If $Ag^{+} + NH_3 \rightleftharpoons [Ag(NH_3)]^+$; $K_1 = 1.6 \times 10^3$ and $[Ag(NH_3)]^+ + NH_3 \rightleftharpoons [Ag(NH_3)_2]^+$; $K_2 = 6.8 \times 10^3$. Then the formation constant of $[Ag(NH_3)_2]^+$ is:

One mole of a compound $AB$ reacts with one mole of a compound $CD$ according to the equation $AB + CD \rightleftharpoons AD + CB$. When equilibrium had been established,it was found that $\frac{3}{4} \ mol$ each of reactant $AB$ and $CD$ had been converted to $AD$ and $CB$. There is no change in volume. The equilibrium constant for the reaction is:

If the equilibrium constant for the reaction ${N_2}_{(g)} + {O_2}_{(g)} \rightleftharpoons 2NO_{(g)}$ is $K$,then what is the equilibrium constant for the reaction $\frac{1}{2}{N_2}_{(g)} + \frac{1}{2}{O_2}_{(g)} \rightleftharpoons NO_{(g)}$?

What is $K_{c}$ for the following equilibrium when the equilibrium concentration of each substance is: $[SO_{2}] = 0.60 \, M, [O_{2}] = 0.82 \, M$ and $[SO_{3}] = 1.90 \, M?$
$2SO_{2(g)} + O_{2(g)} \longleftrightarrow 2SO_{3(g)}$

For the system $A_{(g)} + 2B_{(g)} \rightleftharpoons C_{(g)}$,the equilibrium concentrations are $[A] = 0.06 \ mol/L$,$[B] = 0.12 \ mol/L$,and $[C] = 0.216 \ mol/L$. The $K_{eq}$ for the reaction is:

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. If the equilibrium constant $(K_C)$ for the above reaction at $500 \ K$ is $1.79$ and the equilibrium concentrations of $PCl_5$ and $PCl_3$ are $1.41 \ M$ and $1.59 \ M$,respectively,then the concentration of $Cl_2$ is approximately: (in $M$)