Chemistry: Molecular Approach (4th Edition)

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

Chapter 20 - Exercises - Page 973: 50

Answer

23.4 days.

Work Step by Step

Recall: $\ln \frac{A_{t}}{A_{0}}=-\frac{0.693}{t_{1/2}}t$ $\implies \ln \frac{287/s}{5.88\times10^{4}/s}=-\frac{0.693}{3.042\,d}\times t$ $\implies -5.322415=- 0.22781\,d^{-1}\times t $ $\implies t= \frac{-5.322415}{-0.22781\,d^{-1}}$ $= 23.4\,d$
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