Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 20 - Exercises - Page 973: 46

Answer

34 hours.

Work Step by Step

Rate constant $k= \frac{0.693}{t_{1/2}}=\frac{0.693}{6.0\,h}$ $=0.1155\,h^{-1}$ We can obtain the time taken using integrated rate law which is $\ln \frac{N_{t}}{N_{0}}=-kt$ $\ln \frac{1.0\times10^{-3}\,\mu g}{0.050\,\mu g}=-0.1155\,h^{-1}\times t$ $\implies -3.912023=-0.1155\,h^{-1}\times t$ $\implies t=\frac{-3.912023}{-0.1155\,h^{-1}}$ $=34\,h$
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