MCQ
Significant figures in $0.00051$ are
- A$5$
- B$3$
- ✓$2$
- D$4$
Thus, the given number has only $2$ significant figures.
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${N_2}(g)\, + \,{O_2}(g)\,\underset{{{k_2}}}{\overset{{{k_1}}}{\longleftrightarrow}}\,2NO(g)$
$C_0=Ce^{-2.1 \times 10^{-3}t}$ for the forward reaction and $C'_0=C'e^{-4.2 \times 10^{-4}t}$ for the backward reaction, hence $K_c$ for the above equilibrium is :-

( $Ba = 137,\,\,Cl = 35.5,\,\,S = 32,\,\,H = 1$ and $O = 16$ )