Chem+160+Handout+Problems+Answers

Chem+160+Handout+Problems+Answers - HANDOUT PROBLEMS...

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HANDOUT PROBLEMS SOLUTIONS 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) Δ vap = 22.2 kJ/mol (b) –185.4 – 22.2 = –207.6 kJ/mol 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 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 H1CH13. P = 2.34 x 10
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