Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 18 - Nuclear Chemistry - Questions for Review and Thought - General Questions - Page 817d: 71


1.5 mg

Work Step by Step

Amount of radionuclide at the beginning: $A_{0}=96\, mg$ Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{10.\,min}=0.0693/min$ Time $t=1\,hour=60\,min$ $\ln(\frac{A_{0}}{A})=kt$ where $A$ is the amount of radionuclide remaining. $\implies \ln(\frac{96\,mg}{A})=0.0693\times60=4.158$ Taking the inverse $\ln$ of both sides, we have $\frac{96\,mg}{A}=e^{4.158}=63.94$ Or $A= \frac{96\,mg}{63.94}=1.5\,mg$
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