A-Level Chemistry Practice Test 10

Correct: (B)

(20 × 90.5 + 21 × 0.27 + 22 × 9.25) / 100 = (1810 + 5.67 + 203.5) / 100 = 20.2.

Correct: (C)

Potassium's outer electron is in the fourth shell; the greater distance and extra shielding outweigh the increased nuclear charge, so it is removed more easily.

Correct: (A)

v = 1.20 / 4.40 × 10^-5 = 2.73 × 10^4 m s^-1. KE = 1/2 mv^2 = 0.5 × (39 × 1.66 × 10^-27) × (2.73 × 10^4)^2 = 2.41 × 10^-17 J.

Correct: (D)

Sulfur dioxide is an acidic oxide. Na2O and MgO are basic oxides, and Al2O3 is amphoteric.

Correct: (B)

The full configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1, so the number of s electrons is 2 + 2 + 2 + 1 = 7.

Correct: (A)

V = nRT / p = (0.500 × 8.31 × 300) / (2.00 × 10^5) = 6.23 × 10^-3 m^3 = 6.23 dm^3.

Correct: (D)

Moles of HCl = 0.100 × 0.0225 = 2.25 × 10^-3. Ca(OH)2 reacts with 2HCl, so moles of Ca(OH)2 = 1.125 × 10^-3. Concentration = 1.125 × 10^-3 / 0.0250 = 0.0450 mol dm^-3.

Correct: (C)

Moles of benzene = 5.30 / 78.0 = 0.0679. Maximum mass of nitrobenzene = 0.0679 × 123 = 8.35 g. Percentage yield = (4.92 / 8.35) × 100 = 58.9%.

Correct: (B)

Atom economy = 46.0 / (109 + 56.1) × 100 = 46.0 / 165.1 × 100 = 27.9%.

Correct: (A)

Carbon: 0.880 × (12/44) = 0.240 g = 0.0200 mol. Hydrogen: 0.540 × (2/18) = 0.0600 g = 0.0600 mol. Oxygen: 0.460 – 0.240 – 0.0600 = 0.160 g = 0.0100 mol. Ratio 2 : 6 : 1, so the empirical formula is C2H6O.

Correct: (D)

Moles of MgCl2 = 0.200 × 0.250 = 0.0500. Each MgCl2 gives two Cl^- ions, so moles of Cl^- = 2 × 0.0500 = 0.100 mol.

Correct: (C)

Oxygen has two bonding pairs and two lone pairs. The lone pairs repel strongly, pushing the O-H bonds together and reducing the angle to 104.5°.

Correct: (A)

All four have similar Mr values (44-46). Ethanol forms hydrogen bonds between its molecules, which need much more energy to break than the van der Waals forces and dipole forces in the others.

Correct: (B)

Lattice enthalpy becomes more exothermic as the charges on the ions increase and their radii decrease; MgO has 2+ and 2- ions that are both small, so the attraction in the lattice is very strong.

Correct: (D)

In NH4^+, the nitrogen atom donates its lone pair to an H^+ ion, forming a dative bond. (Once formed, all four N-H bonds are identical.)

Correct: (C)

The C-Cl bonds in CH2Cl2 are polar and the molecule is not symmetrical, so the dipoles do not cancel. In BF3, CO2 and CCl4 the bond dipoles cancel by symmetry.

Correct: (A)

Diamond is a giant covalent structure in which every atom is held by strong covalent bonds throughout the lattice; an enormous amount of energy is needed to break them. MgO melts at a lower temperature despite its strong ionic lattice.

Correct: (B)

q = 500 × 4.18 × 11.5 = 24,035 J. Moles of ethanol = 1.00 / 46.0 = 0.0217. ΔHc = -24,035 / 0.0217 = -1106 kJ mol^-1 (less exothermic than the data-book value because of heat losses).

Correct: (D)

ΔH = sum of ΔHf(products) – sum of ΔHf(reactants) = [2 × (-286)] – [+51] = -572 – 51 = -623 kJ mol^-1.

