Sulphide ores are converted to oxides before reduction. This process is called roasting. For example:
\[ 2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2 \]
This statement is true.
Oxide ores are generally easier to reduce than sulphide ores. This can be explained by the following:
The reaction for sulphide reduction is:
\[ 2\text{MS} + \text{C} \rightarrow 2\text{M} + \text{CS}_2 \]
Since \( \text{CO}_2 \) is more volatile compared to \( \text{CS}_2 \), oxide reduction is favored. This statement is also true.
Final Answer: Both Statement I and Statement II are correct.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)