solution

4.68 atm × 1.00 atm

= 0.0089 g N2/100 g blood = 3.2 g Cl2

78.

Although phenol, C6H5OH, contains a polar O–H bond, it is only partially miscible in water because it has mostly nonpolar C–H bonds.

80.

Water can hydrogen bond with the polar –COOH in an acid, and draw it into solution. As the nonpolar CxHy– portion of the acid molecule increases, the acid becomes less soluble, and eventually immiscible with water.

82.

(a) (b) (c) (d)

Household Substance grease maple syrup food color gasoline

Better Solvent chloroform water water chloroform

Challenge Exercises 84.

MM of NaClO = 22.99 g/mol + 35.45 g/mol + 16.00 g/mol = 74.44 g/mol 5.00% NaClO = 5.00 g NaClO per 100 g solution 1 mol NaClO Solute: 5.00 g NaClO × 74.44 g NaClO = 0.0672 mol NaClO 1 mL bleach 1 L bleach Solution: 100 g bleach × 1.04 g bleach × 1000 mL bleach = 0.0962 L bleach

Molarity:

0.0672 mol NaClO 0.0962 L bleach

= 0.699 M NaClO

Solutions

107

CHAPTER

Acids and Bases

Section 14.1 Properties of Acids and Bases 2. (a) (b) (c) (d)

General Properties of a Basic Solution True, a basic solution has a bitter taste. True, a basic solution turns litmus paper blue. False, a basic solution has a pH greater than 7. False, a basic solution neutralizes an acid.

(a) (b) (c) (d)

Aqueous Solution battery acid lime juice shampoo ammonia

pH 0 2.7 7.0 11.1

Classification strongly acidic weakly acidic neutral weakly basic

Section 14.2 Arrhenius Acids and Bases 6.

10.

(a) (b) (c) (d)

Base Al(OH)3(aq) Sr(OH)2(aq) Ba(OH)2(aq) Mg(OH)2(aq)

Dissociation ~ 1% ~ 100% ~ 100% ~ 1%

(a) (c)

Formula HNO3(aq) KOH(aq)

(a) (b)

Arrhenius Acid HC2H3O2(aq) H3PO4(aq)

Classification Arrhenius acid Arrhenius base

Strength weak base strong base strong base weak base Formula (b) KNO3(aq) (d) K2CO3(aq)

Classification salt salt

Arrhenius Base LiOH(aq) KOH(aq)

Acids and Bases

109

12. (a) (b) (c) (d)

Salt KBr(aq) BaCl2(aq) CoSO4(aq) Na3PO4(aq)

Acid HBr HCl H2SO4 H3PO4

Base KOH Ba(OH)2 Co(OH)2 NaOH

Neutralization Reactions:

14.

(a) (b) (c) (d)

HBr(aq) + KOH(aq) → KBr(aq) + H2O(l) 2 HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2 H2O(l) H2SO4(aq) + Co(OH)2(s) → CoSO4(aq) + 2 H2O(l) H3PO4(aq) + 3 NaOH(aq) → Na3PO4(aq) + 3 H2O(l)

(a) (b)

2 HC2H3O2(aq) + Sr(OH)2(aq) → Sr(C2H3O2)2(aq) + 2 H2O(l) H2SO4(aq) + 2 NH4OH(aq) → (NH4)2SO4(aq) + 2 H2O(l)

Section 14.3 Brønsted–Lowry Acids and Bases 16.

(a)

Acid HNO3(aq)

Base NaHCO3(aq)

(b)

Acid HBr(aq)

Base NaCN(aq)

(a)

Acid HCO3–(aq)

Base OH–(aq)

(b)

Acid HClO4(aq)

Base NO2–(aq)

(a) (b)

H3O+(aq) + SO42–(aq) → HSO4–(aq) + H2O(l) H2PO4–(aq) + NH3(aq) → NH4+(aq) + HPO42–(aq)

18.

20.

Section 14.4 Acid–Base Indicators 22.

The pH of deionized water is 7. Thus, the color of phenolphthalein indicator is colorless.

24.

At pH 5, methyl red has both the red and yellow forms of the indicator. Therefore, the color of the indicator is orange (red + yellow).

26.

At pH 7, bromthymol blue has both the yellow and blue forms of the indicator. Therefore, the color of the indicator is green (yellow + blue).

110

Chapter 14

Section 14.5 Acid–Base Titrations 28.

2 HNO2(aq) + Ba(OH)2(aq) → 27.55 mL solution

0.101 mol Ba(OH)2 1000 mL solution

0.00278 mol Ba(OH)2 0.00557 mol HNO2 25.0 mL solution

Ba(NO2)2(aq) + 2 H2O(l)

2 mol HNO2 1 mol Ba(OH)2

= 0.00278 mol Ba(OH)2

1000 mL solution 1 L solution

0.00557 mol HNO2

0.223 mol HNO2 1 L solution

Molarity of nitrous acid: 0.223 M HNO2 30.

H2SO3(aq)

2 NH4OH(aq)

27.05 mL solution

Na2SO4(aq)

→

2 H2O(l)

0.105 mol NH4OH 1000 mL solution = 0.00284 mol NH4OH

1 mol H2SO3 0.00284 mol NH4OH × 2 mol NH OH 4 0.00142 mol H2SO3 10.0 mL solution

1000 mL solution 1 L solution

0.00142 mol H2SO4 0.142 mol H2SO3 1 L solution

Molarity of sulfurous acid: 0.142 M H2SO3 32.

H3PO3(aq)

3 LiOH(aq)

→

Li3PO3(aq)

45.05 mL solution

0.488 mol LiOH × 1000 mL solution

0.0220 mol LiOH

0.00733 mol H3PO3 50.0 mL solution

1 mol H3PO3 3 mol LiOH

3 H2O(l)

= 0.0220 mol LiOH

= 0.00733 mol H3PO3

1000 mL solution 1 L solution

0.147 mol H3PO3 1 L solution

Molarity of phosphoric acid: 0.147 M H3PO3

Acids and Bases

111

34.

(a)

MM of NaOH = 40.00 g/mol 3.00 mol NaOH 40.00 g NaOH 1 mL solution 1000 mL solution × 1 mol NaOH × 1.12 g solution × 100% = 10.7% NaOH

(b)

MM of KOH = 56.11 g/mol 0.500 mol KOH 56.11 g KOH 1 mL solution 1000 mL solution × 1 mol KOH × 1.02 g solution × 100% = 2.75% KOH

(c)

MM of NH3 = 17.04 g/mol 6.00 mol NH3 17.04 g NH3 1 mL solution 1000 mL solution × 1 mol NH3 × 0.954 g solution × 100% = 10.7% NH3

(d) MM of Na2CO3 = 105.99 g/mol 1.00 mol Na2CO3 105.99 g Na2CO3 1 mL solution × × 1000 mL solution 1 mol Na2CO3 1.10 g solution × 100% = 9.64% Na2CO3 Section 14.6 Acid–Base Standardization 36.

2 HCl(aq) + Na2CO3(aq) → NaCl(aq) + H2O(l) + CO2(g) MM of Na2CO3 = 105.99 g/mol 1 mol Na2CO3 2 mol HCl 0.515 g Na2CO3 × 105.99 g Na CO × 1 mol Na CO = 0.00972 mol HCl 2 3 2 3 0.00972 mol HCl 29.75 mL solution

1000 mL solution 0.327 mol HCl = 1 L solution 1 L solution

Molarity of hydrochloric acid: 0.327 M HCl

112

Chapter 14

38.

H2C2O4(aq) + 2 KOH(aq) → K2C2O4(aq) + 2 H2O(l) MM of H2C2O4 = 90.04 g/mol 1 mol H2C2O4 2 mol KOH 0.627 g H2C2O4 × 90.04 g H C O × 1 mol H C O = 0.0139 mol KOH 2 2 4 2 2 4 0.0139 mol KOH 29.05 mL solution

1000 mL solution 1 L solution

0.479 mol KOH 1 L solution

Molarity of potassium hydroxide: 0.479 M KOH 40.

2 KHC8H4O4(aq) + Ba(OH)2(aq)

→

BaK2(C8H4O4)2(aq + 2 H2O(l)

MM of KHC8H4O4 = 204.23 g/mol 1 mol KHC8H4O4 1 mol Ba(OH)2 1.655 g KHC8H4O4 × 204.23 g KHC H O × 2 mol KHC H O 8 4 4 8 4 4 = 0.004052 mol Ba(OH)2 1000 mL solution 0.004052 mol Ba(OH)2 × 0.150 mol Ba(OH) = 27.0 mL solution 2 42.

H2Tart(aq) + 2 NaOH(aq) → 28.15 mL solution

Na2Tart(aq) + 2 H2O(l)

0.295 mol NaOH 1000 mL solution ×

1 mol H2Tart 2 mol NaOH = 0.00415 mol H2Tart

0.623 g H2Tart MM of tartaric acid = 0.00415 mol H Tart = 1.50 × 102 g/mol 2 Section 14.7 Ionization of Water 44.

