A-Level Chemistry Practice Test 13

Correct: (B)

(35 × 0.758) + (37 × 0.242) = 26.53 + 8.95 = 35.48, which is approximately 35.5.

Correct: (C)

There is a large jump between the 2nd and 3rd ionisation energies, so the third electron is removed from an inner shell; the element has 2 outer electrons and is in Group 2.

Correct: (B)

Fe is [Ar] 4s^2 3d^6. Three electrons are removed, the two 4s electrons first and then one 3d electron, giving [Ar] 3d^5.

Correct: (D)

Ionisation energy generally increases across the period because nuclear charge increases while shielding stays similar, so chlorine, furthest to the right, has the highest value.

Correct: (C)

Flight time is proportional to sqrt(m/z): t = 2.0 × 10^-5 × sqrt(90/40) = 2.0 × 10^-5 × 1.5 = 3.0 × 10^-5 s.

Correct: (A)

V = nRT/p = (0.50 × 8.31 × 300)/100 000 = 0.0125 m^3 = 12.5 dm^3.

Correct: (D)

Moles of HCl = 0.0200 × 0.100 = 2.00 × 10^-3. The reacting ratio is 1:1, so concentration of NaOH = 2.00 × 10^-3 / 0.0250 = 0.080 mol dm^-3.

Correct: (A)

Percentage yield = (9.0/12.0) × 100 = 75%.

Correct: (B)

Amounts: C = 40.0/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33. The ratio is 1 : 2 : 1, so the empirical formula is CH2O.

Correct: (C)

Moles = 0.500 × 0.100 = 0.0500 mol. Mass = 0.0500 × 106 = 5.30 g.

Correct: (D)

Moles of Mg = 0.12/24 = 5.0 × 10^-3. Mg + 2HCl → MgCl2 + H2, so moles of H2 = 5.0 × 10^-3. Volume = 5.0 × 10^-3 × 24 000 = 120 cm^3.

Correct: (C)

Water has two bonding pairs and two lone pairs; the lone pairs repel most strongly, pushing the O-H bonds together to a tetrahedral-based angle of 104.5°.

Correct: (D)

Methanol molecules form hydrogen bonds, which are much stronger than the van der Waals and dipole-dipole forces in the other substances, so more energy is needed to separate the molecules.

Correct: (D)

From the Born-Haber cycle, lattice enthalpy = -411 – (107 + 496 + 121 – 349) = -411 – 375 = -786 kJ mol^-1.

Correct: (A)

The doubly charged ions attract each other strongly in all directions in the giant ionic lattice, so a very large amount of thermal energy is needed to break the lattice up.

Correct: (D)

CO2, BF3 and CCl4 are symmetrical, so their bond dipoles cancel. CHCl3 is asymmetrical, so its dipoles do not cancel and the molecule has an overall dipole.

Correct: (C)

Each carbon atom forms only three covalent bonds, so the fourth outer electron is delocalised within the layers and can carry charge along them.

Correct: (C)

q = 100 × 4.18 × 6.8 = 2842 J. Moles of water formed = 0.0500, so ΔH = -2842/0.0500 = -56 848 J mol^-1, approximately -56.8 kJ mol^-1.

Correct: (A)

ΔH = [(-394) + 2(-286)] – [(-75)] = -966 + 75 = -891 kJ mol^-1.

Correct: (B)

Bonds broken = 436 + 242 = 678 kJ mol^-1. Bonds formed = 2 × 431 = 862 kJ mol^-1. ΔH = 678 – 862 = -184 kJ mol^-1.

Correct: (A)

A gas is produced from a solid, so the disorder of the system increases; in each of the other reactions the system becomes more ordered.

Correct: (B)

At the threshold temperature, T = ΔH/ΔS = 100 000/200 = 500 K. The reaction is feasible below 500 K; above it, the -TΔS term makes ΔG positive.

Correct: (A)

ΔS = [2 × 193] – [192 + (3 × 131)] = 386 – 585 = -199 J K^-1 mol^-1.

Correct: (D)

Doubling A multiplies the rate by 2; doubling B multiplies it by 2^2 = 4. The overall factor is 2 × 4 = 8.

Correct: (C)

The reaction is first order with respect to X and second order with respect to Y, so rate = k[X][Y]^2.

Correct: (D)

k = rate/concentration^3, so the units are (mol dm^-3 s^-1)/(mol dm^-3)^3 = dm^6 mol^-2 s^-1.

Correct: (A)

Gradient = -Ea/R, so Ea = 6000 × 8.31 = 49 860 J mol^-1, approximately 49.9 kJ mol^-1.

Correct: (B)

A catalyst provides an alternative reaction route of lower activation energy; it is not used up and does not change the energy of the molecules.

Correct: (B)

Kc = [NO2]^2/[N2O4] = (0.80)^2/0.60 = 0.64/0.60 = 1.07 mol dm^-3.

Correct: (A)

Kp = p(SO3)^2 / [p(SO2)^2 × p(O2)] = 60^2/(20^2 × 10) = 3600/4000 = 0.90 kPa^-1.

Correct: (A)

Increasing the temperature shifts the equilibrium in the endothermic direction, so more products are present at equilibrium and Kc increases. A catalyst never changes Kc.

Correct: (C)

Kc = [products]/[reactants], with each concentration raised to the power of its stoichiometric coefficient: Kc = [C]/([A]^2[B]).

