The solubility of $AgCl$ in its solution is $1.25 \times 10^{-5} \ mol \ dm^{-3}$. What is the solubility product of $AgCl$?

If the solubility of $MX_2$ in water is $0.0002 \ mol/L$,then its $K_{sp}$ is:

Concentration of $H_2SO_4$ and $Na_2SO_4$ in a solution is $1 \ M$ and $1.8 \times 10^{-2} \ M$,respectively. Molar solubility of $PbSO_4$ in the same solution is $X \times 10^{-Y} \ M$ (expressed in scientific notation). The value of $Y$ is. . . .
[Given: Solubility product of $PbSO_4$ $(K_{sp})=1.6 \times 10^{-8}$. For $H_2SO_4$,$K_{a1}$ is very large and $K_{a2}=1.2 \times 10^{-2}$]

$A$ $1.0 \ L$ of aqueous solution contains $1 \times 10^{-8} \ M \ NaBr$,$1 \times 10^{-8} \ M \ NaCl$ and $1 \times 10^{-8} \ M \ NaI$. To this solution,$1 \times 10^{-10} \ M$ aqueous $AgNO_3$ solution is added dropwise. The order of precipitation of $AgX$ $(X = Cl, Br, I)$ is:
$(K_{sp}(AgCl) = 1.8 \times 10^{-10}; K_{sp}(AgBr) = 5 \times 10^{-13}; K_{sp}(AgI) = 8.3 \times 10^{-17})$

Which of the following reagents can be used to separate $AgCl$ and $AgI$?

Let the solubility of an aqueous solution of $Mg(OH)_2$ be $x$,then its $K_{sp}$ is: