MCQ
Conjugate base of $HPO_4^{2 - }$ is
- ✓$PO_4^{3 - }$
- B${H_2}PO_4^ - $
- C${H_3}P{O_4}$
- D${H_4}P{O_3}$
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${N_2}\left( g \right) + {O_2}\left( g \right)\underset{{{k_2}}}{\overset{{{k_1}}}{\longleftrightarrow}}2NO\left( g \right)$
$C_0 = Ce^{-2.1×10^{-3}\ t}$ for the forward reaction and
$C_0'= C'e^{-4.2×10^{-4}\ t}$ for the backward reaction, hence $K_c$ for the above equilibrium is
