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Page AP-1
Answers
To Odd-Numbered Problems
Chapter 1

1.5 (a) Law. (b) Theory. (c) Hypothesis.

1.7 (a) C and O. (b) F and H. (c) N and H. (d) O.

1.13 (a) K. (b) Sn. (c) Cr. (d) B. (e) Ba. (f) Pu. (g) S. (h) Ar. (i) Hg.

1.15 (a) Homogeneous mixture. (b) Element. (c) Compound. (d) Homogeneous mixture. (e) Heterogeneous mixture. (f) Homogeneous mixture. (g) Heterogeneous mixture.

1.17 (a) Element. (b) Compound. (c) Compound. (d) Element.

1.23 3.12 g/mL.

1.25 (a) 35°C. (b) −11°C. (c) 39°C. (d) 1011°C. (e) −459.67°F.

1.27 2.52 mL.

1.29 (a) 388.36 K. (b) 3.10 × 102 K. (c) 6.30 × 102 K.

1.35 (a) Quantitative. (b) Qualitative. (c) Qualitative. (d) Qualitative. (e) Qualitative.

1.37 (a) Physical change. (b) Chemical change. (c) Physical change. (d) Chemical change. (e) Physical change.

1.39 99.9 g; 20°C; 11.35 g/cm3.

1.45 (a) 0.0152. (b) 0.0000000778. (c) 0.000001. (d) 1600.1.

1.47 (a) 1.8 × 10−2. (b) 1.14 × 1010. (c) −5 × 104. (d) 1.3 × 103.

1.49 (a) One. (b) Three. (c) Three. (d) Four. (e) Three. (f) One. (g) One or two.

1.51 (a) 1.28. (b) 3.18 × 10−3 mg. (c) 8.14 × 107 dm.

1.53 Tailor Z's measurements are the most accurate. Tailor Y's measurements are the least accurate. Tailor X's measurements are the most precise. Tailor Y's measurements are the least precise.

1.55 (a) 1.10 × 108 mg. (b) 6.83 × 10−5 m3. (c) 7.2 × 103 L. (d) 6.24 × 10−8 lb.

1.57 3.1557 × 107 s.

1.59 (a) 81 in/s. (b) 1.2 × 102 m/min. (c) 7.4 km/h.

1.61 88 km/h.

1.63 3.7 × 10−3 g Pb.

1.65 (a) 1.85 × 10−7 m. (b) 1.4 × 1017 s. (c) 7.12 × 10−5 m3. (d) 8.86 × 104 L.

1.67 6.25 × 10−4 g/cm3.

1.69 0.88 s.

1.71 (a) 2.5 cm. (b) 2.55 cm.

1.73 (a) Chemical. (b) Chemical. (c) Physical. (d) Physical. (e) Chemical.

1.75 (a) 8.08 × 104 g. (b) 1.4 × 10−6 g. (c) 39.9 g.

1.77 31.35 cm3.

1.79 10.50 g/cm3.

1.81 11.4 g/cm3.

1.83 %Error (°F) = 0.1%; %Error (°C) = 0.3%.

1.85 −40°F = −40°C.

1.87 5 × 102 mL/breath.

1.89 4.8 × 1019 kg NaCl = 5.3 × 1016 tons NaCl.

1.91 The density of the crucible is equal to the density of pure platinum.

1.93 (a) 75.0 g Au. (b) A troy ounce is heavier than an ounce.

1.95 (a) 0.5%. (b) 3.1%.

1.97 208.0 s = 3 min 28.0 s.

1.99 1.77 × 106 g Cu.

1.101 6.0 × 1012 g Au; $1.3 × 1014.

1.103 7.3 × 1021 kg Si.

1.105 9.5 × 1010 kg CO2.

1.107 (a) homogeneous. (b) heterogeneous.

1.109 1.1 × 102 yr.

1.111 2.3 × 104 kg NaF/yr; 99% NaF wasted.

1.113 4.0 × 10−19 g/L.

1.115 Density = 7.20 g/cm3; r = 0.853 cm.

1.117 It would be more difficult to prove that the unknown substance is an element. Most compounds would decompose on heating, making them easy to identify.

1.119 Gently heat the liquid to see if any solid remains after the liquid evaporates. Also, collect the vapor and then compare the densities of the condensed liquid with the original liquid.

1.121 The volume occupied by the ice is larger than the volume of the glass bottle. The glass bottle would break.

Chapter 2

2.15 0.12 mi.

2.21 145.

2.23 : protons = 7, electrons = 7, neutrons = 8; : protons = 16, electrons = 16, neutrons = 17; : protons = 29, electrons = 29, neutrons = 34; : protons = 38, electrons = 38, neutrons = 46; : protons = 56, electrons = 56, neutrons = 74; : protons = 74, electrons = 74, neutrons = 112; : protons = 80, electrons = 80, neutrons = 122.

2.25 (a) . (b) . (c) . (d) .

2.27 (a) 19. (b) 34. (c) 75. (d) 192.

2.35 Metallic character (a) increases as you progress down a group of the periodic table and (b) decreases from the left to right across the periodic table.

2.37 Na and K; N and P; F and Cl.

2.39 Iron: Fe, period 4, upper-left square of Group 8B; Iodine: I, period 5, Group 7A; Sodium: Na, period 3, Group 1A; Phosphorus: P, period 3, Group 5A; Sulfur: S, period 3, Group 6A; Magnesium: Mg, period 3, Group 2A.

2.45 207.2 amu.

2.47 6Li = 7.5%, 7Li = 92.5%.

2.49 5.1 × 1024 amu.

2.59 (a) Polyatomic, elemental form, not a compound. (b) Polyatomic, compound. (c) Diatomic, compound.

2.61 Elements: N2, S8, H2; Compounds: NH3, NO, CO, CO2, SO2.

2.63 (a) CN. (b) CH. (c) C9H20. (d) P2O5. (e) BH3.

2.65 C3H7NO2.

2.67 (a) Nitrogen trichloride. (b) Iodine heptafluoride. (c) Tetraphosphorus hexoxide. (d) Disulfur dichloride.

2.69 (a) NF3: nitrogen trifluoride. (b) PBr5: phosphorus pentabromide. (c) SCl2: sulfur dichloride.

2.75 Na+: 11 protons, 10 electrons; Ca2+: 20 protons, 18 electrons; Al3+: 13 protons, 10 electrons; Fe2+: 26 protons, 24 electrons; I: 53 protons, 54 electrons; F: 9 protons, 10 electrons; S2−: 16 protons, 18 electrons; O2−: 8 protons, 10 electrons; N3−: 7 protons, 10 electrons.

2.77 (a) Na2O. (b) FeS. (c) Co2(SO4)3. (d) BaF2.

2.79 Ionic: LiF, BaCl2, KCl; Molecular: SiCl4, B2H6, C2H4.

2.81 (a) Potassium dihydrogen phosphate. (b) Potassium hydrogen phosphate. (c) Hydrogen bromide. (d) Hydrobromic acid. (e) Lithium carbonate. (f) Potassium dichromate. (g) Ammonium nitrite. (h) Hydrogen iodate (in water, iodic acid). (i) Phosphorus pentafluoride. (j) Tetraphosphorus hexoxide. (k) Cadmium iodide. (l) Strontium sulfate. (m) Aluminum hydroxide.

2.83 (a) RbNO2. (b) K2S. (c) NaHS. (d) Mg3(PO4)2. (e) CaHPO4. (f) PbCO3. (g) SnF2. (h) (NH4)2SO4. (i) AgClO4. (j) BCl3.

2.85 (a) Mg(NO3)2. (b) Al2O3. (c) LiH. (d) Na2S.

2.87 Acid: compound that produces H+; Base: compound that produces OH; Oxoacids: acids that contain oxygen; Oxoanions: the anions that remain when oxoacids lose H+ ions; Hydrates: ionic solids that have water molecules in their formulas.

2.89 (c) Changing the electrical charge of an atom usually has a major effect on its chemical properties. The two electrically neutral carbon isotopes should have nearly identical chemical properties.

2.91 I.

2.93 NaCl is an ionic compound; it doesn't consist of molecules.

2.95 (a) Molecule and compound. (b) Element and molecule. (c) Element. (d) Molecule and compound. (e) Element. (f) Element and molecule. (g) Element and molecule. (h) Molecule and compound. (i) Compound, not molecule. (j) Element. (k) Element and molecule. (l) Compound, not molecule.

2.97 It establishes a standard mass unit that permits the measurement of masses of all other isotopes relative to carbon-12.

2.99 , protons = 5, neutrons = 6, electrons = 5, net charge = 0; , protons = 26, neutrons = 28, electrons = 24, net charge = +2; , protons = 15, neutrons = 16, electrons = 18, net charge = −3; , protons = 79, neutrons = 117, electrons = 79, net charge = 0; , protons = 86, neutrons = 136, electrons = 86, net charge = 0.

2.101 (a) Li+. (b) S2−. (c) I. (d) N3−. (e) Al3+. (f) Cs+. (g) Mg2+.

2.103 Group 7A, binary: HF, hydrofluoric acid; HCl, hydrochloric acid; HBr, hydrobromic acid; HI, hydroiodic acid. Group 7A, oxoacids: HClO4, perchloric acid; HClO3, chloric acid; HClO2, chlorous acid; HClO, hypochlorous acid; HBrO3, bromic acid; HBrO2, bromous acid; HBrO, hypobromous acid; HIO4, periodic acid; HIO3, iodic acid; HIO, hypoiodous acid. Examples of oxoacids containing other Group A-block elements are: H3BO3, boric acid; H2CO3, carbonic acid; HNO3, nitric acid; HNO2, nitrous acid; H3PO4, phosphoric acid; H3PO3, phosphorous acid; H3PO2, hypophosphorous acid; H2SO4, sulfuric acid; H2SO3, sulfurous acid. Binary acids formed from other Group A-block elements other than Group 7A: H2S, hydrosulfuric acid.

Page AP-2

2.105 : protons = 2, neutrons = 2, neutrons/protons = 1.00; : protons = 10, neutrons = 10, neutrons/protons = 1.00; : protons = 18, neutrons = 22, neutrons/protons = 1.22; : protons = 36, neutrons = 48, neutrons/protons = 1.33; : protons = 54, neutrons = 78, neutrons/protons = 1.44. The neutron/proton ratio increases with increasing atomic number.

2.107 Cu, Ag, and Au are fairly chemically unreactive. This makes them especially suitable for making coins and jewelry that you want to last a very long time.

2.109 MgO and SrO.

2.111 (a) Berkelium (Berkeley, CA); Europium (Europe); Francium (France); Scandium (Scandinavia); Ytterbium (Ytterby, Sweden); Yttrium (Ytterby, Sweden). (b) Einsteinium (Albert Einstein); Fermium (Enrico Fermi); Curium (Marie and Pierre Curie); Mendelevium (Dmitri Mendeleev); Lawrencium (Ernest Lawrence); Meitnerium (Lise Meitner). (c) Arsenic, Cesium, Chlorine, Chromium, Iodine.

2.113 The mass of fluorine reacting with hydrogen and deuterium would be the same. The ratio of F atoms to hydrogen (or deuterium) atoms is 1:1 in both compounds. This does not violate the law of definite proportions. When the law of definite proportions was formulated, scientists did not know of the existence of isotopes.

2.115 (a) Br. (b) Rn. (c) Se. (d) Rb. (e) Pb.

2.117 Mg2+, , Mg(HCO3)2, Magnesium bicarbonate; Sr2+, Cl, SrCl2, Strontium chloride; Fe3+, , Fe(NO2)3, Iron(III) nitrite; Mn2+, , Mn(ClO3)2, Manganese(II) chlorate; Sn4+, Br, SnBr4, Tin(IV) bromide; Co2+, , Co3(PO4)2, Cobalt(II) phosphate; , I, Hg2I2, Mercury(I) iodide; Cu+, , Cu2CO3, Copper(I) carbonate; Li+, N3−, Li3N, Lithium nitride; Al3+, S2−, Al2S3, Aluminum sulfide.

2.119 1.908 × 10−8 g. The predicted change (loss) in mass is too small a quantity to measure. Therefore, for all practical purposes, the law of conservation of mass is assumed to hold for ordinary chemical processes.

2.121

2.123 (a) Yes. (b) Acetylene: any formula with C:H = 1:1 (CH, C2H2, etc.); Ethane: any formula with C:H = 1:3 (CH3, C2H6, etc.).

2.125 Manganese, Mn.

2.127 Chloric acid, nitrous acid, hydrocyanic acid, and sulfuric acid.

Chapter 3

3.3 (a) 50.48 amu. (b) 92.02 amu. (c) 64.07 amu. (d) 84.16 amu. (e) 34.02 amu. (f) 342.3 amu. (g) 17.03 amu.

3.5 (a) 16.04 amu. (b) 46.01 amu. (c) 80.07 amu. (d) 78.11 amu. (e) 149.9 amu. (f) 174.27 amu. (g) 310.2 amu.

3.9 78.77% Sn; 21.23% O.

3.11 (d) Ammonia, NH3.

3.13 39.89% Ca, 18.50% P, 41.41% O, 0.20% H.

3.19 (a) KOH + H3PO4 → K3PO4 + H2O. (b) Zn + AgCl → ZnCl2 + Ag. (c) NaHCO3 → Na2CO3 + H2O + CO2. (d) NH4NO2 → N2 + H2O. (e) CO2 + KOH → K2CO3 + H2O.

3.21 (a) Potassium and water react to form potassium hydroxide and hydrogen. (b) Barium hydroxide and hydrochloric acid react to form barium chloride and water. (c) Copper and nitric acid react to form copper nitrate, nitrogen monoxide and water. (d) Aluminum and sulfuric acid react to form aluminum sulfate and hydrogen. (e) Hydrogen iodide reacts to form hydrogen and iodine.

3.23 (a) 2N2O5 → 2N2O4 + O2. (b) 2KNO3 → 2KNO2 + O2. (c) NH4NO3 → N2O + 2H2O. (d) NH4NO2 → N2 + 2H2O. (e) 2NaHCO3 → Na2CO3 + H2O + CO2. (f) P4O10 + 6H2O → 4H3PO4. (g) 2HCl + CaCO3 → CaCl2 + H2O + CO2. (h) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2. (i) CO2 + 2KOH → K2CO3 + H2O. (j) CH4 + 2O2 → CO2 + 2H2O. (k) Be2C + 4H2O → 2Be(OH)2 + CH4. (l) 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O. (m) S + 6HNO3 → H2SO4 + 6NO2 + 2H2O. (n) 2NH3 + 3CuO → 3Cu + N2 + 3H2O.

3.25 (d) 3A + 2B → 2C + D.

3.31 5.8 × 103 light-yr.

3.33 9.96 × 10−15 mol Co.

3.35 3.01 × 103 g Au.

3.37 (a) 1.244 × 10−22 g/As atom. (b) 9.746 × 10−23 g/Ni atom.

3.39 2.98 × 1022 Cu.

3.41 Two atoms of lead.

3.43 409 g/mol.

3.45 3.01 × 1022 C atoms, 6.02 × 1022 H atoms, 3.01 × 1022 O atoms.

3.47 2.1 × 109 molecules.

3.49 (a) 80.56% C, 7.51% H, 11.93% O. (b) 2.11 × 1021 molecules C9H10O.

3.51 C2H3NO5.

3.53 39.3 g S.

3.55 5.97 g F.

3.57 (a) CH2O. (b) KCN.

3.59 C8H10N4O2.

3.61 6.12 × 1021 molecules.

3.65 C10H20O.

3.67 CH2O.

3.69 Empirical formula: C3H2Cl; Molecular formula: C6H4Cl2.

3.71 C3H7O2NS.

3.75 1.01 mol Cl2.

3.77 2.0 × 101 mol CO2.

3.79 (a) 2NaHCO3 → Na2CO3 + CO2 + H2O. (b) 78.3 g NaHCO3.

3.81 255.9 g C2H5OH; 0.324 L.

3.83 0.294 mol KCN.

3.85 NH4NO3(s) → N2O(g) + 2H2O(g). (b) 2.0 × 101g N2O.

3.87 18.0 g O2.

3.93 6 mol NH3 produced; 1 mol H2 left.

3.95 4NO2 + O2 → 2N2O5. The limiting reactant is NO2.

3.97 HCl is the limiting reactant; 23.4 g Cl2 are produced.

3.99 (a) 7.05 g O2. (b) 92.9%.

3.101 3.48 × 103 g C6H14.

3.103 8.55 g S2Cl2; 76.6%.

3.105 (a) Combustion. (b) Combination. (c) Decomposition.

