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

This preview shows pages 1–2. Sign up to view the full content.

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)]

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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 .

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online