chapter12 - ⋅ atm/K ⋅ mol)(295.7 K) 0.0180 atm = 102...

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Homework Chapter 12 12.2 210 mm Hg · 1 atm 760 mm Hg = 0.28 atm 0.28 atm · 1.013 bar 1 atm = 0.28 bar 210 mm Hg · 101.325 kPa 760 mm Hg = 28 kPa 12.14 () () () ( ) L V V K mmHg K L mmHg T V P T V P 7 2 2 7 2 2 2 1 1 1 10 2 . 1 240 600 2 . 289 10 2 . 1 737 × = = × = The volume of gas is nearly the same (to 2 significant figures) at the higher altitude. 12.20 1.50 g 1 mol C 2 H 5 OH 46.07 g = 0.0326 mol C 2 H 5 OH 251 cm 3 1 mL 1 cm 3 1 L 10 3 mL = 0.251 L P = nRT V = (0.0326 mol C 2 H 5 OH)(0.082057 L atm/K mol)(520 K) 0.251 L
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12.28 d = 0.0125 g 0.165 L = 0.0758 g/L 13.7 mm Hg 1 atm 760 mm Hg = 0.0180 atm M = dRT P = (0.0758 g/L)(0.082057 L
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Unformatted text preview: ⋅ atm/K ⋅ mol)(295.7 K) 0.0180 atm = 102 g/mol 102 g/mol 51 g/mol = 2 The molecular formula is (CHF 2 ) 2 or C 2 H 2 F 4 12.36 767 mm Hg ⋅ 1 atm 760 mm Hg = 1.01 atm n = PV RT = (1.01 atm)(8.90 L) (0.082057 L ⋅ atm/K ⋅ mol)(295.2 K) = 0.371 mol CO 2 0.371 mol CO 2 ⋅ 4 mol KO 2 2 mol CO 2 ⋅ 71.10 g 1 mol KO 2 = 52.7 g KO 2 12.56 C or K mL K mL V T V T o 30 310 5 . 25 360 5 . 21 1 1 2 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =...
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This note was uploaded on 11/16/2011 for the course CHEM 1211k taught by Professor Wood during the Fall '07 term at Valdosta State University .

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chapter12 - ⋅ atm/K ⋅ mol)(295.7 K) 0.0180 atm = 102...

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