Your Answer : (Not Answered)

Correct Answer : D

The pH is close to 1.70.

\[ \begin{aligned} n(\text{H}^{+}) &= 0.800 \times 0.1 = 0.080\ \text{mol} \\ n(\text{OH}^{-}) &= 0.200 \times 0.3 = 0.060\ \text{mol} \\ \text{excess H}^{+} &= 0.020\ \text{mol in } 1.0\ \text{L} \end{aligned} \]

\[ \text{pH} = -\log(0.020) = 1.70 \]

The original 0.1 M HCl had pH 1.00, so 1.10 is too acidic once 0.060 mol of the acid has been neutralised. 13.3 would mean excess base, and 6.90 near-exact neutralisation, but the acid is in excess by 0.020 mol in the combined 1 L.

Your Answer : (Not Answered)

Correct Answer : A

The pH is 5.

Ammonium sulfate is the salt of a weak base and a strong acid, so its solution is acidic through hydrolysis of NH₄⁺. 0.05 M (NH₄)₂SO₄ gives 0.1 M NH₄⁺. Taking K_b = 10⁻⁵ for ammonia (printed '(NH)'):

\[ \begin{aligned} K_{a} &= \frac{K_{w}}{K_{b}} = \frac{10^{-14}}{10^{-5}} = 10^{-9} \\ [\mathrm{H^{+}}] &= \sqrt{K_{a}C} = \sqrt{10^{-9}\times 0.1} = 10^{-5} \\ \text{pH} &= 5 \end{aligned} \]

Forgetting that each formula unit gives two NH₄⁺ and using 0.05 M gives pH 5.15, still closest to 5. pH 6 would need a far weaker acid, and pH 3 or 4 a much stronger one than NH₄⁺.

Your Answer : (Not Answered)

Correct Answer : A

The sketch that shows the primary salt effect is A, in which the barrier is lower after potash alum is added.

The reaction \(\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \to 2\text{SO}_4^{2-} + \text{I}_2\) is between ions of like charge. The activated complex carries a charge of 3− (2− plus 1−), more than either reactant, so the ionic atmosphere created by added electrolyte stabilises it more than it stabilises the reactants. The free energy of the complex falls relative to the reactants, the activation barrier shrinks and the rate rises with ionic strength.

In B and D the barrier grows, which would be the effect for ions of opposite charge in the reverse sense, and in C it is unchanged, which is the case of no salt effect.

Your Answer : (Not Answered)

Correct Answer : B

The correct statement is that K_M has the unit mol dm⁻³.

In the Michaelis-Menten equation K_M is added to the substrate concentration [s] in the denominator, so it must have the same unit as [s], a concentration. It equals the substrate concentration at which the rate is half its maximum value.

Setting K_M = [s] shows why the third option fails:

\[ \nu = \frac{V_{max}[s]}{[s] + [s]} = \frac{V_{max}}{2} \]

so the rate is half of V_max, not equal to it. At low substrate concentration K_M >> [s], and the equation reduces to ν = (V_max/K_M)[s]: the rate varies linearly with [s] (first order), not exponentially. dm³ mol⁻¹ time⁻¹ is the unit of a second-order rate constant, not of K_M.

Your Answer : (Not Answered)

Correct Answer : A

The ratio is 1 : 4.

For a one-electron atom or ion of nuclear charge Z, the energy of level n is

\[ E_{n} = -13.6\,\frac{Z^{2}}{n^{2}}\ \text{eV} \]

Hydrogen ground state: Z = 1, n = 1. Be³⁺ is one-electron beryllium, Z = 4, and its first excited state is n = 2:

\[ \begin{aligned} E_{H} &= -13.6\ \text{eV} \\ E_{Be^{3+}} &= -13.6 \times \tfrac{16}{4} = -54.4\ \text{eV} \end{aligned} \]

The ratio is 13.6 : 54.4 = 1 : 4. 1 : 16 forgets the n² = 4 in the denominator (it is the ground-state ratio), and 1 : 8 and 2 : 9 come from using Z or n instead of their squares.

