EO_440_PS3_Sol_F11 - Chemistry 440 Problem Set 3 Solutions...

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Chemistry 440 Problem Set 3 – Solutions 1. a) P 2 = P 1 T 2 T 1 = (28.0 + 14.7)psi (305.15) 291.15 = 44.8psi The gauge pressure is (44.8 – 14.7) = 30.1 psi b) The excess pressure is (30.1 – 28.0) = 2.1 psi and this corresponds to (2.1/44.8) x 100 = 4.7% of the air in the tire which has to be let out. 2. b = RT c 8P c = (0.08206)(474.8) 8(40.6) = 0.12 a = 27R 2 T c 2 64P c = 27(0.08206) 2 (474.8) 2 64(40.6) = 15.8 liter 2 atm mol -2 3. V c = 3b = 3 x 0.115 = 0.345 liter m T c = 8a 27Rb = 8(18) 27(0.08206)(0.115) = 565K P c = a 27b 2 = 18 27(0.115) 2 = 50.4atm 4. P T = P H 2 + P H 2 O P H 2 O = 23.8 760 0.0313 , P H 2 = (0.98 – 0.0313) = 0.949 atm n H 2 = PV RT = (0.949)(7.80) (0.08206)(298) = 0.303 mol This is also the amount of Zn consumed. Thus mass Zn = (0.303) (65.37) = 19.8g .
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Chemistry 440 Problem Set 3 – Solutions 5. a) Assume ideal behavior: PV = nRT 0.0020 mm 760 mm x 1 liter = n(0.08206 liter atm K -1 mol -1 ) (300 K) n = 1.07 x 10 -7 mol b) molar mass Hg = 200.59 g mol
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This note was uploaded on 11/07/2011 for the course CHEM 440 taught by Professor Gennis during the Fall '08 term at University of Illinois, Urbana Champaign.

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EO_440_PS3_Sol_F11 - Chemistry 440 Problem Set 3 Solutions...

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