$K_C$ for the reaction, $A_2(g) \stackrel{T(\mathrm{~K})}{\rightleftharpoons} B_2(g)$ is 99.0 . In a 1 L closed flask two moles of $B_2(g)$ is heated to $T(\mathrm{~K})$. What is the concentration of $B_2(g)\left(\right.$ in $\left.\mathrm{mol} \mathrm{L}^{-1}\right)$ at equilibrium?

$K_{\mathrm{c}}$ for the following reaction is 99.0

$$ A_2(g) \stackrel{T(K)}{\rightleftharpoons} B_2(g) $$

In a one litre flask, 2 moles of $A_2$ was heated to $T(\mathrm{~K})$ and the above equilibrium is reached. The concentration at equilibrium of $A_2$ and $B_2$ are $C_1\left(A_2\right)$ and $C_2\left(B_2\right)$ respectively. Now, one mole of $A_2$ was added to flask and heated to $T(\mathrm{~K})$ to established the equilibrium again. The concentration of $A_2$ and $B_2$ are $C_3\left(A_2\right)$ and $C_4\left(B_2\right)$ respectively. what is the value of $C_3\left(A_2\right)$ in $\mathrm{mol} \mathrm{L}^{-1}$ ?

At $T(\mathrm{~K}), K_c$ for the reaction $A_2(g) \rightleftharpoons B_2(g)$ is 99.0 . Two moles of $A_2(s)$ was heated to $T(\mathrm{~K})$ in a 1 L . closed flask to reach the above equilibrium. What are the concentrations (in mol $\mathrm{L}^{-1}$ ) of $A_2(g)$ and $B_2(g)$ respectively at equilibrium?

At $27^{\circ} \mathrm{C}$, the degree of dissociation of weak acid ( $\mathrm{H} A$ ) in its 0.5 M aqueous solution is $1 \%$. Its $K_e$, value is approximately