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The Gibbs free energy change for the above reaction at $298\, K$ is $x \times 10^{-1} \,k\,J\, mol ^{-1}$;
The value of $x$ is ..... [nearest integer]$\left [\text { Given : } E _{ Cu ^{2} / / Cu }=0.34\, V ; E _{ Sn ^{2} / Sn }^{\ominus}=-0.14 \,V ; F=96500\, C\, mol ^{-1}\right]$
$\mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{Conc} .) \rightarrow(\mathrm{A})+$ Side products
$(\mathrm{A})+\mathrm{NaOH} \rightarrow(\mathrm{B})+$ side product
$(\mathrm{B})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{dilute})+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow(\mathrm{C})+$ Side product
The sum of the total number of atoms in one molecule each of $(A), (B)$ and $(C)$ is