The correct statement among the following is/are

Solubility product of CuS is very less than $ZnS.$ When $ZnS$ is dissolved in $HCl$
$ZnS+2HCl\stackrel{ }{\to }ZnC{l}_2+H_{2}S$
Since the sulphide ions combine with $H^{+}$ ions forming $H_{2}S,$ which escapes out on heating, $ZnS$ is soluble . As the solubility product of $CuS$ is very less, the sulphide ions present in acid medium are sufficient to precipitate $CuS.$ So it is insoluble in $HCl$ but dissolves in $\text{ dil.}HN{O}_3$ due to oxidizing nature of nitric acid.
$3CuS+8HN{O}_3\stackrel{ }{\to }3Cu{\left(N{O}_3\right)}_2+4H_{2}O+2NO+3S$
When oxalate is added to $CaC{l}_2,$ the HCl formed in the reaction 
$CaC{l}_2+2H_{2}O\stackrel{ }{\to }Ca{\left(OH\right)}_2+2HCl$
Will convert the oxalate ion to oxalic acid, thus decreases the oxalate ion concentration. So calcium oxalate dissolves and cannot be precipitated. But if calcium acetate is used, acetic acid will be formed which is weaker than oxalic acid and will not effect the concentration of oxalate ion. So calcium oxalate will be precipitated.
Due to similar size of $B{a}^{2+}$ and $Cr{O}_4^{2-}$ ions $BaCr{O}_4$ will be precipitated on adding mineral acid. $Cr{O}_4^{2-}$ ion converts into larger $C{r}_2{O}_7^{2-}$ ion in acid medium. Due to large difference in the sizes of $B{a}^{2+}$ and $C{r}_2{O}_7^{2-}$ ions $BaC{r}_2{O}_7^{ }$ dissolves. Barium oxalate is soluble in hot dilute acetic acid but calcium oxalate is insoluble using ammonium oxalate.