(a) (b) (c)

Equation for collision reaction: H2O(l) + H2O(l) → H3O+(aq) + OH–(aq) Molar hydrogen ion concentration at 25 °C: [H+] = 1.0 × 10–7 M Molar hydroxide ion concentration at 25 °C: [OH–] = 1.0 × 10–7 M

Acids and Bases

113

46.

48.

Kw = [H+] [OH–] =

1.0 × 10–14

(a)

1.0 × 10–14 [OH–] = 6.2 × 10–7 = 1.6 × 10–8 M

(b)

1.0 × 10–14 [OH–] = 4.6 × 10–12 = 2.2 × 10–3 M

Kw = [H+] [OH–] =

1.0 × 10–14

(a)

1.0 × 10–14 [H+] = 8.8 × 10–8 = 1.1 × 10–7 M

(b)

1.0 × 10–14 [H+] = 4.6 × 10–13 = 2.2 × 10–2 M

Section 14.8 The pH Concept 50.

52.

pH = – log [H+] (a)

pH = – log 0.000 000 01 = – log 10–8 pH = – (– 8) = 8

(b)

pH = – log 0.000 001 = – log 10–6 pH = – (– 6) = 6

[H+] = 10–pH (a)

[H+] = 10–9 = 0.000 000 001 M (1 × 10–9 M)

(b)

[H+] = 10–11 = 0.000 000 000 01 M (1 × 10–11 M)

Section 14.9 Advanced pH Calculations 54.

114

(a)

pH = – log 0.000 007 9 = – log 7.9 × 10–6 pH = – log 7.9 – log 10–6 pH = – 0.90 – (– 6) = 5.10

(b)

pH = – log 0.000 000 39 = – log 3.9 × 10–7 pH = – log 3.9 – log 10–7 pH = – 0.59 – (– 7) = 6.41

Chapter 14

56.

58.

(a)

[H+] = 10–1.80 = 100.20 × 10–2 [H+] = 1.6 × 10–2 M = 0.016 M

(b)

[H+] = 10–4.75 = 100.25 × 10–5 [H+] = 1.8 × 10–5 M = 0.000 018 M

(a)

[OH–] = 0.000 031 M = 3.1 × 10–5 M [H+ ] =

1.0 × 10–14 1.0 × 10–14 –10 = [ OH– ] 3.1 × 10–5 = 3.2 × 10 M

pH = – log (3.2 × 10–10) pH = – log 3.2 – log 10–10 pH = – 0.51 – (– 10) = 9.49 (b)

[OH–] = 0.000 000 000 66 M = 6.6 × 10–10 M [H+] =

1.0 × 10–14 1.0 × 10–14 –5 = [ OH– ] 6.6 × 10–10 = 1.5 × 10 M

pH = – log (1.5 × 10–5) pH = – log 1.5 – log 10–5 pH = – 0.18 – (– 5) = 4.82 60.

(a)

pH = 0.90 [H+] = 10–0.90 = 100.10 × 10–1 [H+] = 1.3 × 10–1 M = 0.13 M [OH–] =

(b)

1.0 × 10–14 1.0 × 10–14 = = 7.7 × 10–14 M + 0.13 [H ]

pH = 1.62 [H+] = 10–1.62 = 100.38 × 10–2 [H+] = 2.4 × 10–2 M = 0.024 M [OH–] =

1.0 × 10–14 1.0 × 10–14 = = 4.2 × 10–13 M + 0.024 [H ]

Section 14.10 Strong and Weak Electrolytes 62. (a) (b) (c)

Solution weak acids weak bases slightly soluble ionic compounds

Ionization slightly ionized slightly ionized slightly ionized

Acids and Bases

115

64.

66.

(a) (b) (c) (d)

Solution KOH(aq) NH4OH(aq) Sr(OH)2(aq) Al(OH)3(s)

Electrolyte strong weak strong weak

(a) (b) (c) (d)

Solution Fe(C2H3O2)3 AlPO4(s) Ag2CrO4(s) CdSO4(aq)

Electrolyte strong weak weak strong

(a) (b) (c) (d)

Solution NaOH(aq) NH4OH(aq) Ba(OH)2(aq) Al(OH)3(s)

Electrolyte strong weak strong weak

Aqueous Solution Na+(aq) and OH–(aq) NH4OH(aq) Ba2+(aq) and 2 OH–(aq) Al(OH)3(s)

(a) (b) (c) (d)

Solution AlPO4(s) Co(C2H3O2)3(aq) MnSO4(aq) PbSO4(s)

Electrolyte weak strong strong weak

Aqueous Solution AlPO4(s) Co3+(aq) and 3 C2H3O2–(aq) Mn2+(aq) and SO42–(aq) PbSO4(s)

68.

70.

Section 14.11 Net Ionic Equations 72.

The ions that appear in the total ionic equation, but not in the net ionic equation, are termed spectator ions.

74.

(a)

Nonionized equation: 2 HF(aq) + Li2CO3(aq) → 2 LiF(aq) + H2O(l) + CO2(g) Total ionic equation: 2 HF(aq) + 2 Li+(aq) + CO32–(aq) → 2 Li+(aq) + 2 F–(aq) + H2O(l) + CO2(g) Net ionic equation: 2 HF(aq) + CO32–(aq) → 2 F–(aq) + H2O(l) + CO2(g)

116

Chapter 14

(b)

Nonionized equation: H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s) + 2 H2O(l) Total ionic equation: 2 H+(aq) + SO42–(aq) + Ba2+(aq) + 2 OH–(aq) → BaSO4(s) + 2 H2O(l) Net ionic equation: 2 H+(aq) + SO42–(aq) + Ba2+(aq) + 2 OH–(aq) → BaSO4(s) + 2 H2O(l)

76.

(a)

Nonionized equation: Zn(NO3)2(aq) + 2 NaOH(aq) → Zn(OH)2(s) + 2 NaNO3(aq) Total ionic equation: Zn2+(aq) + 2 NO3–(aq) + 2 Na+(aq) + 2 OH–(aq) → Zn(OH)2(s) + 2 Na+(aq) + 2 NO3–(aq) Net ionic equation: Zn2+(aq) + 2 OH–(aq) → Zn(OH)2(s)

(b)

Nonionized equation: MgSO4(aq) + 2 NH4OH(aq) → Mg(OH)2(s) + (NH4)2SO4(aq) Total ionic equation: Mg2+(aq) + SO42–(aq) + 2 NH4OH(aq) → Mg(OH)2(s) + 2 NH4+(aq) + SO42–(aq) Net ionic equation: Mg2+(aq) + 2 NH4OH(aq) → Mg(OH)2(s) + 2 NH4+(aq)

General Exercises 78.

The color of bromthymol blue is green at pH 7, while phenolphthalein is colorless. Therefore, the color of the water with both indicators is green.

80.

The color of bromcresol green is yellow at pH 3.8 and blue at pH 5.4. Therefore, the color of a solution at pH 4.6 is green (yellow + blue).

82.

Notice that K2HPO4 is a proton acceptor in the first reaction and a proton donor in the second reaction. Thus, K2HPO4 is amphiprotic.

Acids and Bases

117

84.

NaOH(aq) → Na+(aq) + OH–(aq) 0.50 M NaOH → 0.50 M OH– [H+] =

1.0 × 10–14 1.0 × 10–14 = = 2.0 × 10–14 0.50 [ OH– ]

pH = – log [H+] pH = – log (2.0 × 10–14) pH = – log 2.0 – log 10–14 pH = – 0.30 – (– 14) = 13.70 3 × 1023 molecules H2O 2 OH– – 14 × 1 mL water 1 × 109 molecules H2O = 6 × 10 OH

86.

10 mL water ×

88.

Nonionized equation: HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l) Total ionic equation: H+(aq) + NO3–(aq) + K+(aq) + OH–(aq) → NO3–(aq) + K+(aq) + H2O(l) Net ionic equation: H+(aq) + OH–(aq) → H2O(l)

Challenge Exercises 90.

KHTart(aq)

KOH(aq)

→

K2Tart(aq)

H2O(l)

1 mole KHTart = 1 mole KOH 0.100 mol KOH 1 mol KHTart 42.10 mL solution × 1000 mL solution × 1 mol KOH = 0.00421 mol KHTart 0.791 g KHTart MM cream of tartar = 0.00421 mol KHTart = 188 g/mol

118

Chapter 14

CHAPTER

Advanced Problem Solving

Section 15.1 Advanced Problem-Solving Strategies 2.

