10-17-11 - Chemistry 1A F11 10.17.11 Dr. Shaka Exam...

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112 Chemistry 1A F11 10.17.11 Dr. Shaka Course Code 40000 Exam Postmortem • Many mistakes! • Units conversions, no units, and arithmetic errors topped the list. • Assigned problems, which we went over in the Discussion Sections, were often missed. • Apparently very few could determine the amount of molecular iodine in the gas. • Also the sodium/strontium carbonate problem proved difficult for many. • The van der Waals problems were mostly okay.
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113 Chemistry 1A F11 10.17.11 Dr. Shaka Course Code 40000 Exam Postmortem • An example. How many ozone molecules are present in one cubic meter of air at 250 K and 1.3 × 10 3 atm? (1 m = 100 cm, 1 L = 1000 cm 3 ). n = pV RT = 1.0 " 10 # 3 atm ( ) 1 m 3 100 cm / m ( ) 3 1 l /1000 cm 3 ( ) 0.0820575 latmmol # 1 K # 1 ( ) 250 K = 0.0487463 moles $ N = N 0 n = 6.0221415 " 10 23 0.0487463 ( ) = 2.93557 " 10 22 molecules • N.B. If 1 m = 100 cm, (1m) 3 = (100 cm) 3 , and not 100 cm 3 . Always put parentheses around the number and the units .
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114 Chemistry 1A F11 10.17.11 Dr. Shaka Course Code 40000 Exam Postmortem • Another example. If 0.0457 moles of I 2 is put into a 2.30 liter container at 1410 K, what would the pressure be if all the gas were I 2 ? • This rounds to 2.30 atm to three digits. But putting 2.29(89) is also completely acceptable. • We cannot assume that any “simple rules” for sig. figs. will actually work in all situations. p = nRT V = 0.0457 mol " 0.0820575 latmmol # 1 K # 1 " 1410 K 2.30 l = 2.29893 atm
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115 Chemistry 1A F11 10.17.11 Dr. Shaka Course Code 40000 Exam Postmortem • Now supposing all the I 2 converted to I atoms according to I 2 ( g ) = 2I( g )? Well, there are just twice as many moles now, and the temperature and volume are the same, so it must be that p = 4.60 atm. No calculation
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10-17-11 - Chemistry 1A F11 10.17.11 Dr. Shaka Exam...

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