some_problem_answers_160

some_problem_answers_160 - CORRECTIONS TO ASSIGNED PROBLEMS...

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CORRECTIONS TO ASSIGNED PROBLEMS Chapter 17 12. This question should specify to use the stoichiometric coefficients that are the smallest integers that balance the equations. Any set of correct stoichiometric coefficients can be used, but the expression for Q c will be different if the coefficients are changed. Chapter 20 75, part(a): assume T = 298 K. 75, part(b), (c), and (d): The phrase “assuming standard conditions” should be added. CORRECTIONS TO SOLUTIONS TO ASSIGNED HOMEWORK PROBLEMS ANSWERED IN THE STUDENT SOLUTIONS MANUAL Chapter 18 104. The solutions manual assumes that the van’t Hoff factor i is 1, and then calculates a molality of 1.0376 mol/kg. Instead, the given molality 1.000 mol/kg should be used to calculate i = 1.0376. Note that i = 1.000 + x. The answer K a = 0.00147 is correct. Chapter 19 37, part(f). 100.10 mL = 0.10010 L, and (OH ) = 9.99 x 10 –5 M; pH = 10.00 ANSWERS TO ASSIGNED HOMEWORK PROBLEMS NOT ANSWERED IN THE STUDENT SOLUTIONS MANUAL (listed in the order given on the syllabus) Chapter 12 15. q total = –4859 J – 13486 J – 2010 J = –20.4 kJ H1CH12. Start problems of this kind with a statement of conservation of energy in words: heat lost by the hot coffee = heat needed to warm the ice to 0°C + heat needed to melt the ice + heat gained by the melted ice. Convert this statement into symbols: –q coffee = q ice + q fusion + q water –q coffee = ms∆T = m coffee (4.184 J/g°C)(T final – 98.4)°C q ice = m ice s ice (0 –(–2.2))°C q water = m water s water (T final – 0)°C m ice = m water s ice ≠ s water q fusion = 7685.39(=7690) J Putting all of this together and solving for T final gives T final = 83.2°C. H2CH12. (a) vap = 22.2 kJ/mol (b) –185.4 – 22.2 = –207.6 kJ/mol 78. (a) A solid E solid liquid F liquid gas H liquid B solid, liquid, and gas, all in equilibrium C gas (b) critical point = D and triple point = B (c) GB curve (d) solid; solid liquid; liquid; liquid gas; gas. (e) liquid; liquid gas; gas. (f) Liquid X is more dense than solid X. H3CH12. (a) The vapor pressure depends on the temperature, which has not changed. Therefore, the partial pressure of water in the compressed gas will be 19.8 torr: some liquid must condense. (b) 0.0388 g (calculated from n = PV/RT). H4CH12. (a) 0.0252 atm (calculated from P = nRT/V). (b) 292 mol 5.26 L 71. (a) 4 (b) 9.23 x 10 –23 cm 3 (c) 1.34 x 10 –22 g (d) 20.2 g/mol = molar mass; 20.2 amu/atom = atomic mass; conversion: N A amu = 1 g. H5CH12. 1.68 x 10 –8 cm H6CH12. r = 1.28 x 10 –8 cm H7CH12. (a) 0.5236 (b) 0.7405 (c) 0.6802 (d) body-centered d = 8.3 g/cm 3 and simple cubic d = 6.4 g/cm 3 83. AZ 3 86. There are four formula units of NaCl or KF in a unit cell; density = 2.46 g/cm 3 .
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Chapter 13 86. Xe is a much larger atom than He, so it is much more polarizable. This would increase the dipole-induced dipole forces when Xe is placed in water, increasing the solubility relative to He. 16.
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some_problem_answers_160 - CORRECTIONS TO ASSIGNED PROBLEMS...

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