When calomel reacts with $NH_4OH$,we get

$A$ is:
$\text{Salt } A \text{ (Colourless)} \xrightarrow{Cu} \text{Blue solution}$
$\text{Salt } A \text{ (Colourless)} \xrightarrow{HCl_{(aq)}} \text{White ppt}$
$\text{Salt } A \text{ (Colourless)} \xrightarrow{\text{Conc. } H_2SO_4} \text{Brown fumes}$

Which ion will not be precipitated by aqueous $(NH_4)_2S$?

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

In the group $III$ radicals,in place of $NH_4Cl$,which of the following can be used?

The $V^{-}A$ group precipitate was dissolved in $HNO_3$ and treated with an excess of $NH_4OH$. It gives a white precipitate because of: