Answers+to+Ch3-2

00atm498k00821latmmolk 0183

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Unformatted text preview: mm Hg; PH2O = (0.23/2.5)(752 mm Hg) = 69 mm Hg 45. total moles =(0. 986 L)(1. 00 atm)/(315 K)(0. 0821 L ∙ atm/ mol ∙ K) = 0.0381 mol dry air =(1. 04 L)(1. 00 atm)(363 K)(0. 0821 L ∙ atm/ mol ∙ K) = 0.0349 3 mol H2 O = 0.0381 mol – 0.0349 mol = 0.0032 PH2O =(0. 0032 mol)(315 K)(0. 0821 L ∙ atm/ mol ∙ K)/1. 00 L = 0.083 atm = 63 mm Hg 47. mol Ne =2. 50×1. 09/RT=2. 72/RT; mol CO =2. 00×0. 773/RT=1. 55/RT; total mol =2. 72/RT+1. 55/RT=4. 27/RT P =(4. 27/RT )(RT )/(2. 50 + 2. 00) L= 0.949 atm 49. (a) 2C2 H2 (g ) + 5O2 (g ) → 4CO2 (g ) + 2H2 O(g ) (b) mol C2 H2 =(7. 50 L)(1. 00 atm)/(498 K)(0. 0821 L ∙ atm/ mol ∙ K) = 0.183 4 mol CO2 + 2 mol H2 O = 6 mol products/2 mol...
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This note was uploaded on 02/13/2014 for the course CHEM 1024 taught by Professor Dr.zhifengding during the Winter '12 term at UWO.

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