(a) concept map

(b)

visualization

(a) meter (m) (c) cubic meter (m3) (e) Kelvin unit (K)

(b) kilogram (kg) (d) second (s) (f) joule (J)

(a) length (c) volume (e) temperature

(b) mass (d) time (f) heat energy

(a) velocity (c) specific heat

(b) density (d) molar concentration (M)

0.1 mol AlCl3 3 mol PbCl2 300 g PbCl2 10. (a) 40 mL AlCl3 × 1000 mL AlCl × 2 mol AlCl × 1 mol PbCl 3 3 2 Estimated answer: ~2 g PbCl2

= g PbCl2

Calculated answer: 1.895 g PbCl2

0.2 mol HCl 1 mol Ba(OH)2 1000 mL Ba(OH)2 (b) 40 mL HCl x 1000 mL HCl × 2 mol HCl × 0.1 mol Ba(OH) = mL Ba(OH)2 2

Estimated answer: ~40 mL Ba(OH)2

Calculated answer: 39.0 mL Ba(OH)2

Advanced Problem Solving

119

Section 15.2 Concept Maps—Mole Relationships 12. (a)

mass CHCl3

moles CHCl3

molecules CHCl3

volume HCl(aq)

moles HCl

mass HCl

14. (a)

volume of gas reactant

volume of gas product

(b)

volume of gas product

volume of gas reactant

(b)

16. (a) 0.0755 mol NH3 ×

17.04 g NH3 1 mol NH3

= 1.29 g NH3

(b) 0.0755 mol NH3 ×

22.4 L NH3 1 mol NH3

= 1.69 L NH3

6.02 × 1023 molecules NH3 = 4.55 × 1022 molecules NH3 1 mol NH3

0.0755 mol NH3 (d) 0.155 L solution

= 0.487 M NH3

1 mol N2H4 18. (a) 2.00 g N2H4 × 32.06 g N H 2 4 1mol N2H4 (b) 2.00 g N2H4 × 32.06 g N H 2 4

22.4 L N2H4 × 1 mol N H 2 4 ×

= 1.40 L N2H4

6.02 × 1023 molecules N2H4 1 mol N2H4 = 3.76 × 1022 molecules N2H4

2.00 g N2H4 (c) 0.250 L solution

120

Chapter 15

1 mol N2H4 × 32.06 g N H 2 4

= 0.250 M N2H4

20. (a) 555 mL SO2 ×

1 mol SO2 22400 mL SO2

64.07 g SO2 1 mol SO2

(b) 555 mL SO2 ×

1 mol SO2 22400 mL SO2

6.02 × 1023 molecules SO2 1 mol SO2

= 1.59 g SO2

= 1.49 × 1022 molecules SO2 555 mL SO2 (c) 0.250 L solution

1 mol SO2 22400 mL SO2

0.0991 M H2SO3

1 mol SO3 22.4 L SO3 22. (a) 1.12 × 1023 molecules SO3 × 6.02 × 1023 molecules SO × 1 mol SO 3 3 = 4.17 L SO3 1 mol SO3 80.07 g SO3 (b) 1.12 × 1023 molecules SO3 × 6.02 × 1023 molecules SO × 1 mol SO 3 3 = 14.9 g SO3 1.12 × 1023 molecules SO3 (c) 0.325 L solution

1 mol SO3 6.02 × 1023 molecules SO3

0.572 M H2SO4

Section 15.3 Concept Maps—Stoichiometry Pt/825 °C

24. 4 NH3(g)

5 O2(g)

(a) 25.0 L NO ×

→

4 NO(g)

6 H2O(l)

4 mol NH3 4 mol NO = 25.0 L NH3

6 mol H2O 18.02 g H2O 1 mol NO (b) 25.0 L NO × 22.4 L NO × 4 mol NO × 1 mol H O 2

= 30.2 g H2O

26. 2 KHCO3(s) → K2CO3(s) + H2O(g) + CO2(g) 1 L CO2 1 mol CO2 2 mol KHCO3 100.12 g KHCO3 (a) 255 mL CO2 × 1000 mL CO × 22.4 L CO × 1 mol CO × 1 mol KHCO 2 2 2 3 = 2.28 g KHCO3 1 L CO2 1 mol CO2 1 mol K2CO3 138.21 g K2CO3 (b) 255 mL CO2 × 1000 mL CO × 22.4 L CO × 1 mol CO × 1 mol K CO 2

= 1.57 g K2CO3

Advanced Problem Solving

121

28. Hg(NO3)2(aq) + 2 NaI(aq)

→ HgI2(s) +

2 NaNO3(aq)

1 mol Hg(NO3)2 1000 mL Hg(NO3)2 0.170 mol NaI (a) 24.0 mL NaI × 1000 mL NaI × × 2 mol NaI 0.209 mol Hg(NO3)2 = 9.76 mL Hg(NO3)2 1 mol HgI2 0.170 mol NaI (b) 24.0 mL NaI × 1000 mL NaI × 2 mol NaI

454.39 g HgI2 1 mol HgI2

= 0.927 g HgI2

Section 15.4 Multiple-Reaction Stoichiometry 30. 8 H2S(aq) + 8 Cl2(g)

→ 16 HCl(aq) +

S8(s)

S8(s) + 24 F2(g) → 8 SF6(g) 1 mol Cl2 1 mol S8 8 mol SF6 146.07 g SF6 (a) 0.950 L Cl2 × 22.4 L Cl × 8 mol Cl × 1 mol S × 1 mol SF = 6.19 g SF6 2 2 8 6

(b) 0.950 L Cl2 ×

(c)

122

1 L S8 8 L SF6 × 8 L Cl2 1 L S8 = 0.950 L SF6

1 mol Cl2 8 mol H2S 1 L H2 S 0.950 L Cl2 × 22.4 L Cl × 8 mol Cl × 0.0265 mol H S = 1.60 L H2S 2 2 2

Chapter 15

200°C

32. 3 Fe2O3(s) + CO(g)

2 Fe3O4(s) + CO2(g)

→ 700°C

Fe3O4(s) + CO(g)

3 FeO(s) + CO2(g)

→ 1200°C

FeO(s) +

CO(g)

→

Fe(l) + CO2(g)

1000 g Fe2O3 1 mol Fe2O3 2 mol Fe3O4 3 mol FeO (a) 1.00 kg Fe2O3 × 1 kg Fe O × 159.70 g Fe O × 3 mol Fe O × 1 mol Fe O 2 3 2 3 2 3 3 4 ×

71.85 g FeO 1 kg FeO 1 mol FeO × 1000 g FeO

= 0.900 kg FeO

1000 g Fe2O3 1 mol Fe2O3 2 mol Fe3O4 3 mol FeO (b) 1.00 kg Fe2O3 × 1 kg Fe O × 159.70 g Fe O × 3 mol Fe O × 1 mol Fe O 2 3 2 3 2 3 3 4 1 mol Fe 55.85 g Fe 1 kg Fe × 1 mol FeO × 1 mol Fe × 1000 g Fe

= 0.699 kg Fe

1000 g Fe2O3 1 mol Fe2O3 2 mol Fe3O4 3 mol FeO (c) 1.00 kg Fe2O3 × 1 kg Fe O × 159.70 g Fe O × 3 mol Fe O × 1 mol Fe O 2 3 2 3 2 3 3 4 1 mol Fe 55.85 g Fe 1 kg Fe × 1 mol FeO × 1 mol Fe × 1000 g Fe × 70.0% = 0.490 kg Fe 34. C3H7OH(l) + O2(g)

→ H2O2(l) + C3H6O(l)

2 H2O2(l) + N2H4(l)

→ N2(g) + 4 H2O(g)

0.786 g C3H7OH 1 mol C3H7OH 1 mol H2O2 (a) 50.0 mL C3H7OH × 1 mL C H OH × 60.11 g C H OH × 1 mol C H OH 3 7 3 7 3 7 1 mol N2 28.02 g N2 × 2 mol H O × 1 mol N = 9.16 g N2 2 2 2 0.786 g C3H7OH 1 mol C3H7OH 1 mol H2O2 (b) 50.0 mL C3H7OH × 1 mL C H OH × 60.11 g C H OH × 1 mol C H OH 3 7 3 7 3 7 1 mol N2 22.4 L N2 × 2 mol H O × 1 mol N = 7.32 L N2 2 2 2 0.786 g C3H7OH 1 mol C3H7OH 1 mol H2O2 (c) 50.0 mL C3H7OH × 1 mL C H OH × 60.11 g C H OH × 1 mol C H OH 3 7 3 7 3 7 4 mol H2O 18.02 g H2O × 2 mol H O × 1 mol H O = 23.6 g H2O 2 2 2

Advanced Problem Solving

123

Section 15.5 Advanced Problem-Solving Examples PV 36. PV = nRT; n = R T P = 0.974 atm; V = 2.50 L; T = 35 °C + 273 = 308 K; R = n =