Your Answer : (Not Answered)

Correct Answer : C

The formula is XY.

An atom at a corner of a cubic unit cell is shared by eight cells, and an atom at the body centre belongs wholly to one cell:

\[ \begin{aligned} \mathrm{X} &= 8 \times \tfrac{1}{8} = 1 \\ \mathrm{Y} &= 1 \times 1 = 1 \end{aligned} \]

One X and one Y per cell gives XY, the arrangement of the cesium chloride structure.

X₂Y and X₄Y come from counting corner atoms as more than one-eighth each, and XY₄ reverses the roles of the two atoms. The same counting with face centres (each shared by two cells) is what gives XY₃ in a face-centred arrangement.

Your Answer : (Not Answered)

Correct Answer : B

The weight of the drop equals 2πrγ.

The drop hangs from the tip of the capillary, held by surface tension acting along the circumference of the tip. Surface tension γ is a force per unit length, and the length of the contact line is the circumference 2πr. At the moment of detachment the upward force just balances the weight (Tate's law):

\[ mg = 2\pi r\gamma \]

This is the basis of the drop-weight (stalagmometer) method. πr²γ uses an area where a length is needed, and the other two forms are not dimensionally a force.

Your Answer : (Not Answered)

Correct Answer : C

The shift reagents are complexes of Eu³⁺.

Lanthanide shift reagents such as Eu(dpm)₃ and Eu(fod)₃ are β-diketonate complexes of europium(III). Eu³⁺ is paramagnetic but has a very short electron-spin relaxation time, so when it coordinates to a Lewis-basic site in the substrate it shifts nearby proton signals (by the pseudocontact mechanism) without broadening them badly, spreading out a crowded spectrum.

La³⁺ (4f⁰) is diamagnetic and induces no paramagnetic shift. Ce³⁺ and Ho³⁺ complexes are paramagnetic, but europium (and praseodymium, which shifts upfield) are the ions used routinely.

Your Answer : (Not Answered)

Correct Answer : B

The Lennard-Jones function is \(E_{LJ} = \sum (A_{ik}/r_{ik}^{12} - C_{ik}/r_{ik}^{6})\).

The Lennard-Jones 12-6 potential has two terms. The positive r⁻¹² term is the steep short-range repulsion that arises when electron clouds overlap. The negative r⁻⁶ term is the London dispersion attraction, which falls off as the sixth power of the distance. In force fields the sum runs over all nonbonded atom pairs, with A_ik and C_ik constants for each pair.

Option A puts the powers on the wrong terms, making the attraction short-range. Options C and D add the two terms, which would make the interaction purely repulsive and leave no van der Waals minimum.

Your Answer : (Not Answered)

Correct Answer : A

The incorrect statement is (A): the two compounds do not show equal quadrupole splitting.

Quadrupole splitting measures the electric field gradient at the iron nucleus, and in both compounds the I = 3/2 excited state splits into two levels, so each gives a doublet. FeCl₃.6H₂O contains high-spin Fe(III), d⁵, whose half-filled t_2g³e_g² shell is spherical, so the valence electrons add no field gradient and only the lattice (ligand) contribution acts: the splitting is small. FeCl₂.4H₂O contains high-spin Fe(II), d⁶, whose extra electron sits in a non-spherical d orbital and adds a large valence contribution, so its splitting is much larger. The splittings are therefore unequal, which makes (B) true, and (D) states exactly this valence versus lattice difference.

Statement (C) is true in substance: both salts are high-spin, and the pair is one Fe(III) and one Fe(II) complex. As printed, 'respectively' pairs Fe(II) with the first formula, FeCl₃.6H₂O, which is the wrong way round; the question evidently means only the pair of oxidation states, so (A), false on any reading, is the answer. The splitting values themselves are not quoted here; the argument rests on the d⁵ versus d⁶ electronic structure.

