A-Level Chemistry Practice Test 14

Correct: (C)

(24 × 0.790) + (25 × 0.100) + (26 × 0.110) = 18.96 + 2.50 + 2.86 = 24.32, which is approximately 24.3.

Correct: (B)

A large jump after five electrons have been removed means there are five electrons in the outer shell, so the element is in Group 15.

Correct: (A)

Fe is [Ar] 4s^2 3d^6. The 4s electrons are removed first when the ion forms, so Fe^2+ is [Ar] 3d^6.

Correct: (D)

Across the period, protons are added but the electrons enter the same outer shell, so shielding stays similar and the effective nuclear charge increases, drawing the electrons inwards.

Correct: (B)

Flight time is proportional to sqrt(m/z): t = 1.6 × 10^-5 × sqrt(128/32) = 1.6 × 10^-5 × 2 = 3.2 × 10^-5 s.

Correct: (D)

n = pV/RT = (101 000 × 250 × 10^-6)/(8.31 × 298) = 25.25/2476 = 0.0102 mol. Mass = 0.0102 × 44 = 0.45 g.

Correct: (A)

Moles of NaOH = 0.0300 × 0.100 = 3.00 × 10^-3. H2SO4 reacts with NaOH in a 1 : 2 ratio, so moles of H2SO4 = 1.50 × 10^-3. Concentration = 1.50 × 10^-3 / 0.0250 = 0.060 mol dm^-3.

Correct: (C)

Amounts: C = 85.7/12 = 7.14; H = 14.3/1 = 14.3, a ratio of 1 : 2, so the empirical formula is CH2 (Mr = 14). 84/14 = 6, so the molecular formula is C6H12.

Correct: (B)

Atom economy = (total Mr of desired product / total Mr of all reactants) × 100 = (2 × 46)/180 × 100 = 92/180 × 100 = 51.1%.

Correct: (D)

Moles = 5.85/58.5 = 0.100 mol. Concentration = 0.100/0.250 = 0.400 mol dm^-3.

Correct: (A)

Moles of Mg = 0.48/24 = 0.0200. From the equation, moles of O2 = 0.0100. Volume = 0.0100 × 24 000 = 240 cm^3.

Correct: (C)

BF3 has three bonding pairs and no lone pairs around the central boron atom; the pairs repel into a trigonal planar arrangement with 120° bond angles.

Correct: (B)

Oxygen is very electronegative and the O-H bond is highly polar, so water forms hydrogen bonds; sulfur is not electronegative enough for hydrogen bonding in H2S, so far less energy is needed to separate its molecules.

Correct: (D)

From the Born-Haber cycle, lattice enthalpy = -437 – (89 + 418 + 121 – 349) = -437 – 279 = -716 kJ mol^-1.

Correct: (A)

Aluminium atoms release three electrons per atom into the delocalised sea and form small 3+ ions, giving the strongest attraction between the ions and the electrons, so aluminium has the highest melting point.

Correct: (C)

Fluorine is the most electronegative element, so the difference in electronegativity across the H-F bond is the largest and the bond is the most polar.

Correct: (B)

Within each layer, every carbon atom is covalently bonded to three others, but only weak van der Waals forces act between the layers, so the layers slide easily and graphite is soft.

Correct: (A)

q = 100 × 4.18 × 32.7 = 13 669 J. Moles of ethanol = 0.46/46 = 0.0100, so ΔH = -13 669/0.0100 = -1367 kJ mol^-1.

Correct: (D)

Using a Hess cycle with combustion data, ΔH = [(-1411) + (-286)] – (-1560) = -1697 + 1560 = -137 kJ mol^-1.

Correct: (C)

Bonds broken = (4 × 413) + (2 × 498) = 1652 + 996 = 2648 kJ mol^-1. Bonds formed = (2 × 805) + (4 × 464) = 1610 + 1856 = 3466 kJ mol^-1. ΔH = 2648 – 3466 = -818 kJ mol^-1.

Correct: (B)

ΔG = ΔH – TΔS. With both ΔH and ΔS positive, ΔG becomes negative only when TΔS exceeds ΔH, which happens at high temperature.

Correct: (D)

ΔG = ΔH – TΔS = -50 – (298 × -0.100) = -50 + 29.8 = -20.2 kJ mol^-1, so the reaction is feasible at 298 K.

Correct: (A)

Strong acids and strong alkalis are fully ionised in solution, so the only chemical change in every such neutralisation is hydrogen ions combining with hydroxide ions to form water.

Correct: (C)

Halving A multiplies the rate by (1/2)^2 = 1/4; doubling B multiplies the rate by 2. The overall factor is 1/4 × 2 = 1/2, so the rate halves.

Correct: (B)

k = rate/[A]^2 = (4.0 × 10^-3)/(0.10)^2 = (4.0 × 10^-3)/0.010 = 0.40 mol^-1 dm^3 s^-1.

Correct: (A)

For a first order reaction, k = rate/concentration, so the units are (mol dm^-3 s^-1)/(mol dm^-3) = s^-1.

Correct: (D)

Gradient = -Ea/R, so Ea = 9000 × 8.31 = 74 790 J mol^-1, approximately 74.8 kJ mol^-1.

Correct: (C)

A constant half-life that does not depend on the starting concentration is the characteristic feature of a first order reaction.

Correct: (B)

Kc = (0.70 × 0.70)/(0.30 × 0.30) = 0.49/0.09 = 5.4. Because all the amounts are in the same volume, the amounts can be used directly, and Kc has no units here because there are equal numbers of molecules on each side of the equation.

Correct: (A)

Partial pressures: p(N2O4) = 0.25 × 200 = 50 kPa; p(NO2) = 0.75 × 200 = 150 kPa. Kp = p(NO2)^2/p(N2O4) = 150^2/50 = 22 500/50 = 450 kPa.

