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$NO(g) + Br_2 (g) \rightleftharpoons NOBr_2 (g)$
$NOBr_2(g)+ NO(g) \longrightarrow 2NOBr(g)$
If the second step is the rate determining step, the order of the reaction with respect to $NO(g)$ is
$Zn_{(s)} + Ag_2O_{(s)} + H_2O_{(l)} \rightleftharpoons $$2Ag_{(s)} + Zn^{2+}_{(aq)}+ 2OH^-_{(aq)}$
If half cell potentials are
$Zn^{2+}_{(aq)} + 2e^- \rightarrow Zn_{(s)}\,;\,\, E^o = - 0.76\, V$
$Ag_2O_{(s)} + H_2O_{(l)} + 2e^- \rightarrow 2Ag_{(s)} + 2OH^-_{(aq)}\,,$$E^o = 0.34\, V$
The cell potential will be ........... $V.$
$9\, mol$ $O_2$ and $14\, mol$ $N_2$ here allowed to react. When $3\, mol$ $O_2$ remains unreacted, till then how many moles of $N_2O_3$ would have been produced?
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\mathop C\limits_7 {H_3} - \mathop C\limits_6 - \mathop C\limits_5 H = \mathop C\limits_4 H - \mathop C\limits_3 H - \mathop C\limits_2 \equiv \mathop C\limits_1 H} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
is in the following sequence