Chemistry: Molecular Approach (4th Edition)

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

Chapter 20 - Exercises - Page 973: 52

Answer

$1.22\times10^{4}\,y$

Work Step by Step

Recall: $\ln \frac{N_{t}}{N_{0}}=-\frac{0.693}{t_{1/2}}\times t$ $\ln \frac{22.8}{100}=-\frac{0.693}{5730\,y}\times t$ $\implies -1.4784=-1.2094\times10^{-4}\,y^{-1}\times t$ $\implies t=\frac{-1.4784}{-1.2094\times10^{-4}\,y^{-1}}$ $=1.22\times10^{4}\,y$
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