31. (a) Using Eq. 42-4, the energy released would beE=NEavg=(3.1g)(63.54 g/mol)/(6.02×1023/mol)µ35¶(7.0 eV)(1.6×10−19J/eV)=1.98×104J≈20 kJ.(b) Keeping in mind that a Watt is a Joule per second, we have
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