Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 16 - A Macroscopic Description of Matter - Exercises and Problems - Page 464: 10


The diameter of the sphere is 2.7 cm

Work Step by Step

The mass of 1 mole of gold is $196.967~g$. We can find the volume of this mass of gold as: $V = \frac{m}{\rho}$ $V = \frac{0.196967~kg}{19.3\times 10^3~kg/m^3}$ $V = 1.02\times 10^{-5}~m^3$ We can find the radius of the sphere as: $\frac{4}{3}\pi~R^3 = V$ $R^3 = \frac{3V}{4\pi}$ $R = (\frac{3V}{4\pi})^{(1/3)}$ $R = [\frac{(3)(1.02\times 10^{-5}~m^3)}{4\pi}]^{(1/3)}$ $R = 0.0135~m = 1.35~cm$ Since the diameter is $2R$, the diameter of the sphere is 2.7 cm.
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