Solution Manual For Chemistry And Chemical Reactivity, 9th Edition

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iiiTable of ContentsPreface........................................................................................................................................................................... vChapter 1 ....................................................................................................................................................................... 1Basic Concepts of ChemistryLet’s Review.................................................................................................................................................................. 9The Tools of Quantitative ChemistryChapter 2 ..................................................................................................................................................................... 20Atoms, Molecules, and IonsChapter 3 ..................................................................................................................................................................... 48Chemical ReactionsChapter 4 ..................................................................................................................................................................... 65Stoichiometry: Quantitative Information about Chemical ReactionsChapter 5 ..................................................................................................................................................................... 94Principles of Chemical Reactivity: Energy and Chemical ReactionsChapter 6 ................................................................................................................................................................... 120The Structure of AtomsChapter 7 ................................................................................................................................................................... 137The Structure of Atoms and Periodic TrendsChapter 8 ................................................................................................................................................................... 152Bonding and Molecular StructureChapter 9 ................................................................................................................................................................... 174Bonding and Molecular Structure: Orbital Hybridization and Molecular OrbitalsChapter 10 ................................................................................................................................................................. 190Gases and Their PropertiesChapter 11 ................................................................................................................................................................. 216Intermolecular Forces and LiquidsChapter 12 ................................................................................................................................................................. 229The Chemistry of SolidsChapter 13 ................................................................................................................................................................. 245Solutions and Their BehaviorChapter 14 ................................................................................................................................................................. 270Chemical Kinetics: The Rates of Chemical ReactionsChapter 15 ................................................................................................................................................................. 295Principles of Chemical Reactivity: EquilibriaChapter 16 ................................................................................................................................................................. 316The Chemistry of Acids and BasesChapter 17 ................................................................................................................................................................. 338Principles of Reactivity: Other Aspects of Aqueous EquilibriaChapter 18 ................................................................................................................................................................. 369Principles of Chemical Reactivity: Entropy and Free Energy

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ivChapter 19..................................................................................................................................................................394Principles of Chemical Reactivity: Electron Transfer ReactionsChapter 20..................................................................................................................................................................419Environmental ChemistryChapter 21..................................................................................................................................................................431The Chemistry of the Main Group ElementsChapter 22..................................................................................................................................................................457The Chemistry of the Transition ElementsChapter 23..................................................................................................................................................................470Carbon: More Than Just Another ElementChapter 24..................................................................................................................................................................494BiochemistryChapter 25..................................................................................................................................................................504Nuclear Chemistry

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Chapter 11Chapter 1Basic Concepts of ChemistryINSTRUCTOR’SNOTESThis chapter discusses a wide range of basic subjects needed for the study of chemistry. How you use the materialwill depend on the preparation level of your students. Essentially all of the topics in this chapter would be found in atypical high school chemistry class. However, if students took that course two or three years earlier, or if theirperformance at that time was inconsistent with their present goal of a professional study for science or engineering,an extensive review might be necessary. Such a review could take two or three lecture periods. If your class consistsof students who have been screened by a placement exam or other process to reasonably ensure that they have goodabilities in high school chemistry, you may assign most or even all of the chapter for outside reading with only asingle class to review and stress key points. The chapter can stand on its own without direct instruction if yourstudents learned the topics previously. For an intermediate approach you could assign sections 1.1-1.4 for studentreading then take one lecture for the rest of the chapter.A pre-quiz selected from Chapter 1 Study Questions could help establish what degree of review is needed for yourstudents.SUGGESTEDDEMONSTRATIONS1.DensityKolb, K. E.; Kolb, D. K. “Method for Separating or Identifying Plastics,”Journal of Chemical Education1991,68, 348.Franz, D. A. “Densities and Miscibilities of Liquids and Liquid Mixtures,”Journal of Chemical Education1991,68, 594.Checkai, G.; Whitsett, J. “Density Demonstration Using Diet Soft Drinks,”Journal of Chemical Education1986,63, 515.Shakhashiri, B. Z. “Density and Miscibility of Liquids,”Chemical Demonstrations: A Handbook forTeachers of Chemistry;University of Wisconsin Press, 1989; Vol. 3, pp. 229–233.2.Illustration of “Physical Change”Liquid nitrogen is always a favorite of students. We freeze a banana, a hot dog, a flower, or similar object.3.Illustration of “Chemical Change”The first lecture in the course is often begun with a “bang” by setting off several hydrogen-filled balloonsin a darkened lecture room. The demonstration is described inChemical Demonstrations: A Handbook forTeachers of Chemistry;University of Wisconsin Press, 1983; Vol. 1, pp. 106–112.