Your Answer : (Not Answered)

Correct Answer : D

The rate law is Rate = k[I⁻][OCl⁻].

Rows 1 and 2 keep [OCl⁻] at 0.05 M while [I⁻] doubles from 0.1 to 0.2 M, and the rate doubles from 3.05 to 6.10 × 10⁻⁴ M s⁻¹, so the reaction is first order in I⁻. Rows 1 and 3 then fix the order in OCl⁻:

\[ \begin{aligned} k &= \frac{3.05 \times 10^{-4}}{0.1 \times 0.05} = 0.061 \\ k &= \frac{1.83 \times 10^{-4}}{0.3 \times 0.01} = 0.061 \end{aligned} \]

The same constant from both rows confirms first order in OCl⁻, so the reaction is second order overall. A rate independent of I⁻ (B) or second order in I⁻ (A, C) contradicts the doubling between rows 1 and 2. The fourth row, as printed, gives a somewhat higher constant (0.077), but the first three rows fix the rate law.

Your Answer : (Not Answered)

Correct Answer : C

The empirical formula is M₂O₃.

Divide each percentage by the atomic mass to get the mole ratio:

\[ \begin{aligned} \mathrm{M} &: \frac{54}{27} = 2 \\ \mathrm{O} &: \frac{46}{16} = 2.875 \end{aligned} \]

The ratio M : O is 2 : 2.875, or 1 : 1.44, which is closest to 1 : 1.5, that is 2 : 3. (A metal of atomic mass 27 is aluminium, and Al₂O₃ contains about 53 % metal, matching the data.)

M₂O and MO₂ give ratios of 1 : 0.5 and 1 : 2, and M₃O₂ and M₂O₅ are further from 1 : 1.44.

Your Answer : (Not Answered)

Correct Answer : B

The mass of ammonia formed is 3.4 g.

\(\text{N}_{2} + 3\text{H}_{2} \rightarrow 2\text{NH}_{3}\). The moles are \(\frac{2.8}{28} = 0.1\) of nitrogen and \(\frac{0.8}{2} = 0.4\) of hydrogen.

0.1 mol of nitrogen needs only 0.3 mol of hydrogen, so nitrogen is the limiting reagent.

\[ \begin{aligned} n(\text{NH}_{3}) &= 2 \times 0.1 = 0.2\ \text{mol} \\ m &= 0.2 \times 17 = 3.4\ \text{g} \end{aligned} \]

3.6 g would be the total mass of the reactants, which is more than can react because 0.1 mol of hydrogen is left over.

Your Answer : (Not Answered)

Correct Answer : A

The amount of methane needed is 0.25 mol.

Combustion of methane gives one CO₂ per CH₄: CH₄ + 2O₂ → CO₂ + 2H₂O.

\[ \begin{aligned} n(\mathrm{CO_{2}}) &= \frac{11}{44} = 0.25\ \text{mol} \\ &= n(\mathrm{CH_{4}}) \end{aligned} \]

0.5 mol would give 22 g of CO₂, and the larger values more still. The answer equals the moles of CO₂ because the mole ratio is 1 : 1.

Your Answer : (Not Answered)

Correct Answer : A

A 4s orbital has three nodal surfaces.

The total number of nodes in an orbital is \(n - 1\), made up of \(l\) angular nodes and \(n - l - 1\) radial (spherical) nodes. For 4s, \(n = 4\) and \(l = 0\):

\[ \begin{aligned} \text{radial nodes} &= n - l - 1 = 4 - 0 - 1 = 3 \\ \text{angular nodes} &= l = 0 \end{aligned} \]

So all three nodal surfaces are spheres. Four is the value of \(n\), not the node count, and one or two would belong to the 2s and 3s orbitals.

