Chemistry 9th Edition

$26\text{ kJ/mol}$
$q=mc\Delta T$ $q=75.0*4.18*(23.34-25.00)=-520.41\text{ J}$ $1.60\text{ grams$NH_4NO_3$}*\frac{1\text{ mol$NH_4NO_3$}}{80.043\text{ grams$NH_4NO_3$}}=0.0200$ We switch the signs of the enthalpy change as the system is the $NH_4OH$ which is absorbing energy. $\frac{520.14\text{ J}}{0.200\text{ mols}}=26\text{ kJ/mol}$