0.974 atm × 2.50 L x 308 K

1 mol · K 0.0821 atm · L = 0.0963 mol NO2

0.0963 mol NO2 x

38. 100.0 mL LiCl

0.0821 atm · L 1 mol · K

46.01 g NO2 1 mol NO2

0.156 mol LiCl 1000 mL LiCl

= 4.43 g NO2

1 mol Cl– × 1 mol LiCl

= 0.0156 mol Cl–

0.225 mol BaCl2 2 mol Cl– 150.0 mL BaCl2 × 1000 mL BaCl × 1 mol BaCl = 0.0675 mol Cl– 2 2 (0.0156 + 0.0675) mol Cl– 1000 mL solution Molarity of Cl–: (100.0 + 150.0) mL solution × = 0.332 M Cl– 1 L solution

40. Pb(NO3)2(aq) + 2 HCl(aq) → 2 PbCl2(s) + 2 HNO3(g) 1 mol PbCl2 2 mol HCl 1000 mL HCl 1.88 g PbCl2 × 278.1 g PbCl × 1 mol PbCl × 0.150 mol HCl = 90.1 mL HCl 2 2

124

Chapter 15

42. PV = nRT; V =

nRT P 0.0821 atm · L 1 mol · K

P = 0.750 atm; T = 75 °C + 273 = 348 K; R =

1 mol CO2 n = 6.55 g CO2 × 44.01 g CO = 0.149 mol CO2 2 V =

0.149 mol × 348 K 0.750 atm

0.149 mol CO2 ×

0.0821 atm · L 1 mol · K

= 5.68 L CO2

6.02 × 1023 molecules CO2 1 mol CO2

= 8.97 × 1022 molecules CO2

44. Cl2(g) + 2 NaBr(aq) → Br2(l) + 2 NaCl(aq) 1 mol Cl2 1 mol Br2 159.80 g Br2 1 mL Br2 10.0 g Cl2 × 70.90 g Cl × 1 mol Cl × 1 mol Br × 3.12 g Br = 7.22 mL Br2 2 2 2 2 1 mol Br2 159.80 g Br2 1 mL Br2 1 mol NaBr 10.0 g NaBr × 102.89 g NaBr × 2 mol NaBr × 1 mol Br × 3.12 g Br = 2.49 mL Br2 2

Since NaBr is the limiting reactant, 2.49 mL Br2 are produced.

46. 2 KClO3(s)

2 KCl(s)

3 O2(g)

Ptotal = Poxygen + Pwater vapor Poxygen = 766 mm Hg – 21 mm Hg = 745 mm Hg 1.00 atm P = 745 mm Hg × 760 mm Hg = 0.980 atm V = 455 mL = 0.455 L; T = 23 °C + 273 = 296 K; R =

0.0821 atm · L 1 mol · K

PV PV = nRT; n = R T n =

0.980 atm × 0.455 L 1 mol · K × 296 K 0.0821 atm · L =

0. 0.0183 mol O2 ×

0.0183 mol O2

2 mol KClO3 122.55 g KClO3 × = 1.50 g KClO3 3 mol O2 1 mol KClO3 ..

Advanced Problem Solving

125

General Exercises 1 ream 48. 4,125,365 sheets x 500 sheets

= 8250.73 reams

Rounded to a whole number, the number is 8250 reams. 50. 2 SO2(g) + O2(g) → 2 SO3(g) 2 L SO3 25.0 L SO2 x 2 L SO = 25.0 L SO3 2 25.0 L O2 x

2 L SO3 1 L O2 = 50.0 L SO3

(a) SO2 is the limiting reactant (b) 25.0 L SO3 52. Ca(C2H3O2)2(aq) + Na2CO3(aq) → CaCO3(s) + 2 NaC2H3O2(aq) 0.266 mol ––––––––––––––– Ca(C2H3O2)2 1 mol –––––––––– CaCO3 36.5 mL –––––––––––––– Ca(C2H3O2)2 × 1000 mL –––––––––––––– Ca(C2H3O2)2 × 1 mol ––––––––––––––– Ca(C2H3O2)2 × 100.1 g CaCO3 1 mol –––––––––– CaCO3 = 0.972 g CaCO3 0.385 mol ––––––––––– Na2CO3 1 mol ––––––––– CaCO3 100.1 g CaCO3 25.0 mL –––––––––– Na2CO3 × 1000 mL × × –––––––––– Na2CO3 1 mol ––––––––––– Na2CO3 1 mol ––––––––– CaCO3 = 0.963 g CaCO3 (a) Na2CO3 is the limiting reactant (b) 0.963 g CaCO3

54. 1.4 × 1021 –––––––––– L seawater ×

1000 mL ––––––––––– seawater 1.05 –––––––––– g seawater 1.90 g Cl– × × 1 –––––––––– L seawater 1 mL ––––––––––– seawater 100 –––––––––– g seawater = 2.8 × 1022 g chlorine

126

Chapter 15

0.100 mol NaCl 1 mol Cl– 56. 50.0 mL NaCl solution × 1000 mL NaCl solution × 1 mol NaCl = 0.00500 mol Cl– from NaCl 0.100 mol KCl 1 mol Cl– 25.0 mL KCl solution × 1000 mL KCl solution × 1 mol KCl = 0.00250 mol Cl– from KCl Total mol Na+: 0.00500 mol + 0.00250 mol = 0.00750 mol Cl– 0.00750 mol Cl– 1000 mL solution × (50.0 + 25.0) mL solution 1 L solution

58. BaCl(s) + 2 AgNO3(aq)

= 0.100 M Cl–

→ 2 AgCl(s) + Ba(NO3)2 (aq)

2 AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + KNO3(aq) 0.100 mol AgNO3 Total AgNO3: 50.0 mL AgNO3 × 1000 mL AgNO = 0.00500 mol AgNO3 3 Excess AgNO3: 0.125 mol K2CrO4 2 mol AgNO3 17.0 mL K2CrO4 × 1000 mL K CrO × 1 mol K CrO 2 4 2 4

= 0.00425 mol AgNO3

BaCl2 precipitated: (0.00500– 0.00425) mol AgNO3 = 0.00075 mol AgNO3 1 mol BaCl2 208.23 g BaCl2 0.00075 mol AgNO3 × 2 mol AgNO × 1 mol BaCl = 0.078 g BaCl2 3 2

Advanced Problem Solving

127

Challenge Exercises 60. A concept map for the software application program is: Word

Application

Processor

Program

Spreadsheet

128

Chapter 15

Database

CHAPTER

Chemical Equilibrium

Section 16.1 Collision Theory 2.

(a) (b) (c)

Theoretical Factors that Increase the Rate of a Reaction Increasing collision frequency of molecules increases the rate. Increasing collision energy of molecules increases the rate. Effective collision orientation of molecules increases the rate.

Ineffective Collision Orientation

Reaction Conditions increase the concentration of a reactant

Effect on the Rate of Reaction The rate of reaction increases because more collisions occur.

(b)

increase the temperature of the reaction

The rate of reaction increases because both collision frequency and collision energy decrease.

(c)

add a metal catalyst

The rate of reaction increases because molecular collisions are more effective.

(a)

No, the catalyst speeds up the rate at which ammonia is produced, but does not increase the amount of ammonia produced.

Chemical Equilibrium

129

Section 16.2 Energy Profiles of Chemical Reactions 10.

2 B(s) + 3 F2(g)

2 BF3(g) + heat

→ ←

2 O3

Energy

3 O2 Progress of reaction

12.

H2(g) + I2(g) + heat

→ ←

2 HI(g) transition state

Energy

Eact

2 HI ΔΗ

H2 + I2 Progress of reaction

14.

130

The heat of reaction (ΔH) is independent from the effects of a catalyst.

Chapter 16

Section 16.3 The Chemical Equilibrium Concept 16.

(a)

True. The equilibrium expression for a given reaction can be determined experimentally.

(b)

True. The equilibrium expression for a given reaction can be determined theoretically.

Section 16.4 General Equilibrium Constant, Keq 18.

20.

Equilibrium Reaction

Equilibrium Constant Expression

(a)

→ 2 HBr(g) H2(g) + Br2(g) ←

[HBr]2 Keq = [H ] [Br ]

(b) (c)

4 HCl(g) + O2(g)

→ ←

2 Cl2(g) + 2 H2O(g)

[Cl2]2 [H2O]2 Keq = [HCl]4 [O ] 2

CO(g) + 2 H2(g)

→ ←

CH3OH(l)

1 Keq = [CO] [H ]2

N2(g) + 3 H2(g)

2 NH3(g)

→ ←

[N2] = 0.400 M

[H2] = 1.20 M

[NH3]2 Keq = [N ] [H ]3 2

[NH3] = 0.195 M

(0.195)2 (0.400) (1.20)3

= 0.0550

Section 16.5 Equilibria Shifts for Gases 22.