Correct: (D)

A catalyst speeds up the forward and reverse reactions equally, so it does not change Kc or the equilibrium composition; it only shortens the time needed to reach equilibrium.

Correct: (C)

Kc = [PCl3][Cl2]/[PCl5], so the units are (mol dm^-3 × mol dm^-3)/(mol dm^-3) = mol dm^-3.

Correct: (B)

At equilibrium, [CO] = [H2O] = 0.40 mol dm^-3 and [CO2] = [H2] = 0.60 mol dm^-3. Kc = (0.60 × 0.60)/(0.40 × 0.40) = 0.36/0.16 = 2.25.

Correct: (D)

2(+1) + 2x + 3(-2) = 0, so 2x = +4 and x = +2.

Correct: (A)

EMF = E(more positive half-cell) – E(more negative half-cell) = 0.80 – 0.34 = 0.46 V. Copper is oxidised at the negative electrode.

Correct: (C)

Bromine has the more positive E value, so Br2 is the stronger oxidising agent and oxidises iodide ions to iodine; the EMF for this reaction is 1.09 – 0.54 = +0.55 V, so it is feasible.

Correct: (B)

Q = It = 2.0 × 965 = 1930 C = 1930/96 500 = 0.0200 mol of electrons. Ag^+ + e^- → Ag, so moles of Ag = 0.0200 and mass = 0.0200 × 107.9 = 2.16 g.

Correct: (D)

Standard conditions are a temperature of 298 K, a pressure of 100 kPa and all solutions at a concentration of 1.00 mol dm^-3.

Correct: (A)

pH = -log(2.5 × 10^-3) = 2.60.

Correct: (C)

pH 3.00 means [H^+] = 1.0 × 10^-3 mol dm^-3. Ka = [H^+]^2/[HA] = (1.0 × 10^-3)^2/0.100 = (1.0 × 10^-6)/0.100 = 1.0 × 10^-5 mol dm^-3.

Correct: (B)

The large vertical section of the curve means that any indicator changing colour within about pH 3 to pH 11 will give an accurate end point.

Correct: (D)

pH = pKa + log([salt]/[acid]) = 4.74 + log(0.300/0.100) = 4.74 + log 3 = 4.74 + 0.48 = 5.22.

Correct: (A)

pOH = -log(0.020) = 1.70, so pH = 14.00 – 1.70 = 12.30.

Correct: (C)

Group 2 sulfate solubility decreases down the group; BaSO4 is insoluble, which is why acidified barium chloride solution is used as a test for sulfate ions.

Correct: (B)

Silver chloride is a white precipitate that dissolves in dilute ammonia solution; silver bromide is cream and dissolves only in concentrated ammonia, and silver iodide is yellow and does not dissolve in ammonia.

Correct: (D)

Fe^3+(aq) + 3OH^-(aq) → Fe(OH)3(s). Iron(III) hydroxide forms as an orange-brown (rust-coloured) precipitate. Fe^2+ ions give a green precipitate instead.

Correct: (A)

Ethane-1,2-diamine has two nitrogen atoms, each with a lone pair, so it forms two coordinate bonds to a central metal ion and is a bidentate ligand.

Correct: (C)

The vanadium oxidation states have characteristic colours: +5 (VO2^+) is yellow, +4 (VO^2+) is blue, +3 is green and +2 is violet.

Correct: (B)

In initiation, ultraviolet light supplies the energy for homolytic fission of the Cl-Cl bond, forming two chlorine radicals. Steps (A) and (C) are propagation steps and step (D) is a termination step.

Correct: (D)

The cyanide ion acts as a nucleophile and replaces the bromine: CH3CH2Br + CN^- → CH3CH2CN + Br^-. The nitrile product has one more carbon atom than the starting halogenoalkane.

Correct: (A)

Ethene reacts with steam at about 300 °C and high pressure over a phosphoric(V) acid catalyst to give ethanol. Conditions (B) describe fermentation, which is a different route to ethanol.

Correct: (C)

Addition polymers are made from alkene monomers; in poly(propene) the double bonds of propene molecules open up and the molecules join into a long chain.

Correct: (B)

2-methylpropan-2-ol is a tertiary alcohol; the carbon atom carrying the -OH group has no hydrogen atom attached, so it resists oxidation and the mixture stays orange. Primary alcohols are oxidised to carboxylic acids and secondary alcohols to ketones.

Correct: (D)

The alcohol is dehydrated in an elimination reaction; a molecule of water is removed and the alkene ethene is formed.

Correct: (A)

Aldehydes reduce the blue Cu^2+ ions in Fehling's solution to brick-red copper(I) oxide, Cu2O; ketones give no reaction. A silver mirror is formed with Tollens' reagent instead.

Correct: (C)

The negative inductive effect of the chlorine atom spreads out (disperses) the negative charge on the ethanoate ion, making the anion more stable and shifting the ionisation equilibrium to the right.

Correct: (B)

In the name of an ester, the first part comes from the alcohol and the second part from the carboxylic acid, so methanol and propanoic acid give methyl propanoate (plus water).

Correct: (D)

Addition to benzene would destroy the stable delocalised pi system, so benzene undergoes electrophilic substitution (with bromine it needs an FeBr3 catalyst), whereas the localised double bond in cyclohexene reacts by addition.

Correct: (A)

Glycine normally exists as the zwitterion H3N^+CH2COO^-. In strong acid, the hydrogen ions protonate the -COO^- group, giving the cation H3N^+CH2COOH.

Correct: (C)

The two CH3 groups are in identical environments and give one peak, and the carbonyl carbon gives a second peak, so there are two peaks in total.