Correct: (D)

Forming 1.60 mol of HI uses 0.80 mol each of H2 and I2, so [H2] = [I2] = 0.20 mol dm^-3 and [HI] = 1.60 mol dm^-3. Kc = (1.60)^2/(0.20 × 0.20) = 2.56/0.04 = 64.

Correct: (B)

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

Correct: (A)

EMF = E(more positive half-cell) – E(more negative half-cell) = 0.77 – (-0.76) = 1.53 V.

Correct: (D)

The strongest oxidising agent is the species in the half-equation with the most positive E value; chlorine has the most positive value, +1.36 V.

Correct: (B)

Q = It = 0.50 × 1930 = 965 C = 965/96 500 = 0.0100 mol of electrons. Cu^2+ + 2e^- → Cu, so moles of Cu = 0.00500 and mass = 0.00500 × 63.5 = 0.318 g.

Correct: (A)

In NaCl the chlorine has oxidation number -1 (it has been reduced) and in NaClO it is +1 (it has been oxidised). The same element is both oxidised and reduced, so this is a disproportionation reaction.

Correct: (B)

Nitric acid is a strong acid and is fully ionised, so [H^+] = 0.010 mol dm^-3 and pH = -log(0.010) = 2.00.

Correct: (C)

[H^+] = sqrt(Ka × [acid]) = sqrt(1.8 × 10^-5 × 0.100) = sqrt(1.8 × 10^-6) = 1.34 × 10^-3 mol dm^-3. pH = -log(1.34 × 10^-3) = 2.87.

Correct: (C)

pOH = -log(0.050) = 1.30, so pH = 14.00 – 1.30 = 12.70.

Correct: (A)

pH = pKa + log([salt]/[acid]) = 4.74 + log(0.100/0.200) = 4.74 – 0.30 = 4.44.

Correct: (D)

Each Ca(OH)2 unit gives two OH^- ions, so [OH^-] = 2.0 × 10^-3 mol dm^-3. pOH = -log(2.0 × 10^-3) = 2.70, so pH = 14.00 – 2.70 = 11.30.

Correct: (B)

Group 2 hydroxide solubility increases down the group; Mg(OH)2 is only sparingly soluble. The carbonates become more stable to heating down the group, and magnesium reacts with steam.

Correct: (C)

Oxidising power decreases down Group 7, so iodine is a weaker oxidising agent than bromine and cannot displace bromine from its salts.

Correct: (D)

Chloride ions act as ligands and replace the water ligands in a ligand substitution reaction: [Cu(H2O)6]^2+ + 4Cl^- ⇌ [CuCl4]^2- + 6H2O. The tetrachlorocuprate(II) ion is yellow-green.

Correct: (A)

Calcium gives a brick red (orange-red) flame; lithium gives crimson, potassium gives lilac and copper(II) gives blue-green.

Correct: (C)

Vanadium(V) oxide catalyses the oxidation of sulfur dioxide to sulfur trioxide: 2SO2 + O2 ⇌ 2SO3. Iron is the catalyst in the Haber process.

Correct: (B)

In a propagation step a radical reacts with a molecule to give a product and a new radical, keeping the chain going. Step (A) is initiation, and steps (C) and (D) are termination steps.

Correct: (D)

The C-I bond is the weakest carbon-halogen bond, so it breaks most easily and 1-iodobutane reacts fastest. Reactivity increases in the order C-F < C-Cl < C-Br < C-I.

Correct: (A)

By Markovnikov's rule, the hydrogen adds to the carbon of the double bond that already has the greater number of hydrogens, so the bromine ends up on the middle carbon, giving 2-bromopropane.

Correct: (C)

The C=C double bond of an alkene reacts with bromine in an addition reaction, so orange bromine water is decolourised; alkanes do not react in this way.

Correct: (B)

Under reflux with excess oxidising agent, a primary alcohol is oxidised fully to the carboxylic acid, propanoic acid. Propanal is obtained only if it is distilled off as it forms.

Correct: (C)

Hot ethanolic KOH acts as a base and removes HBr in an elimination reaction, giving ethene. Aqueous KOH would give substitution to ethanol instead.

Correct: (B)

Tollens' reagent is reduced by aldehydes to give a silver mirror; ketones are not oxidised by it. 2,4-dinitrophenylhydrazine gives an orange precipitate with both aldehydes and ketones.

Correct: (D)

The carbon of the C=O group is trigonal planar; attack from either side of the plane is equally likely, so a racemic (equal) mixture of the two optical isomers of the hydroxynitrile is produced.

Correct: (A)

Acyl chlorides react vigorously with water, hydrolysing to the carboxylic acid and hydrogen chloride, which forms misty fumes in moist air.

Correct: (C)

Concentrated sulfuric acid protonates nitric acid and generates the nitronium ion: HNO3 + H2SO4 → NO2^+ + HSO4^- + H2O. The NO2^+ ion is the electrophile.

Correct: (B)

The alkyl group in ethylamine pushes electrons towards the nitrogen, making its lone pair more available to accept a proton. In phenylamine the lone pair is delocalised into the benzene ring, making it much less available, so phenylamine is the weakest base.

Correct: (D)

There are two different proton environments: the three CH3 protons give one peak (a singlet) and the COOH proton gives one peak (also a singlet, as it has no neighbouring protons).