N2O4(g) (a) (b)

24.

heat

2 NO2(g)

→ ←

On hot days, the increase in heat shifts the equilibrium to the right producing more NO2. Thus, the NO2 concentration increases. On cool days, the decrease in heat shifts the equilibrium to the left producing less NO2. Thus, the NO2 concentration decreases. 2 C2H4(g)

(a) (c) (e) (g) (i)

2 O3(g)

increase [C2H4] increase [CH2O] increase temp. increase volume add Ne inert gas

→ ←

4 CH2O(g) + O2(g)

shifts right shifts left shifts left shifts right no shift

(b) (d) (f) (h) (j)

Chemical Equilibrium

+ heat

decrease [O3] decrease [O2] decrease temp. decrease volume ultraviolet light

131

shifts left shifts right shifts right shifts left no shift

Section 16.6 Ionization Equilibrium Constant, Ki 26.

28.

Equilibrium Reaction

Equilibrium Constant Expression NH3OH+(aq) + OH–(aq) [NH3OH+] [OH–] Ki = [NH2OH]

(a)

NH2OH(aq) + H2O(l)

(b)

→ C6H5NH3+(aq) + OH–(aq) C6H5NH2(aq) + H2O(l) ← [C6H5NH3+] [OH–] Ki = [C6H5NH2]

(c)

(CH3)2NH(aq) + H2O(l)

NH4OH(aq)

→ ←

(CH3)2NH+(aq) + OH–(aq) [(CH3)2NH2+] [OH–] Ki = [(CH3)2NH] + – NH4 (aq) + OH (aq) → ←

[NH4+] = [OH–] = 2.1 × 10–3 M

[NH4OH] = 0.245 M

[NH4+] [OH–] (2.1 × 10–3) (2.1 × 10–3) Ki = = [NH4OH] (0.245) 30.

N2H4(aq) + H2O(l)

→ ←

= 1.8 × 10–5

N2H5+(aq) + OH–(aq)

[H+] = 10–pH = 10–11.00 = 1.0 × 10–11 M Kw 1.0 × 10–14 [OH–] = [H+] = 1.0 × 10–11 = 1.0 × 10–3 = 0.0010 M [N2H5+] = [OH–] = 0.0010 M Ki =

[N2H4] = 0.139 M

[N2H5+] [OH–] (0.0010) (0.0010) = = 7.2 × 10–6 [N2H4] (0.139)

Section 16.7 Equilibria Shifts for Weak Acids and Bases 32.

HNO2(aq) (a) (c) (e) (g)

132

→ ←

decrease [HNO2] increase [HNO2] add solid KNO2 add solid KOH

Chapter 16

H+(aq)

NO2–(aq)

shift left shift right shift left shift right

(b) (d) (f) (h)

decrease [H+] increase [NO2–] add solid KCl increase pH

shift right shift left no shift shift right

34.

NH4OH(aq) (a) (c) (e) (g)

NH4+(aq)

→ ←

increase [NH4+] increase [NH4OH] add gaseous NH3 add solid KOH

shift left shift right shift right shift left

OH–(aq) (b) (d) (f) (h)

decrease [OH–] decrease pH add solid KCl add solid NH4Cl

shift right shift right no shift shift left

Section 16.8 Solubility Product Equilibrium Constant, Ksp 36.

38.

Reaction

Solubility Product Expression

(a)

→ 2 Cu+(aq) + CO32–(aq) Cu2CO3(s) ←

Ksp = [Cu+]2 [CO32–]

(b)

ZnCO3(s)

Zn2+(aq) + CO32–(aq)

Ksp = [Zn2+] [CO32–]

(c)

Al2(CO3)3(s)

MgF2(s)

→ ←

2 Al3+(aq) + 3 CO3 2–(aq)

→ ←

Ksp = [Al3+]2 [CO32–]3

Mg2+(aq) + 2 F–(aq)

Because two moles of F– are produced for every mole of Mg2+, 1

[Mg2+] = 2 [F–] =

1 –3 2 (2.3 × 10 )

= 1.2 × 10–3 M.

Ksp = [Mg2+] [F–]2 = (1.2 × 10–3) (2.3 × 10–3)2 = 6.3 × 10–9

Section 16.9 Equilibria Shifts for Slightly Soluble Compounds 40.

SrCO3(s) (a) (c) (e) (g)

42.

increase [Sr2+] decrease [Sr2+] add solid SrCO3 add solid KNO3 CdS(s)

(a) (c) (e) (g)

Sr2+(aq)

→ ←

shift left shift right no shift no shift Cd2+(aq)

increase [Cd2+] decrease [Cd2+] add solid CdS add solid NaNO3

CO32–(aq) (b) (d) (f) (h)

S2–(aq)

shift left shift right no shift no shift

(b) (d) (f) (h)

Chemical Equilibrium

increase [CO32–] decrease [CO32–] add solid Sr(NO3)2 add H+

shift left shift right shift left shift right

increase [S2–] decrease [S2–] add solid Cd(NO3)2 add H+

shift left shift right shift left shift right

133

General Exercises 44.

N2O4(g)

46. (a) (c)

2 NO2(g)

→ ←

PN2O4 = 0.0014 atm

PNO2 = 0.092 atm

Kp =

(PNO2)2 PN 2 O 4 =

(0.092 atm)2 0.0014 atm = 6.0 atm

H2O(l)

→ ←

H+(aq)

add gaseous HCl add liquid H2SO4

OH–(aq)

shift left shift left

(b) add solid NaOH (d) add solid NaF

shift left shift right

Challenge Exercises 48.

Zn(OH)2(s)

Zn2+(aq) + 2 OH–(aq)

→ ←

pH = 8.44 [H+] = 3.6 × 10–9 M [OH–] =

1.0 × 10–14 = [H+]

1.0 × 10–14 –6 3.6 × 10–9 = 2.8 × 10 M

Because two moles of OH– are produced for every mole of Zn2+, [Zn2+] =

1 – 2 [OH ]

1 –6 2 (2.8 × 10 )

= 1.4 × 10–6 M.

therefore, Ksp = [Zn2+] [OH –]2 = (1.4 × 10–6) (2.8 × 10–6)2

134

Chapter 16

= 1.1 × 10–17

CHAPTER

Oxidation and Reduction

Section 17.1 Oxidation Numbers 2.

Element Ox. no. Element Ox. no. (a) Au 0 (b) Mn 0 (c) P 0 (d) O2 0 (Note: All elements in the free state have an oxidation number of zero.)

(a) (c)

Ion Au3+ P3–

Ox. no. +3 –3

(a)

Compound Si2H6

Oxidation Number of Silicon 2(ox. no. Si) + 6(ox. no. H) = 0 2(ox. no. Si) + 6(–1) = 0 2(ox. no. Si) + (–6) = 0 2(ox. no. Si) = +6 ox. no. Si = +3

(b)

SiO2

ox. no. Si + 2(ox. no. O) = 0 ox. no. Si + 2(–2) = 0 ox. no. Si + (–4) = 0 ox. no. Si = +4

(c)

Si3N4

3(ox. no. Si) + 4(ox. no. N) = 0 3(ox. no. Si) + 4(–3) = 0 3(ox. no. Si) + (–12) = 0 3(ox. no. Si) = +12 ox. no. Si = +4

(d) CaSiO3

Ion (b) Mn2+ (d) O2–

Ox. no. +2 –2

ox. no. Ca + ox. no. Si + 3(ox. no. O) = 0 +2 + ox. no. Si + 3(–2) = 0 +2 + ox. no. Si + (–6) = 0 ox. no. Si = +4

Oxidation and Reduction

137

8. (a)

Ion CO32–

Oxidation Number of Carbon ox. no. C + 3(ox. no. O) = –2 ox. no. C + 3(–2) = –2 ox. no. C + (–6) = –2 ox. no. C = –2 +6 ox. no. C = +4

(b)

HCO3–

ox. no. H + ox. no. C + 3(ox. no. O) = –1 +1 + ox. no. C + 3(–2) = –1 +1 + ox. no. C + (–6) = –1 ox. no. C –5 = –1 ox. no. C = –1 +5 ox. no. C = +4

(c)

CN–

ox. no. C + ox. no. N = –1 ox. no. C + (–3) = –1 ox. no. C = –1 +3 ox. no. C = +2

(d) CNO–

ox. no. C + ox. no. N + ox. no. O = –1 ox. no. C + (–3) + (–2) = –1 ox. no. C + (–5) = –1 ox. no. C = –1 +5 ox. no. C = +4

Section 17.2 Oxidation–Reduction Reactions 10.

(a) (b)

A redox process characterized by electron loss is termed oxidation. A substance undergoing oxidation in a redox reaction is an reducing agent.

12.

(a) (b)

Oxidized: Mg Oxidized: Mn

Reduced: S Reduced: Br2 +4

14.

(a)

H2O(l) + 2 Fe3+(aq) + SO 32–(aq) → 2 Fe 2+(aq) + SO 42–(aq) + 2 H +(aq)

Oxidized: SO32–

Reduced: Fe3+

–1 0

(b)

0 –1

Cl 2(g) + 2 I – (aq) → 2 Cl–(aq) + I 2 (s) Oxidized: I– 138

Chapter 17

Reduced: Cl2 ..

