Zinc reacts with dilute `H_(2)SO_(4)` to give `H_(2)`. It also reacts with conc. `H_(2)SO_(4)` to give `SO_(2)` . How will you account for this difference ?
Dilute `H_(2)SO_(4)` behaves as a proton acid and zinc metal displaces hydrogen from the acid.
`Zn+H_(2)SO_(4)("dil")to ZnSO_(4)+H_(2)`
Concentrated `H_(2)O_(4)` behaves as a strong oxidising agent and itself is reduced to `SO_(2)`.
`Zn +2H_(2)SO_(4) ("conc.")to ZnSO_(4)+SO_(2)2H_(2)O`.
Correct Answer - C
`NaNO_2` gives `NO_2` (brown vapours) both with dil. as well as conc. `H_2SO_4`. However NaBr gives `Br_2` (brown vapours) only with conc. `H_2SO_4`. This is because halides...
Correct Answer - B
`underset(1mol)(SO_(2)Cl_(2))+underset(2mol)(2H_(2)O)tounderset(1mol)(H_(2)SO_(4))+underset(2mol)(2HCl)`
`underset(1mol)(H_(2)SO_(4))+underset(2mol)(2NaOH)toNa_(2)SO_(4)+H_(2)O`
`underset(1mol)(HCl)+underset(1mol)(NaOH)toNaCl+H_(2)xx2`
`:.` 1 mol of `SO_(2)Cl_(2)` in aq. solution requires
4 mol of NaOH or 2 mol of `Ca(OH)_(2)`
Correct Answer - C
Since `HNO_(3)` is also a strong oxidising agent, therefore, aniline forms tarry oxidation products along with some oxidation products.
i. This statement is True.
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