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Answers to Selected Exercises Answers to Selected Exercises The answers listed here are from the Complete Solutions Guide, in transition metal (Cr) that generally exhibits two or more stable which rounding is carried out at each intermediate step in a calcu- charges when in ionic compounds. We need to indicate which lation in order to show the correct number of significant figures charged ion we have in the compound. This is generally true when- for that step. Therefore, an answer given here may differ in the last ever the metal in the ionic compound is a transition metal. ICl3 is digit from the result obtained by carrying extra digits throughout made from only nonmetals and is a covalent compound. Predicting the entire calculation and rounding at the end (the procedure you formulas for covalent compounds is extremely difficult. Because of should follow). this, we need to indicate the number of each nonmetal in the binary covalent compound. The exception is when there is only one of the Chapter 2 first species present in the formula; when this is the case, mono- is not used (it is assumed). 45. a. sulfur difluoride; b. dinitrogen 19. ClF3 21. All the masses of hydrogen in these three compounds can be expressed as simple whole-number ratios. The g H/g N in tetroxide; c. iodine trichloride; d. tetraphosphorus hexoxide hydrazine, ammonia, and hydrogen azide are in the ratios 6:9:1. 47. a. copper(I) iodide; b. copper(II) iodide; c. cobalt(II) iodide; 23. O, 7.94; Na, 22.8; Mg, 11.9; O and Mg are incorrect by a d. sodium carbonate; e. sodium hydrogen carbonate or sodium Ϸ bicarbonate; f. tetrasulfur tetranitride; g. sulfur hexafluoride; factor of 2; correct formulas are H2O, Na2O, and MgO. 25. d(nucleus) ϭ 3 ϫ 1015 g/cm3; d(atom) ϭ 0.4 g/cm3 27. Since h. sodium hypochlorite; i. barium chromate; j. ammonium nitrate Ϫ ϫ 13 49. a. SO2;b.SO3;c.Na2SO3; d. KHSO3;e.Li3N; f. Cr2(CO3)3; all charges are whole-number multiples of 6.40 10 zirkombs, ϩ Ϫ ϫ 13 g. Cr(C2H3O2)2; h. SnF4;i.NH4HSO4 (composed of NH4 and then the charge on one electron could be 6.40 10 zirkombs. Ϫ Ϫ However, 6.40 ϫ 10 13 zirkombs could be the charge of two elec- HSO4 ions); j. (NH4)2HPO4; k. KClO4; l. NaH; m. HBrO; trons (or three electrons, etc.). All one can conclude is that the n. HBr 51. a. lead(II) acetate; b. copper(II) sulfate; c. calcium charge of an electron is 6.40 ϫ 10Ϫ13 zirkombs or an integer frac- oxide; d. magnesium sulfate; e. magnesium hydroxide; f. calcium tion of 6.40 ϫ 10Ϫ13. 29. If the plum pudding model were correct sulfate; g. dinitrogen monoxide or nitrous oxide (common) (a diffuse positive charge with electrons scattered throughout), then 53. a. nitric acid, HNO3; b. perchloric acid, HClO4; c. acetic acid, ␣ particles should have traveled through the thin foil with very HC2H3O2; d. sulfuric acid, H2SO4; e. phosphoric acid, H3PO4 minor deflections in their path. This was not the case because a few 55. 116 g S; 230. g O 57. a. True; b. False. The isotope has 34 of the ␣ particles were deflected at very large angles. Rutherford protons; c. False. The isotope has 45 neutrons; d. False. The iden- 2Ϫ 2Ϫ reasoned that the large deflections of these ␣ particles could be tity is selenium, Se. 59. Ra; 142 n 61. SeO4 : selenate; SeO3 : 2Ϫ 2Ϫ caused only by a center of concentrated positive charge that con- selenite; TeO4 : tellurate; TeO3 : tellurite 63. InO, atomic mass tains most of the atom’s mass (the nuclear model of the atom). of In 76.54; In2O3, atomic mass of In 114.8 65. SbCl3; 31. The atomic number of an element is equal to the number of antimony(III) chloride 67. chlorine; 18 electrons 69. a. The com- protons in the nucleus of an atom of that element. The mass num- pounds have the same number and types of atoms (same ber is the sum of the number of protons plus neutrons in the formula), but the atoms in the molecules are bonded together nucleus. The atomic mass is the actual mass of a particular isotope differently. Therefore, the two compounds are different compounds (including electrons). As we will see in Chapter 3, the average mass with different properties. The compounds are called isomers of each of an atom is taken from a measurement made on a large number other. b. When wood burns, most of the solid material in wood is of atoms. The average atomic mass value is listed in the periodic converted to gases, which escape. The gases produced are most table. 