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.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32