3.107 Diagram (b).

3.109 C3H2ClF5O; 184.50 g/mol.

3.111 Cl2O7.

3.113 (a) 0.212 mol O. (b) 0.424 mol O.

3.115 39.6% NO.

3.117 30.20% C, 5.069% H, 44.57% Cl, 20.16% S.

3.119 700 g.

3.121 (a) Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g). (b) 64.2%. (c) We assume that the impurities are inert and do not react with the sulfuric acid to produce hydrogen.

3.123 1.7 × 1015 g CO2/yr.

3.125 0.0011 mol chlorophyll.

3.127 (a) 6.532 × 104 g. (b) 7.6 × 102 g HG.

3.129 (a) 4.24 × 1022 K+ ions, 4.24 × 1022 Br ions. (b) 4.58 × 1022 Na+ ions, 2.29 × 1022 ions. (c) 4.34 × 1022 Ca2+ ions, 2.89 × 1022 ions.

3.131 (a) 0.307 g salicylic acid. (b) 0.410 g salicylic acid. (c) 12.08 g aspirin; 90.2%.

3.133 (a) C3H7NO. (b) C6H14N2O2.

3.135 6.022 × 1023 amu = 1 g.

3.137 16.00 amu.

3.139 (e) 0.50 mol Cl2.

3.141 PtCl2 and PtCl4.

3.143 (a) Compound X: MnO2; Compound Y: Mn3O4. (b) 3MnO2 → Mn3O4 + O2.

3.145 6.1 × 105 tons H2SO4.

3.147 Mg3N2, magnesium nitride.

3.149 PbC8H20.

3.151 (a) 4.3 × 1022 Mg atoms. (b) 1.6 × 102 pm.

3.153 28.97 g/mol.

3.155 3.1 × 1023 molecules/mol.

3.157 (a) C3H8(g) + 3H2O(g) → 3CO(g) + 7H2(g). (b) 909 kg H2.

3.159 (a) $0.47/kg. (b) 0.631 kg K2O.

3.161 BaBr2.

3.163 32.17% NaCl, 20.09% Na2SO4, 47.75% NaNO3.

3.165 0.709 g NO2 are produced. O3 is the limiting reactant. 0.0069 mol NO are left.

3.167 (a) NH4NO2 is the only reactant. When a reaction involves only a single reactant, we usually do not describe that reactant as the “limiting reactant” because it is understood that this is the case. (b) In principle, two or more reactants can be limiting reactants if they are exhausted simultaneously. In practice, it is virtually impossible to measure the reactants so precisely that two or more run out at the same time.

Page AP-3
Chapter 4

4.7 Diagram (c).

4.9 (a) Strong electrolyte. (b) Nonelectrolyte. (c) Weak electrolyte. (d) Strong electrolyte.

4.11 (a) Nonconducting. (b) Conducting. (c) Conducting.

4.13 Since HCl dissolved in water conducts electricity, HCl(aq) must actually exist as H+(aq) cations and Cl(aq) anions. Since HCl dissolved in benzene solvent does not conduct electricity, then we must assume that the HCl molecules in benzene solvent do not ionize, but rather exist as un-ionized molecules.

4.17 Diagram (c).

4.19 (a) Insoluble. (b) Insoluble. (c) Soluble. (d) Soluble.

4.21 (a) 2Ag+(aq) + 2 (aq) + 2Na+(aq) + (aq) → Ag2SO4(s) + 2Na+(aq) + 2(aq); 2Ag+(aq) + (aq) → Ag2SO4(s). (b) Ba2+(aq) + 2Cl(aq) + Zn2+(aq) + (aq) → BaSO4(s). (c) 2(aq) + (aq) + Ca2+(aq) + 2Cl(aq) → CaCO3(s) + 2(aq) + 2Cl(aq); Ca2+(aq) + (aq) → CaCO3(s).

4.23 (a) No precipitate forms. (b) Ba2+(aq) + (aq) → BaSO4(s).

4.31 (a) Brønsted base. (b) Brønsted base. (c) Brønsted acid. (d) Brønsted acid and Brønsted base.

4.33 (a) HC2H3O2(aq) + KOH(aq) → HC2H3O2(aq) + H2O(l); Ionic: HC2H3O2(aq) + K+(aq) + OH(aq) → + K+(aq) + H2O(l); Net ionic: HC2H3O2(aq) + OH(aq) → + H2O(l). (b) H2CO3(aq) + 2NaOH(aq) → Na2CO3(aq) + 2H2O(l), Ionic: H2CO3(aq) + 2Na+(aq) + 2OH(aq) → 2Na+(aq) + (aq) + 2H2O(l); Net ionic: H2CO3(aq) + 2OH(aq) → (aq) + 2H2O(l). (c) 2HNO3(aq) + Ba(OH)2(aq) → Ba(NO3)2(aq) + 2H2O(l), Ionic: 2H+(aq) + 2(aq) + Ba2+(aq) + 2OH(aq) → Ba2+(aq) + 2(aq) + 2H2O(l); Net ionic: 2H+(aq) + 2OH(aq) → 2H2O(l) or H+(aq) + OH(aq) → H2O(l).

4.41 (a) 2Sr → 2Sr2+ + 4e, Sr is the reducing agent; O2 + 4e → 2O2−, O2 is the oxidizing agent. (b) 2Li → 2Li+ + 2e, Li is the oxidizing agent; H2 + 2e → 2H, H2 is the oxidizing agent. (c) 2Cs → 2Cs+ + 2e, Cs is the reducing agent; Br2 + 2e → 2Br, Br2 is the oxidizing agent. (d) 3Mg → 3Mg2+ + 6e Mg is the reducing agent; N2 + → 6e → 2N3−, N2 is the oxidizing agent.

4.43 H2S(−2), S2− (−2), HS (−2) < S8 (0) < SO2 (+4) < SO3 (+6), H2SO4 (+6).

4.45 (a) +1. (b) +7. (c) −4. (d) −1. (e) −2. (f) +6. (g) +6. (h) +7. (i) +4. (j) 0. (k) +5. (l) −½. (m) +5. (n) +3.

4.47 (a) +1. (b) −1. (c) +3. (d) +3. (e) +4. (f) +6. (g) +2. (h) +4. (i) +2. (j) +3. (k) +5.

4.49 If nitric acid is a strong oxidizing agent and zinc is a strong reducing agent, then zinc metal will probably reduce nitric acid when the two react; that is, N will gain electrons and the oxidation number of N must decrease. Since the oxidation number of nitrogen in nitric acid is +5, then the nitrogen-containing product must have a smaller oxidation number for nitrogen. The only compound in the list that doesn't have a nitrogen oxidation number less than +5 is N2O5. This is never a product of the reduction of nitric acid.

4.51 Molecular oxygen is a powerful oxidizing agent. In SO3, the oxidation number of the element bound to oxygen (S) is at its maximum value (+6); the sulfur cannot be oxidized further. The other elements bound to oxygen in this problem have less than their maximum oxidation number and can undergo further oxidation. Only SO3 does not react with molecular oxygen.

4.53 (a) Decomposition. (b) Displacement. (c) Decomposition. (d) Combination.

4.59 232 g KI.

4.61 6.00 × 10−3 mol MgCl2.

4.63 (a) 1.16 M. (b) 0.608 M. (c) 1.78 M.

4.65 (a) 136 mL. (b) 62.2 mL. (c) 47 mL.

4.67 Dilute 323 mL of the 2.00 M HCl solution to a final volume of 1.00 L.

4.69 Dilute 3.00 mL of the 4.00 M HNO3 solution to a final volume of 60.0 mL.

4.71 (a) BaCl2: 0.300 M Cl; NaCl: 0.566 M Cl; AlCl3: 3.606 M Cl. (b) 1.28 M Sr(NO3)2.

4.73 2.325 M.

4.81 0.215 g AgCl.

4.83 0.165 g NaCl; Ag+(aq) + Cl(aq) → AgCl(s).

4.85 NaOH: Diagram (d); Ba(OH)2: Diagram (c).

4.87 (a) 42.78 mL. (b) 158.5 mL. (c) 79.23 mL.

4.89 (a) Redox. (b) Precipitation. (c) Acid-base. (d) Combination. (e) Redox. (f ) Redox. (g) Precipitation. (h) Redox. (i) Redox. (j) Redox.

4.91 (d) 0.20 M Mg(NO3)2 (greatest concentration of ions).

4.93 773 mL.

4.95 (a) Weak electrolyte. (b) Strong electrolyte. (c) Strong electrolyte. (d) Nonelectrolyte.

4.97 (a) C2H5ONH2 molecules. (b) K+ and F ions. (c) and ions. (d) C3H7OH molecules.

4.99 1146 g/mol.

4.101 1.28 M.

4.103 43.4 g BaSO4.

4.105 (a) Mix a solution containing Mg2+ ions such as MgCl2(aq) or Mg(NO3)2(aq) with a solution containing hydroxide ions such as NaOH(aq). Mg(OH)2 will precipitate and can be collected by filtration. (b) 0.78 L.

4.107 1.72 M.

4.109 (1) Electrolysis to ascertain if hydrogen and oxygen were produced, (2) The reaction with an alkali metal to see if a base and hydrogen gas were produced, and (3) The dissolution of a metal oxide to see if a base was produced (or a nonmetal oxide to see if an acid was produced).

4.111 Diagram (a) showing Ag+ and . The reaction is AgOH(aq) + HNO3(aq) → H2O(l) + AgNO3(aq).

4.113 (a) Check with litmus paper, combine with carbonate or bicarbonate to see if CO2 gas is produced, combine with a base and check for neutralization with an indicator. (b) Titrate a known quantity of acid with a standard NaOH solution. (c) Visually compare the conductivity of the acid with a standard NaCl solution of the same molar concentration.

4.115 (a) Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq); net ionic: Pb2+(aq) + (aq) → PbSO4(s). (b) 6.34 × 10-5 M.

4.117 (a) HI(aq) + KOH(aq) → KI(aq) + H2O(l), evaporate to dryness. (b) 2HI(aq) + K2CO3(aq) → 2KI(aq) + CO2(g) + H2O(l), evaporate to dryness.

4.119 (a) Combine any soluble magnesium salt with a soluble hydroxide, filter the precipitate. (b) Combine any soluble silver salt with any soluble iodide salt, filter the precipitate. (c) Combine any soluble barium salt with any soluble phosphate salt, filter the precipitate.

4.121 (a) Add Na2SO4. (b) Add KOH. (c) Add AgNO3. (d) Add Ca(NO3)2. (e) Add Mg(NO3)2.

4.123 Reaction 1: (aq) + H2O2(aq) → (aq) + H2O(l); Reaction 2: (aq) + Ba2+(aq) → BaSO4(s).

4.125 Cl2O (+1), Cl2O3 (+3), ClO2 (+4), Cl2O6 (+6), Cl2O7 (+7).

4.127 4.99 grains.

4.129 (a) CaF2(s) + H2SO4(aq) → CaSO4(s) + 2HF(g); 2NaCl(s) + H2SO4(aq) → Na2SO4(aq) + 2HCl(g). (b) The sulfuric acid would oxidize the Br and I ions to Br2 and I2. (c) PBr3(l) + 3H2O(l) → 3HBr(g) + H3PO3(aq).

4.131 Electric furnace method: P4(s) + 5O2(g) → P4O10(s) (redox), P4O10(s) + 6H2O(l) → 4H3PO4(aq) (acid-base); Wet process: Ca5(PO4)3F(s) + 5H2SO4(aq) → HF(aq) + 3H3PO4(aq) + 5CaSO4(s) (acid-base and precipitation).

4.133 (a) 4KO2(s) + 2CO2(g) → 2K2CO3(s) + 3O2(g). (b) −½. (c) 34.4 L air.

4.135 No. The oxidation number of all oxygen atoms is zero.

4.137 (a) Acid: H3O+, base: OH.

(b) Acid ; base .

4.139 When a solid dissolves in solution, the volume of the solution usually changes.

4.141 (a) The precipitate CaSO4 formed over Ca preventing the Ca from reacting with the sulfuric acid. (b) Aluminum is protected by a tenacious oxide layer with the composition Al2O3. (c) These metals react more readily with water: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g). (d) The metal should be placed below Fe and above H. (e) Any metal above Al in the activity series will react with Al3+. Metals from Mg to Li will work.

4.143 (a) (1) Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) (redox); (2) Cu(NO3)2(aq) + 2NaOH(aq) → Cu(OH)2(s) + 2NaNO3(aq) (precipitation); (3) Cu(OH)2(s) → CuO(s) + H2O(g) (decomposition); (4) CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l) (acid-base); (5) CuSO4(aq) + Zn(s) → Cu(s) + ZnSO4(aq) (redox); (6) Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) (redox). (b) (1) 194 g Cu(NO3)2; (2) 101 g Cu(OH)2; (3) 82.1 g CuO; (4) 165 g CuSO4; (5) 65.6 g Cu. (c) All of the reaction steps are clean and almost quantitative; therefore, the recovery yield should be high.

4.145 [Na+] = 0.5295 M, [] = 0.4298 M, [OH] = 0.09968 M, [Mg2+] ≈ 0 M.

4.147 1.41 M KMnO4.

4.149 0.00666 M C6H12O6.

4.151 Cu2+(aq) + S2−(aq) → CuS(s), 2.31 × 10−4 M Cu2+.

Page AP-4
Chapter 5

5.13 48 J.

5.15 577 J (work done on the system).

5.23 (a) 0. (b) −9.5 J. (c) −18 J.

5.25 4.51 kJ/g.

5.27 4.80 × 102 kJ.

5.29 For processes occurring in solution (cellular fluids), there is usually almost no volume change, so PΔV ≈ 0 and ΔU + PΔV ≈ ΔU (see Equation 5.10).

5.35 728 kJ.

5.37 50.7°C.

5.39 26.3°C.

5.41 47.8°C/day; 4.1 kg water/day.

5.45 0.30 kJ/mol.

5.47 −238.7 kJ/mol.

5.53 CH4(g) and H(g).

5.55 [H2O(l)].

5.57 [H2O(l)] is more negative than [H2O2(l)].

5.59 177.8 kJ/mol.

5.61 (a) −571.6 kJ/mol. (b) −2599 kJ/mol.

5.63 (a) −724 kJ/mol. (b) −1.37 × 103 kJ/mol. (c) −2.01 × 103 kJ/mol.

5.65 −3924 kJ/mol.

5.67 −3.43 × 104 kJ.

5.69 In a chemical reaction the same elements and the same numbers of atoms are always on both sides of the equation. This provides a consistent reference that allows the energy change in the reaction to be interpreted in terms of the chemical or physical changes that have occurred. In a nuclear reaction the same elements are not always on both sides of the equation and no common reference point exists.

5.71 −44.35 kJ/mol.

5.73 −175.3 kJ.

5.75 (a) Although we cannot measure for this reaction, the reverse process is the combustion of glucose. We could easily measure for this combustion by burning a mole of glucose in a bomb calorimeter. (b) 1.l × 1019 kJ.

5.77 −350.7 kJ/mol.

5.79 0.492 J/g · °C.

5.81 The first reaction, which is exothermic, can be used to promote the second reaction, which is endothermic. Thus, the two gases are produced alternately.

5.83 446 mol H2.

5.85 0.167 mol ethane.

5.87 5.60 kJ/mol.

5.89 (a).

5.91 (a) 0 J. (b) −9.1 J.

5.93 (a) The more closely packed, the greater the mass of food. Heat capacity depends on both the mass and specific heat: C = ms. The heat capacity of the food is greater than the heat capacity of air; hence, the cold in the freezer will be retained longer. (b) Tea and coffee are mostly water; whereas, soup might contain vegetables and meat. Water has a higher heat capacity than the other ingredients in soup; therefore, coffee and tea retain heat longer than soup.

5.95 −1.84 × 103 kJ.

5.97 3.0 × 109 atomic bombs.

5.99 (a) 2LiOH(aq) + CO2(g) → Li2CO3(aq) + H2O(l). (b) 1.1 kg CO2, 1.2 kg LiOH.

5.101 −5.2 × 106 kJ.

5.103 −3.60 × 102 kJ/mol Zn or −3.60 × 102 kJ/2 mol Ag+.

5.105 We assume that when the car is stopped, its kinetic energy is completely converted into heat (friction of the brakes and friction between the tires and the road). Thus, q = (½)mu2, and the amount of heat generated must be proportional to the braking distance, d: d α q or d α u2. Therefore, as u increases to 2u, d increases to (2u)2 = 4u2, which is proportional to 4d.

