To convert \(K_p\) to \(K_c\):
\[ K_p = K_c \cdot (RT)^{\Delta n_g} \]
For the reaction \( \text{N}_2\text{O}_4(g) \leftrightharpoons 2\text{NO}_2(g) \),
\[ \Delta n_g = 2 - 1 = 1 \]
Therefore:
\[ K_c = \frac{K_p}{RT} = \frac{0.492}{0.082 \times 300} = 2 \times 10^{-2} \]
So, the correct answer is: $2 \times 10^{-2}$
x mg of Mg(OH)$_2$ (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of x is ____ mg. (Nearest integer) (Given: Mg(OH)$_2$ is assumed to dissociate completely in H$_2$O)
Match List-I with List-II: List-I