Cinnabar (HgS) and galena (PbS) are converted into their respective metals on roasting but zinc blende (ZnS) does not. Explain with reason.

Cinnabar (HgS) and galena (PbS) on roasting often yield their respective metals but zinc blende (ZnS) does not. Explain with reason.

Explain why is roasting of zinc blende carried out in controlled supply of air.

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

If galena (PbS) is contaminated with zinc blende (ZnS), how can these be separated by froth-floatation?

A cell is prepared by dipping a zinc rod in 1(M) ZnSo_4 solution. The standard electrode potential for Pb^(2+)/ Pb and Zn^(2+) /Zn electrodes are -0.126V and -0.763 V respectively. Write the half cell reactions and overall cell reaction.

A cell is prepared by dipping a zinc rod in 1(M) ZnSo_4 solution. The standard electrode potential for Pb^(2+)/ Pb and Zn^(2+) /Zn electrodes are -0.126V and -0.763 V respectively. How will you represent the cell?

Why does roasting of cinnabar HgS, produce the metal, mercury, rather than an oxide of mercury?

The crystal structure of Zinc blende (ZnS) consists of cubic close-packed (ccp) array of S^(2-) ions. If the radii of Zn^(2+)andS^(2-) ions are 0.74 and 1.84Å respectively, then which type of voids (tetrahedral or octahedral) formed in ccp array of S^(2-) ions are occupied by Zn^(2+) ions? What fraction of total number of this void remains unoccupied?