MCQ
In $LiAlH_4,$ metal $Al$ is present in
- ✓anionic part
- Bcationic part
- Cin both anionic and cationic part
- Dneither in cationic nor in anionic part
Therefore metal Al is present in anionic part.
Hence correct option is ( $\mathrm{A}$ ).
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$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to 2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$
Here $20\, mL$ of $0.1\, M\, KMnO_4$ is equivalent to
$(A)$ $\Delta U = q + p \Delta V$
$(B)$ $\Delta G =\Delta H - T \Delta S$
$(C)$ $\Delta S =\frac{ q _{ rev }}{ T }$
$(D)$ $\Delta H =\Delta U -\Delta nRT$
Choose the most appropriate answer from the options given below :