Find the equivalent weight of ${\mathrm{Br}}_2$ in each of the following reactions:

(a) ${\mathrm{Br}}_2+2{\mathrm{e}}^{-}=2{\mathrm{Br}}^{-}$ or $5{\mathrm{Br}}_2+10{\mathrm{e}}^{-}=10{\mathrm{Br}}^{-}$

(b) ${\mathrm{Br}}_2+12{\mathrm{OH}}^{-}=2{\mathrm{BrO}}_3^{-}+6{\mathrm{H}}_2\mathrm{O}+10{\mathrm{e}}^{-}$

(c) $6{\mathrm{Br}}_2+12{\mathrm{OH}}^{-}=10{\mathrm{Br}}^{-}+2{\mathrm{BrO}}_3^{-}+6{\mathrm{H}}_2\mathrm{O}$

Equation (c) is the sum of equations (a) and (b). What is the relationship between the answer to (c) and the answers to (a) and (b)?

 

Find the equivalent weight of ${\mathrm{KMnO}}_4$ in the reaction:

${\mathrm{Mn}}^{2+}+{\mathrm{MnO}}_4^{-}+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{MnO}}_2+{\mathrm{H}}^{+}\left(\text{unbalanced}\right)$

What mass of ${\mathrm{MnSO}}_4$ is oxidised by $1.25 \mathrm{grams}$ of ${\mathrm{KMnO}}_4$?

The time required for $10\%$ completion of a first-order reaction at $298 \mathrm{K}$ is equal to that required for its $25\%$ completion at $308 \mathrm{K}$. If the pre-exponential factor for the reaction is $3.56 \times {10}^9 {\mathrm{s}}^{-1}$, calculate its rate constant at $318 \mathrm{K}$ and also the energy of activation.

In a $\mathrm{Cu}$-voltameter, the mass deposited in $30$ seconds is $\mathrm{m}$ grams. If the time-current graph is as shown in the figure, calculate the electrochemical equivalent of $\mathrm{Cu}$.

$1 \mathrm{L}$ of ${\mathrm{CO}}_2$ at $15°\mathrm{C}$ and $1 \mathrm{atm}$ pressure dissolves in $1 \mathrm{L}$ of water at the same temperature, when the pressure of ${\mathrm{CO}}_2$ is $1 \mathrm{atm}$. Calculate the molal concentration of ${\mathrm{CO}}_2$ in a solution over which the partial pressure of ${\mathrm{CO}}_2$ is $150 \mathrm{mm} \mathrm{Hg}.$ Report your answer after multiplying by $100$.

The voltage of the cell, $\mathrm{Pb}\left(\mathrm{s}\right) \left| {\mathrm{PbSO}}_4\left(\mathrm{s}\right) \right| {\mathrm{NaHSO}}_4 \left(0.6\mathrm{M}\right) \left|\right| {\mathrm{Pb}}^{2+} \left(2.5\times {10}^{-5}\right) | \mathrm{Pb}\left(\mathrm{s}\right)$ $\text{is} 0.061 \mathrm{V}. \text{Calculate }{\mathrm{K}}_2=\frac{\left[{\mathrm{H}}^{+}\right] \left[{\mathrm{SO}}_4^{2-}\right]}{\left[{\mathrm{HSO}}_4^{-}\right]}.$ If ${\mathrm{K}}_2 =\mathrm{a}\times {10}^{-2}$, report $\text{'}\mathrm{a}\text{'}$ correct up to one place of decimal.

$\text{Given,} {\mathrm{PbSO}}_4\left(\mathrm{s}\right)+2{\mathrm{e}}^{-}\to \mathrm{Pb}\left(\mathrm{s}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right); {\mathrm{E}}^0=-0.356 \mathrm{V}$ and ${\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Pb}\left(\mathrm{s}\right); {\mathrm{E}}^0=-0.126 \mathrm{V}$.