5.107 Water has a larger specific heat than air. Thus cold, damp air can extract more heat from the body than cold, dry air. By the same token, hot, humid air can deliver more heat to the body.

5.109 5.8 × 102 m.

5.111 This law does not hold for Hg because it is a liquid.

5.113 (C2H2) = 226.6 kJ/mol, (C6H6) = 49.0 kJ/mol, = −630.8 kJ/mol.

5.115 4.1 cents.

5.117 (a) CaC2(s) + 2H2O(l) → Ca(OH)2(s) + C2H2(g). (b) 1.51 × 106 J.

5.119 (a) glucose: 31 kJ, sucrose: 33 kJ. (b) glucose: 15 m, sucrose: 16 m.

5.121 w = 0, ΔU = −5153 kJ/mol.

5.123 −564.2 kJ/mol.

5.125 96.21%.

5.127 58.1°C.

5.129 −9.78 kJ/mol.

5.131

  q w ΔU ΔH
(a) 0
(b)
(c) + ? +
(d) + 0 + +
(e) + 0 0

5.133 43.5°C/day; 4.1 kg water/day.

5.135 23.6°C.

Chapter 6

6.5 (a) 3.5 × 103 nm. (b) 5.30 × 1014 Hz.

6.7 7.0 × 102 s.

6.9 3.26 × 107 nm, microwave.

6.15 2.82 × 10−19 J.

6.17 (a) 4.6 × 107 nm, not in the visible region. (b) 4.3 × 10−24 J/photon. (c) 2.6 J/mol.

6.19 1.29 × 10−15 J.

6.21 Infrared photons have insufficient energy to cause the chemical changes.

6.23 ν = 3.23 × 1020 s−1, λ = 9.29 × 10−13 m.

6.29 Analyze the emitted light by passing it through a prism.

6.31 Excited atoms of the chemical elements emit the same characteristic frequencies or lines in a terrestrial laboratory, in the sun, or in a star many light-years distant from Earth.

6.33 3.027 × 10−19 J.

6.35 ν = 1.60 × 1014 Hz, λ = 1.88 × 103 nm.

6.37 5.

6.41 0.565 nm.

6.43 9.96 × 10−32 cm.

6.51 Δu≥ 4.38 × 10−26 m/s. This uncertainty is far smaller than can be measured.

6.55 ℓ = 1, m = −1, 0, and 1; ℓ = 0, m = 0.

6.57 4s, 4p, 4d, and 4f subshells; 1, 3, 5, and 7 orbitals, respectively.

6.63 (a) n = 2, ℓ = 1, m = 1, 0, or −1. (b) n = 3, ℓ = 0, m = 0. (c) n = 5, ℓ = 2, m = 2, 1, 0, −1, or −2.

6.65 A 2s orbital is larger than a 1s orbital and exhibits a node. Both have the same spherical shape. The 1s orbital is lower in energy than the 2s.

6.67 In H, energy depends only on n, but for all other atoms, energy depends on n and ℓ.

6.69 (a) 2s. (b) 3. (c) equal. (d) equal. (e) 5s.

6.75 In the nth shell, there are n2 orbitals and 2n2 electrons.

6.77 (a) 3. (b) 6. (c) 0.

6.79 Al: two 2p electrons missing, 1s22s22p63s23p1; B: too many 2p electrons, 1s22s22p1; F: too many 2p electrons, 1s22s22p5.

6.81 1s: (1, 0, 0, + ½) and (1, 0, 0, – ½), 2s: (2, 0, 0, +½) and (2, 0, 0, −½), 2p (2, 1, −1, +½), (2, 1, −1, −½), (2, 1, 0, +½), (2, 1, 0, −½), (2, 1, +1, +½), and (2, 1, +1, −½), 3s (3, 0, 0, +½) and (3, 0, 0, −½). The element is Mg.

6.91 [Kr]5s24d5.

6.93 Ge: [Ar]4s23d104p2; Fe: [Ar]4s23d6; Zn: [Ar]4s23d10; Ni: [Ar]4s23d8; W: [Xe]6s24f145d4; Tl: [Xe]6s24f145d106p1.

6.95 Part (b) is correct in the view of contemporary quantum theory. Bohr's explanation of emission and absorption line spectra appears to have universal validity. Parts (a) and (c) are artifacts of Bohr's early planetary model of the hydrogen atom and are not considered to be valid today.

6.97 (a) 4. (b) 6. (c) 10. (d) 1. (e) 2.

6.99 In the photoelectric effect, light of sufficient energy shining on a metal surface causes electrons to be ejected (photoelectrons). Since the electrons are charged particles, the metal surface becomes positively charged as more electrons are lost. After a long enough period of time, the positive surface charge becomes large enough to start attracting the ejected electrons back toward the metal with the result that the kinetic energy of the departing electrons becomes smaller.

6.101 (a) 1.20 × 1018 photons. (b) 3.76 × 108 W.

6.103 λ = 419 nm. This wavelength is in the visible region of the electromagnetic spectrum. Since water is continuously being struck by visible radiation without decomposition, it seems unlikely that photodissociation of water by this method is feasible.

6.105 3.0 × 1019 photons/s.

6.107 He+: n = 3 → 2: λ = 164 nm; n = 4 → 2: λ = 121 nm; n = 5 → 2: λ = 108 nm; n = 6 → 2: λ = 103 nm. H: n = 3 → 2:λ = 656 nm; n = 4 → 2: λ = 486 nm; n = 5 → 2: λ = 434 nm; n = 6 → 2:λ = 410 nm. All the Balmer transitions for He+ are in the ultraviolet region; whereas, the transitions for H are all in the visible region.

6.109

6.111 (a) He, 1s2. (b) N, 1s22s22p3.

(c) Na, 1s22s22p63s1. (d) As, [Ar]4s23d104p3. (e) Cl, [Ne]3s23p5.

6.113 (a) (b) [Ne] (c)

6.115 λ = 0.596 m; Microwave/radio region.

6.117 (a) False. (b) False. (c) True. (d) False. (e) True.

6.119 4.9 × 1023 photons; 34 H2O molecules/photon.

6.121 2.2 × 105 J.

6.123 (b) and (d) are allowed transitions. Any of the transitions in Figure 6.11 is possible as long as ℓ for the final state differs from ℓ of the initial state by 1.

6.125 17.4 pm.

6.127 λ = 0.382 pm, ν = 7.86 × 1020 s−1.

6.129 ni = 1: ΔE = 1.96 × 10−17 J; ni = 5: ΔE = 7.85 × 10−19 J; transition from n = 5 to n = 1: ΔE = −1.88 × 10−17 J; λ = 10.6 nm.

6.131 3.9 × 105 m/s.

6.133 (a) We note that the maximum solar radiation centers around 500 nm. Thus, over billions of years, organisms have adjusted their development to capture energy at or near this wavelength. The two most notable cases are photosynthesis and vision. (b) Astronomers record blackbody radiation curves from stars and compare them with those obtained from objects at different temperatures in the laboratory. Because the shape of the curve and the wavelength corresponding to the maximum depend on the temperature of an object, astronomers can reliably determine the temperature at the surface of a star from the closest matching curve and wavelength.

6.135 1.06 nm.

Page AP-5
Chapter 7

7.17 Sulfur, 1s22s22p63s23p4.

7.19 (a) and (d); (b) and (e); (c) and (f).

7.21 (a) Group 1A or 1. (b) Group 5A or 15. (c) Group 8A or 18. (d) Group 8B or 10.

7.25 (a) σ = 2 and Zeff = +4. (b) 2s, Zeff = +3.22; 2p, Zeff = +3.14. The values are lower than those in part (a) because the 2s and 2p electrons actually do shield each other somewhat.

7.33 Na > Mg > Al > P > Cl.

7.35 Fluorine.

7.37 Left to right: S, Se, Ca, K.

7.39 The atomic radius is largely determined by how strongly the outer-shell electrons are held by the nucleus. The larger the effective nuclear charge, the more strongly the electrons are held and the smaller the atomic radius. For the second period, the atomic radius of Li is largest because the 2s electron is well shielded by the filled 1s shell. The effective nuclear charge that the outermost electrons feel increases across the period as a result of incomplete shielding by electrons in the same shell. Consequently, the orbital containing the electrons is compressed and the atomic radius decreases.

7.41 K < Ca < P < F < Ne.

7.43 The Group 3A elements (such as Al) all have a single electron in the outermost p subshell, which is well shielded from the nuclear charge by the inner electrons and the ns2 electrons. Therefore, less energy is needed to remove a single p electron than to remove a paired s electron from the same principal energy level (such as for Mg).

7.45 496 kJ/mol is paired with 1s22s22p63s1. 2080 kJ/mol is paired with 1s22s22p6, a very stable noble gas configuration.

7.47 8.40 × 106 kJ/mol.

7.49 Cl.

7.51 Alkali metals have a valence electron configuration of ns1 so they can accept another electron in the ns orbital. On the other hand, alkaline earth metals have a valence electron configuration of ns2. Alkaline earth metals have little tendency to accept another electron, as it would have to go into a higher energy p orbital.

7.57 Fe.

7.59 (a) [Ne]. (b) [Ne]. (c) [Ar]. (d) [Ar]. (e) [Ar]. (f) [Ar]3d6. (g) [Ar]3d9. (h) [Ar]3d10.

7.61 (a) Cr3+. (b) Sc3+. (c) Rh3+. (d) Ir3+.

7.63 Be2+ and He; N3− and F; Fe2+ and Co3+; S2− and Ar.

7.67 (a) Cl. (b) Na+. (c) O2−. (d) Al3+. (e) Au3+.

7.69 The Cu+ ion is larger than Cu2+ because it has one more electron.

7.71 The binding of a cation to an anion results from electrostatic attraction. As the +2 cation gets smaller (from Ba2+ to Mg2+), the distance between the opposite charges decreases and the electrostatic attraction increases.

7.75 −199.7°C.

7.77 Since ionization energies decrease going down a column in the periodic table, francium should have the lowest first ionization energy of all the alkali metals. As a result, Fr should be the most reactive of all the Group 1A elements toward water and oxygen. The reaction with oxygen would probably be similar to that of K, Rb, or Cs.

7.79 The Group 1B elements are much less reactive than the Group 1A elements. The 1B elements are more stable because they have much higher ionization energies resulting from incomplete shielding of the nuclear charge by the inner d electrons. The ns1 electron of a Group 1A element is shielded from the nucleus more effectively by the completely filled noble gas core. Consequently, the outer s electrons of 1B elements are more strongly attracted by the nucleus.

7.81 (a) Li2O(s) + H2O(l) → 2LiOH(aq). (b) CaO(s) + H2O(l) → Ca(OH)2(aq). (c) SO3(g) + H2O(l) → H2SO4(aq).

7.83 BaO. As we move down a column, the metallic character of the elements increases.

7.85 (a) Br. (b) N. (c) Rb. (d) Mg.

7.87 O2− < F < Na+ < Mg2+.

7.89 M = K, × = Br.

7.91 O+ and N; S2− and Ar; N3− and Ne; As3+ and Zn; Cs+ and Xe.

7.93 (a) and (d).

7.95 Fluorine is a yellow-green gas that attacks glass; chlorine is a pale yellow gas; bromine is a fuming red liquid; iodine is a dark, metallic-looking solid.

7.97 F.

7.99 H. Since H has only one proton compared to two protons for He, the nucleus of H will attract the two electrons less strongly compared to He.

7.101 Li2O, lithium oxide, basic; BeO, beryllium oxide, amphoteric; B2O3, diboron trioxide, acidic; CO2, carbon dioxide, acidic; N2O5, dinitrogen pentoxide, acidic.

7.103 Hydrogen has just one valence electron but it is also just one electron short of a noble gas configuration. Thus, it can behave like a member of Group 1A or like a member of Group 7A.

7.105 0.66.

7.107 77.5%.

7.109 6.94 × 10−19 J/electron. If there are no other electrons with lower kinetic energy, then this is the electron from the valence shell. UV light of the longest wavelength (lowest energy) that can still eject electrons should be used.

7.111 X must belong to Group 4A; it is probably Sn or Pb because it is not a very reactive metal (it is certainly not reactive like an alkali metal). Y is a nonmetal since it does not conduct electricity. Since it is a light yellow solid, it is probably phosphorus (Group 5A). Z is an alkali metal since it reacts with air to form a basic oxide or peroxide.

7.113 (a) IE1 = 3s1 electron, IE2 = 2p6 electron, IE3 = 2p5 electron, IE4 = 2p4 electron, IE5 = 2p3 electron, IE6 = 2p2 electron, IE7 = 2p1 electron, IE8 = 2s2 electron, IE9 = 2s1 electron, IE10 = 1s2 electron, IE11 = 1s1 electron. (b) Each break (IE1IE2 and IE9IE10) represents the transition to another shell (n = 3 → 2 and n = 2 → 1).

7.115 Considering electron configurations, Fe2+[Ar]3d6 → Fe3+[Ar]3d5, Mn2+[Ar]3d5 → Mn3+[Ar]3d4, a half-filled shell has extra stability. In oxidizing Fe2+, the product is a d5-half-filled shell. In oxidizing Mn2+, a d5-half-filled shell electron is being lost, which requires more energy.

7.117 The electron configuration of titanium is [Ar]4s23d2. K2TiO4 is unlikely to exist because of the oxidation state of Ti of +6. Ti in an oxidation state greater than +4 is unlikely because of the very high ionization energies needed to remove the fifth and sixth electrons.

7.119 (a) 2KClO3(s) → 2KCl(s) + 3O2(g). (b) N2(g) + 3H2(g) → 2NH3(g) (industrial); NH4Cl(s) + NaOH(aq) → NH3(g) + NaCl(aq) + H2O(l). (c) CaCO3(s) → CaO(s) + CO2(g) (industrial); CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g). (d) Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g). (e) Same as (c), (first equation).

7.121 Examine a solution of Na2SO4, which is colorless. This shows that the ion is colorless. Thus the blue color is due to Cu2+(aq).

7.123 Zeff increases from left to right across the table, so electrons are held more tightly. (This explains the electron affinity values of C and O.) Nitrogen has a zero value of electron affinity because of the stability of the half-filled 2p subshell (that is, N has little tendency to accept another electron).

7.125 Once an atom gains an electron forming a negative ion, adding additional electrons is typically an unfavorable process due to electron-electron repulsions. Second and third electron affinities do not occur spontaneously and are therefore difficult to measure.

7.127 (a) It was determined that the periodic table was based on atomic number, not atomic mass. (b) Argon: 39.95 amu; Potassium: 39.10 amu.

7.129 Element 119: [Rn]7s25f146d107p68s1.

7.131 2A. There is a large jump from the second to the third ionization energy, indicating a change in the principal quantum number n.

7.133 (a) SiH4, GeH4, SnH4, PbH4. (b) RbH should be more ionic than NaH. (c) Ra(s) + 2H2O(l) → Ra(OH)2(aq) + H2(g). (d) Be (diagonal relationship).

7.135 See sections 7.1 and 7.2.

7.137 (c) Carbon.

7.139 Li: Zeff = 1.26, Zeff/n = 0.630; Na: Zeff = 1.84, Zeff/n = 0.613; K: Zeff = 2.26, Zeff/n = 0.565. As we move down a group, Zeff increases. This is what we would expect because shells with larger n values are less effective at shielding the outer electrons from the nuclear charge. The Zeff/n values are fairly constant, meaning that the screening per shell is about the same.

7.141 Nitrogen. Lithium forms a stable nitride (Li3N).

Page AP-6
Chapter 8

8.19 (a) Decreases the ionic bond energy. (b) Triples the ionic bond energy. (c) Increases the bond energy by a factor of 4. (d) Increases the bond energy by a factor of 2.

8.21

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

8.23 860 kJ/mol.

8.33 (a) BF3, boron trifluoride, covalent. (b) KBr, potassium bromide, ionic.

8.37 C—H (ΔEN = 0.4) < Br—H (ΔEN = 0.7) < F—H (ΔEN = 1.9) < Li—Cl (ΔEN = 2.0) < Na—Cl (ΔEN = 2.1) < K—F (ΔEN = 3.2).

8.39 Cl—Cl < Br—Cl < Si—C < Cs—F.

8.41 (a) Covalent. (b) Polar covalent. (c) Ionic. (d) Polar covalent.

8.43

(a)

(b)

(c)

(d)

(e)

(f)

8.45

(a)

(b)

(c)

(d)

(e)

(f)

(g)

8.47 (a) (b) or (c) (d)

8.49 (a) Neither oxygen atom has a complete octet, and the left-most hydrogen atom shows two bonds (4 electrons). Hydrogen can hold only two electrons in its valence shell.

