When $2 \ mol$ of $HI$ is heated in a closed vessel at $440 \ ^\circ C$,$22\%$ of $HI$ dissociates until equilibrium is reached. The equilibrium constant $K_c$ for the reaction is ..........

Write the equilibrium constant $K$ for the reaction $CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}$

$HI$ was heated in a closed tube at $440\,^{\circ}C$ until equilibrium was obtained. At this temperature,$22\%$ of $HI$ was dissociated. The equilibrium constant for this dissociation will be:

State the law of chemical equilibrium for an equilibrium process.

For the reaction $CO_{(g)} + 2H_{2_{(g)}} \rightleftharpoons CH_3OH_{(g)}$,the equilibrium constant $K_c$ is $0.5$. If the concentrations of $CO$ and $H_2$ at equilibrium are $0.18 \ M$ and $0.22 \ M$ respectively,what is the concentration of $CH_3OH$?

Given the reaction between $2$ gases represented by $A_2$ and $B_2$ to give the compound $AB_{(g)}$:
$A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$
At equilibrium,the concentrations are $[A_2] = 3.0 \times 10^{-3} \, M$,$[B_2] = 4.2 \times 10^{-3} \, M$,and $[AB] = 2.8 \times 10^{-3} \, M$.
If the reaction takes place in a sealed vessel at $527^{\circ}C$,then the value of $K_c$ will be: