If $C$ (diamond) $\rightarrow C$ (graphite) $+ X \ kJ \ mol^{-1}$,$C$ (diamond) $+ O_{2(g)} \rightarrow CO_{2(g)} + Y \ kJ \ mol^{-1}$,and $C$ (graphite) $+ O_{2(g)} \rightarrow CO_{2(g)} + Z \ kJ \ mol^{-1}$,at constant temperature,then which of the following relations is correct?

Find the standard enthalpy of formation of ammonia from the following reaction:
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} ; \Delta_{r}H^0 = -92.0 \ kJ$

$C + 2S \rightarrow CS_2$; $\Delta H^o = +117 \, kJ \, mol^{-1}$; $C + O_2 \rightarrow CO_2$; $\Delta H^o = -393 \, kJ \, mol^{-1}$; $S + O_2 \rightarrow SO_2$; $\Delta H^o = -297 \, kJ \, mol^{-1}$. The heat of combustion of $CS_2$ in $kJ \, mol^{-1}$ is:

The lowest value of heat of neutralization is obtained for

The heat of combustion of $C_xH_y$,carbon,and hydrogen are $a, b$,and $c \ cal/mole$ respectively. The heat of formation of $C_xH_y$ will be:

When $600 \; mL$ of $0.2 \; M \; HNO_3$ is mixed with $400 \; mL$ of $0.1 \; M \; NaOH$ solution in a flask,the rise in temperature of the flask is $\dots \times 10^{-2} \; ^{\circ}C$. (Enthalpy of neutralisation $= 57 \; kJ \; mol^{-1}$ and Specific heat of water $= 4.2 \; J \; K^{-1} \; g^{-1}$) (Neglect heat capacity of flask)