$=120+198-140$
$=178\, \mathrm{S}\, \mathrm{cm}^{2} \mathrm{eq}^{-1}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$CoCl _{3} . xNH _{3}+ AgNO _{3}( aq ) \rightarrow$
If two equivalents of $AgCl$ precipitate out, then the value of $x$ will be $....$
$(I)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,}
\end{array}$
$(II)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$(III)$ $\begin{array}{*{20}{c}}
\,\,{C{H_3}} \\
{|\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - COOH} \\
{|\,\,\,\,\,\,} \\
\,\,{C{H_3}\,}
\end{array}$
$(IV)$ $(CH_3-CH_2)_3C-COOH$
$\begin{array}{|l|l|l|l|} \hline & Column\,I & & Column\,II \\ & (Reduction\,process) & & (Charge\,required) \\ \hline (a) & 1\,mol\,of\,MnO_4^-\,to\,Mn^{2+} & (p) & 193000\,\,C \\ \hline (b) & 1\,\,mole\,of\,Cr_2O_7^{2-}\,to\,Cr^{3+} & (q) & 289500\,\,C \\ \hline (c) & 1\,\,mole\,of\,Sn^{4+}\,to\,Sn^{2+} & (r) & 482500\,\,C \\ \hline (d) & 1\,mole\,of\,Al^{3+}\,to\,Al & (s) & 579000\,\,C \\ \hline \end{array}$
