Chemistry: Molecular Approach (4th Edition)

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

Chapter 20 - Exercises - Page 973: 51

Answer

$2.66\times10^{3}\,y$

Work Step by Step

Recall: $\ln \frac{N_{t}}{N_{0}}=-\frac{0.693}{t_{1/2}}t$ $\implies \ln \frac{72.5}{100}=-\frac{0.693}{5730\,y}\times t$ $\implies -0.32158=-1.2094\times10^{-4}\,y^{-1}\times t $ $\implies t= \frac{-0.32158}{-1.2094\times10^{-4}\,y^{-1}}$ $= 2.66\times10^{3}\,y$
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