(b)

8.53 (a)

(b)

8.55

8.57

8.59

8.67 No. To make an octet on Be is not plausible,

8.69 No.

8.71 Coordinate covalent bond,

8.73 Completed octet on S: ; zero formal charge on S:

8.77 303.0 kJ/mol.

8.79 (a) −2759 kJ/mol. (b) −2855.4 kJ/mol.

8.81 −651 kJ/mol.

8.83 Ionic: RbCl, KO2. Covalent: PF5, BrF3, CI4.

8.85 Ionic: NaF, MgF2, AlF3. Covalent: SiF4, PF5, SF6, ClF3.

8.87 KF is an ionic compound. It is a solid at room temperature made up of K+ and F ions. It has a high melting point, and it is a strong electrolyte. Benzene, C6H6, is a covalent compound that exists as discrete molecules. It is a liquid at room temperature. It has a low melting point, is insoluble in water, and is a nonelectrolyte.

8.89

8.91 (a) (b) . (c) AlCl3.

8.93 CF2 would be very unstable because carbon does not have an octet. LiO2 would not be stable because the lattice energy between Li+ and superoxide would be too low to stabilize the solid. CsCl2 requires a Cs2+ cation. The second ionization energy is too large to be compensated by the increase in lattice energy. PI5 appears to be a reasonable species. However, the iodine atoms are too large to have five of them “fit” around a single P atom.

8.95 (a) False. (b) True. (c) False. (d) False.

8.97 −67 kJ/mol.

8.99 N2, since it has a triple bond.

8.101 CH4 and ; N2 and CO; C6H6 and B3N3H6.

8.103

8.105

8.107 The central iodine atom in has ten electrons surrounding it: two bonding pairs and three lone pairs. The central iodine has an expanded octet. Elements in the second period such as fluorine cannot have an expanded octet as would be required for .

8.109

8.111 Form (a) is the most important structure with no formal charges and all satisfied octets. (b) is likely not as important as (a) because of the positive formal charge on O. Forms (c) and (d) do not satisfy the octet rule for all atoms and are likely not important.

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

8.115 (a) −9.2 kJ/mol. (b) −9.2 kJ/mol.

Page AP-7

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

8.119 True. Each noble gas atom already has completely filled ns and np subshells.

8.121 (a) 114 kJ/mol. (b) The bond in is weaker.

8.123 (a) The first structure is the most important. (b) No.

8.125 The arrows indicate coordinate covalent bonds. This dimer does not possess a dipole moment.

8.127 (a) (b) To complete its octet, the OH radical has a strong tendency to form a bond with a H atom. (c) −46 kJ/mol. (d) 260 nm.

8.129 347 kJ/mol.

8.131 EN(O) = 3.2 (Pauling 3.5); EN(F) = 4.4 (Pauling 4.0); EN(Cl) = 3.5 (Pauling 3.0).

8.133 (1) You could estimate the lattice energy of the solid by trying to measure its melting point. Mg+O would have a lattice energy (and, therefore, a melting point) similar to that of Na+Cl. This lattice energy and melting point are much lower than those of Mg2+O2−. (2) You could determine the magnetic properties of the solid. An Mg+O solid would be paramagnetic while Mg2+O2− solid is diamagnetic. See Chapter 9 of the text.

8.135

(a) (b) −413 kJ/mol.

8.137

8.139 C—C: 347 kJ/mol; N—N: 193 kJ/mol; O—O: 142 kJ/mol. Lone pairs appear to weaken the bond.

Chapter 9

9.7 (a) Trigonal pyramidal. (b) Tetrahedral. (c) Tetrahedral. (d) See-saw.

9.9 (a) Tetrahedral. (b) Trigonal planar. (c) Trigonal pyramidal. (d) Bent. (e) Bent.

9.11 (a) Linear. (b) Tetrahedral. (c) Trigonal bipyramidal. (d) Trigonal pyramidal. (e) Tetrahedral.

9.13 Carbon at the center of H3C—: electron domain geometry = tetrahedral, molecular geometry = tetrahedral; Carbon at center of —CO—OH: electron-domain geometry = trigonal planar, molecular geometry = trigonal planar; Oxygen in —O—H: electron domain geometry = tetrahedral, molecular geometry = bent.

9.17 (a) Polar. (b) Nonpolar.

9.19 Only (c) is polar.

9.29 (a) sp3. (b) sp3.

9.31 In BF3, B is sp2 hybridized. In NH3, N is sp3 hybridized. In F3B—NH3, B and N are both sp3 hybridized.

9.33 sp3d.

9.37 (a) sp. (b) sp. (c) sp.

9.39 sp.

9.41 Nine σ bonds and nine π bonds.

9.43 36 σ bonds and 10 π bonds.

9.49 In order for the two hydrogen atoms to combine to form a H2 molecule, the electrons must have opposite spins. If two H atoms collide and their electron spins are parallel, no bond will form.

9.51 = < Li2. (Both and have bond order of ½. Li2 has a bond order of 1.)

9.53 , with a bond order of ½. (B2 has a bond order of 1.)

9.55 The Lewis diagram has all electrons paired (incorrect) and a double bond (correct). The MO diagram has two unpaired electrons (correct), and a bond order of 2 (correct).

9.57 O2: bond order = 2, paramagnetic; : bond order = 2.5, paramagnetic; : bond order = 1.5, paramagnetic; : bond order = 1, diamagnetic.

9.59 The two shared electrons that make up the single bond in B2 both lie in pi molecular orbitals and constitute a pi bond. The four shared electrons that make up the double bond in C2 all lie in pi molecular orbitals and constitute two pi bonds.

9.63 The left symbol shows three delocalized double bonds (correct). The right symbol shows three localized double bonds and three single bonds (incorrect).

9.65 (a) (b) sp2. (c) Sigma bonds join the nitrogen atom to the fluorine and oxygen atoms. There is a pi molecular orbital delocalized over the N and O atoms.

9.67 The central oxygen atom is sp2 hybridized. The unhybridized 2pz orbital on the central oxygen overlaps with the 2pz orbitals on the two terminal atoms.

9.69 Linear. You could establish the geometry of HgBr2 by measuring its dipole moment.

9.71 Bent; sp3.

9.73 (a) C6H8O6. (b) Five central O atoms sp3; hybridization of the sixth (double bonded) peripheral O atom is sp2. Three C atoms sp2; three C atoms sp3. (c) The five central O atoms bent. The three sp2 C atoms are trigonal planar. The three sp3 C atoms are tetrahedral.

9.75 (a) 180°. (b) 120°. (c) 109.5°. (d) 109.5°. (e) 180°. (f) 120°. (g) 109.5°. (h) 109.5°.

9.77 (a) , planar. (b) , nonplanar.

(c) polar. (d) polar. (e) greater than 120°. Experimental value is around 135°.

9.79 The S—S bond is a normal 2-electron shared pair covalent bond. Each S is sp3 hybridized, so the X—S—S angle is about 109°.

9.81 Only and CdBr2 are linear.

9.83 (a) sp2. (b) The molecule on the right is polar.

9.85 Rotation about the sigma bond in 1,2-dichloroethane does not destroy the bond, so the bond is free to rotate. Thus, the molecule is nonpolar because the C—Cl bond moments cancel each other because of the averaging effect brought about by rotation. The π bond between the C atoms in cis-dichloroethylene prevents rotation (in order to rotate, the π bond must be broken). Therefore, the molecule is polar.

9.87 All except homonuclear diatomic molecules N2 and O2.

9.89

The carbon atoms and nitrogen atoms marked with an asterisk (C* and N*) are sp2 hybridized; unmarked carbon atoms and nitrogen atoms are sp3 hybridized; and the nitrogen atom marked with (#) is sp hybridized.

9.91 C has no d orbitals but Si does (3d). Thus, H2O molecules can add to Si in hydrolysis (valence-shell expansion).

9.93 The carbons are all sp2 hybridized. The nitrogen double bonded to carbon in the ring is sp2 hybridized. The other nitrogens are sp3 hybridized.

9.95 F2 has 8 electrons in bonding orbitals and 6 electrons in antibonding orbitals, giving it a bond order of 1. has 8 electrons in bonding orbitals and 7 electrons in antibonding orbitals, giving it a bond order of ½.

9.97 43.6%.

9.99 The second (“asymmetric stretching mode”) and third (“bending mode”).

9.101

The four carbons marked with an asterisk are sp2 hybridized. The remaining carbons are sp3 hybridized.

Page AP-8

9.103 (a) [Ne2](σ3s)2(. (b) 3. (c) Diamagnetic.

9.105 In the Lewis structure, all the electrons are paired. From molecular orbital theory, the electrons would be arranged as follows: . This shows two unpaired electrons. To pair these electrons, energy is required to flip the spin of one, thus making the Lewis structure an excited state.

9.107 (a) Although the O atoms are sp3 hybridized, they are locked in a planar structure by the benzene rings. The molecule is symmetrical and therefore is not polar. (b) 20 σ bonds and 6 π bonds.

9.109 (a) The electronegativity difference between O and C suggests that electron density should concentrate on the O atom, but assigning formal charges places a negative charge on the C atom. Therefore, we expect CO to have a small dipole moment. (b) CO is isoelectronic with N2, bond order 3. This agrees with the triple bond in the Lewis structure. (c) Since C has a negative formal charge, it is more likely to form bonds with Fe2+. (OC—Fe2+ rather than CO—Fe2+).

9.111 Tetrahedral.

9.113 No.

Chapter 10

10.7 (a) Amine. (b) Aldehyde. (c) Ketone. (d) Carboxylic acid. (e) Alcohol.

10.9 (a) 3-ethyl-2,4,4-trimethylhexane. (b) 6,6-dimethyl-2-heptanol. (c) 4-chlorohexanal.

10.11 3,5-dimethyloctane.

10.13 (a) (CH3)3CCH2CH(CH3)2. (b) HO(CH2)2CH(CH3)2. (c) CH3(CH2)4C(O)NH2. (d) Cl3CCHO.

10.15

10.17

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

10.21 (a) Kekule: ,

(b) Condensed: (C2H5)2CHCH2CO2C(CH3)3,

(c) Condensed: (CH3)2CHCH2NHCH(CH3)2,

10.23 (a) Condensed structural: (CH3)2NCHO,

(b) Condensed structural: (CH2COOH)2C(OH)COOH,

(c) Condensed structural: (CH3)2CH(CH2)2OC(O)CH3,

10.25 (a) (b) (c)

10.27 (a)

(b)

(c)

10.29 (a)

(b)

(c)

10.37

Page AP-9

10.39

10.41

10.43(a) (b)

10.45

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

10.55

10.57 (a)

(b)

10.59 ; elimination reaction.

10.61 This is not an oxidation-reduction reaction. There are no changes in oxidation states for any of the atoms.

10.63 (a) Sulfuric acid is a catalyst.

(b) n-propanol: (c) No.

10.71

10.73 (a)

(b)

10.75

10.77

10.79

10.81 (a)

(b)

10.83 (a) Cis/trans stereoisomers. (b) Constitutional isomers. (c) Resonance structures. (d) Different representations of the same structure.

10.85

10.87 (b).

10.89

10.91 Since N is less electronegative than O, electron donation in the amide would be more pronounced.

10.93 (a) The more negative ΔH implies stronger alkane bonds; branching decreases the total bond enthalpy (and overall stability) of the alkane. (b) The least highly branched isomer (n-octane).

10.95 (a) 15.81 mg C, 1.32 mg H, 3.49 mg O.

(b) C6H6O.

(c)

Page AP-10

10.97

10.99

10.101 8 different tripeptides: Lys—Lys–Lys, Lys—Lys–Ala, Lys—Ala–Lys, Ala—Lys–Lys, Lys—Ala–Ala, Ala—Lys–Ala, Ala—Ala–Lys, Ala—Ala–Ala.

Chapter 11

11.13 0.493 atm, 0.500 bar, 375 torr, 5.00 × 104 Pa.

11.15 13.1 m.

11.17 7.3 atm.

11.21 (a) Diagram (d). (b) Diagram (b).

11.23 45.9 mL.

11.25 587 mmHg.

11.27 31.8 L.

11.29 1 volume of NO.

11.35 6.6 atm.

11.37 1590°C.

11.39 1.8 atm.

11.41 0.70 L.

11.43 63.31 L.

11.45 6.1 × 10−3 atm.

11.47 35.0 g/mol.

11.49 2.1 × 1022 N2 molecules, 5.6 × 1022 O2 molecules, 2.7 × 1020 O2 molecules.

11.51 2.98 g/L.

11.53 SF4.

11.55 3.70 × 102 L.

11.57 88.9%.

11.59 M(s) + 3HCl(aq) → 1.5H2(g) + MCl3(aq), M2O3; M2(SO4)3.

11.61 94.7%. Assuming the impurity (or impurities) do not produce CO2.

11.63 C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l); 1.44 × 103L air.

11.67 (a) 0.89 atm. (b) 1.4 L.

11.69 349 mmHg.

11.71 19.8 g Zn.

11.73 = 217 mmHg, = 650 mmHg.

11.75 (a) Box 2. (b) Box 2.

11.83 urms(N2) = 472 m/s, urms(O2) = 441 m/s, urms(O3) = 360 m/s.

11.85 RMS = 2.8 m/s, Average speed = 2.7 m/s. The root-mean-square value is always greater than the average value, because squaring favors the larger values compared to just taking the average value.

11.87 43.8 g/mol, CO2.

11.93 18.0 atm; Ideal: 18.5 atm.

11.95 (a) Neither the amount of gas in the tire nor its volume changes appreciably. The pressure is proportional to the temperature. Therefore, as the temperature rises, the pressure increases. (b) As the paper bag is hit, its volume decreases so that its pressure increases. The popping sound occurs when the bag is broken. (c) As the balloon rises, the pressure outside decreases steadily, and the balloon expands. (d) The pressure inside the bulb is less than 1 atm.

11.97 1.7 × 102 L. = 0.49 atm, = 0.41 atm, = 0.25 atm, and = 0.041 atm.

11.99 (a) NH4NO2(s) → N2(g) + 2H2O(l ). (b) 0.273 g.

11.101 (a) Pii = 4.0 atm, Piii = 2.67 atm. (b) 2.67 atm, PA = 1.33 atm, PB = 1.33 atm.

11.103 (a) 9.54 atm. (b) If the tetracarbonylnickel gas starts to decompose significantly above 43°C, then for every mole of tetracarbonylnickel gas decomposed, four moles of carbon monoxide would be produced.

11.105 1.3 × 1022 molecules. CO2, O2, N2 and H2O.

11.107 (a) 2KClO3(s) → 2KCl(s) + 3O2(g). (b) 6.21 L.

11.109 0.0701 M.

11.111 PHe = 0.16 atm, PNe = 2.0 atm.

11.113 When the water enters the flask from the dropper, some hydrogen chloride dissolves, creating a partial vacuum. Pressure from the atmosphere forces more water up the vertical tube.

11.115 7.

11.117 (a) 61.2 m/s. (b) 4.58 × 10−4 s. (c) 328 m/s. urms = 366 m/s. The magnitudes of the speeds are comparable, but not identical. The rms speed is derived for 3-dimensional motion and is not applicable to the speed distributions in directed molecular beams. Also, the experiment is measuring a most-probable speed, not an rms speed.

11.119 20.9 kg O2; 1.58 × 104 L O2.

11.121 The fruit ripens more rapidly because the quantity (partial pressure) of ethylene gas inside the bag increases.

11.123 As the pen is used the amount of ink decreases, increasing the volume inside the pen. As the volume increases, the pressure inside the pen decreases. The hole is needed to equalize the pressure as the volume inside the pen increases.

11.125 (a) NH4NO3(s) → N2O(g) + 2H2O(l). (b) 0.0821 L · atm/K · mol.

11.127 C6H6.

11.129 The gases inside the mine were a mixture of carbon dioxide, carbon monoxide, methane, and other harmful compounds. The low atmospheric pressure caused the gases to flow out of the mine (the gases in the mine were at a higher pressure), and the man suffocated.

11.131 (a) 4.90 L. (b) 6.0 atm. (c) 1 atm.

11.133 (a) 4 × 10−22 atm. (b) 6 × 1020 L/g of H.

11.135 91%.

11.137 1.7 × 1012.

11.139 NO2.

11.141 1.4 g.

11.143 = 8.7 × 106 Pa, = 3.2 × 105 Pa, = 1.4 × 103Pa.

