Cadmium amalgam is prepared by the electrolysis of a solution of ${\mathrm{CdCl}}_2$ using $\mathrm{Hg}$ cathode. For how long $\left(\mathrm{in} \sec .\right)$ should the electrolysis be carried out, in order to prepare $12\%$ by weight of cadmium amalgam using $22.0 \mathrm{g}$ of $\mathrm{Hg}$ as cathode and a current strength of $5$ ampere? $(\mathrm{Cd}=112)$

The tungsten used in the filaments for the light bulbs can be prepared from tungsten $\left(\mathrm{VI}\right)$ oxide by reduction with hydrogen at $1200°\mathrm{C}$. ${\mathrm{WO}}_3\left(\mathrm{s}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to \mathrm{W}\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right); \mathrm{\Delta H}=114.9 \mathrm{kJ}$

${\mathrm{\Delta H}}_{\mathrm{f}}$ for ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)=-241.8 \mathrm{kJ} {\mathrm{mole}}^{-1}.$ Calculate ${\mathrm{\Delta H}}_{\mathrm{f}}$ for ${\mathrm{WO}}_3\left(\mathrm{s}\right) \mathrm{in} \mathrm{kJ} {\mathrm{mole}}^{-1}$

Water gas is produced by the action of superheated steam on red hot coke. ${\mathrm{C}}_{\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+131.4 \mathrm{kJ}=\underset{\mathrm{water}gas\mathrm{ }}{\underbrace{{\mathrm{CO}}_{\left(\mathrm{g}\right)}+{{\mathrm{H}}_2}_{\left(\mathrm{g}\right)}}}$ ${\mathrm{\Delta H}}_{\mathrm{f}}$ for ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)=-241.8 \mathrm{kJ} {\mathrm{mole}}^{-1}.$ Calculate ${\mathrm{\Delta H}}_{\mathrm{f}}$ for $\mathrm{CO}\left(\mathrm{g}\right)$ in kJ/mole.

$$\begin{gathered} \mathrm{A} : -120.4 \\ \mathrm{B} : -110.4 \\ \mathrm{C}: -100.4 \end{gathered}$$

Enter your correct answer as A, B or C.

Calculate $\mathrm{\Delta H}{°}_f$ for chloride ion (in KJ) from the following data.

$\begin{array}{c}\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{Cl}}_2\left(\mathrm{g}\right)\to \mathrm{HCl}\left(\mathrm{g}\right); \mathrm{\Delta H}{°}_{\mathrm{f}}=-92.4 \mathrm{kJ}\end{array}$

$\mathrm{HCl}\left(\mathrm{g}\right)+{\mathrm{xH}}_2\mathrm{O}\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right); {\mathrm{\Delta H}}_{298}=-74.8 \mathrm{kJ}$

$\mathrm{\Delta }H{°}_f\left[{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\right]=0.0 \mathrm{kJ}$

The data given below is after the reaction of $\mathrm{NO}$ and ${\mathrm{Cl}}_2$ to form $\mathrm{NOCl}$ at $295 \mathrm{K}$.

$\begin{array}{|ccc|}\hline \left[{\mathrm{Cl}}_2\right]& \left[\mathrm{NO}\right]& \text{ Initial rate }\left({\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}\right)\\ 0.05\mathrm{M}& 0.05\mathrm{M}& 1\times {10}^{-3}\\ 0.15\mathrm{M}& 0.05\mathrm{M}& 3\times {10}^{-3}\\ 0.05\mathrm{M}& 0.15\mathrm{M}& 9\times {10}^{-3}\\ \hline\end{array}$

The data below is after the reaction of $\mathrm{NO}$ and ${\mathrm{Cl}}_2$ to form $\mathrm{NOCl}$ at $295 \mathrm{K}$:

$\begin{array}{|ccc|}\hline \left[{\mathrm{Cl}}_2\right]& \left[\mathrm{NO}\right]& \text{ Initial rate }\left({\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}\right)\\ 0.05\mathrm{M}& 0.05\mathrm{M}& 1\times {10}^{-3}\\ 0.15\mathrm{M} & 0.05\mathrm{M} & 3\times {10}^{-3}\\ 0.05\mathrm{M}& 0.15\mathrm{M} & 9\times {10}^{-3}\\ \hline\end{array}$

The data given below is after the reaction of $\mathrm{NO}$ and ${\mathrm{Cl}}_2$, to form $\mathrm{NOCl}$ at $295 \mathrm{K}$:

$\begin{array}{|ccc|}\hline \left[{\mathrm{Cl}}_2\right]& \left[\mathrm{NO}\right]& \text{ Initial rate }\left({\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}\right)\\ 0.05\mathrm{M}& 0.05\mathrm{M}& 1\times {10}^{-3}\\ 0.15\mathrm{M}& 0.05\mathrm{M}& 3\times {10}^{-3}\\ 0.05\mathrm{M}& 0.15\mathrm{M}& 9\times {10}^{-3}\\ \hline\end{array}$

The data below is after the reaction of $\mathrm{NO}$ and ${\mathrm{Cl}}_2$, to form $\mathrm{NOCl}$ at $295 \mathrm{K}$ :

$\begin{array}{|ccc|}\hline \left[{\mathrm{Cl}}_2\right]& \left[\mathrm{NO}\right]& \text{ Initial rate }\left({\mathrm{molL}}^{-1}{\mathrm{s}}^{-1}\right)\\ 0.05\mathrm{M}& 0.05\mathrm{M}& 1\times {10}^{-3}\\ 0.15\mathrm{M}& 0.05\mathrm{M}& 3\times {10}^{-3}\\ 0.05\mathrm{M}& 0.15\mathrm{M}& 9\times {10}^{-3}\\ \hline\end{array}$