Chemistry: Molecular Approach (4th Edition)

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

Chapter 20 - Exercises - Page 974: 54

Answer

$2.9\times10^{4}$ years ago.

Work Step by Step

Recall: $\ln\frac{A_{t}}{A_{0}}=-\frac{0.693}{t_{1/2}}\times t$ $\implies \ln \frac{0.48}{15.3}=-\frac{0.693}{5715\,y}\times t$ $\implies -3.461822=-1.212598\times10^{-4}\,y^{-1}\times t$ $\implies t= \frac{-3.461822}{-1.212598\times10^{-4}\,y^{-1}}$ $=2.9\times10^{4}\,y$
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