11.145 (a) (i) He atoms in both flasks have the same rms speed. (ii) The atoms in the larger volume strike the walls of their container with the same average force as those in the smaller volume, but pressure in the larger volume will be less. (b) (i) u2 is greater than u1 by a factor of . (ii) Both the frequency of collision and the average force of collision of the atoms at T2 will be greater than those at T1. (c) (i) False. The rms speed of He is greater than that of Ne because He is less massive. (ii) True. The average kinetic energies of the two gases are equal. (iii) The rms speed of the whole collection of atoms is 1.47 × 103 m/s, but the speeds of the individual atoms are random and constantly changing.

11.147 Vapor: 3.7 nm; Liquid: 0.31 nm. Water molecules are packed very closely together in the liquid, but much farther apart in steam.

11.149 9.98 atm.

11.151 1.41 atm.

11.153 χmethane = 0.789, χethane = 0.211.

11.155 (a) 0.154%. (b) 1.5%. (c) 7.7%.

11.157 NO.

11.159 Reaction (b).

11.161 1.80 × 10−4 g/mL.

11.163 The kinetic energy of a particle depends on its mass and its speed (KE = mu2/2). At a given temperature, the more massive UF6 molecules move more slowly than do the H2 molecules.

11.165 Intermolecular attractions tend to lower Z at low pressures and excluded volume tends to increase Z at high pressures.

Chapter 12

12.9 Butane would be a liquid in winter (boiling point −44.5°C), and on the coldest days even propane would become a liquid (boiling point −0.5°C). Only methane would remain gaseous (boiling point −161.6°C).

12.11 (a) Dispersion. (b) Dispersion and dipole-dipole. (c) Dispersion and dipole-dipole. (d) Dispersion and ionic. (e) Dispersion.

12.13 (e) CH3COOH.

12.15 1-butanol has greater intermolecular forces because it can form hydrogen bonds.

12.17 (a) Xe, it is larger and therefore has stronger dispersion forces. (b) CS2, it is larger and therefore has stronger dispersion forces. (c) Cl2, it is larger and therefore has stronger dispersion forces. (d) LiF, it is an ionic compound, and the ion-ion attractions are much stronger than the dispersion forces between F2 molecules. (e) NH3, it can form hydrogen bonds and PH3 cannot.

12.19 (a) Dispersion and dipole-dipole, including hydrogen bonding. (b) Dispersion only. (c) Dispersion only. (d) Covalent bonds.

12.21 The compound with —NO2 and —OH groups on adjacent carbons can form hydrogen bonds with itself (intramolecular hydrogen bonds). Such bonds do not contribute to intermolecular attraction and do not help raise the melting point of the compound. The other compound, with the –NO2 and –OH groups on opposite sides of the ring, can form only intermolecular hydrogen bonds; therefore, it will take a higher temperature to escape into the gas phase.

Page AP-11

12.33 Ethylene glycol has two –OH groups, allowing it to exert strong intermolecular forces through hydrogen bonding. Its viscosity should fall between ethanol (1 OH group) and glycerol (3 OH groups).

12.35 331 mm Hg.

12.37 29.9 kJ/mol.

12.45 Simple cubic: one sphere; body-centered cubic: two spheres; face-centered cubic: four spheres.

12.47 6.20 × 1023 atoms/mol.

12.49 458 pm.

12.51 XY3.

12.53 0.220 nm.

12.55 ZnO.

12.59 Molecular solid.

12.61 Molecular: Se8, HBr, CO2, P4O6, and SiH4; covalent: Si and C.

12.63 Diamond: each carbon atom is covalently bonded to four other carbon atoms. Because these bonds are strong and uniform, diamond is a very hard substance. Graphite: the carbon atoms in each layer are linked by strong bonds, but the layers are bound by weak dispersion forces. As a result, graphite may be cleaved easily between layers and is not hard. In graphite, all atoms are sp2 hybridized; each atom is covalently bonded to three other atoms. The remaining unhybridized 2p orbital is used in pi bonding, forming a delocalized molecular orbital. The electrons are free to move around in this extensively delocalized molecular orbital, making graphite a good conductor of electricity in directions along the planes of carbon atoms.

12.85 2.72 × 103 kJ.

12.87 47.03 kJ/mol.

12.89 Two phase changes occur in this process. First, the liquid is turned to solid (freezing), then the solid ice is turned to gas (sublimation).

12.91 When steam condenses to liquid water at 100°C, it releases a large amount of heat equal to the enthalpy of vaporization. Thus, steam at 100°C exposes one to more heat than an equal amount of water at 100°C.

12.95 Initially, the ice melts because of the increase in pressure. As the wire sinks into the ice, the water above the wire refreezes. Eventually the wire actually moves completely through the ice block without cutting it in half.

12.97 (a) Ice would melt. (If heating continues, the liquid water would eventually boil and become a vapor.) (b) Liquid water would vaporize. (c) Water vapor would solidify without becoming a liquid.

12.99 (d).

12.101 Covalent.

12.103 CCl4.

12.105 0.26 cm; 7.69 × 106 base pairs.

12.107 760 mmHg.

12.109 It has reached the critical point; the point of critical temperature (Tc) and critical pressure (Pc).

12.111 Crystalline SiO2. Its regular structure results in a more efficient packing.

12.113 (a) and (b).

12.115 8.3 × 10−3 atm.

12.117 (a) K2S. Ionic forces are much stronger than the dipole-dipole forces in (CH3)3N. (b) Br2. Both molecules are nonpolar; but Br2 has a larger mass.

12.119 SO2 will behave less ideally because it is polar and has greater intermolecular forces.

12.121 62.4 kJ/mol.

12.123 Smaller ions can approach polar water molecules more closely, resulting in larger ion-dipole interactions. The greater the ion-dipole interaction, the larger is the heat of hydration.

12.125 (a) 30.7 kJ/mol. (b) 192.5 kJ/mol. It requires more energy to break the bond than to vaporize the molecule.

12.127 (a) Decreases. (b) No change. (c) No change.

12.129 CaCO3(s) → CaO(s) + CO2(g). Initial state: one solid phase, final state: two solid phase components and one gas phase component.

12.131 The molecules are all polar. The F atoms can form H-bonds with water and other —OH and —NH groups in the membrane, so water solubility plus easy attachment to the membrane would allow these molecules to pass the blood-brain barrier.

12.133 The time required to cook food depends on the boiling point of the water in which it is cooked. The boiling point of water increases when the pressure inside the cooker increases.

12.135 (a) Extra heat produced when steam condenses at 100°C. (b) Avoids extraction of ingredients by boiling in water.

12.137 (a) ~2.3 K. (b) ~10 atm. (c) ~5 K. (d) No.

12.139 (a) Pumping allows Ar atoms to escape, thus removing heat from the liquid phase. Eventually the liquid freezes. (b) The slope of the solid-liquid line of cyclohexane is positive. Therefore, its melting point increases with pressure. (c) These droplets are super-cooled liquids. (d) When the dry ice is added to water, it sublimes. The cold CO2 gas generated causes nearby water vapor to condense, hence the appearance of fog.

12.141 (a) Two triple points: Diamond/graphite/liquid and graphite/liquid/vapor. (b) Diamond. (c) Apply high pressure at high temperature.

12.143 Ethanol mixes well with water. The mixture has a lower surface tension and readily flows out of the ear channel.

12.145 When water freezes it releases heat, helping keep the fruit warm enough not to freeze. Also, a layer of ice is a thermal insulator.

12.147 The fuel source for the Bunsen burner is most likely methane gas. When methane burns in air, carbon dioxide and water are produced. The water vapor produced during the combustion condenses to liquid water when it comes in contact with the outside of the cold beaker.

12.149 6.019 × 1023

12.151 127 mmHg.

12.153 55°C.

12.155 0.833 g/L. The hydrogen-bonding interactions in HF are relatively strong, and since the ideal gas equation ignores intermolecular forces, it underestimates significantly the density of HF gas near its boiling point.

12.157 Fluoromethane. Of the three compounds, only fluoromethane has a permanent dipole moment.

Chapter 13

13.9 “Like dissolves like.” Naphthalene and benzene are nonpolar, whereas CsF is ionic.

13.11 O2 < Br2 < LiCl < CH3OH.

13.15 (a) 8.47%. (b) 17.7%. (c) 11%.

13.17 (a) 0.0610 m. (b) 2.04 m.

13.19 (a) 1.7 m. (b) 0.87 m. (c) 7.0 m.

13.21 3.0 × 102 g.

13.23 18.3 M; 27.4 m.

13.25 χ(N2) = 0.677, χ(O2) = 0.323. Due to the greater solubility of oxygen, it has a larger mole fraction in solution than it does in the air.

13.33 45.9 g.

13.35 According to Henry's law, the solubility of a gas in a liquid increases as the pressure increases (c = kP). The soft drink tastes flat at the bottom of the mine because the carbon dioxide pressure is greater and the dissolved gas is not released from the solution. As the miner goes up in the elevator, the atmospheric carbon dioxide pressure decreases and dissolved gas is released from his stomach.

13.37 1.0 × 10−5 mol/L.

13.39 3.3 atm. This pressure is only an estimate since we ignored the amount of CO2 that was present in the unopened container in the gas phase.

13.57 1.3 × 103 g.

13.59 Pethanol = 30 mmHg; P1-propanol = 26.3 mmHg.

13.61 187 g.

13.63 0.59 m.

13.65 −5.4°C.

13.67 (a) CaCl2. (b) Urea. (c) CaCl2. CaCl2 is an ionic compound and is therefore an electrolyte in water. Assuming that CaCl2 completely dissociates, the total ion concentration will be 3 × 0.35 = 1.05 m, which is larger than the urea (nonelectrolyte) concentration of 0.90 m.

13.69 0.15 m C6H12O6 > 0.15 m CH3COOH > 0.10 m Na3PO4 > 0.20 m MgCl2 > 0.35 m NaCl.

13.71 Boiling point: 102.8°C, Freezing point: −10.0°C. (b) Boiling point: 102.0°C, Freezing point: −7.14°C.

13.73 Both NaCl and CaCl2 are strong electrolytes. Urea and sucrose are nonelectrolytes. The NaCl or CaCl2 will yield more particles per mole of the solid dissolved, resulting in greater freezing point depression. Also, sucrose and urea would make a mess when the ice melts.

13.75 2.47.

13.77 9.16 atm.

13.81 282.5 g/mol; C19H38O.

13.83 4.3 × 102 g/mol; C24H20P4.

13.85 1.75 × 104g/mol.

13.87 342 g/mol.

13.89 15.7%.

13.93 ΔP = 2.05 × 10−5 mmHg; ΔTf = 8.9 × 10−5 °C; ΔTb = 2.5 × 10−5 °C; π = 0.889 mmHg.

13.95 Before: 7.77 × 10−6 mol glucose/mL blood; 3.89 × 10−3 mol glucose; 7.00 g glucose. After: 1.33 × 10−5 mol glucose/mL blood; 6.66 × 10−2 mol glucose; 12.0 g glucose.

13.97 Water migrates through the semipermeable cell walls of the cucumber into the concentrated salt solution.

13.99 3.5.

13.101 × = water, Y = NaCl, Z = urea.

13.103 The pill is in a hypotonic solution. Consequently, by osmosis, water moves across the semipermeable membrane into the pill. The increase in pressure pushes the elastic membrane to the right, causing the drug to exit through the small holes at a constant rate.

13.105 Reverse osmosis involves no phase changes and is usually cheaper than distillation or freezing. 34 atm.

13.107 0.295 M; −0.55°C.

13.109 (a) 2.14 × 103 g/mol. (b) 4.50 × 104 g/mol.

13.111 (a) 0.099 L. (b) 9.9.

13.113 (a) At reduced pressure, the solution is supersaturated with CO2. (b) As the escaping CO2 expands it cools, condensing water vapor in the air to form fog.

13.115 (a) Runoff of the salt solution into the soil increases the salinity of the soil. If the soil becomes hypertonic relative to the tree cells, osmosis would reverse, and the tree would lose water to the soil and eventually die of dehydration. (b) Assuming the collecting duct acts as a semipermeable membrane, water would flow from the urine into the hypertonic fluid, thus returning water to the body.

13.117 33 mL, 67 mL.

13.119 12.3 M.

13.121 14.2%.

13.123 (a) Solubility decreases with increasing lattice energy. (b) Ionic compounds are more soluble in a polar solvent. (c) Solubility increases with enthalpy of hydration of the cation and anion.

Page AP-12

13.125 1.43 g/mL; 37.0 m.

13.127 NH3 can form hydrogen bonds with water; NCl3 cannot.

13.129 3%.

13.131 Egg yolk contains lecithins that solubilize oil in water (See Figure 13.18 of the text). The nonpolar oil becomes soluble in water because the nonpolar tails of lecithin dissolve in the oil, and the polar heads of the lecithin molecules dissolve in polar water (like dissolves like).

13.133 Yes.

13.135 As the water freezes, dissolved minerals in the water precipitate from solution. The minerals refract light and create an opaque appearance.

13.137 1.8 × 102 g/mol.

13.139 1.9 m.

13.141 = 4.0 × 102 mmHg; = 1.9 × 102 mmHg.

13.143 −0.737°C.

13.145 About 0.4 molal.

Chapter 14

14.5 (a) (b)

14.7 (a) 0.066 M/s. (b) 0.033 M/s.

14.15 8.1 × 10−6 M/s.

14.17 The reaction is first order in A and first order overall; k = 0.213 s−1.

14.19 (a) 2. (b) 0. (c) 2.5. (d) 3.

14.21 First order; k = 1.19 × 10−4 s−1.

14.27 30 min.

14.29 (a) 0.034 M. (b) 17 s; 23 s.

14.31 4.5 × 10−10 M/s; 5.4 × 106 s.

14.33 (a) 4:3:6. (b) The relative rates would be unaffected, each absolute rate would decrease by 50%. (c) 1:1:1.

14.41 103 kJ/mol.

14.43 3.0 × 103 s−1.

14.45 51.8 kJ/mol.

14.47 1.3 × 102 kJ/mol. For maximum freshness, fish should be frozen immediately after capture and kept frozen until cooked.

14.49 Diagram (a).

14.59 (a) Second-order. (b) The first step is the slower (rate-determining) step.

14.61 Mechanism I can be discarded. Mechanisms II and III are possible.

14.71 (i) and (iv).

14.73 Temperature, energy of activation, concentration of reactants, and a catalyst.

14.75 Temperature must be specified.

14.77 0.035 s−1.

14.79 272 s.

14.81 Since the methanol contains no oxygen-18, the oxygen atom must come from the phosphate group and not the water. The mechanism must involve a bond-breaking process like:

14.83 Most transition metals have several stable oxidation states. This allows the metal atoms to act as either a source or a receptor of electrons in a broad range of reactions.

14.85 (a) rate = k[CH3COCH3][H+]. (b) 3.8 × 10−3 M−1 · s−1. (c) k = k1k2/k−1.

14.87 (I) Fe3+ oxidizes I: 2Fe3+ + 2I → 2Fe2+ + I2; (II) Fe2+ reduces : 2Fe2+ + → 2Fe3+ + 2; Overall Reaction: 2I + → I2 + 2. (Fe3+ undergoes a redox cycle: Fe3+ → Fe2+ → Fe3+). The uncatalyzed reaction is slow because both I and are negatively charged, which makes their mutual approach unfavorable.

14.89 (a) (i) rate = k[A]0 = k, (ii) [A] = –kt + [A]0, (b) . (c) t = 2t½.

14.91 There are three gases present and we can measure only the total pressure of the gases. To measure the partial pressure of azomethane at a particular time, we must withdraw a sample of the mixture, analyze and determine the mole fractions. Then, Pazomethane = PTχazomethane.

14.93

14.95 (a) A catalyst works by changing the reaction mechanism, thus lowering the activation energy. (b) A catalyst changes the reaction mechanism. (c) A catalyst does not change the enthalpy of reaction. (d) A catalyst increases the forward rate of reaction. (e) A catalyst increases the reverse rate of reaction.

14.97 At very high [H2]: k2[H2] ≫ 1, rate . At very low [H2]: k2[H2] ≫ 1, rate . The result from Problem 14.80 agrees with the rate law determined for low [H2].

14.99 Rate = k[N2O5]; k = 1.0 × 10−5 s−1.

14.101 The red bromine vapor absorbs photons of blue light and dissociates to form bromine atoms: Br2 → 2. The bromine atoms collide with methane molecules and abstract hydrogen atoms: + CH4 → HBr + . The methyl radical then reacts with Br2, giving the observed product and regenerating a bromine atom to start the process over again: + Br2 → CH3Br + , + CH4 → HBr + . and so on.

14.103 Lowering the temperature would slow all chemical reactions, which would be especially important for those that might damage the brain.

