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- B

- C

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Reason : Hydrogen is present above $Cu$ in the reactivity series.
In the above reaction, $3.9\, g$ of benzene on nitration gives $4.92\, g$ of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is............. $\%$. (Round off to the Nearest Integer).
(Given atomic mass: $C : 12.0\, u , H : 1.0\, u$$O : 16.0\, u , N : 14.0\, u )$
$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq}),(1 \mathrm{M}) \| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})$
The fraction of total iron present as $\mathrm{Fe}^{3+}$ ion at the cell potential of $1.500\, \mathrm{~V}$ is $\mathrm{X} \times 10^{-2}$. The value of $x$ is $.....$ (Nearest integer).
$\left(\right.$ Given $\left.E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=0.77\, \mathrm{~V}, \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \,\mathrm{~V}\right)$
