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: $2 \times 10^{-2}$
An ideal massless spring \( S \) can be compressed \( 1 \) m by a force of \( 100 \) N in equilibrium. The same spring is placed at the bottom of a frictionless inclined plane inclined at \( 30^\circ \) to the horizontal. A \( 10 \) kg block \( M \) is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring by \( 2 \) m. If \( g = 10 \) m/s\( ^2 \), what is the speed of the mass just before it touches the spring?
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32