14.105 (a) rate = k[X][Y]2. (b) 0.019 M−2s−1.

14.107 Second-order; k = 2.4 × 107 M−1s−1.

14.109 During the first five minutes or so the engine is relatively cold, so the exhaust gases will not fully react with the components of the catalytic converter. Remember, for almost all reactions, the rate of reaction increases with temperature.

14.111 (a) The activation energy of reaction B is larger than that of reaction A. (b) Ea ≈ 0. Orientation factor is not important.

14.113 5.7 × 105 yr.

14.115 (a) Catalyst: Mn2+; intermediates: Mn3+, Mn4+; first step is rate-determining. (b) Without the catalyst, the reaction would be a termolecular one involving 3 cations! (TI+ and two Ce4+). The reaction would be slow. (c) The catalyst is a homogeneous catalyst because it has the same phase (aqueous) as the reactants.

14.117 0.45 atm.

14.119 (a) Δ[B]/Δt = k1[A] – k2[B]. (b) [B] = (k1/k2)[A].

14.121 (a) k = 0.0247 yr−1. (b) 9.8 × 10−4. (c) 187 yr.

14.123 (a) Three. (b) Two. (c) The third step. (d) Exothermic.

14.125 1.8 × 103 K.

14.127 (a) 2.5 × 10−5 M/s. (b) 2.5 × 10−5 M/s. (c) 8.3 × 10−6 M.

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14.129

14.131 (a) 1.13 × 10−3 M/min. (b) 6.83 × 10−4 M/min; 8.8 × 10−3 M.

14.133 Second-order, k = 0.42/M · min.

14.135 60% increase. The result shows the profound effect of an exponential dependence.

14.137 4.1 × 102 kJ/mol.

Chapter 15

15.9 (a) . (b) 7.2 × 102.

15.11 1.08 × 107.

15.21 (1) Diagram (a). (2) Diagram (d). Volume will cancel from the Kc expression.

15.23 2.40 × 1033.

15.25 3.5 × 10−7.

15.27 (a) 8.2 × 10−2. (b) 0.29.

15.29 KP = 0.105, Kc = 2.05 × 10−3.

15.31 KP = 7.09 × 10−3.

15.33 KP = 9.6 × 10−3, Kc = 3.9 × 10−4.

15.35 4.0 × 10−6.

15.37 5.6 × 1023.

15.39 The equilibrium pressure is less than the original pressure.

15.41 0.173 mol H2.

15.43 [H2] = [Br2] = 1.80 × 10−4 M; [HBr] = 0.267 M.

15.45 = 0.408 atm; PCO = = 0.352 atm.

15.47 PCO = 1.96 atm; = 2.54 atm.

15.49 The forward reaction will not occur.

15.55 (a) The equilibrium would shift to the right. (b) The equilibrium would be unaffected. (c) The equilibrium would be unaffected.

15.57 (a) No effect. (b) No effect. (c) Shift to the left. (d) No effect. (e) Shift to the left.

15.59 (a) Shift to the right. (b) Shift to the left. (c) Shift to the right. (d) Shift to the left. (e) A catalyst has no effect on equilibrium position.

15.61 No change.

15.63 (a) Shift to the right. (b) No effect. (c) No effect. (d) Shift to the left. (e) Shift to the right. (f) Shift to the left. (g) Shift to the right.

15.65 (a) 2O3(g) ⇄ 3O2(g), = −284.4 kJ//mol. (b) Equilibrium would shift to the left. The number of O3 molecules would increase and the number of O2 molecules would decrease.

15.67 (a) PNO = 0.24 atm; = 0.12 atm. (b) KP = 0.017.

15.69 (a) No effect. (b) More CO2 and H2O.

15.71 (a) Kc = 8 × 10−44. (b) A mixture of H2 and O2 can be kept at room temperature because of a very large activation energy.

15.73 (a) KP = 1.7. (b) PA = 0.69 atm, PB = 0.81 atm.

15.75 KP = 1.5 × 105.

15.77 = 0.28 atm; = 0.051 atm; PHCl = 1.67 atm.

15.79 5.0 × 101 atm.

15.81 0.038.

15.83 3.3 × 102 atm.

15.85 Kc = 6.28 × 10−4.

15.87 = 0.860 atm, = 0.366 atm, = 4.40 × 10−3 atm.

15.89 (a) Kc = 1.16. (b) 53.7%.

15.91 (a) KP = 0.49. (b) 0.23 (23%). (c) 0.037 (3.7%). (d) Greater than 0.037 mol.

15.93 [H2] = 0.07 M, [I2] = 0.18 M, [HI] = 0.83 M.

15.95 (c). N2O4(colorless) → 2NO2(brown) is consistent with the observations. The reaction is endothermic so heating darkens the color. Above 150°C, the NO2 breaks up into colorless NO and O2: 2NO2(g) → 2NO(g) + O2(g). An increase in pressure shifts the equilibrium back to the left, restoring the color by producing NO2.

15.97 (a) KP = 4.2 × 10−4. (b) 0.83. (c) 1.1. (d) 2.3 × 103; 2.1 × 10−2.

15.99 (a) Color deepens. (b) Increases. (c) Decreases. (d) Increases. (e) Unchanged.

15.101 Potassium is more volatile than sodium. Therefore, its removal shifts the equilibrium from left to right.

15.103 3.5 × 10−2.

15.105 (a) Shifts to the right. (b) Shifts to the right. (c) No change. (d) No change. (e) No change. (f) Shifts to the left.

15.107 (a) 1.8 × 10−16. (b) [H+][OH] = 1.0 × 10−14; [H+] = [OH] = 1.0 × 10−7 M.

15.109 = 0.09 atm; = 0.10 atm.

15.111 (a) 1.03 atm. (b) 0.39 atm. (c) 1.67 atm. (d) 0.620 (62.0%).

15.113 (a) KP = 2.6 × 10−6; Kc = 1.1 × 10−7. (b) 2.2 g; 22 mg/m3; Yes.

15.115 There is a temporary dynamic equilibrium between the melting of ice cubes and the freezing of water between the ice cubes.

15.117 [NH3] = 0.042 M; [N2] = 0.086 M; [H2] = 0.26 M.

15.119 (a) . (b) If P increases, the fraction (and therefore x) must decrease. Equilibrium shifts to the left to produce less NO2 and more N2O4 as predicted.

15.121 = 3.58 atm, = = 2.71 atm.

15.123 4.0.

15.125 −3.

15.127 KP = Kc: (d), (g); cannot write a KP: (c), (f).

Chapter 16

16.3 (a) Both. (b) Base. (c) Acid. (d) Base. (e) Acid. (f) Base. (g) Base. (h) Base. (i) Acid. (j) Acid.

16.5 (a) . (b) . (c) HS. (d) CN. (e) HCOO.

16.7 (a) CH2ClCOO. (b) . (c) . (d) . (e) . (f) . (g) . (h) . (i) . (j) NH3. (k) HS. (l) S2−. (m) OCl.

16.15 (a) 4.0 × 10−13 M. (b) 6.0 × 10−10 M. (c) 1.2 × 10−12 M. (d) 5.7 × 10−3 M.

16.17 (a) 2.05 × 10−11 M. (b) 3.07 × 10−8 M. (c) 5.95 × 10−11 M. (d) 2.93 × 10−1 M.

16.21 7.1 × 10−12 M.

16.23 (a) 3.00. (b) 13.89.

16.25 (a) 3.8 × 10−3 M. (b) 6.2 × 10−12 M. (c) 1.1 × 10−7 M. (d) 1.0 × 10−15 M.

16.27 pH < 7: [H+] > 1.0 × 10−7 M, solution is acid; pH > 7: [H+] < 1.0 × 10−7 M, solution is basic; pH = 7: [H+] = 1.0 × 10−7 M, solution is neutral.

16.29 2.5 × 10−5 M.

16.31 2.2 × 10−3 g.

16.37 (a) −0.009. (b) 1.46. (c) 5.82.

16.39 (a) 6.2 × 10−5 M. (b) 2.8 × 10−4 M. (c) 0.10 M.

16.41 (a) pOH = −0.093, pH = 14.09. (b) pOH = 0.36, pH = 13.64. (c) pOH = 1.07, pH = 12.93.

16.43 (a) 1.1 × 10−3 M. (b) 5.5 × 10−4 M.

16.51 (a) Strong base. (b) Weak base. (c) Weak base. (d) Weak base. (e) Strong base.

16.53 2.59.

16.55 5.17.

16.57 (a) 10%. (b) 84%. (c) 0.97%.

16.59 1.3 × 10−6.

16.61 4.8 × 10−9.

16.63 2.3 × 10−3 M.

16.65 6.80.

16.69 6.97 × 10−7.

16.71 11.98.

16.73 (a) A. (b) B.

16.77 Kb(CN) = 2.0 × 10−5; Kb(F) = 1.4 × 10−11; Kb(CH3COO) = 5.6 × 10−10; = 2.4 × 10−8.

16.79 (1) Diagram (c). (2) Diagrams (b) and (d).

16.81 pH (0.040 M HCl) = 1.40; pH (0.040 M H2SO4) = 1.31.

16.83 [H+] = [] = 1.0 × 10−4 M, [] = 4.8 × 10−11 M.

16.85 pH = 1.00.

16.89 (a) H2SO4 > H2SeO4. (b) H3PO4 > H3AsO4.

16.91 The conjugate bases are C6H5O from phenol and CH3O from methanol. The C6H5O is stabilized by resonance:

The CH3O ion has no such resonance stabilization. A more stable conjugate base means an increase in the strength of the acid.

16.97 (a) Neutral. (b) Basic. (c) Acidic. (d) Acidic.

16.99 HZ < HY < HX.

16.101 pH > 7.

16.103 4.82.

16.105 5.39.

16.109 (a) Al2O3 < BaO < K2O. (b) CrO3 < Cr2O3 < CrO.

16.111 Al(OH)3(s) + OH(aq) → . This is a Lewis acid-base reaction.

16.115 AlCl3 is a Lewis acid with an incomplete octet of electrons and Cl is the Lewis base donating a pair of electrons.

16.117 CO2, SO2, and BCl3.

16.119 (a) AlBr3 is the Lewis acid; Br is the Lewis base. (b) Cr is the Lewis acid; CO is the Lewis base. (c) Cu2+ is the Lewis acid; CN is the Lewis base.

16.121 A strong acid: Diagram (b); a weak acid: Diagram (c); a very weak acid: Diagram (d).

16.123 CH3COO(aq) and HCl(aq). This reaction will not occur to any measurable extent.

16.125 pH = 1.70; % ionization = 2.3%.

16.127 (c).

16.129 (a) For the forward reaction and NH3 are the acid and base, respectively. For the reverse reaction NH3 and are the acid and base, respectively. (b) H+ corresponds to ; OH corresponds to . For the neutral solution, [] = [].

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16.131 , [HA] ≈ 0.1 M, [A] ≈ 0.1 M. Therefore,

16.133 1.7 × 1010.

16.135 (a) H (base1) + H2O (acid2) → OH (base2) + H2 (acid1). (b) H is the reducing agent and H2O is the oxidizing agent.

16.137 2.8 × 10−2.

16.139 PH3 is a weaker base than NH3.

16.141 (a) HNO2. (b) HF. (c) BF3. (d) NH3. (e) H2SO3. (f) and .

16.143 (a) Trigonal pyramidal. (b) H4O2+ does not exist because the positively charged H3O+ has no affinity to accept the positive H+ ion. If H4O2+ existed, it would have a tetrahedral geometry.

16.145 Cl2(g) + H2O(l) ⇄ HCl(aq) + HClO(aq); HCl(aq) + AgNO3(aq) ⇄ AgCl(s) + HNO3(aq). In the presence of OH ions, the first equation is shifted to the right: H+ (from HCl) + OH → H2O. Therefore, the concentration of HClO increases. (The ‘bleaching action' is due to ClO ions.)

16.147 11.80.

16.149 (a) 0.181 or 18.1%. (b) 4.63 × 10−3.

16.151 4.26.

16.153 7.2 × 10−3 g.

16.155 1.000.

16.157 (a) The pH of the solution of HA would be lower. (b) The electrical conductance of the HA solution would be greater. (c) The rate of hydrogen evolution from the HA solution would be greater.

16.159 1.4 × 10−4.

16.161 2.7 × 10−3 g.

16.163 (a) + H2O (acid) → NH3 + OH; N3− (base) + 3H2O (acid) → NH3 + 3OH. (b) N3−.

16.165 When the smelling salt is inhaled, some of the powder dissolves in the basic solution. The ammonium ions react with the base as follows: + OH(aq) → NH3(aq) + H2O. It is the pungent odor of ammonia that prevents a person from fainting.

16.167 (c).

16.169 21 mL.

16.171 Magnesium.

16.173 Both NaF and SnF2 provide F ions in solution. The F ions replace OH ions during the remineralization process 5Ca2+ + 3 + F → Ca5(PO4)3F (fluorapatite). Because F is a weaker base than OH, fluorapatite is more resistant to attacks by acids compared to hydroxyapatite.

16.175 Ka = 5.2 × 10−10.

Chapter 17

17.5 (a) 2.57. (b) 4.44.

17.9 (c) and (d).

17.11 8.89.

17.13 0.024.

17.15 0.58.

17.17 9.25; 9.18.

17.19 Na2A/NaHA.

17.21 (1) Solutions (a), (b), and (c); (2) Solution (a).

17.27 202 g/mol.

17.29 0.25 M.

17.31 (a) 1.10 × 102 g/mol. (b) 1.6 × 10−6.

17.33 5.82.

17.35 (a) 2.87. (b) 4.56. (c) 5.34. (d) 8.78. (e) 12.10.

17.37 (a) Cresol red or phenolphthalein. (b) Most of the indicators in Table 17.3, except thymol blue and, to a lesser extent, bromophenol blue and methyl orange. (c) Bromophenol blue, methyl orange, methyl red, or chlorophenol blue.

17.39 Red.

17.41 (1) Diagram (c), (2) Diagram (b), (3) Diagram (d), (4) Diagram (a). The pH at the equivalence point is below 7 (acidic).

17.49 (a) [I] = 9.1 × 10−9 M. (b) [Al3+] = 7.4 × 10−8 M.

17.51 1.8 × 10−11.

17.53 3.3 × 10−93.

17.55 9.52.

17.57 Yes.

17.63 (a) 0.013 M or 1.3 × 10−2 M. (b) 2.2 × 10−4 M. (c) 3.3 × 10−3 M.

17.65 (a) 1.0 × 10−5 M. (b) 1.1 × 10−10 M.

17.67 (b), (c), (d), and (e)

17.69 (a) 0.016 M or 1.6 × 10−2 M. (b) 1.6 × 10−6 M.

17.71 A precipitate of Fe(OH)2 will form.

17.73 [Cd2+] = 1.1 × 10−18 M, [] = 4.2 × 10−3 M, [CN] = 0.48 M.

17.75 3.5 × 10−5 M.

17.77 Cu2+(aq) + 4NH3(aq) ⇄ [Cu(NH3)4]2+(aq). (b) Ag+(aq) + 2CN(aq) ⇄ [Ag(CN)2](aq). (c) Hg2+(aq) + 4Cl(aq) ⇄ [HgCl4]2−(aq).

17.81 Greater than 2.68 but less than 8.11.

17.83 0.011 M.

17.85 Chloride ion will precipitate Ag+ but not Cu2+. So, dissolve some solid in H2O and add HCl. If a precipitate forms, the salt was AgNO3. A flame test will also work. Cu2+ gives a green flame test.

17.87 2.51 to 4.41.

17.89 1.3 M.

17.91 [Na+] = 0.0835 M, [HCOO] = 0.0500 M; [OH] = 0.0335 M; [H+] = 3.0 × 10−13 M; [HCOOH] = 8.8 × 10−11 M.

17.93 Cd(OH)2(s) + 2OH(aq) ⇄ . This is a Lewis acid-base reaction.

17.95 (d).

17.97 [Ag+] = 2.0 × 10−9 M; [Cl] = 0.080 M; [Zn2+] = 0.070 M; = 0.060 M.

17.99 0.035 g/L.

17.101 2.4 × 10−13.

17.103 [Ba2+] = 1.0 × 10−5 M. Ba(NO3)2 is too soluble to be used for this purpose.

17.105 (a) AgBr. (b) 1.8 × 10−7 M. (c) 0.0018%.

17.107 3.0 × 10−8.

17.109 (a) 1.0 × 1014. (b) 1.8 × 109. (c) 1.8 × 109. (d) 3.2 × 104.