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Basic Concepts of Chemistry2Other reactions could be done as well, depending on the facilities available. Possibilities include placingsmall pieces of potassium in water, the thermite reaction (Shakhashiri, Volume 1, page 85), or the reactionof zinc and ammonium nitrate (Shakhashiri, Volume 1, page 51). The latter reaction gives off a largeamount of ZnO dust and other irritating fumes. It is not suitable for a room without good ventilation.For fun, and to give some color, as well as talking about our future study of acid-base reactions, we addaqueous NH3to separate flasks containing (i) very dilute acid with phenolphthalein, (ii) Al(NO3)3, and (iii)dilute CuSO4.We often have a student contribute a penny to put into concentrated HNO3. The reaction brings up a briefdiscussion of oxidation–reduction processes. The NO2gas generated prompts a discussion of air pollutionproblems, as well as the fuels used in the Lunar Lander and in the Space Shuttle. (CAUTION: NO2is avery corrosive gas. Use only in a well ventilated room. We do the reaction by putting the penny in a fewmilliliters of acid in a 2-L Erlenmeyer flask that is lightly stoppered. This effectively contains the gas.)

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Chapter 13SOLUTIONS TOSTUDYQUESTIONS1.1(a)CCarbon(c)ClChlorine(e)MgMagnesium(b)KPotassium(d)PPhosphorus(f)NiNickel1.2(a)MnManganese(c)NaSodium(e)XeXenon(b)CuCopper(d)BrBromine(f)FeIron1.3(a)bariumBa(d)leadPb(b)titaniumTi(e)arsenicAs(c)chromiumCr(f)zincZn1.4(a)silverAg(d)tinSn(b)aluminumAl(e)technetiumTc(c)plutoniumPu(f)kryptonKr1.5(a)NaCl is a compound; sodium is an element(b)Sugar is a compound; carbon is an element.(c)Gold chloride is a compound; gold is an element.1.6(a)Pt(NH3)2Cl2is a compound; Pt is an element(b)Copper is an element; copper(II) oxide is a compound(c)Silicon is an element; sand is a compound1.7(a)physical property(b)chemical property(c)chemical property(d)physical property(e)physical property(f)physical property1.8(a)chemical change(b)physical change(c)chemical change(d)physical change1.9(a)Physical properties: color (colorless), physical state (liquid)Chemical property: reactivity (burns in air)(b)Physical properties: color (shiny metal, orange), physical state (liquid)Chemical property: reactivity (aluminum reacts readily with bromine)1.10(a)Physical properties: color (white), physical state (solid), density (2.71 g/cm3)Chemical properties: reactivity towards acid (reacts to produce gaseous carbon dioxide)(b)Physical property: color (gray zinc, purple iodine, white compound)Chemical property: reactivity (zinc and iodine react to give a white compound)

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Basic Concepts of Chemistry41.11.Mechanical energy is used to move the lever, producing electrical and radiant energy1.12.Mechanical energy, electrical energy, radiant energy1.13(a)Kinetic energy(b)Potential energy(c)Potential energy(d)Potential energy1.14(a)Potential energy converted to kinetic energy(b)Kinetic energy converted to potential energy(c)Potential energy converted to kinetic energy(d)Kinetic energy converted to potential energy1.15(a)Qualitative observations: blue-green color, solid physical stateQuantitative observations: density of 2.65 g/cm3, mass of 2.5 g(b)Mass is an extensive property; color, physical state, and density are intensive properties(c)2.5 g ·31 cm2.65 g= 0.94 cm31.16(a)Qualitative observations: shiny golden metallic appearance, crystals in form of perfect cubesQuantitative observations: length of 0.40 cm on a side, mass of 0.064 g(b)Mass and length are extensive properties; color, luster, and crystalline form are intensive properties(c)0.064 g/(0.40 cm)3= 1.0 g/cm3= 1.0 g/mL1.17Observation (c) identifies a chemical property.1.18Observations (a) and (b) identify chemical properties.1.19Calcium: CaFluorine: FThe shape of the fluorite crystals can be described as interwoven cubes.The overall shape of the crystals indicates that the ions in the solid matrix arrange themselves withalternating calcium and fluoride ions to produce the crystal appearance.1.20(a)Copper, CuCarbon, COxygen: O(b)Oxygen is a gas, while copper, carbon, and azurite are solids at room temperature. Oxygen iscolorless, while copper has a reddish color and carbon is gray. The gemstone is a bluish color.1.21The water can be removed by boiling the solution until all the water is converted to water vapor; the solidleft behind will be NaCl.