Correct: (C)

ΔS = 2(193) – [192 + 3(131)] = 386 – 585 = -199 J K^-1 mol^-1. The value is negative because four moles of gas become two moles of gas.

Correct: (B)

Feasibility requires ΔG = ΔH – TΔS to be negative. At ΔG = 0, T = 95,000 / 210 = 452 K. Above 452 K, the -TΔS term makes ΔG positive.

Correct: (A)

Bonds broken: 436 + 193 = 629 kJ. Bonds formed: 2 × 366 = 732 kJ. ΔH = 629 – 732 = -103 kJ mol^-1.

Correct: (D)

When the lattice breaks up and the ions disperse through the water, entropy rises. The TΔS term outweighs the small positive ΔH, so ΔG is negative and dissolving is feasible.

Correct: (C)

Doubling [A] leaves the rate unchanged (zero order in A); doubling [B] quadruples the rate (second order in B).

Correct: (D)

k = rate / ([A]^2[B]) = 8.00 × 10^-4 / (0.100^2 × 0.100) = 8.00 × 10^-4 / 1.00 × 10^-3 = 0.800 mol^-2 dm^6 s^-1.

Correct: (A)

The Arrhenius plot has gradient = -Ea / R, so Ea = 6500 × 8.31 = 54,015 J mol^-1 = 54.0 kJ mol^-1.

Correct: (B)

0.800 → 0.400 → 0.200 is two half-lives, so 2 × t1/2 = 30.0 s and t1/2 = 15.0 s.

Correct: (C)

A small temperature rise only slightly increases collision frequency, but it greatly increases the fraction of collisions with enough energy to react, which is why the rate rises so sharply.

Correct: (A)

Kc = [FeSCN^2+] / ([Fe^3+][SCN^-]) = 0.0800 / (0.0200 × 0.0200) = 0.0800 / 0.000400 = 200 mol^-1 dm^3.

Correct: (D)

Kp = p(SO3)^2 / (p(SO2)^2 × p(O2)) = 60^2 / (20^2 × 10) = 3600 / 4000 = 0.900 kPa^-1.

Correct: (A)

At constant volume, adding an inert gas does not change the partial pressures (or concentrations) of the reacting gases, so the position of equilibrium is unaffected.

Correct: (C)

Kc = [FeSCN^2+] / ([Fe^3+][SCN^-]), so the units are (mol dm^-3) / (mol dm^-3)^2 = mol^-1 dm^3.

Correct: (B)

Cooling shifts the equilibrium in the exothermic (forward) direction, converting brown NO2 into colourless N2O4, so the mixture becomes paler.

Correct: (D)

2(+1) + 2x + 7(-2) = 0, so 2x = +12 and x = +6.

Correct: (A)

E°cell = E°(right) – E°(left) = +0.80 – (-0.44) = +1.24 V.

Correct: (B)

Zinc is readily oxidised (E°(Zn^2+/Zn) = -0.76 V, the most negative value here), so it is the strongest reducing agent. Cl2, Fe^3+ and Cu^2+ all act as oxidising agents.

Correct: (C)

In this reaction copper(I) is simultaneously oxidised (+1 to +2 in Cu^2+) and reduced (+1 to 0 in Cu), which is the definition of disproportionation.

Correct: (D)

Hydrogen is produced at the cathode (2H^+ + 2e^- → H2) and oxygen at the anode. Overall, 2H2O → 2H2 + O2, so the volume ratio H2 : O2 is 2 : 1.

Correct: (B)

[OH^-] = 1.00 × 10^-3, so pOH = 3.00 and pH = 14.00 – 3.00 = 11.00.

Correct: (A)

[H^+] = 10^-3.00 = 1.00 × 10^-3 mol dm^-3. Ka = [H^+]^2 / [HX] = (1.00 × 10^-3)^2 / 0.0500 = 2.00 × 10^-5 mol dm^-3.

Correct: (D)

The added H^+ converts 0.0100 mol of ethanoate into ethanoic acid, giving [acid] = 0.110 and [salt] = 0.090 mol dm^-3. [H^+] = 1.80 × 10^-5 × (0.110 / 0.090) = 2.20 × 10^-5, so pH = 4.66.

