In order to decompose $9 \ g$ of water,$142.5 \ kJ$ of heat is required. Hence,the enthalpy of formation of water is ...... $kJ/mol$.

If the heat of neutralization of an acid-base reaction is $56 \ kJ \ mol^{-1}$,then the substances could be:

Calculate the enthalpy change for the reaction
$H_2 + F_2 \longrightarrow 2HF$
given that
Bond energy of $H-H$ bond $= 434 \ kJ/mol$
Bond energy of $F-F$ bond $= 158 \ kJ/mol$
Bond energy of $H-F$ bond $= 565 \ kJ/mol$
Result in $kJ$.

Given that $C_{(g)} + 4H_{(g)} \longrightarrow CH_{4(g)}$,$\Delta H^{\circ} = -1665 \ kJ$. What is the bond energy per mole of $C-H$ bond?

$H_2 + \frac{1}{2} O_2 \to H_2O; \Delta H = -68.39 \ kcal$
$K + H_2O + \text{water} \to KOH_{(aq)} + \frac{1}{2} H_2; \Delta H = -48 \ kcal$
$KOH + \text{water} \to KOH_{(aq)}; \Delta H = -14 \ kcal$
The heat of formation of $KOH$ is (in $kcal$):

The heat evolved in the combustion of benzene is given by $C_6H_6 + 7.5 O_2 \to 6CO_2(g) + 3H_2O(l); \Delta H = -3264.6 \, kJ$. Which of the following quantities of heat energy will be evolved when $39 \, g$ of $C_6H_6$ is burnt? (in $kJ$)