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)
Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.
Match List I with List II:
Choose the correct answer from the options given below:
For the AC circuit shown in the figure, $ R = 100 \, \text{k}\Omega $ and $ C = 100 \, \text{pF} $, and the phase difference between $ V_{\text{in}} $ and $ (V_B - V_A) $ is 90°. The input signal frequency is $ 10^x $ rad/sec, where $ x $ is:
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
A point particle of charge \( Q \) is located at \( P \) along the axis of an electric dipole 1 at a distance \( r \) as shown in the figure. The point \( P \) is also on the equatorial plane of a second electric dipole 2 at a distance \( r \). The dipoles are made of opposite charge \( q \) separated by a distance \( 2a \). For the charge particle at \( P \) not to experience any net force, which of the following correctly describes the situation?