Correct: (C)

This is a strong acid-weak base titration, so the equivalence point is acidic (the solution contains NH4^+). Methyl orange changes colour within the sharp pH fall at the end point.

Correct: (A)

NH4Cl is the salt of a weak base (NH3) and a strong acid (HCl), so NH4^+ hydrolyses to give an acidic solution. CH3COONa and Na2CO3 give alkaline solutions, and KNO3 is neutral.

Correct: (B)

Reactivity of Group 2 metals increases down the group, so barium reacts most vigorously with cold water, forming Ba(OH)2 and hydrogen.

Correct: (C)

Iodide ions are the largest and lose electrons most easily, so reducing power increases down Group 7: I^- is the strongest reducing agent.

Correct: (D)

Platinum(II) forms four-coordinate square planar complexes. [Fe(H2O)6]^3+ is octahedral, [Zn(NH3)4]^2+ is tetrahedral and [Ag(NH3)2]^+ is linear.

Correct: (C)

Cu^2+ + 2OH^- → Cu(OH)2, a blue precipitate. Unlike Al(OH)3 or Zn(OH)2, copper(II) hydroxide does not dissolve in excess NaOH (answers A, B and D describe Fe^2+, Fe^3+ and Al^3+/Zn^2+ behaviour respectively).

Correct: (B)

Bromide ions are strong enough reducing agents to reduce concentrated H2SO4 to SO2 while being oxidised to Br2 (orange vapour). Chloride ions cannot do this, so NaCl gives only HCl.

Correct: (A)

In initiation, the Br-Br bond breaks by homolytic fission under UV light to form two bromine radicals. Options B and C are propagation steps, and D is termination.

Correct: (D)

It is a tertiary halogenoalkane with the weakest carbon-halogen bond (C-I), so it reacts fastest. Reactivity increases from primary to tertiary and from C-Cl to C-I.

Correct: (C)

Hot ethanolic KOH gives elimination. Removal of H from either side is possible, but the more substituted alkene, but-2-ene, is the major product.

Correct: (B)

Industrial hydration of alkenes uses steam with an H3PO4 catalyst. Electrophilic addition of water follows Markovnikov's rule, so the OH group ends up on the central carbon, giving propan-2-ol.

Correct: (D)

2-methylpropan-2-ol is a tertiary alcohol: the carbon carrying the OH group has no hydrogen atoms, so it cannot be oxidised. Primary alcohols oxidise to carboxylic acids and secondary alcohols to ketones.

Correct: (A)

In pent-2-ene each double-bond carbon carries two different groups (H and CH3 on one, H and C2H5 on the other), so E-Z isomers exist. In the other alkenes at least one double-bond carbon has two identical groups attached.

Correct: (C)

LiAlH4 is a powerful reducing agent that reduces carboxylic acids to primary alcohols. NaBH4 reduces aldehydes and ketones but not carboxylic acids.

Correct: (B)

Alkaline hydrolysis of an ester gives the sodium salt of the carboxylic acid and the alcohol: CH3COOC2H5 + NaOH → CH3COONa + C2H5OH.

Correct: (D)

This Friedel-Crafts acylation involves attack on the ring by the CH3CO^+ electrophile; the aromatic system is preserved because a proton is then lost, making it substitution rather than addition.

Correct: (A)

The -NH2 group strongly activates the benzene ring, so phenylamine reacts rapidly with bromine water at the 2, 4 and 6 positions without needing a catalyst, decolourising it and giving a white precipitate.

Correct: (C)

Amino acids form zwitterions (^+H3N-CHR-COO^-) with ionic character, so their solids have strong ionic attractions and high melting points. They react with both acids and alkalis because they contain both basic and acidic groups.

Correct: (B)

There are three proton environments: CH3CO (3H), OCH2 (2H) and CH2CH3 (3H), giving three signals with an integration ratio of 3 : 2 : 3.