$\left[ Co \left( NH _{3}\right)_{3}\left( NO _{2}\right)_{3}\right]=$ out of the coordination sphere no ions (non ionisable) $K_{4}\left[F e(C N)_{6}\right]=5$ ions
$\left[ Co \left( NH _{3}\right)_{4}\right] . SO _{4}=2$ ions
More ions $\propto$ more conductivity
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$K = \left( {4.5 \times {{10}^{11}}{S^{ - 1}}} \right){e^{ - \left( {28000\frac{{{K^2}mol}}{J}} \right)\frac{R}{T}}}$
The activation energy of reaction (in $J/mol$) is
$C{H_3} - C{H_2} - Br\xrightarrow{{{\text{alc}}{\text{.KCN}}}}$ $C{H_3}C{H_2}CN$$\xrightarrow{{{\text{HOH}}}}X$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{C{H_2} = CH - C{H_2} - C - H}
\end{array} \to C{H_3} - C{H_2} - C{H_2} - C{H_2}OH$
