- A$ScO$
- B${V_2}{O_3}$
- ✓$CuCN$
- D$C{r_2}{(S{O_4})_3}$

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$6 OH ^{-}+ Cl ^{-} \rightarrow ClO _{3}^{-}+3 H _{2} O +6 e ^{-}$
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 $\left. KClO _{3}=122\,gmol ^{-1}\right)$
$\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$
is $\mathrm{K}_{\mathrm{C}}=4.9 \times 10^{-2}$. The value of $\mathrm{K}_{\mathrm{C}}$ for the reaction given below is
$2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$ is