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Chapter 151.22The non-uniform appearance of the mixture indicates that samples taken from different regions of themixture would be different—a characteristic of a heterogeneous mixture. Recalling that iron is attracted toa magnetic field while sand is generally not attracted in this way suggests that passing a magnet throughthe mixture would separate the sand and iron.1.23Physical changes: a, b, dChemical changes: c1.24Physical changes: a, dChemical changes: b, c1.25The large colorless block of salt represents the macroscopic view. The spheres represent the microscopic orparticulate view. If one can imagine producing multiple “copies” of the particulate view, the macroscopicview will result.1.26The orange solid and liquid in the bowl and the orange liquid and gas in the round bottom flask representthe macroscopic views. The spheres represent the particulate view. The particulate view of the soliddisplays atoms of bromine tightly packed to produce the solid. The liquid view displays molecules of Br2with space separating the individual molecules. The gas view displays molecules of Br2; however, themolecules are further apart than in the liquid view.1.27The plastic (with a much lower density than CCl4) will float. The aluminum (which is more dense thanCCl4) will sink.1.28Liquids: mercury and waterSolid: copperOf the substances shown, mercuryis most dense and water is least dense.1.29(a)mixture(b)mixture(c)element(d)compound1.30(a)mixture(b)compound(c)mixture(d)mixture

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Basic Concepts of Chemistry61.31(a)solid iron(b)liquid water(c)water vapor1.32(a)water vapor, helium(b)liquid water, solid aluminum(c)brass1.33The three liquids will be layered, with hexane floating on top of water, which in turn floats on top of theperfluorohexane. HDPE will lie at the top of the water layer. PVC will lie on top of the perfluorohexanelayer. Teflon will sink to the bottom of the perfluorohexane layer.1.34Melting point. Sugar melts around 160-186 °C while salt melts at 800 °C.1.35HDPE will float in liquids with a density higher than 0.97 g/mL, ethylene glycol, water, acetic acid, andglycerol.1.36The sample's density and melting point could be compared with that of silver to prove whether or not thesample is silver.1.37Milk is mostly water. When water freezes its volume increases (its density decreases). When the milkfroze, the increase in volume was so great that it pushed out of the bottle.1.38The mass of the object is determined and then the volume is determined by submersion in a known volumeof liquid. The increase in volume would be equal to the volume of the irregularly shaped object. Thedensity could be calculated by dividing its mass by its volume.1.39If too much sugar is excreted, the density of urine will increase. If too much water is excreted, the densityof urine will decrease.1.40One could check for an odor, check the boiling or freezing point, or determine the density. If the density isapproximately 1 g/cm3at room temperature, the liquid could be water. If it boils at about 100 °C andfreezes at about 0 °C, that would be consistent with water. To check for the presence of salt, boil the liquidaway. If a substance remains, it could be salt, but further testing would be required.

