
- A

- B

- C

- ✓








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$(a)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - Br} \\
{|\,\,} \\
{\,\,\,\,\,{C_2}{H_5}\,}
\end{array}$ $(b)$ $\begin{array}{*{20}{c}}
{\,\,Br\,} \\
{|\,} \\
{C{H_3} - C - C{H_3}} \\
{|\,} \\
{\,\,\,\,\,{C_2}{H_5}}
\end{array}$ $(c)$ $\begin{array}{*{20}{c}}
{{C_2}{H_5} - CH - C{H_2}Br} \\
{\,\,|\,\,\,\,\,\,\,\,\,\,\,} \\
{{C_2}{H_5}\,\,}
\end{array}$
$6 OH ^{-}+ Cl ^{-} \rightarrow ClO _{3}^{-}+3 H _{2} O +6 e ^{-}$
If only $60 \%$ of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce $10\, g$ of $KCIO _{3}$ using a current of $2\, A$ is..........
(Given : $F =96,500\, C\, mol ^{-1}$ molar mass of $\left. KClO _{3}=122\,gmol ^{-1}\right)$