Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 1 - Chemistry: The Study of Change - Questions & Problems - Page 33: 1.73


Please see the work below.

Work Step by Step

We know that $volume=\frac{mass}{density}$ We plug in the known values to obtain: $volume \space of \space crucible =\frac{860.2}{21.45}=40.10cm^3$ Now $Volume \space of\space water \space displaced=\frac{(860.2-820.2)}{0.9986}=40.1cm^3$
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