16.

(a)

Ca (s) + 2 H2O(l) → Ca(OH)2(aq) + H 2(g) Oxidizing agent: H2O

Reducing agent: Ca

(b)

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H 2(g) Oxidizing agent: HCl

Reducing agent: Mg

Section 17.3 Balancing Redox Equations: Oxidation Number Method 18.

Yes, the total ionic charge on the reactant side always equals the total ionic charge on the product side of a balanced redox reaction.

20.

(a) –1 0

gains 2 e–

Cl2(g)

2 KI (aq)

loses 2 e– –1

2 KCl(aq)

→

I2(s)

(b) +2 +3

loses 2 e–

gains 3 e–

Fe2O3(s) + 3 CO (g)

22.

→

2 Fe(s) + 3 CO2(g)

(a) gains 1 e–

–1 +2 2+

2 Fe

loses 1 e–

–2

+3 +

(aq) + H 2O2(aq) + 2 H (aq)

2 Fe

→

(aq) + 2 H2O(l)

(b) –1 +6

loses 1 e–

gains 3 e–

Cr2O72–(aq) + 6 Br –(aq) + 14 H +(aq) → 2 Cr 3+(aq) + 3 Br 2(l) + 7 H 2O(l)

Oxidation and Reduction

139

Section 17.4 Balancing Redox Equations: Half–Reaction Method 24.

(a) (b)

Ni(OH)2(s) + 2 OH–(aq) → NiO2(s) + 2 H2O(l) + 2 e– 2 NO2–(aq) + 3 H2O(l) + 4 e– → N2O(g) + 6 OH–(aq)

26.

(a)

Oxidation: Reduction:

3 (S2– → S + 2 e–) 2 (MnO4– + 2 H2O + 3 e– → MnO2 + 4 OH–)

3 S2–(aq) + 2 MnO4–(aq) + 4 H2O(l) → 3 S(s) + 2 MnO2(s) + 8 OH–(aq) (b)

Oxidation: Reduction:

2 (Cu → Cu 2+ + 2 e–) 2 ClO– + 2 H2O + 4 e– → 2 Cl– + 4 OH–

2 Cu(s) + 2 ClO –(aq) + 2 H2O(l) → 2 Cu2+(aq) + 2 Cl–(aq) + 4 OH–(aq) 28.

Oxidation: Reduction:

Cl2 + 8 OH– → 2 ClO2– + 4 H2O + 6 e– 3 (Cl2 + 2 e– → 2 Cl–)

4 Cl2(g) + 8 OH–(aq) → 2 ClO2–(aq) + 6 Cl–(aq) + 4 H2O(l)

Section 17.5 Predicting Spontaneous Redox Reactions 30.

32.

Substances (a) Pb2+(aq) or Zn2+(aq) (b) Fe3+(aq) or Al3+(aq) (c) Ag+(aq) or I2(s) (d) Cu2+(aq) or Br2(l)

Greater Tendency to be Reduced Pb2+(aq) Fe3+(aq) Ag+(aq) Br2(l)

(a) (b) (c) (d)

Substances F2(g) or Cl2(g) Ag+(aq) or Br2(l) Cu2+(aq) or H+(aq) Mg2+(aq) or Mn2+(aq)

34.

(a)

spontaneous

(b)

spontaneous

36.

(a)

nonspontaneous

(b)

spontaneous

140

Chapter 17

Stronger Oxidizing Agent F2(g) Br2(l) Cu2+(aq) Mn2+(aq)

Section 17.6 Voltaic Cells 38.

Ag (aq)

(aq)

NO3 (aq)

Anode: Ni Cathode: Ag Ni(s) + 2 AgNO3(aq) → 2 Ag(s) + Ni(NO3)2(aq)

40.

electron flow

e– →

e– → e– ↑

e– ←

↓

NO 3–

ion flow

←

Ag ↑

2+ (aq)

↓

Ag (aq)

NO3 (aq)

Ni(s) + 2 AgNO3(aq) → 2 Ag(s) + Ni(NO3)2(aq)

Oxidation and Reduction

141

Section 17.7 Electrolytic Cells 42.

Sn(s) + CoSO4(aq)

Co(s)

→

SnSO4(aq)

(a)

oxidation half-cell reaction: Sn (s) → Sn 2+(aq) + 2 e–

(b)

reduction half-cell reaction: Co 2+(aq) + 2 e– → Co(s)

(c)

anode: Sn electrode

cathode: Co electrode

(d) direction of e– flow: Sn anode → Co cathode (e) 44.

direction of SO42– in the salt bridge: Co half-cell → Sn half-cell

Br2(l) + 2 NaCl(aq) →

2 NaBr(aq) + Cl2(g)

(a)

oxidation half-cell reaction: 2 Cl–(aq) → Cl2(g) + 2 e–

(b)

reduction half-cell reaction: Br2(g) + 2 e– → 2 Br–(aq)

(c)

anode: Pt(Cl2) electrode

cathode: Pt(Br2) electrode

(d) direction of e– flow: Pt(Cl2) anode → Pt(Br2) cathode (e)

direction of Na+ in the salt bridge: Pt(Cl2) half-cell → Pt(Br2) half-cell

General Exercises 46.

Calcium Thiosulfate CaS2O3

48.

Net ionic equation: 6 Hg(l) + 2 Au3+(aq)

50.

Net ionic equation: 2 Al(s) + 6 H+(aq)

Oxidation Number of Sulfur (ox. no. Ca) + 2(ox. no. S) + 3(ox. no. O) = 0 (+2) + 2(ox. no. S) + 3(–2) = 0 (+2) + 2(ox. no. S) + (– 6) = 0 2(ox. no. S) = –2 +6 = +4 ox. no. S = +2

→

3 Hg22+(aq) + 2 Au(s)

2 Al3+(aq) + 3 H2(g)

Challenge Exercises 70.

142

Net ionic equation: 3 C2H5OH(aq) + 2 Cr2O72–(aq) + 16 H+(aq) → 3 HC2H3O2(aq) + 4 Cr3+(aq) + 11 H2O(l)

Chapter 17

CHAPTER

Nuclear Chemistry

Section 18.1 Natural Radioactivity 2.

An alpha particle (α) cannot penetrate human skin, and requires heavy paper or clothing as minimum protective shielding.

A beta particle (β) can penetrate about 1 cm of human flesh, and requires wood or aluminum as minimum protective shielding.

A gamma ray (γ) can pass through the human body, and requires thick lead or concrete as minimum protective shielding.

Section 18.2 Nuclear Equations 8.

Radiation (a) alpha particle (α) (b) beta particle (β–) (c) gamma ray (γ) (d) positron (β+) (e) neutron (n0) (f) proton (p+)

10.

(a)

160 W 74

→

156 Hf + 72

(b)

32 P 15

→

32 S + 16

(c)

55 Co 27

→

55 Fe + +10e 26

(d)

44 Ti 22

0 e -1

→

Relative Charge 2+ 1– 0 1+ 0 1+ 4 He 2

0 e –1

44 Sc 21

Nuclear Chemistry

145

12.

(a)

222 X 90

→

218 Ra + 88

4 He 2

(b)

56 X 25

→

56 Fe + 26

(c)

19 X 10

→

19 0 F + +1 e 9

(d)

37 X + -10 e → 18

222 222 X = 90 Th 90

0 e -1

56 X = 56 Mn 25 25 19 19 X = 10 Ne 10

37 Cl 17

37 X = 37 Ar 18 18

Section 18.3 Radioactive Decay Series 14.

Let X = the parent nuclide: 210 X 84

16.

→

The parent nuclide is:

207 Pb + 82

0 e -1

210 Po. 84

207

The parent nuclide is: 81 Tl.

Decay Series for Neptunium–237 237 Np 93

225 Ac 89

⎯⎯→

←⎯⎯

233 Pa 91

225 Ra 88

⎯⎯→

←⎯⎯

Answers 233 U 92

(a)

233 Pa 91

(b)

233 U 92

↓α

(c)

(d)

229 Th 90

(e)

229 Th 90 225 Ac 89

225 Ra 88 221 Fr 87

217 At 85 213 Po 84

(h) 213 Bi 83

↓

221 Fr 87

209 Bi 83

146

4 He 2

Let X = the parent nuclide: 207 X 81

18.

206 Pb + 82

→

Chapter 18

(g) α

⎯⎯→ β

←⎯⎯

217 At 85

209 Pb 82

⎯⎯→ α

←⎯⎯

213 Bi 83

↓β 213 Po 84

(i)

(f)

(j)

209 Pb 82

Section 18.4 Radioactive Half–Life 20.

100% × 2 × 2 × 2 = 12.5% 12.5% of the U-238 radionuclides remain after three half-lives.

