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One gram sample of $\mathrm{NH}_4 \mathrm{NO}_3$ is decomposed in a bomb calorimeter. The temperature of the calorimeter increases by $6.12 \mathrm{~K}$. The heat capacity of the system is $1.23 \mathrm{~kJ} / \mathrm{g} / \mathrm{deg}$. What is the molar heat of decomposition for $\mathrm{NH}_4 \mathrm{NO}_3$ ?
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Verified Answer
The correct answer is:
$-602 \mathrm{~kJ} / \mathrm{mol}$
Heat of decomposition, $\Delta E=m s \Delta T$
$$
=1 \times 1.23 \times 6.12=7.5276 \mathrm{~kJ} \text {. }
$$
Molar heat of decomposition for $\mathrm{NH}_4 \mathrm{NO}_3$
$$
=7.5276 \times 80=602.2 \mathrm{~kJ} / \mathrm{mol}
$$
$$
=1 \times 1.23 \times 6.12=7.5276 \mathrm{~kJ} \text {. }
$$
Molar heat of decomposition for $\mathrm{NH}_4 \mathrm{NO}_3$
$$
=7.5276 \times 80=602.2 \mathrm{~kJ} / \mathrm{mol}
$$
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