33. a. The noble gases are He, Ne, Ar, Kr, Xe, and Rn likely CO2 and H2O. c. The atom is not an indivisible particle, but (helium, neon, argon, krypton, xenon, and radon). Radon has is instead composed of other smaller particles—electrons, neutrons, only radioactive isotopes. In the periodic table, the whole number and protons. d. The two hydride samples contain different isotopes enclosed in parentheses is the mass number of the longest-lived of either hydrogen or lithium. Although the compounds are com- isotope of the element. b. promethium (Pm) and technetium (Tc) posed of different isotopes, their properties are similar because 35. a. Five; F, Cl, Br, I, and At; b. Six; Li, Na, K, Rb, Cs, Fr (H is different isotopes of the same element have similar properties not considered an alkali metal.) c. 14; Ce, Pr, Nd, Pm, Sm, Eu, Gd, (except, of course, their mass). 71. The ratio of the masses of Ϻ Tb, Dy, Ho, Er, Tm, Yb, and Lu; d. 40; all elements in the block R that combine with 1.00 g Q is 3 1, as expected by the law of defined by Sc, Zn, Uub, and Ac are transition metals. 37. a. 12 p, multiple proportions. R3Q 73. C:H ratio 8:18 or 4:9 12 n, 12 e; b. 12 p, 12 n, 10 e; c. 27 p, 32 n, 25 e; d. 27 p, 32 Chapter 3 n, 24 e; e. 27 p, 32 n, 27 e; f. 34 p, 45 n, 34 e; g. 34 p, 45 n, 151 3ϩ 36 e; h. 28 p, 35 n, 28 e; i. 28 p, 31 n, 26 e 39. 63Eu ; 23. 47.88 amu; Ti 25. 185 amu 27. There are three peaks in the 118 2ϩ Ϫ 2ϩ Ϫ 50Sn 41. a. Lose 2 e to form Ra ; b. Lose 3 e to form mass spectrum, each two mass units apart. This is consistent with In3ϩ; c. Gain 3 eϪ to form P3Ϫ; d. Gain 2 eϪ to form Te2Ϫ; e. Gain two isotopes, differing in mass by two mass units. The peak at Ϫ Ϫ Ϫ ϩ 1 e to form Br ; f. Lose 1 e to form Rb 43. AlCl3, aluminum 157.84 corresponds to a Br2 molecule composed of two atoms of ͞ chloride; CrCl3, chromium(III) chloride; ICl3, iodine trichloride; the lighter isotope. This isotope has mass equal to 157.84 2 or 79 81 AlCl3 and CrCl3 are ionic compounds following the rules for nam- 78.92, which corresponds to Br. The second isotope is Br with ͞ ing ionic compounds. The major difference is that CrCl3 contains a mass equal to 161.84 2 80.92. The peaks in the mass spectrum A41 A42 Answers to Selected Exercises 79 79 81 81 ϩ Ϫ correspond to Br2, Br Br, and Br2, in order of increasing exists as separate hydrated Rb ions and hydrated Cl ions in solu- mass. The intensities of the highest and lowest masses tell us the tion. AgCl is an insoluble ionic compound, so the ions stay together two isotopes are present at about equal abundance. The actual in solution and fall to the bottom of the container as a precipitate. 79 81 ϩ abundance is 50.68% Br and 49.32% Br. 29. GaAs can be d. HNO3 is a strong acid and exists as separate hydrated H ions 69 71 Ϫ either GaAs or GaAs. The mass spectrum for GaAs will have and hydrated NO3 ions in solution. CO is a polar covalent mole- 2 peaks at 144 (69 ϩ 75) and 146 (71 ϩ 75) with intensities in the cule and is hydrated as explained in part a. 13. a. picture iv; b. Ϻ Ϻ 69 69 71 ratio of 60 40 or 3 2. Ga2As2 can be Ga2As2, Ga GaAs2, picture ii; c. picture iii; d. picture i; 15. a. Place 20.0 g NaOH in 71 or Ga2As2. The mass spectrum will have 3 peaks at 288, 290, a 2-L volumetric flask; add water to dissolve the NaOH, and fill to and 292 with intensities in the ratio of 36Ϻ48Ϻ16 or 9Ϻ12Ϻ4. the mark with water, mixing several times along the way. b. Add 31. a. 1.03 ϫ 10Ϫ4 mol; b. 4.52 ϫ 10Ϫ3 mol; c. 3.41 ϫ 10Ϫ2 mol 500. mL of 1.00 M NaOH stock solution to a 2-L volumetric flask; Ͻ Ͻ Ͻ Ͻ 33. 4.0 g He 1.0 mol F2 44.0 g CO2 4.0 g H2 146 g SF6 fill to the mark with water, mixing several times along the way. 35. a. 165.39 g/mol; b. 3.023 mol; c. 3.3 g; d. 5.5 1022 atoms; c. Similar to the solution made in part a, instead using 38.8 g Ϫ19 e.
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