22.

Activity of C-14

Elapsed Time

100% 50% 25% 12.5% 6.25%

0 t1/2 1 t1/2 2 t1/2 3 t1/2 4 t1/2

4 t 1/2 24.

5730 years ≈ 22,900 years old 1 t 1/2

1 t1/2 24 hours × 6 hours = 4 t1/2 160 mg Tc–99 ×

1 2

= 10 mg Tc–99

Section 18.5 Applications of Radionuclides 26.

Uranium–lead dating can be used to estimate geological events occurring billions of years ago.

28.

The γ-emitting radionuclide cobalt-60 can be used to sterilize male insects for agricultural pest control.

30.

The γ–emitting radionuclide iridium–192 can be inserted directly into breast tissue for the treatment of tumors.

Section 18.6 Induced Radioactivity 32.

6 1 Li + 0 n 3

34.

238 2 U + 1H 92

36.

249 Cf + 157 N → 98

260 Db + 4 10 n 105

The Americans created: 260 Db. 105

38.

54 Cr + 209 Bi 24 83

262 Bh + 107

The Germans created: 262 Bh. 107

→

3 H + 1

4 He 2

The radionuclide is: 1 H.

238 1 0 Pu + 2 0n + –1 e 94

→

The radionuclide is:

1 n 0

238 Pu. 94

Nuclear Chemistry

147

Section 18.7 Nuclear Fission 1st

40.

1 n 0

2nd 1 3 0n

→

3rd 1 9 0n

→

27 0 n

After the third step, 27 neutrons are produced. 42.

Uranium-235 can fission in many different ways to give various products. The number of neutrons released from a single fission is usually 2 or 3, but the average number of neutrons is 2.4.

44.

235 1 U + 0n 92

→

142 91 1 56 Ba + 36 Kr + 3 0n

The number of neutrons is 3.

46.

233 1 U+ 0n 92

→

142 90 1 Xe + 38 Sr + 2 0 n 54

The fissionable nuclide is: 92 U.

233

Section 18.8 Nuclear Fusion 48.

3 3 He + 2 He 2

50.

7 Li + 3

52.

1 X + 11 X 1

2 X 1

→ 2 1H + 2X

The unknown particle is: 2 He.

→ 2 42 He + 10 n

The unknown particle is: 21 H.

→

2 H + +10 e 1

The nuclide X is: 11 H.

General Exercises 54.

Since Pb–206 and U–238 are present in equal proportions, one half-life has expired. Thus, the approximate age of the lunar rocks is ~ 4.5 billion years because 1 t1/2 = 4.5 billion years.

56.

0 0 e + -1e +1

→ 2 0γ

Challenge Exercises 58.

148

6 1 Li + 0n 3

Chapter 18

→

4 3 He + 1X + energy 2

The nuclide X is tritium: 1H.

CHAPTER

Organic Chemistry

Section 19.1 Hydrocarbons 2.

The alkene and alkyne classes of hydrocarbons are unsaturated.

The alkene class of compounds has a double bond.

The arene class of compounds contains a benzene ring.

Section 19.2 Alkanes 8.

Molecular Formula

Class of Compound

(a)

C8 H ?

The molecular formula of the alkanes is CnH2n+2. Given n = 8, the formula of the alkane is C8H18.

(b)

C10H?

The molecular formula of the alkanes is CnH2n+2. Given n = 10, the formula of the alkane is C10H22.

Alkane

Structural Formula

propane pentane heptane nonane

CH3–CH2–CH3 CH3–CH2–CH2–CH2–CH3 CH3–CH2–CH2–CH2–CH2–CH2–CH3 CH3–CH2–CH2–CH2–CH2–CH2–CH2–CH2–CH3

10. (a) (b) (c) (d)

Organic Chemistry

151

12.

Five Isomers of Hexane (C6H14) CH3 CH3 – CH2 – CH2 – CH2 – CH2 – CH3

CH3 – CH – CH 2 – CH2 – CH3

CH3

CH3 CH3

CH3 – CH2 – CH – CH2 – CH3

CH3 – CH – CH – CH 3

CH3 CH3 – C – CH2 – CH3 CH3

14.

Two Isomers of Dibromoethane (C2H4Br2) Br–CH2–CH2–Br

16.

18.

CH3–CHBr2

Alkyl Group

IUPAC Name

(a) (b)

propyl isopropyl

CH3CH2CH2– (CH3)2CH–

(a)

2-methylpropane

CH3 (b)

2,2-dimethylbutane

CH3 – CH – CH3 CH3 CH3 – C – CH2 – CH 3

(c)

CH3

3-ethylheptane

CH 2CH 3 CH 3 – CH 2 – CH – CH 2 – CH 2 CH 2 – CH 3 (d) 3,3-diethylpentane CH 2 CH 3 CH3 – CH 2 – C

CH2 – CH 3

CH 2 CH 3

152

Chapter 19

20.

(a)

2 C2H6 + 7 O2

→

4 CO2 + 6 H2O

(b)

2 C4H10 + 13 O2

→

8 CO2 + 10 H2O

(c)

C5H12 + 8 O2 →

5 CO2 + 6 H2O

(d) C7H16 + 11 O2 →

7 CO2 + 8 H2O

Section 19.3 Alkenes and Alkynes 22.

24.

26.

(a)

Molecular Formula C8 H ?

Class of Compound The molecular formula of the alkenes is CnH2n. Given n = 8, the formula of the alkene is C8H16.

(b)

C10H?

The molecular formula of the alkenes is CnH2n. Given n = 10, the formula of the alkene is C10H20.

Alkene (a) 2-pentene (b) 3-heptene (c) 1-propene (d) 4-nonene

Structural Formula CH3–CH=CH–CH2–CH3 CH3–CH2–CH=CH–CH2–CH2–CH3 CH2=CH–CH3 CH3–CH2–CH2–CH=CH–CH2–CH2–CH2–CH3

(a)

Molecular Formula C8 H ?

Class of Compound The molecular formula of the alkynes is CnH2n–2. Given n = 8, the formula of the alkyne is C8H14.

(b)

C10H?

The molecular formula of the alkynes is CnH2n–2. Given n = 10, the formula of the alkyne is C10H18.

Alkyne

Structural Formula

(a)

1-propyne

CH≡C–CH3

(b)

4-octyne

CH3–CH2–CH2–C≡C–CH2–CH2–CH3

(c)

2-butyne

CH3–C≡C–CH3

28.

(d) 3-decyne

CH3–CH2–C≡C–CH2–CH2–CH2–CH2–CH2–CH3

Organic Chemistry

153

30.

Two Isomers of Straight-Chain Pentyne (C5H8) CH≡C–CH2–CH2–CH3

32.

(a)

CH3–C≡C–CH2–CH3 CH 3 C CH 3

CH 2

2-methyl-1-propene

CH3 CH C – C – CH3 (b)

CH3

3,3-dimethyl-1-butyne

CH 2 CH 3 34.

(a)

CH 3 – CH C – CH 2 – CH 2 – CH 2 – CH 3

3-ethyl-2-heptene

CH 3 CH 3 – C

36.

C – C – CH 3 CH 3

(b)

4,4-dimethyl-2-pentyne

(a)

CH≡C–CH3 + 4 O2

→

3 CO2 + 2 H2O

(b)

CH3–C≡CH + 2 H2

→

CH3–CH2–CH3

Section 19.4 Arenes 38.

Three Isomers of Dinitrobenzene [C6H4(NO2)2] NO2

NO2

NO2 NO2 ortho-

NO2 para-

neta-

spark

40.

2 C6H6

154

Chapter 19

15 O2

→

12 CO2 + 6 H2O

Section 19.5 Hydrocarbon Derivatives 42.

General Formula

Class of Compound

(a)

carboxylic acid

C OH

O (b)

C H

aldehyde

(c)

C O R´

ester

O (d) 44.

C NH2

amide

Chemical Formula

Class of Compound

(a)

CH3–OH

alcohol

(b)

CH3–CH2–O–CH2–CH3

ether

O (c)

amide

CH3 C NH2 O

(d)

ketone

CH3 C CH2CH3

O (e)

(f)

(g)

(h)

CH3 C O

CH3

ester OH

phenol

organic halide

CH2 OH

alcohol

Organic Chemistry

155

Section 19.6 Organic Halides 46.

48.

Organic Halide

Structural Formula

(a)

dichloromethane

CH2Cl2

(b)

iodoethane

CH3–CH2–I

(c)

2-fluoropropane

CH3–CHF–CH3

(d) 1,2-dibromopropane

CH2Br–CHBr–CH3

1,1,1-trichloroethane

CH3–CCl3

Section 19.7 Alcohols, Phenols, and Ethers 50.

