无机及分析化学课后重点习题答案详解(高等教育出版社)（Explanation of key exercises in inorganic and Analytical Chemistry (Higher Education Press)）.doc

无机及分析化学课后重点习题答案详解(高等教育出版社)（Explanation of key exercises in inorganic and Analytical Chemistry (Higher Education Press)） Inorganic and analytical chemistry after-school key exercises answer detailed explanation (Higher Education Press).Txt, a person, a box of cigarettes, a computer, one day a person, a bottle of wine, a dish of beans, a night. Never carry a womans little temper, women can never withstand the sweet words of men. Chapter 1 gases and Solutions 2. solution: according to ideal gas equation of state: Available: Then: The relative molecular mass is 16 4.: the solution by the nitrogen isothermal change, oxygen volume change According to Daltons law of partial pressure: 7. solution: T = (273+15) K = 288K; P total =100kPa; V=1.20L 288K, P (H2O) =1.71kPa M (Zn) =65.39 Then p hydrogen = (100-1.71) kPa = 98.29kPa According to: Zn (s) + 2HCl, ZnCl2 + H2 (g) 65.39g 1mol M (Zn) = 0.0493mol The solution is m (Zn) =3.22g Then the mass fraction of impurities is w (impurity) = (3.45-3.22) / 3.45 = 0.067 The 14. solution: for very dilute solution, with P = 1 g - mL-1 (1) (2) The relative molecular mass of heme is (3) (4) the measurement error is very large because the boiling point elevation and the freezing point drop are too small, so the two methods are not applicable. The second chapter is preliminary chemical thermodynamics 15. answers: (1) wrong; (2) wrong; (3) wrong, can be changed to: the response can be spontaneous; (4) wrong when the temperature is higher than 0K; (5) a mistake, as of reaction 17. answers (1) spontaneous at high temperature (2) spontaneous at low temperature (3) spontaneous at any temperature 22. solutions: (1) -277.6 -235.3 161282 (measured boiling point 78.3 centigrade) (2) 30.91 152.2, 245.35 (measured boiling point 58.76 centigrade) 24. solutions: (1) the standard state = -2286+ (-1267) - - (-3644) = 91kJ/mol 0 Positive, non spontaneous, not weathering (1) under 60% RH conditions, pH2O = 0.6*3.17kPa=, 1.90 kPa = (91-98.2) kJ/mol = -7.2, kJ