When $H_2S$ is passed through a mixture containing $Cu^{2+}$,$Ni^{2+}$,and $Zn^{2+}$ in an acidic solution,which ion will precipitate?

In which of the following is $NH_{3}$ not used?

What is formed when calomel $(Hg_2Cl_2)$ reacts with $NH_4OH$?

Conc. $HNO_3$ is added before proceeding to test for group $III$ members. This is to

$Pb^{2+}$,$Cu^{2+}$,$Zn^{2+}$,and $Ni^{2+}$ ions are present in a given acidic solution. On passing hydrogen sulphide gas through this solution,the precipitate formed will contain:

$A$ mixture of two salts is used to prepare a solution $S$,which gives the following results:
White precipitate$(s)$ $\xleftarrow{\text{Dilute } NaOH(aq.) \text{ at Room temperature}}$ $S$ (aq. solution of the salts) $\xrightarrow{\text{Dilute } HCl(aq.) \text{ at Room temperature}}$ White precipitate$(s)$ only
The correct option$(s)$ for the salt mixture is(are):
$A$. $Pb(NO_3)_2$ and $Zn(NO_3)_2$
$B$. $Pb(NO_3)_2$ and $Bi(NO_3)_3$
$C$. $AgNO_3$ and $Bi(NO_3)_3$
$D$. $Pb(NO_3)_2$ and $Hg(NO_3)_2$