Chapter6 - Chapter 6 Homework 103 J 1 kJ 6.4 1670 kJ 6.14 1...

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Chapter 6 Homework 6.4 1670 kJ · 10 3 J 1 kJ · 1 cal 4.184 J · 1 Cal 10 3 cal = 399 Cal 6.14 q metal + q water = 0 [(182 g)(0.128 J/g·K)(300.7 K – T i )] + [(22.1 g)(4.184 J/g·K)(300.7 K – 298.2 K)] = 0 T i = 311 K (37 ºC) 6.20 16 cubes · 62.0 g 1 cube · 333 J/g = 3.30 × 10 5 J 6.24 q total = energy to heat metal + energy to change phase from solid to liquid q heat metal = (454 g)(0.227 J/g·K)(505.1 K – 298.2 K) = 2.13 × 10 4 J q phase change = (454 g)(59.2 J/g) = 2.69 × 10 4 J q total = 2.13 × 10 4 J + 2.69 × 10 4 J = 4.82 × 10 4 J 6.28 endothermic 10.0 g CaO · 1 mol CaO 56.08 g · 464.8 kJ 1 mol CaO = 82.9 kJ heat required 6.36 Assume C solution = 4.2 J/g·K q solution = (140.2 g)(4.2 J/g·K)(302.0 K – 293.4 K) = 5100 J 5.2 g H 2 SO 4 · 1 mol H 2 SO 4 98.1 g = 0.053 mol H 2 SO 4 q dissolving = – q solution = – 5100 J 0.053 mol H 2 SO 4 · 1 kJ 10 3 J = –96 kJ/mol H 2 SO 4 6.40 q = q water + q calorimeter q = [(575 g)(4.184 J/g·K)(298.37 K – 294.85 K)] + [(650 J/K)(298.37 K – 294.85 K)]
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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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Chapter6 - Chapter 6 Homework 103 J 1 kJ 6.4 1670 kJ 6.14 1...

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