17.111 (a) Mix 500 mL of 0.40 M CH3COOH with 500 mL of 0.40 M CH3COONa. (b) Mix 500 mL of 0.80 M CH3COOH with 500 mL of 0.40 M NaOH. (c) Mix 500 mL of 0.80 M CH3COONa with 500 mL of 0.40 M HCl.

17.113 (a) Figure (b). (b) Figure (a).

17.115 (a) Add sulfate. (b) Add sulfide. (c) Add iodide.

17.117 They are insoluble in water.

17.119 The ionized polyphenols have a dark color. In the presence of citric acid from lemon juice, the anions are converted to the lighter-colored acids.

17.121 Yes.

17.123 (c).

17.125 Decreasing the pH would increase the solubility of calcium oxalate and should help minimize the formation of calcium oxalate kidney stones.

17.127 At pH = 1.0: +NH3—CH2—COOH. At pH = 7.0: +NH3—CH2–COO. At pH = 12.0: NH2—CH2—COO.

17.129 (a) Before dilution: pH = 4.74. After a 10-fold dilution: pH = 4.74. (b) Before dilution: pH = 2.52. After dilution, pH = 3.02.

17.131 AgBr(s) will precipitate first. Ag2SO4(s) will precipitate last.

17.133 (c).

17.135 8.8 × 10−12.

17.137 4.75.

Chapter 18

18.7 (a) With barrier: 16; Without barrier: 64. (b) 16; 16; 32. Both particles on one side: S = 3.83 × 10−23 J/K; particles on opposite sides: S = 4.78 × 10−23. The most probable state is the one with the larger entropy; that is, the state in which the particles are on opposite sides.

18.13 (a) ΔSsys = −0.031 J/K. (b) ΔSsys = −0.29 J/K. (c) ΔSsys = 1.5 × 102 J/K.

18.15 (c) < (d) < (e) < (a) < (b).

18.17 (a) 47.5 J/K · mol. (b) −12.5 J/K · mol. (c) −242.8 J/K · mol.

18.21 (a) 291 J/K · mol; spontaneous. (b) 2.10 × 103 J/K · mol; spontaneous. (c) ΔSsurr = 2.99 × 103 J/K · mol; spontaneous.

18.23 (a) ΔSsys = −75.6 J/K · mol; ΔSsurr = 185 J/K · mol; spontaneous. (b) ΔSsys = 215.8 J/K · mol; ΔSsurr = −304 J/K · mol; not spontaneous. (c) ΔSsys = 98.2 J/K · mol; ΔSsurr = −1.46 × 103 J/K · mol; not spontaneous. (d) ΔSsys = −282 J/K · mol; ΔSsurr = 7.20 × 103 J/K · mol; spontaneous.

18.29 (a) −1139 kJ/mol. (b) −140.0 kJ/mol. (c) −2935 kJ/mol.

18.31 (a) All temperatures. (b) Below 111 K.

18.33 ΔSfus = 99.9 J/K · mol; ΔSvap = 93.6 J/K · mol.

18.35 = −226.6 kJ/mol.

18.37 75.9 kJ.

18.41 0.35.

18.43 79 kJ/mol.

18.45 (a) = 35.4 kJ/mol; KP = 6.2 × 10−7. (b) 44.6 kJ/mol.

18.47 (a) 1.6 × 10−23 atm. (b) 0.535 atm.

18.49 3.1 × 10−2 atm or −3.6 mmHg.

18.53 93 ATP molecules.

18.55 ΔHfus > 0. ΔSfus > 0. (a) ΔGfus < 0. (b) ΔGfus = 0. (c) ΔGfus > 0.

18.57 U and H.

18.59 42°C.

18.61 ΔS must be positive (ΔS > 0).

18.63 (a) Benzene: ΔSvap = 87.8 J/K · mol; hexane: ΔSvap = 90.1 J/K · mol; mercury: ΔSvap = 93.7 J/K · mol; toluene: ΔSvap = 91.8 J/K · mol. Trouton's rule is a statement about . In most substances, the molecules are in constant and random motion in both the liquid and gas phases, so ≈ 90 J/K · mol. (b) Ethanol: ΔSvap = 111.9 J/K · mol; water: ΔSvap = 109.4 J/K · mol. In ethanol and water, there are fewer possible arrangements of the molecules due to the network of H-bonds, so is greater.

18.65 (a) 2CO + 2NO → 2CO2 + N2. (b) The oxidizing agent is NO; the reducing agent is CO. (c) KP = × 3 10120. (d) QP = 1.2 × 1014. To the right. (e) No.

18.67 2.6 × 10−9.

18.69 976 K = 703°C.

18.71 38 kJ.

18.73 174 kJ/mol.

18.75 (a) Positive. (b) Negative. (c) Positive. (d) Positive.

18.77 625 K. We assume that ΔH° and ΔS° do not depend on temperature.

18.79 No. A negative ΔG° tells us that a reaction has the potential to happen, but gives no indication of the rate.

18.81 (a) ΔG° = −106.4 kJ/mol; KP = 4 × 1018. (b) ΔG° = −53.2 kJ/mol; KP = 2 × 109. The KP in (a) is the square of the KP in (b). Both ΔG° and KP depend on the number of moles of reactants and products specified in the balanced equation.

18.83 Talking involves various biological processes (to provide the necessary energy) that lead to an increase in the entropy of the universe. Since the overall process (talking) is spontaneous, the entropy of the universe must increase.

18.85 (a) 86.7 kJ/mol. (b) 4 × 10−31. (c) 3 × 10−6. (d) Lightning supplies the energy necessary to drive this reaction, converting the two most abundant gases in the atmosphere into NO(g). The NO gas dissolves in the rain, which carries it into the soil where it is converted into nitrate and nitrite by bacterial action. This “fixed” nitrogen is a necessary nutrient for plants.

Page AP-15

18.87 T > 673.2 K.

18.89 (a) 7.6 × 1014. (b) 4.1 × 10−12. The activity series is correct. The very large value of K for reaction (a) indicates that products are highly favored; whereas, the very small value of K for reaction (b) indicates that reactants are highly favored.

18.91 (a) Disproportionation redox reaction. (b) 8.2 × 1015. This method is feasible for removing SO2. (c) Less effective.

18.93 1.8 × 1070. Due to the large magnitude of K, you would expect this reaction to be spontaneous in the forward direction. However, this reaction has a large activation energy, so the rate of reaction is extremely slow.

18.95 ΔSsys = 91.1 J/K; ΔSsurr = −91.1 J/K; ΔSuniv = 0. The system is at equilibrium.

18.97 8.5 kJ/mol.

18.99 (a) CH3COOH: 27 kJ/mol; CH2ClCOOH: 16 kJ/mol. (b) The system's entropy change dominates. (c) The breaking and making of specific O—H bonds. Other contributions include solvent separation and ion solvation. (d) The CH3COO ion, which is smaller than CH2ClCOO, can participate in hydration to a greater extent, leading to solutions with fewer possible arrangements.

18.101 = 0.45; = 0.55. We assumed that ΔG° calculated from values was temperature independent.

18.103 249 J/K.

18.105 ΔSsys = −327 J/K · mol; ΔSsurr = 1918 J/K · mol; ΔSuniv = 1591 J/K · mol.

18.107 q, and w.

18.109 ΔG must be negative. ΔS must be negative. ΔH must be negative.

18.111 (a) S = 5.76 J/K · mol. (b) The fact that the actual residual entropy is 4.2 J/K · mol means that the orientation is not totally random.

18.113 ΔH° = 33.89 kJ/mol; ΔS° = 96.4 J/K · mol; ΔG° = 5.2 kJ/mol. ΔH° is positive because this is an endothermic process. We also expect ΔS° to be positive because this is a liquid → vapor phase change. ΔG° is positive because we are at a temperature that is below the boiling point of benzene (80.1°C).

Chapter 19

19.1 a) 2H+ + H2O2 + 2Fe2+ → 2Fe3+ + 2H2O. (b) 6H+ + 2HNO3 + 3Cu → 3Cu2+ + 2NO + 4H2O. (c) 3CN + 2 + H2O → 3CNO + 2MnO2 + 2OH. (d) 6OH + 3Br2. (e) .

19.11 Al(s) + 3Ag+(1.0 M) → Al3+(1.0 M) + 3Ag(s); = 2.46 V.

19.13 Cl2(g) and (aq).

19.15 (a) Spontaneous. (b) Not spontaneous. (c) Not spontaneous. (d) Spontaneous.

19.17 (a) Li. (b) H2. (c) Fe2+. (d) Br.

19.21 × 3 1054.

19.23 (a) 2 × 1018. (b) 3 × 108. (c) 3 × 1062.

19.25 Ce4+(aq) + Fe2+(aq) → Ce3+(aq) + Fe3+(aq); ΔG° = −81 kJ/mol; K = 2 × 1014.

19.29 1.09 V.

19.31 = 0.76 V; Ecell = 0.78 V.

19.33 6.0 × 10−38.

19.39 1.09 V.

19.43 12.2 g.

19.45 Sodium.

19.47 0.012 F.

19.49 5.33 g Cu; 13.4 g Br2.

19.51 7.70 × 103 C.

19.53 1.84 kg Cl2/h.

19.55 63.3 g/mol.

19.57 27.0 g/mol.

19.63 (a) Half-reactions: H2(g) → 2H+(aq) + 2e, Ni2+(aq) + 2e → Ni(s); Balanced equation: H2(g) + Ni2+(aq) → 2H+(aq) + Ni(s). The reaction will proceed to the left. (b) Half-reactions: 5e + 8H+(aq) + (aq) → Mn2+(aq) + 4H2O; 2Cl(aq) → Cl2(g) + 2e. Balanced equation: 16H+(aq) + 2(aq) + 10Cl(aq) → 2Mn2+(aq) + 8H2O(l) + 5Cl2(g). The reaction will proceed to the right. (c) Half-reactions: Cr(s) → Cr3+(aq) + 3e, Zn2+(aq) + 2e → Zn(s). Balanced equation: 2Cr(s) + 3Zn2+(aq) → 2Cr3+(aq) + 3Zn(s). The reaction will proceed to the left.

19.65 0.00944 g SO2.

19.67 (a) 2 + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2. (b) 0.0602 M.

19.69 0.232 mg Ca2+/mL blood.

19.71 5 × 10−13.

19.73 (a) 3.14 V. (b) 3.13 V.

19.75 0.035 V.

19.77 Mercury(I) is .

19.79 1.44 g Mg; [Ag+] = 7 × 10−55 M; [Mg2+] = 0.0500 M.

19.81 (a) H2, 0.206 L. (b) 6.09 × 1023 e/mol e.

19.83 (a) −1356.8 kJ/mol. (b) 1.17 V.

19.85 + 3.

19.87 ΔG° = 6.8 kJ/mol; K = 0.064.

19.89 1.4 A.

19.91 + 4.

19.93 1.60 × 10−19 C/e.

19.95 Cells of higher voltage require very reactive oxidizing and reducing agents, which are difficult to handle. (From Table 19.1 of the text, we see that 5.92 V is the theoretical limit of a cell made up of Li+/Li and F2/F electrodes under standard-state conditions.) Batteries made up of several cells in series are easier to use.

19.97 Kf = 2 × 1020.

19.99 (a) for X is negative. for Y is positive. (b) 0.59 V.

19.101 (a) Gold does not tarnish in air because the reduction potential for oxygen is not sufficiently positive to result in the oxidation of gold. (b) Yes. (c) 2Au + 3F2 → 2AuF3.

19.103 [Fe2+] = 0.0920 M; [Fe3+] = 0.0680 M.

19.105 H2O2(aq) + 2H+(aq) + 2e → 2H2O(l), = 1.77 V. H2O2(aq) → O2(g) + 2H+(aq) + 2e, = 0.68 V. = = 1.09 V. The decomposition is spontaneous.

19.107 (a) Unchanged. (b) Unchanged. (c) Squared. (d) Doubled. (e) Doubled.

19.109 As [H+] increases, F2 does become a stronger oxidizing agent.

19.111 4.4 × 102 atm.

19.113 (a) Anode: Zn → Zn2+ + 2e; Cathode: + 2e → O2− 1.65 V. (b) 1.63 V. (c) 4.86 × 103 kJ/kg Zn. (d) 64 L of air.

19.115 −3.05 V.

19.117 1 × 10−14.

19.119 (a) 1A·h = 1A × 3600s = 3600 C. (b) 105 A·h. This ampere·hour cannot be fully realized because the concentration of H2SO4 keeps decreasing. (c) = 2.01 V; ΔG° = −388 kJ/mol.

19.121 $217.

19.123 −0.037 V.

19.125 2 × 1037.

19.127 5 mol ATP/mol .

19.129 A small non-zero emf will appear if the temperatures of the two half-cells are different.

Chapter 20

20.5 (a). (b) or . (c). (d). (e).

20.13 2.72 3 × 1014 g/cm3.

20.15 (a) Ni. (b) Se. (c) Cd.

20.17 −4.85 × 10−12 kg/mol H2.

20.19 (a) 6.30 × 10−12 J; 9.00 × 10−13 J/nucleon. (b) 4.78 × 10−11 J; 1.37 × 10−12 J/nucleon.

20.21 7.963 × 10−26 kg.

20.25

  1. .

  2. .

  3. .

20.27 4.88 × 1019 atoms.

20.29 3.09 × 103 yr.

20.31 No A remains, 0.25 mole of B, no C is left, 0.75 mole of D.

20.33 5.5 dpm.

20.35 43:1.

20.39 (a) 14N(α,p)17O. (b) 9Be(α,n)12C. (c) 238U(d,2n)238Np.

20.41 (a) 40Ca(d,p)41Ca. (b) 32S(n,p)32P. (c) 239Pu(α,n)242Cm.

20.43 .

20.55 The fact that the radioisotope appears only in the I2 shows that the is formed only from the .

20.57 Add iron-59 to the person's diet, and allow a few days for the iron-59 isotope to be incorporated into the person's body. Isolate red blood cells from a blood sample and monitor radioactivity from the hemoglobin molecules present in the red blood cells.

20.63 65.3 yr.

20.65 (a) (b) 70.5 dpm.

20.67 (a). (b). (c). (d).

20.69 (a). (b). (c). (d).

20.71 Because both Ca and Sr belong to Group –A, radioactive strontium that has been ingested into the human body becomes concentrated in bones (replacing Ca) and can damage blood cell production. 2.12 × 103 mCi.

20.73 Normally the human body concentrates iodine in the thyroid gland. The purpose of the large doses of KI is to displace radioactive iodine from the thyroid and allow its excretion from the body.

20.75 .

20.77 2.77 × 103 yr.

20.79 0.070%.

20.81 (a) 5.59 × 10−15 J; 2.84 × 10−13 J. (b) 0.024 mol. (c) 4.26 × 106 kJ.

20.83 2.8 × 1014 iodine-131 atoms.

20.85 6.1 × 1023 atoms/mol.

20.87 (a) 1.83 × 10−12 J. (b) The α particle will move away faster because it is smaller.

20.89 U-238, t½ = 4.5 × 109 yr and Th-232, t½ = 1.4 × 1010 yr. They are still present because of their long half-lives.

20.91 8.3 × 10−4 nm.

20.93 Only 3H has a suitable half-life. The other half-lives are either too long or too short to determine the time span of 6 years accurately.

20.95 A small-scale chain reaction (fission of 235U) took place. Copper played the crucial role of reflecting neutrons from the splitting uranium-235 atoms back into the uranium sphere to trigger the chain reaction. Note that a sphere has the most appropriate geometry for such a chain reaction. In fact, during the implosion process prior to an atomic explosion, fragments of uranium-235 are pressed roughly into a sphere for the chain reaction to occur (see Section 20.5 of the text).

20.97 2.1 × 102 g/mol.

20.99 (a) r = r0, where r0 is a proportionality constant. (b) 1.7 × 10−42 m3.

20.101 0.49 rem.

20.103 3.4 mL.

Page AP-16
Chapter 21

21.5 3.30 × 10−4; 330 ppm.

21.7 In the stratosphere, the air temperature rises with altitude. This warming effect is the result of exothermic reactions triggered by UV radiation from the sun.

21.11 260 nm.

21.21 4.0 × 1037 molecules; 3.2 × 1012 kg O3.

21.23 CCl4 + HF → HCl + CFCl3 (Freon-11); CFCl3 + HF → HCl + CF2Cl2 (Freon-12).

21.25 E = 479 kJ/mol. Solar radiation preferentially breaks the C—Cl bond. There is not enough energy to break the C—F bond.

21.27 .

