Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Let's Review: The Tools of Quantitative Chemistry - Study Questions - Page 43e: 60


$90.\ m$

Work Step by Step

Mass of the ingot in g: $57\ kg \times 10^3\ g/kg = 57\times10^3\ g$ Given the density of $8.96\ g/cm^3$, the volume is: $V=m/\rho=6.4\times10^3\ cm^3$ With the volume of the wire given by $V=\pi r^2L$, and the radius (half of the diameter) being $4.75\ mm = 0.475\ cm$, the length is: $L=9.0\times10^3\ cm=90.\ m$
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