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Chapter 171.41(a)Solid potassium reacts with liquid water to produce gaseous hydrogen and a homogeneous mixture ofpotassium hydroxide in liquid water.(b)The reaction is a chemical change.(c)Potassium and water are reactants, hydrogen and potassium hydroxide are products.(d)Among the qualitative observations are (i) the reaction is violent; and (ii) heat and light (a purpleflame) are produced.1.42Least dense liquid: waterMost dense liquid: mercuryMedium density: carbon tetrachloride1.43Any balloon filled with a gas having a density less than 1.12 g/L will float in air. Helium and neon balloonswill float.1.44A copper-colored metal could be copper, but it may also be an alloy of copper, for example, brass orbronze. Testing the material’s density and melting temperature would be one way to find out if it is copper.1.45Physical change1.46(a)Mass CHCl3= 10.0 mL·1.492 g/mL = 14.9 gMass CHBr3= 5.0 mL·2.890 g/mL = 14 gMass of solution = 14.9 g + 14 g = 29 gDensity = mass of solution/volume of solution = 29 g/15.0 mL = 1.9 g/mL(b)Mass of solution = 2.07 g/mL·20.0 mL = 41.4 gAlso, mass of solution = mass CHCl3+ mass CHBr3Let X = volume CHCl3and (20.0 mL−X) = volume CHBr3Then, mass of solution = (1.492 g/mL)X + (2.890 g/mL)(20.0 mL – X) = 41.4 gX = 11.7 mL CHCl3(20.0 – X) = 8.3 mL CHBr3To obtain 20.0 mL of solution withd= 2.07 g/cm3, mix 11.7 mL CHCl3and 8.3 mL CHBr3.

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Basic Concepts of Chemistry81.47(a) Au, Cu, Ag(b) (1 K/1 °C)(1064 °C + 273 °C) = 1337 K(c) Gold is not the most dense element. Iridium is the most dense element.(d) 0.75 × 5.58 g = 4.2 g(e) 56 × 106couples × (1 ring/person) × (6.15 mg/ring) × (1 g/1000 mg) = 3.4 × 105g3.4 × 105g × 1 troy ounce/31.1 g × $1620/troy ounce = $ 1.8 × 107

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Let’s Review9Let’s ReviewThe Tools of Quantitative ChemistryINSTRUCTOR’SNOTESThis section reviews a wide range of mathematical topics that will be needed by the student in the course, somastery of this material should be stressed. The need for accurate measurement of physical quantities and thecorrect recording of this information can be characterized as necessary for reasons of safety using medical (drugdosage) and engineering examples (material composition and properties). Mistakes in this area can mean thedifference between life and death.A pre-quiz selected from the Study Questions at the end of this section could help establish what degree of review isneeded for your students.SUGGESTEDDEMONSTRATIONS1.Units of MeasurementAdd a few drop of bromcresol green to a 2-L flask before the lecture. On filling with water during lecture,the water becomes blue. When the flask is almost full, it is topped off with dilute HCl, and the solutionturns yellow. Next, the water is poured into an ordinary 1-L flask or beaker that already contains somedilute base, and the solution turns blue again. This is then poured into a graduated cylinder containingsome phenolphthalein. We use this sequence to comment on the relative accuracies of the different types ofglassware. We take some containers such as soda cans so students can connect metric units with familiarobjects.A weighed piece of fruit or some other solid gives some meaning to mass expressed in grams.Earley, C. W. “A Simple Demonstration for Introducing the Metric System to Introductory ChemistryClasses,”Journal of Chemical Education1999,76, 1215.2.Demonstrations Using Significant FiguresKirksey, H. G. “Significant Figures: A Classroom Demonstration,”Journal of Chemical Education1992,69, 497.Abel, K. B.; Hemmerlin, W. M. “Significant Figures,”Journal of Chemical Education1990,67, 213.

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The Tools of Quantitative Chemistry10SOLUTIONS TOSTUDYQUESTIONSLR.1(25 °C + 273.15 °C)1 K1C= 298 KLR.2(5.5103°C + 273.15 °C)1 K1C= 5.8103KLR.3(a)289 K(b)97 °C(c)310 KLR.4(a)–196 °C(b)336 K(c)1177 °CLR.542.195 km ·1000 m1 km= 4.2195104m42.195 km · 0.62137 miles1 km= 26.219 milesLR.619 cm ·10 mm1 cm= 190 mm19 cm ·1 m100 cm= 0.19 mLR.72.5 cm2.1 cm = 5.3 cm25.3 cm2·21 m100 cm= 5.310–4m2LR.822211.8 cmArea =r== 109 cm2 22–221 m109 cm= 1.0910m100 cmLR.9250. mL ·31 cm1 mL= 250. cm3250. mL ·31 L10mL= 0.250 L250. mL ·31 cm1 mL·31 m100 cm= 2.5010–4m3250. mL ·31 L10mL·31 dm1 L= 0.250 dm3

