At $20^{\circ}C$,the concentration of $Ag^{+}$ ions in a saturated solution of $Ag_2CrO_4$ is $1.5 \times 10^{-4} \ mol/L$. The solubility product $(K_{sp})$ of $Ag_2CrO_4$ at $20^{\circ}C$ is:

$5.6 \ L$ of helium gas at $STP$ is adiabatically compressed to $0.7 \ L$. Taking the initial temperature to be $T_1$,the work done in the process is

$\sqrt{12-\sqrt{68+48 \sqrt{2}}}$ is equal to :

One mole of $A_{(g)}$ is heated to $200^{\circ} C$ in a one litre closed flask,till the following equilibrium is reached.
$A_{(g)} \rightleftharpoons B_{(g)}$
The rate of forward reaction at equilibrium is $0.02 \ mol \ L^{-1} \ min^{-1}$. What is the rate (in $mol \ L^{-1} \ min^{-1}$) of the backward reaction at equilibrium?

The $PV$ diagram shows four different possible reversible processes performed on a monatomic ideal gas. Process $A$ is isobaric (constant pressure). Process $B$ is isothermal (constant temperature). Process $C$ is adiabatic. Process $D$ is isochoric (constant volume). For which process(es) does the temperature of the gas decrease?

If $|x| < 1$,then the sum of the series $1 + 2x + 3x^2 + 4x^3 + \dots \infty$ will be