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 one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: