A-Level Chemistry Practice Test 20

Correct: (C)

Ar = (10 × 0.20) + (11 × 0.80) = 2.0 + 8.8 = 10.8.

Correct: (D)

The first three electrons are removed from the outer shell relatively easily. The fourth electron comes from an inner shell, much closer to the nucleus, so its ionisation energy is far larger. The element has three outer-shell electrons, so it is in Group 13.

Correct: (A)

A neutral Fe atom is [Ar] 4s^2 3d^6. When Fe^2+ forms, the two 4s electrons are lost first, leaving [Ar] 3d^6.

Correct: (B)

Aluminium oxide reacts with both acids and alkalis, so it is amphoteric. Na2O and MgO are basic oxides and SO2 is an acidic oxide.

Correct: (D)

Mg has one more proton than Na, the electrons are in the same shell so shielding is almost the same, and the Mg atom is smaller. The outer electron is therefore attracted more strongly.

Correct: (A)

Number of molecules = 0.500 × 6.02 × 10^23 = 3.01 × 10^23.

Correct: (C)

n = pV/RT = (100000 × 1.24 × 10^-3) / (8.31 × 300) = 124 / 2493 = 0.0497 mol. Mr = mass/moles = 2.00 / 0.0497 = 40.2, approximately 40.

Correct: (B)

moles of NaOH = 0.200 × 0.0250 = 0.00500 mol. H2SO4 + 2NaOH → Na2SO4 + 2H2O, so moles of H2SO4 = 0.00250 mol. Concentration = 0.00250 / 0.0250 = 0.100 mol dm^-3.

Correct: (D)

Percentage yield = (4.60 / 5.00) × 100 = 92%.

Correct: (A)

The number of moles does not change on dilution: new concentration = 2.00 × (25.0 / 250) = 0.200 mol dm^-3.

Correct: (B)

Mass of water lost = 2.87 − 1.84 = 1.03 g, which is 1.03/18 = 0.0572 mol. Moles of CuSO4 = 1.84/159.5 = 0.0115 mol. Ratio = 0.0572/0.0115 = 4.96, so x = 5.

Correct: (D)

Boron in BF3 has three bonding pairs of electrons and no lone pairs. The bonding pairs repel each other equally, giving a trigonal planar shape with bond angles of 120°.

Correct: (C)

In the ammonium ion, the lone pair on the nitrogen atom of NH3 is donated to an H^+ ion, forming a coordinate bond in which both shared electrons come from nitrogen.

Correct: (A)

Lattice enthalpy depends on the charges and sizes of the ions. The 2+ and 2− ions in MgO are also smaller, so the electrostatic attractions in the lattice are much stronger than in NaCl.

Correct: (B)

In NH3 the hydrogen atoms are bonded to the very electronegative nitrogen atom, so hydrogen bonds form between the H of one molecule and the lone pair on N of another.

Correct: (D)

In Mg the metallic bonding is stronger than in Na because there are twice as many delocalised electrons per atom and the positive ions carry a 2+ charge and are smaller, giving stronger attraction.

Correct: (C)

Water has two bonding pairs and two lone pairs. Lone pair-lone pair repulsion is strongest, so the bonding pairs are pushed together, reducing the angle from 109.5° to 104.5°.

Correct: (A)

q = mcΔT = 100 × 4.18 × 20.0 = 8360 J. Moles of ethanol = 0.460/46 = 0.0100 mol. ΔHc = −8360 / 0.0100 = −836000 J mol^-1 = −836 kJ mol^-1.

Correct: (D)

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

Correct: (B)

ΔS = S(products) − S(reactants) = 350 − 200 = +150 J K^-1 mol^-1.

Correct: (C)

The reaction becomes feasible when ΔG = 0, so T = ΔH/ΔS = 120000 / 300 = 400 K. Above 400 K, ΔG is negative.

Correct: (A)

For C + 0.5O2 → CO, a Hess cycle using the combustion enthalpies gives ΔHf(CO) = ΔHc(C) − ΔHc(CO) = (−394) − (−283) = −111 kJ mol^-1.

Correct: (D)

From the Born-Haber cycle: lattice enthalpy = ΔHf − (atomisation + ionisation + atomisation + electron affinity) = −411 − (108 + 500 + 121 − 364) = −411 − 365 = −776 kJ mol^-1.

Correct: (B)

Doubling [A] multiplies the rate by 4 = 2^2, so the reaction is second order in A. Doubling [B] doubles the rate, so it is first order in B. Rate = k[A]^2[B].

Correct: (C)

k = rate / ([A]^2[B]) = (2.0 × 10^-4) / (0.10^2 × 0.10) = (2.0 × 10^-4) / (1.0 × 10^-3) = 0.20 mol^-2 dm^6 s^-1.

Correct: (A)

The exponential term gives the proportion of molecules (collisions) with enough energy to react. The term A is the frequency factor and represents the collision frequency and orientation.

Correct: (D)

A catalyst provides an alternative route with a lower activation energy. The distribution itself is unchanged, but more molecules now have energy greater than the lowered activation energy.

Correct: (B)

For a first-order reaction, rate = k[A], so k = rate/[A] = (mol dm^-3 s^-1)/(mol dm^-3) = s^-1.

Correct: (C)

Acid reacted = 1.00 − 0.33 = 0.67 mol, so at equilibrium: acid = 0.33, ethanol = 0.33, ester = 0.67, water = 0.67 mol. All are in the same volume, so Kc = (0.67 × 0.67) / (0.33 × 0.33) = 4.12 (no units, as there are equal numbers of moles on each side).

