Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 5 - Gases - Questions & Problems: 5.140

Answer

Please see the work below.

Work Step by Step

We know that $v_{rms}=\sqrt{\frac{3RT}{M}}$ We plug in the known values to obtain: $v_{rms}=\sqrt{\frac{3\times8.314\times 1.7\times 10^{-7}}{0.085}}=0.00706\frac{m}{s}$ Now $KE=\frac{1}{2}Mv_{rms}^2=\frac{1}{2}\times 0.085\times (0.00706)^2=0.0003J$
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