Alcohol

Structural Formula

(a)

“methyl alcohol”

CH3–OH

(b)

“ethyl alcohol”

CH3–CH2–OH

(c)

“propyl alcohol”

CH3–CH2–CH2–OH

(d) “isopropyl alcohol” 52. (a)

Phenol ortho-chlorophenol

(CH3)2–CH–OH Structural Formula OH Cl

(b)

meta-ethylphenol OH

CH2CH3

156

Chapter 19

54.

Ether

Structural Formula CH3

(a)

“diisopropyl ether”

(b)

“diphenyl ether”

(c)

“ethyl propyl ether”

CH3

CH – O – CH CH3 –O–

CH3 – CH2 – O – CH2 – CH2 – CH3 CH3 – O –

(d) “methyl phenyl ether” 56.

Methanol, ethanol, and propanol are soluble in water because the nonpolar portion of the molecule is relatively small. Butanol is only slightly soluble in water because the nonpolar CxHy– portion of the molecule is larger than the polar –OH.

58.

Alcohol (C3H8O) CH3 CH2 CH2 OH

Alcohol (C3H8O) CH3 CH CH3

Ether (C3H8O) CH3 CH2 O CH3

Section 19.8 Amines 60. (a) (b) 62.

Amine

Structural Formula

“methylamine” “isopropylamine”

CH3–NH2 (CH3)2CH–NH2

“Ethylamine” has two N–H bonds and “diethylamine” has one N–H bond, which aid solubility by hydrogen bonding with water. “Triethylamine” does not have an N–H bond, and therefore cannot hydrogen bond and is only slightly soluble in water.

Section 19.9 Aldehydes and Ketones 64.

Aldehyde

Structural Formula O

(a)

H C H

“formaldehyde”

O (b)

CH3 C H

“acetaldehyde”

Organic Chemistry

157

O (c)

CH3CH2 C H

“propionaldehyde”

O C H (d) benzaldehyde 66.

Ketone

Structural Formula

O (a)

CH3 C CH2CH3

“methyl ethyl ketone”

(b)

CH3 CH2 C CH2CH3 O

“diethyl ketone”

CH3 C (c)

“methyl phenyl ketone”

O C (d) “diphenyl ketone”

68.

Aldehyde (C8H8O)

Ketone (C8H8O)

CH2 – C– H

C CH3

Section 19.10 Carboxylic Acids, Esters, and Amides 70.

Carboxylic Acid

Structural Formula O

158

(a)

“formic acid”

H C OH O

(b)

“acetic acid”

CH3 C OH

Chapter 19

72.

Ester

Structural Formula O

(a)

H C O CH2CH2 CH3

“propyl formate”

74.

CH3 C O CH2 CH 3

(b)

“ethyl acetate”

(a)

Ethanol and methanoic acid react to produce ethyl methanoate.

(b)

Phenol and ethanoic acid react to produce phenyl ethanoate.

76.

Amide

Structural Formula O

(a)

CH3 C NH2

“acetamide”

O (b)

CH3 CH2 C NH2

“propionamide”

General Exercises 78. (a) (b)

Compound

Classification

cyclohexene cyclopropane

unsaturated (–ene suffix) saturated (–ane suffix)

80.

Sulfanilamide has an –amide suffix; thus, the functional group present is an amide.

82.

Progesterone has an –one suffix; thus, the functional group present is a ketone.

84.

Ethynyl estradiol has a –ynyl suffix and an –ol suffix; thus, the functional groups present are an alkyne and an alcohol.

Organic Chemistry

159

Challenge Exercises 86.

Compounds

Higher Boiling Point

(a)

CH3COOH or HCOOCH3

CH3COOH

(b)

CH3CH2CONH2 or CH3COOCH3

CH3CH2CONH2

In each of the above pairs, the compound with the higher boiling point can hydrogen bond. (a) CH3COOH has an –OH and can hydrogen bond; HCOOCH3 cannot. (b) CH3CH2CONH2 has an –NH2 and can hydrogen bond; CH3COOCH3 cannot. Hydrogen bonds increase the effective molar mass of the compound in the liquid, and raise the boiling point by increasing the energy required to separate molecules as they leave the liquid and enter the vapor state.

160

Chapter 19

CHAPTER

Biochemistry

Section 20.1 Biological Compounds 2.

(a) (b)

A lipid is a compound consisting of glycerol and carboxylic acids. A nucleic acid is a polymer consisting of many nucleotides, which each contain a sugar, a base, and phosphoric acid.

(a) (b)

Glycerol and a carboxylic acid in a lipid are joined by an ester linkage. Two nucleotides in a nucleic acid are joined by a phosphate linkage.

Section 20.2 Proteins 6.

Hydrogen bonds are responsible for the α–helix in the secondary structure of a protein.

The secondary structure of a protein refers to the twists and bends in the chain; that is, an α–helix. The tertiary structure refers to the overall shape; for example, the chain can be long and extended, or compact and folded.

Biochemistry

163

10.

The dipeptide formed from two molecules of cysteine is named cystylcysteine (Cys–Cys):

12.

The dipeptide formed from a molecule of tyrosine and a molecule of serine is named tyrosylserine (Tyr–Ser):

14.

Tripeptide

Amino Acid Sequence

1 2 3 4 5 6

Gly–His–Trp Gly–Trp–His His–Gly–Trp His–Trp–Gly Trp–Gly–His Trp–His–Gly

16.

The two amino acids in NutraSweet are aspartic acid and the methyl ester of phenylalanine.

18.

Owing to the two different amino acids in vasopressin and oxytocin, the polypeptides have different primary and secondary structures. Thus, they have different chemical properties and differ in their biological activity.

164

Chapter 20

Section 20.3 Enzymes 20.

In the lock-and-key model, the lock represents the substrate (S).

22.

The term for a location on an enzyme that conforms to the shape of the substrate molecule is called an active site.

24.

The term for a molecule that blocks an enzyme and prevents a substrate reaction is called an inhibitor.

Section 20.4 Carbohydrates 26.

A hexose sugar molecule has six carbon atoms.

28.

A ketose sugar has a ketone functional group and one or more alcohol groups.

30.

A disaccharide has two sugar molecules joined by a glycoside linkage, whereas a polysaccharide has hundreds or thousands of simple sugar molecules joined in a long chain by glycoside linkages.

32.

Galactose and glucose are joined by a glycoside linkage in lactose.

Section 20.5 Lipids 34. (a) (b) (c) (d) 36. (a) (b) (c)

Constituents

Lipid

saturated fatty acids unsaturated fatty acids glycerol ester linkages

more typical of a fat more typical of an oil fat and oil fat and oil

Name of Fatty Acid

Structure of Fatty Acid

stearic acid HOOC–(CH2)16–CH3 linoleic acid HOOC–(CH2)7–CH=CH–CH2–CH=CH–(CH2)4–CH3 linolenic acid HOOC–(CH2)7–CH=CH–CH2–CH=CH–CH2–CH=CH–CH2–CH3

38.

The phospholipid used to emulsify chocolate is lecithin, which contains choline.

40.

The saponification of lanolin with aqueous NaOH produces the sodium salt of a fatty acid and a long-chain alcohol.

42.

The steroid structures of testosterone and progesterone are similar; however, testosterone has a hydroxyl group where progesterone has a ketone group.

Biochemistry

165

Section 20.6 Nucleic Acids 44.

The three general components of an RNA nucleotide are: ribose sugar, a nitrogen base, and phosphoric acid.

46.

The four RNA bases are adenine (A), cytosine (C), guanine (G), and uracil (U).

48.

An RNA molecule has a single strand of nucleotides.

50.

There are three hydrogen bonds between cytosine and guanine (C≡G).

52.

During DNA replication, a cytosine base (C) on the template strand codes for a guanine base (G) on the complementary strand.

54.

During RNA transcription, a cytosine base (C) on the template strand codes for a guanine base (G) on the growing strand.

General Exercises 56.

A protein that serves a metabolic role, such as oxygen transport, has a globular shape.

58.

When a polysaccharide undergoes hydrolysis in aqueous acid, glycoside linkages are broken, which release the monosaccharide sugars in the polymer.

60.

Vitamins A, C, and E (and traces of Se) help eliminate free radicals.

62.

There are three possible codons having one adenine base (A) and two cytosine bases (C): ACC, CAC, and CCA.

Challenge Exercises 64.

Transcription is the process by which a DNA molecule synthesizes a complementary single strands of RNA; whereas, translation is the process by which an RNA codon (trinucleotide) designates a specific amino acid to be added to a growing protein chain.

66.

The first step in transcription is for a DNA molecule to unwind by breaking the hydrogen bonds holding the double helix together. Next, a single strand of DNA acts as a template to synthesize a complementary strand of RNA. For example, a nucleotide in the DNA template strand that contains the base thymine (T) codes for adenine (A) on the RNA strand; cytosine (C) codes for guanine (G); and guanine (G) codes for cytosine (C). However, adenine (A) codes for uracil (U)— not thymine—on the growing RNA chain during transcription.

166

Chapter 20