Lithium aluminium hydride can be prepared from the reaction of:
\(LiCl \ and \ Al_2H_6\)
\(LiH \ and \ Al_2Cl_6\)
\(LiCl, Al \ and \ H_2\)
\(LiH \ and \ Al(OH)_3\)
Step 1: Reaction for Preparation of \(\text{LiAlH}_4\)
Lithium aluminium hydride \(\text{LiAlH}_4\) is prepared by the reaction of lithium hydride (\(\text{LiH}\)) with aluminium chloride \(\text{Al}_2\text{Cl}_6\). The reaction is as follows:
\[8\text{LiH} + \text{Al}_2\text{Cl}_6 \rightarrow 2\text{LiAlH}_4 + 6\text{LiCl}.\]
Conclusion: The correct reactants for preparing \(\text{LiAlH}_4\) are \(\text{LiH}\) and \(\text{Al}_2\text{Cl}_6\). Therefore, the correct answer is (2)
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: