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Cu(2)S is subjected to partial roasting...

`Cu_(2)S` is subjected to partial roasting

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Both Cus and ZnS are precipitated if H_(2)S gas is passed through an alkaline solution of Cu^(2+) and Zn^(2+) . Explain.

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Roasting converts Cu_(2)S, HgS or PbS to the corresponding metal. However, roasting cannot convert ZnS to Zn - Why?

Cu_(2)S is black and Cu_(2)O is red through Cu^(+) has no unpaired electrons and is colourless.

Cu_(2)O is reduced easily by C but CaO is not - explain.

When H_(2)S gas is passed through an acidified solution of Cu^(2+) and Zn^(2+) , only CuS is precipitated-why?