## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 25 Nuclear Chemistry - Study Questions - Page 1007b: 27

#### Answer

$7.700\times10^{8}$ kJ/mol nucleons

#### Work Step by Step

Mass defect $\Delta m=[m_{p}Z+(A-Z)m_{n}]-M$ $=[1.00783\,g/mol\times8+(16-8)1.00867\,g/mol]-15.99492\,g/mol$$=0.13708\,g/mol=1.3708\times10^{-4}\,kg/mol Binding energy per mol E_{b}=(\Delta m)c^{2}=(1.3708\times10^{-4}\,kg/mol)(2.99792\times10^{8}\,m/s^{2})^{2}$$=1.23201\times10^{13}\,J/mol$ Binding energy per mole of nucleons $\frac{E_{b}}{n}=\frac{1.23201\times10^{13}\,J/mol}{16\,nucleons}=7.700\times10^{11}$ J/mol nucleons $=7.700\times10^{8}$ kJ/mol nucleons

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