Chemistry Test - 7

Due to delocalisation of electrons of the N-atom over benzene ring in aniline (3) and involvement of these electrons in aromatic sextet formation in pyrrole (1), both these amines are weaker bases than pyridine (2), in which such a delocalisation does not exist since the lone pair of electrons is present in an sp² orbital, which lies in the plane of the ring.
Hence, 2 is more basic than 1 and 3. Further, 1 (pyrrole) is also weakly acidic at the N-H position, with a pKa of 17.5.

Hence option B is the correct answer.

For first order reaction,

\(A \longrightarrow B\)

Rate \(=\mathrm{k} \times[\mathrm{A}]\)

Given, rate \(=2.0 \times 10^{-5} \mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}\)

\([\mathrm{A}]=\) Conc. of \(\mathrm{A}=0.01 \mathrm{M}\)

\(\mathrm{So}, 2.0 \times 10^{-5}=\mathrm{k} \times 0.01\)

\(\mathrm{k}=\frac{2.0 \times 10^{-5}}{0.01} \mathrm{s}^{-1}\)

\(=2.0 \times 10^{-3} \mathrm{s}^{-1}\)

For first order reaction,

\(\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}}=\frac{0.693}{2.0 \times 10^{-3}}\)

\(=346.5=347 \mathrm{s}\)

Hence option D is the correct answer.

At 1500 ^oC, Al can be used to reduce Fe₂O₃, but not MgO.

MgO can be reduced by Al above 1750 ^oC, where the MgO curve intersects the Al₂O₃ curve.

Hence option A is the correct answer.

Before removing face centred atoms:

Number of 'A' atoms = (8 x 1/8) + (6 x 1/2) = 1 + 3 = 4 [There are 8 'A' atoms at the corners and 6 'A' atoms at the face centres].

Number of 'B' atoms = (12 x 1/4) + (1 x 1) = 3 + 1 = 4 [There are 12 'B' atoms at the edges and 1 'B' atom at the centre of the unit cell].

After removing two face centred 'A' atoms along one of the axes:

Number of 'A' atoms = (8 x 1/8) + (4 x 1/2) = 1 + 2 = 3 [There are now only 4 'A' atoms at the face centres]

Number of 'B' atoms = (12 x 1/4) + (1 x 1) = 3 + 1 = 4

Therefore, the stoichiometry after removing these atoms is A₃B₄.

Hence option D is the correct answer.