21.39 2.6 × 104 tons.

21.41 1.6 × 1019 kJ; 4.8 × 1016 kg.

21.49 5.2 × 108 L.

21.57 (a) rate = k[NO]2[O2]. (b) rate = k′[NO]2, where k′ = k[O2]. (c) 1.3 × 103 min.

21.59 4.1 × 10−7 atm; 1 × 1016 molecules/L.

21.65 378 g.

21.67 O3: greenhouse gas, toxic to humans, attacks rubber; SO2: toxic to humans, forms acid rain; NO2: forms acid rain, destroys ozone; CO: toxic to humans; PAN: a powerful lachrymator, causes breathing difficulties; Rn: causes lung cancer.

21.69 (b) 4.7 × 10−2.

21.71 (a) N2O + O → 2NO; 2NO + 2O3 → 2NO2 + 2O2; overall: N2O + O + 2O3 → 2NO2 + 2O2. (b) N2O is a more effective greenhouse gas than CO2 because it has a permanent dipole. (c) 3.0 × 1010 mol.

21.73 1.8 × 1019 Rn atoms; 6.4 × 1016 Rn atoms.

21.75 (a) Its small concentration is the result of the high reactivity of the OH radical. (b) OH has an unpaired electron; free radicals are always good oxidizing agents. (c) OH + NO2 → HNO3. (d) OH + SO2 → HSO3; HSO3 + O2 + H2O → H2SO4 + HO2.

21.77 The blackened bucket has a large deposit of elemental carbon. When heated over the burner, it forms poisonous carbon monoxide. C + CO2 → 2CO. A smaller amount of CO is also formed as follows: 2C + O2 → 2CO.

21.79 The use of the aerosol liberates CFCs that destroy the ozone layer.

21.81 Yes. Light of wavelength 409 nm (visible) or shorter will break the C—Br bond.

21.83 (a) 6.2 × 108. (b) The CO2 liberated from limestone contributes to global warming.

21.85 Most water molecules contain oxygen-16, but a small percentage of water molecules contain oxygen-18. The ratio of the two isotopes in the ocean is essentially constant, but the ratio in the water vapor evaporated from the oceans is temperature-dependent, with the vapor becoming slightly enriched with oxygen-18 as temperature increases. The water locked up in ice cores provides a historical record of this oxygen-18 enrichment, and thus ice cores contain information about past global temperatures.

Chapter 22

22.11 (a) +3. (b) 6. (c) Oxalate ion ().

22.13 (a) Na: + 1; Mo: +6. (b) Mg: +2; W: +6. (c) Fe: 0.

22.15 (a) cis-dichlorobis(ethylenediamine)cobalt(III). (b) pentamminechloroplatinum(IV) chloride. (c) pentamminechlorocobalt(III) chloride.

22.17 (a) [Cr(en)2Cl2]+. (b) Fe(CO)5. (c) K2[Cu(CN)4]. (d) [Co(NH3)4(H2O)Cl]Cl2.

22.23 (a) Two. (b) Two.

22.25 (a) (b)

22.31 Cyanide ion is a very strong-field ligand. It causes a larger crystal field splitting than water, resulting in the absorption of higher energy (shorter wavelength) radiation when a d electron is excited to a higher energy d orbital.

22.33 (a) Orange. (b) 255 kJ/mol.

22.35 Two moles. [Co(NH3)4Cl2]Cl. Refer to Problem 22.25 (a) for a diagram of the structure of the complex ion.

22.37 Δ would be greater for the higher oxidation state.

22.41 Use a radioactive label such as 14CN (in NaCN). Add NaCN to a solution of K3Fe(CN)6. Isolate some of the K3Fe(CN)6 and check its radioactivity. If the complex shows radioactivity, then it must mean that the CN ion has participated in the exchange reaction.

22.43 Cu(CN)2 is the white precipitate. It is soluble in KCN(aq), due to formation of [Cu(CN)4]2−, so the concentration of Cu2+ is too small for Cu2+ ions to precipitate with sulfide.

22.45 1.4 × 102.

22.47 The purple color is caused by the build-up of deoxyhemoglobin. When either oxyhemoglobin or deoxyhemoglobin takes up CO, the carbonylhemoglobin takes on a red color, the same as oxyhemoglobin.

22.49 Mn3+ is 3d4 and Cr3+ is 3d5. Therefore, Mn3+ has a greater tendency to accept an electron and is a stronger oxidizing agent. The 3d5 electron configuration of Cr3+ is a stable configuration.

22.51 Ti3+; Fe3+.

22.53 1.6 × 104 g hemoglobin/mol Fe. The discrepancy between our minimum value and the actual value can be explained by realizing that there are four iron atoms per mole of hemoglobin.

22.55 (a) [Cr(H2O)6]Cl3, number of ions: 4. (b) [Cr(H2O)5Cl]Cl2·H2O, number of ions: 3. (c) [Cr(H2O)4Cl2]Cl·2H2O, number of ions: 2. Compare the compounds with equal molar amounts of NaCl, MgCl2, and FeCl3 in an electrical conductance experiment. The solution that has similar conductance to the NaCl solution contains (c); the solution with the conductance similar to MgCl2 contains (b); and the solution with conductance similar to FeCl3 contains (a).

22.57 ΔG° = −1.8 × 102 kJ/mol; K = 6 × 1030.

22.59 Iron is much more abundant than cobalt.

22.61 Oxyhemoglobin absorbs higher energy light than deoxyhemoglobin. Oxyhemoglobin is diamagnetic (low spin), while deoxyhemoglobin is paramagnetic (high spin). These differences occur because oxygen (O2) is a strong-field ligand.

22.63 Complexes are expected to be colored when the highest occupied orbitals have between one and nine d electrons. Zn2+, Cu+, and Pb2+ are d10 ions. V5+, Ca2+, and Sc3+ are d0 ions.

22.65 Dipole moment measurement. Only the cis isomer has a dipole moment.

22.67 EDTA sequesters metal ions (like Ca2+ and Mg2+), which are essential for the growth and function of bacteria.

22.69 3.

22.71 2.2 × 10−20 M.

22.73 (a) 2.7 × 106. (b) Free Cu+ ions are unstable in solution. Therefore, the only stable compounds containing Cu+ ions are insoluble.

22.75 Y.

22.77 0.0 M.

22.79 [Mn(CN)6]5−: Mn is +1, one unpaired d electron. [Mn(CN)6]4−: Mn is +2, one unpaired d electron. [Mn(CN)6]3−: Mn is +3, two unpaired d electrons.

Chapter 23

23.3 (a) CaCO3. (b) CaCO3 · MgCO3. (c) CaF2. (d) NaCl. (e) Al2O3. (f) Fe3O4. (g) Be3Al2Si6O18. (h) PbS. (i) MgSO4 · 7H2O. (j) CaSO4.

23.13 KP = 4.5 × 105.

23.15 Ag, Pt, and Au will not be oxidized, but the other metals will.

23.17 (a) 8.9 × 1012 cm3. (b) 4.0 × 108 kg.

23.19 Al, Na, and Ca.

23.33 (a) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g). (b) 2NaOH(aq) + CO2(g) → Na2CO3(aq) + H2O(l). (c) Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l). (d) NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l). (e) 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g). (f) No reaction.

23.35 5.59 L.

23.39 . The magnesium nitrate is recovered from solution by evaporation, dried, and heated in air to obtain magnesium oxide: .

23.41 The electron configuration of magnesium is [Ne]3s2. The 3s electrons are outside the neon core (shielded), so they have relatively low ionization energies. Removing the third electron means separating an electron from the neon (closed shell) core, which requires a great deal more energy.

23.43 Even though helium and the Group 2A metals have ns2 outer electron configurations, helium has a closed shell noble gas configuration and the Group 2A metals do not. The electrons in He are much closer to and more strongly attracted by the nucleus. Hence, the electrons in He are not easily removed. Helium is inert.

23.45 (a) CaO(s). (b) Ca(OH)2(s).

23.49 60.7 h.

23.51 (a) 1.03 V. (b) 3.32 × 104 kJ/mol.

23.53 + 12NO2(g) + 3O2(g).

23.55 The “bridge” bonds in Al2Cl6 break at high temperature: . This increases the number of molecules in the gas phase and causes the pressure to be higher than expected for pure Al2Cl6.

23.57 In Al2Cl6, each aluminum atom is surrounded by 4 bonding pairs of electrons (AB4-type molecule), and therefore each aluminum atom is sp3 hybridized. VSEPR analysis shows AlCl3 to be an AB3-type molecule (no lone pairs on the central atom). The geometry should be trigonal planar, and the aluminum atom should therefore be sp2 hybridized.

Page AP-17

23.59 65.4 g/mol.

23.61 Copper(II) ion is more easily reduced than either water or hydrogen ion. Copper metal is more easily oxidized than water. Water should not be affected by the copper purification process under standard conditions.

23.63 (a) 1482 kJ/mol. (b) 3152.8 kJ/mol.

23.65 Mg(s) reacts with N2(g) at high temperatures to produce Mg3N2(s). Ti(s) also reacts with N2(g) at high temperatures to produce TiN(s).

23.67 (a) In water the aluminum(III) ion causes an increase in the concentration of hydrogen ion (lower pH). This results from the effect of the small diameter and high charge (3+) of the aluminum ion on surrounding water molecules. The aluminum ion draws electrons in the O—H bonds to itself, thus allowing easy formation of H+ ions. (b) Al(OH)3 is an amphoteric hydroxide. It will dissolve in strong base with the formation of a complex ion. . The concentration of OH2 in aqueous ammonia is too low for this reaction to occur.

23.69 H2O(l).

23.71 Metals have closely spaced energy levels and a very small energy gap between filled and empty levels. Consequently, many electronic transitions can take place with absorption and subsequent emission of light continually occurring. Some of these transitions fall in the visible region of the spectrum and give rise to the flickering appearance.

23.73 NaF: cavity prevention. Li2CO3: antidepressant. Mg(OH)2: laxative. CaCO3: calcium supplement; antacid. BaSO4: radiocontrast agent.

23.75 Both Li and Mg form oxides (Li2O and MgO). Other Group 1A metals (Na, K, etc.) also form peroxides and superoxides. In Group 1A, only Li forms nitride (Li3N), like Mg (Mg3N2). Li resembles Mg in that its carbonate, fluoride, and phosphate have low solubilities.

23.77 Zn.

23.79 87.66%; 12.34%.

23.81 727 atm.

Chapter 24

24.11 HCl; CaH2. A water solution of HCl is called hydrochloric acid. Calcium hydride will react according to the equation CaH2(s) + 2H2O(l) → Ca(OH)2(aq) + 2H2(g).

24.13 NaH: Ionic compound, reacts with water as follows: NaH(s) + H2O(l) → NaOH(aq) + H2(g); CaH2: Ionic compound, reacts with water as follows: CaH2(s) + 2H2O(l) → Ca(OH)2(aq) + 2H2(g); CH4: Covalent compound, unreactive, burns in air or oxygen: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l); NH3: Covalent compound, weak base in water: NH3(aq) + H2O(l) ⇄ (aq) + OH(aq); H2O: Covalent compound, forms strong intermolecular hydrogen bonds, good solvent for both ionic compounds and substances capable of forming hydrogen bonds; HCl: Covalent compound (polar), acts as a strong acid in water: HCl(g) + H2O(l) → H3O+(aq) + Cl(aq).

24.15 CaH2(s) + 2H2O(l) → Ca(OH)2(aq) + 2H2(g); 22.7 g.

24.17 (a) CuO(s) + H2(g) → Cu(s) + H2O(l). (b) No reaction.

24.25 .

24.27 (a) 2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g). (b) Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l). The visual proof is the formation of a white precipitate of CaCO3.

24.29 Heat causes bicarbonates to decompose according to the reaction: + H2O + CO2. Generation of carbonate ion causes precipitation of the insoluble MgCO3.

24.31 NaHCO3 plus some Na2CO3.

24.33 Yes.

24.39 (a) 2NaNO3(s) → 2NaNO2(s) + O2(g). (b) NaNO3(s) + C(s) → NaNO2(s) + CO(g).

24.41 NH3(g) + CO2(g) → (NH2)2CO(s) + H2O(l). The reaction should be run at high pressure.

24.43 The oxidation state of N in nitric acid is +5, the highest oxidation state for N. N can be reduced to an oxidation state −3.

24.45 (a) NH4NO3(s) → N2O(g) + 2H2O(l). (b) 2KNO3(s) → 2KNO2(s) + O2(g). (c) Pb(NO3)2(s) → PbO(s) + 2NO2(g) + O2(g).

24.47 KNO3(s) + C(s) → KNO2(s) + CO(g); 48.0 g.

24.49 (a) 86.7 kJ/mol. (b) 4 × 10−31. (c) 4 × 10−31.

24.51 125 g/mol; P4.

24.53 4HNO3(aq) + P4O10(s) → 2N2O5(g) + 4HPO3(l); 60.4 g.

24.55 sp3.

24.63 = −198.3 kJ/mol; KP = 6 × 1034; Kc = KP.

24.65 (a) To exclude light. (b) 0.371 L.

24.67 F: −1; O: 0.

24.69 (a) HCOOH(l) ⇄ CO(g) + H2O(l). (b) 4H3PO4(l) ⇄ P4O10(s) + 6H2O(l). (c) 2HNO3(l) ⇄ N2O5(g) + H2O(l). (d) 2HClO3(l) ⇄ Cl2O5(l) + H2O(l).

24.71 To form OF6 there would have to be six bonds (twelve electrons) around the oxygen atom. This would violate the octet rule. Since oxygen does not have d orbitals, it cannot have an expanded octet.

24.73 35 g.

24.75 9H2SO4(aq) + 8NaI(aq) → 4I2(s) + H2S(g) + 4H2O(l) + 8NaHSO4(aq).

24.79 (a) (b)

24.81 (a) Linear. (b) Tetrahedral. (c) Trigonal bipyramidal. (d) See-saw.

24.83 25.3 L.

24.85 2.81 L.

24.87 I2O5(s) + 5CO(g) → 5CO2(g) + I2(s). Iodine is reduced; carbon is oxidized.

24.89 (a) 2H3PO3(aq) → H3PO4(aq) + PH3(g) + O2(g). (b) Li4C(s) + 4HCl(aq) → 4LiCl(aq) + CH4(g). (c) 2HI(g) + 2HNO2(aq) → I2(s) + 2NO(g) + 2H2O(l). (d) H2S(g) + 2Cl2(g) → 2HCl(g) + SCl2(l).

24.91 (a) SiCl4. (b) F. (c) F. (d) CO2.

24.93 There is no change in oxidation number; it is zero for both compounds.

24.95

24.97 25°C: K = 9.61 × 10−22; 100°C: K = 1.2 × 10−15.

24.99 The glass is etched by the reaction: 6HF(aq) + SiO2(s) → H2SiF6(aq) + 2H2O(l). This process gives the glass a frosted appearance.

24.101 1.18.

24.103 0.833 g/L. The molar mass derived from the observed density is 74.41, which suggests that the molecules are associated to some extent in the gas phase. This makes sense due to strong hydrogen bonding in HF.

Chapter 25

25.3 The monomer must have a triple bond.

25.5

25.9 (1) Produce the alkoxide: Sc(s) + 2C2H5OH(l) → Sc(OC2H5)(alc) + 2H+(alc) (“alc” indicates a solution in alcohol); (2) Hydrolyze to produce hydroxide pellets: Sc(OC2H5)(alc) + 2H2O(l) → Sc(OH)2(s) + 2C2H5OH(alc); (3) Sinter pellets to produce ceramic: Sc(OH)2(s) → ScO(s) + 2H2O(g).

25.11 Bakelite is best described as a thermosetting composite polymer.

25.15 No. These polymers are too flexible, and liquid crystals require long, relatively rigid molecules.

25.19 As shown, it is an alternating condensation copolymer of the polyester class.

25.21 Metal amalgams expand with age; composite fillings tend to shrink.

25.25 sp2.

25.27 Dispersion forces; dispersion forces.

25.31 (a) n-type. (b) p-type.

25.35 Bi2Sr2CuO6.

25.37 Two are +2 ([Ar]3d9), one is 13 ([Ar]3d8). The +3 oxidation state is unusual for copper.

25.39

25.41 Plastic polymer: covalent bonds, disulfide (covalent) bonds, H-bonds and dispersion forces. Ceramics: ionic and network covalent bonds.

25.43 Fluoroapatite is less soluble than hydroxyapatite, particularly in acidic solutions. Dental fillings must also be insoluble.

25.45 The molecule is long, flat and rigid, so it should form a liquid crystal.