1442answerstest4

1442answerstest4 - Answers Test 4 Chapters 20 and 21...

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Answers Test 4 Chapters 20 and 21 Silberberg 1. c 2. (1g CH 4 /1)( 1 mol CH 4 / 16g)(891 kJ/ 1 mole CH 4 )* = 55.7 kJ The conversion factor 891 kJ/ 1 mole comes from the coefficient in the balanced equation. You will notice that the answer is usualy expressed as a positive number. If you are given the choice of -55.7 kJ pick that. 3. c 4. 2, 3 5. ∆G = ∆H - T∆S = -403.13kJ/mol - (273 K)(-.1893 kJ/mol.K) = -351.45 kJ/mole Notice that S will always be in J so you will need to change it to kJ. 6. ∆G = -RT lnK lnK = ∆G / -RT = 217 kJ/mole / -(.008314 kJ/mol.K)(298 K) lnK = -87.586 K = 9.16 x 10 -39 Notice that R is in J and had to be converted to kJ. To find K, use 2nd then ln on the calculator. 7. ∆S = ∑S products -∑S reactants = [2NO 2 + 3 H 2 O] - [2NH 3 + 7/2 O 2 ] = [2(240.40 + 3(188.7)] - [ (2)192.5 + (7/2) 205] = [1046.1] - [1102.5] = -56.4 J 8. Assign oxidation numbers first: XeF 2 + BrO 3 - Xe + 2 HF + BrO 4 - 2+ 1- +5 2- 0 1+ 1- 7+ 2- Xe is the oxidizing agent and Br is oxidized
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1442answerstest4 - Answers Test 4 Chapters 20 and 21...

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