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Practice_problems_Answer

Practice_problems_Answer - Answers 1(a Heat required to...

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Answers 1. (a) Heat required to increase the temperature of 1.00 g of Li metal = g/mol 6.94g 1 × 24.8 J/mol·ºC × 1ºC = 3.57 J Heat required to increase the temperature of 1.00 g of Rb metal = g/mol 85.47g 1 × 31.0 J/mol·ºC × 1ºC = 0.363 J (b) To achieve a temperature change of 1ºC, Rb of g 1Rb of Mass = J 0.363J 3.57 Mass of Rb = 9.83 g of Rb 2. ΔHº = +5.0 kJ 3. Mass of CO 2 = 506 g of CO 2 4. c = 0.14 J/g·ºC 5. (a) N 2 O 4(g) + 3CO (g) → N 2 O (g) + 3CO 2(g) ΔHº rxn = [ΔHº f N 2 O (g) + 3ΔHº f CO 2(g) ] – [ΔHº f N 2 O 4(g) + 3ΔHº f CO (g) ] = [81.6 kJ/mol + 3(–393.5 kJ/mol)] – [11.1 kJ/mol + 3(–110.5 kJ/mol)] = –1441.5 kJ/mol of N 2 O 4 (b) 4FeS 2(s) + 11O 2(g) → 8SO 2(g) + 2Fe 2 O 3(s) ΔHº rxn = [8ΔHº f SO 2(g) + 2ΔHº f Fe 2 O 3(s) ] – [ 4ΔHº f FeS 2(s) – 11ΔHº f O 2(g) ] = [8(–296.8 kJ/mol) + 2(–824.2 kJ/mol)] – [4(–178.2 kJ/mol) – 11(0 kJ/mol)] = –3310 kJ/4 mol of FeS 2 = –827.5 kJ/mol of FeS 2 6. (a) C 6 H 12 O 6(s) + 6O 2(g) → 6CO 2(g) + 6H 2 O (l) (b) Heat gained by water = m·c·ΔT (water) = (200.00 g)(4.184 J/g·ºC)(37.3ºC) = 31 200 J = 31.2 kJ n mol C 6 H 12 O 6 = g/mol 180.18g 000.2 = 0.01110 mol ΔHº comb = mol 0.01110kJ 2.31− = –2.81 × 10 3 kJ/mol (c) ΔHº comb = [6ΔHº f CO
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