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.
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]