Your Answer : (Not Answered)

Correct Answer : C

The kinetic energy is 5 x 10^-22 J.

From \(\lambda = \frac{h}{mv}\) and \(K = \frac{1}{2}mv^{2}\), \(K = \frac{h^{2}}{2m\lambda^{2}}\).

\[ \begin{aligned} v &= \frac{6.3 \times 10^{-34}}{10^{-31} \times 63 \times 10^{-9}} \\ &= 10^{5}\ \text{m s}^{-1} \\ K &= \tfrac{1}{2} \times 10^{-31} \times (10^{5})^{2} \\ &= 5 \times 10^{-22}\ \text{J} \end{aligned} \]

The option \(1 \times 10^{-21}\) J forgets the \(\frac{1}{2}\) in kinetic energy, and \(5 \times 10^{-21}\) J slips a power of ten.

Your Answer : (Not Answered)

Correct Answer : B

The minimum value of \(|W|\) is 600R.

Against a constant external pressure, the work done on the gas is \(P_{ext}(V_{i} - V_{f})\), and at 600 K each volume is \(V = \frac{nRT}{p}\) with \(nRT = 0.5 \times 600 \times R = 300R\). The two steps take the gas from 2 bar to \(P\) bar, then from \(P\) to 8 bar:

\[ \begin{aligned} W &= P\left(\tfrac{nRT}{2} - \tfrac{nRT}{P}\right) + 8\left(\tfrac{nRT}{P} - \tfrac{nRT}{8}\right) \\ &= nRT\left(\tfrac{P}{2} + \tfrac{8}{P} - 2\right) \end{aligned} \]

The bracket is least when \(\frac{P}{2} = \frac{8}{P}\), that is \(P = 4\) bar (the geometric mean of 2 and 8):

\[ W_{min} = 300R\,(2 + 2 - 2) = 600R \]

900R is the work for a single step straight against 8 bar, \(300R(4 - 1)\). 630R is what the two-step path gives at \(P = 5\) or \(P = 3.2\), not at the best \(P\). 207R is below even the reversible work \(300R\ln 4 \approx 416R\), the least any compression from 2 to 8 bar can need, so it is impossible.

Your Answer : (Not Answered)

Correct Answer : B

In a dry cell, chemical energy is converted into electrical energy.

A dry (Leclanche type) cell has a zinc container as the negative electrode and a carbon rod surrounded by manganese dioxide as the positive electrode, with an ammonium chloride paste as electrolyte. The redox reactions at the electrodes release chemical energy as a flow of electrons.

No mechanical, kinetic or potential (gravitational) energy is involved. Mechanical energy is converted into electrical energy in a generator, not a cell.

Your Answer : (Not Answered)

Correct Answer : E

The pre-exponential factor is called the frequency factor.

In the Arrhenius equation \(k = A e^{-E_{a}/RT}\), \(A\) is the pre-exponential or frequency factor. In collision theory it reflects how often molecules collide with suitable orientation; it has the same units as the rate constant.

Activation energy is \(E_{a}\), in the exponent. The probability (steric) factor \(P\) is only one part of \(A\) in collision theory, as is the collision frequency \(Z\). The reaction coordinate is the horizontal axis of an energy profile, not a term in the equation.

Your Answer : (Not Answered)

Correct Answer : D

The incorrect statement is The dipole moment of NF₃ is much greater than that in NH₃.

It is the other way round: NH₃ is about 1.47 D and NF₃ only about 0.23 D. In NH₃ the N-H bond dipoles point toward nitrogen, the same way as the lone-pair moment, so they add. In NF₃ the N-F bond dipoles point away from nitrogen, against the lone-pair moment, so they largely cancel.

The other statements are true: BF₃ is trigonal planar and non-polar; CO and NO⁺ are isoelectronic (14 electrons) with bond order 3; resonance makes the two O-O bonds in ozone equal; and the carbonate ion has three equivalent canonical forms.