- APotassium fumarate
- BCalcium carbide
- CEthylene bromide
- ✓All of these
$\begin{array}{*{20}{l}}
{CH} \\
{|\,|\,|} \\
{\underbrace {CH}_{Anode}}
\end{array} + 2\,C{O_2} + 2\,KOH + \underbrace {{H_2}}_{{\text{Cathode}}}$
$Ca{C_2} + 2\,{H_2}O \to Ca{\left( {OH} \right)_2} + CH \equiv CH$
$\,\mathop {\mathop {C{H_2}}\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{Br\,\,\,\,\,} - \mathop {C{H_2}}\limits_{\mathop {|\,\,\,\,\,\,\,\,\,}\limits_{Br\,\,\,\,} } + 2\,KOH \to CH \equiv CH + 2\,KBr + 2\,{H_2}O$
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The time taken for $A$ to become $1 / 4^{\text {th }}$ of its inital concentration is twice the time taken to become $1 / 2$ of the same. Also, when the change of concentration of $B$ is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is . . . . .