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Let’s Review11LR.101.5 L ·310mL1 L= 1.5103mL1.5 L ·310mL1 L·31 cm1 mL= 1.5103cm31.5103cm3·31 dm10 cm= 1.5 dm3LR.112.52 kg ·310g1 kg= 2.52103gLR.122.265 g ·31 kg10g= 2.26510–3kg2.265 g ·310mg1 g= 2.265103mgLR.13500. mL ·31 cm1 mL·31.11 g1 cm= 555 gLR.1431 cm2.365 g10.5 g= 0.225 cm3LR.15density =2.361 g(2.35 cm1.34 cm0.105 cm)= 7.14 g/cm3The metal is (c) ZincLR.16600 g H2O ·31 cm0.995 g= 600 cm3600 g Pb ·31 cm11.35 g= 50 cm3Given equal masses the less dense substance (water) will have a greater volume.LR.171200 Cal ·310cal1 Cal·4.184 J1 cal= 5.0106JLR.181670 kJ ·310J1 kJ·1 cal4.184 J·31 Cal10cal= 399 CalLR.19170 kcal ·310cal1 kcal·4.184 J1 cal·31 kJ10J= 710 kJThe food product with 170 kcal per serving has a greater energy content per serving.

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The Tools of Quantitative Chemistry12LR.20130 Calories × 4.184 J/cal × 1000 cal/Calorie = 5.4 × 105J = 5.4 × 102kJ. The juice provides more totalenergy.5.4 × 102kJ/335 mL = 1.6 kJ/mL 630 kJ/295 mL = 2.1 kJ/mL. The juice provides more energy permilliliter.LR.21(a)Method A average =2.2 + 2.3 + 2.7 + 2.44= 2.4 g/cm3Method B average = 2.703 + 2.701 + 2.705 + 5.8814= 3.480 g/cm3For method B the reading of 5.811 can be excluded because it is more than twice as large as all otherreadings. Using only the first three readings, average = 2.703 g/cm3.(b)Method A: Percent error =2.4 – 2.702100 %2.702= 10 %Method B: Percent error =2.703 – 2.702100 %2.702= 0.04 % (omitting data point)(c)Method A: Standard deviation =22220.2+ 0.1+ 0.3+ 03= 0.2 g/cm3Method B: Standard deviation =2220+ 0.002+ 0.0022= 0.002 g/cm3(omitting data point)(d)Before excluding a data point for B, method A is more accurate and more precise. After excluding thefourth data point for B, this method gives a more accurate and more precise result.LR.22(a)Student A: Average = 135 °CPercent error =135 – 135100 %135= 0 %Student B: Average = 138 °CPercent error =138 – 135100 %135= 2 %(b)Student B is more precise; Student A is more accurateLR.23(a)5.410–2g (2 significant figures)(c)7.9210–4g (3 significant figures)(b)5.462103g (4 significant figures)(d)1.6103mL (2 significant figures)LR.24(a)1623 (4 significant figures)(c)0.0632 (3 significant figures)(b)0.000257 (3 significant figures)(d)3404 (4 significant figures)LR.25(a)9.4410–3(c)11.9(b)5.694103(d)0.122LR.26(a)2.44108(c)0.133(b)4.8510–2(d)0.0286

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Let’s Review13LR.27y = 0.1637x + 0.09580123456702040Number of kernelsslope = 0.1637 g/kernelThe slope represents the average mass of a kernel.Using the straight-line equation, mass = (0.1637 g/kernel)(20 kernels) + 0.0958 g = 3.370 g20.88 g = (0.1637 g/kernel)(number of kernels) + 0.0958 gnumber of kernels = 127LR.28(a)0.21(b)5.6(c)slope =5.6 – 4.00.30 – 0.21= 18; intercept = 0.20(d)y= 18x+ 0.20 = (18)(1.0) + 0.20 = 18LR.29(a)slope =20.00 – 4.000.00 – 4.00= –4.00, intercept = 20.00,y= –4.00x+ 20.00(b)y= –4.00x+ 20.00 = (–4.00)(6.0) + 20.00 = –4.00LR.30(a)y = 4E-05x - 0.0416012345670.E+001.E+052.E+051/speed(b)y = 4.010–5x – 0.0416y-intercept = –0.0416LR.31C= 0.0823LR.32n= 1.63LR.33T= 295

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