MCQ
$NaOH$ is a strong base because
- ✓It gives $O{H^ - }$ion
- BIt can be oxidised
- CIt can be easily ionised
- DBoth $(a)$ and $(c)$
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$\left[\right.$ Given $, E_{C u^{2+} / C u}^{o}=0.34\, V , E _{ NO _{3}^{-} / NO_2 }^{\circ}=0.96\, V$ (Rounded-off to the nearest integer) $E _{ NO _{3} / NO _{2}}^{\circ}=0.79 \,V$ and at $298 \,K$ $\left.\frac{ RT }{ F }(2.303)=0.059\right]$
$X_2O_4(g) \to 2XO_2(g)$
$\Delta U = 2.1\, kcal, \Delta S = 20\, cal\, K^{-1}$ at $300\, K$.
Hence $\Delta G$ is ....$kcal$