Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 1 - Sections 1.1-1.8 - Exercises - Cumulative Problems - Page 41: 115

Answer

$\approx22$ $ in^{3}$

Work Step by Step

1. Convert the density of Titanium from $g/cm^{3}$ to $lb/in^{3}$ by using the following conversion factors: $1lb=453.5926g$ $1in^{3}=16.3871cm^{3}$ $p = \frac{4.51 g/cm^{3} \times 16.3871 cm^{3}/in^{3} }{453.5926 g/lb} = 0.1629 lb/in^{3}$ 2. Divide the mass over density to get the volume $V=\frac{m}{p}=\frac{3.5lb}{0.1629in^{3}}=21.48 in^{3}\approx 22in^3$
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