Correct: (D)

Kp = (pNO2)^2 / (pN2O4). There is one more mole of gas on the right than on the left, so the units are (kPa^2)/(kPa) = kPa.

Correct: (A)

Raising the temperature shifts the position of equilibrium in the endothermic (reverse) direction, so the concentration of products falls relative to reactants and Kc decreases. Kc changes only when temperature changes.

Correct: (B)

The equilibrium [Co(H2O)6]^2+ + 4Cl^- ⇌ [CoCl4]^2- + 6H2O shifts to the right as the chloride ion concentration increases, replacing the pink octahedral aqua complex with the blue tetrahedral chloro complex.

Correct: (C)

Partial pressure = mole fraction × total pressure = 0.20 × 300 = 60 kPa.

Correct: (D)

Using oxidation states: 2(+1) + 2(S) + 3(−2) = 0, so 2 + 2S − 6 = 0 and S = +2 (average value).

Correct: (A)

In this reaction chlorine (oxidation state 0) is simultaneously reduced to Cl^- (−1) in HCl and oxidised to ClO^- (+1) in HClO, so the same element is both oxidised and reduced.

Correct: (B)

E(cell) = E(more positive) − E(more negative) = +0.80 − (−0.76) = +1.56 V.

Correct: (C)

In the alkaline fuel cell, hydrogen reacts with hydroxide ions at the anode: 2H2 + 4OH^- → 4H2O + 4e^-. The electrons released travel through the external circuit.

Correct: (D)

At the cathode, Cu^2+ + 2e^- → Cu, so 2 mol of electrons deposits 1 mol of copper. Moles of Cu = 0.100/2 = 0.0500 mol; mass = 0.0500 × 63.5 = 3.18 g.

Correct: (A)

HCl is fully ionised, so [H^+] = 0.200 mol dm^-3. pH = −log(0.200) = 0.70.

Correct: (B)

For a weak acid, [H^+] = √(Ka × c) = √(1.3 × 10^-5 × 0.100) = √(1.3 × 10^-6) = 1.14 × 10^-3 mol dm^-3. pH = −log(1.14 × 10^-3) = 2.94.

Correct: (C)

pH = pKa + log([salt]/[acid]) = 4.00 + log(0.100/0.200) = 4.00 + log(0.5) = 4.00 − 0.30 = 3.70.

Correct: (D)

The salt formed is the conjugate base of a weak acid, which hydrolyses to give an alkaline solution, so the pH at equivalence is above 7. Phenolphthalein changes colour in the alkaline range covered by the steep section of the curve.

Correct: (A)

In pure water [H^+] = [OH^-] at all temperatures, so it is always neutral. A larger Kw means [H^+] is greater, so the pH falls below 7 even though the water is neutral.

Correct: (C)

Ca + 2H2O → Ca(OH)2 + H2. Calcium reacts steadily with cold water, forming a (partially soluble) hydroxide and hydrogen gas.

Correct: (B)

Iodide ions reduce concentrated sulfuric acid through SO2 (+4) and S (0) to hydrogen sulfide, H2S, in which sulfur has oxidation state −2. The iodide is oxidised to I2.

Correct: (D)

The Contact process oxidises SO2 to SO3 using a vanadium(V) oxide catalyst at about 450°C. Iron is the catalyst in the Haber process.

Correct: (A)

Cu^2+ ions react with hydroxide ions to form a pale blue precipitate of copper(II) hydroxide, Cu(OH)2. Unlike the reaction with excess ammonia, this precipitate does not dissolve in excess sodium hydroxide.

Correct: (C)

Fe^2+ ions form a green precipitate of iron(II) hydroxide, Fe(OH)2, which slowly turns brown at its surface as it is oxidised by air to iron(III) hydroxide.

Correct: (B)

In propagation, a radical reacts with a molecule to form a product and a new radical, keeping the chain going. Steps (C) and (D) destroy radicals (termination) and step (A) is initiation.

Correct: (D)

2-Bromo-2-methylpropane is a tertiary halogenoalkane. It forms a relatively stable tertiary carbocation intermediate, so it hydrolyses fastest by the SN1 mechanism.

Correct: (A)

Cold, dilute manganate(VII) is a mild oxidising agent that adds two -OH groups across the C=C double bond (dihydroxylation), forming a diol and discharging the purple colour.

Correct: (C)

Concentrated sulfuric acid acts as a dehydrating agent, eliminating water from ethanol to form the alkene ethene.

Correct: (B)

With aqueous alkali, nucleophilic substitution takes place: the -OH group replaces the bromine atom, forming ethanol. (Ethanolic KOH would give elimination to ethene instead.)

Correct: (D)

In the stratosphere, UV light causes homolytic fission of C-Cl bonds in CFCs. The chlorine radicals produced react with ozone, O3, and are regenerated, so each radical can destroy many ozone molecules.

Correct: (A)

The CN^- nucleophile adds to the carbonyl carbon and the H^+ adds to the oxygen, forming a hydroxynitrile (cyanohydrin): CH3CHO + HCN → CH3CH(OH)CN, which is 2-hydroxypropanenitrile.

Correct: (C)

Esterification is a reversible reaction catalysed by concentrated sulfuric acid, which also acts as a dehydrating agent, and the mixture is warmed.

Correct: (B)

Benzene is stabilised by its delocalised ring of pi electrons (its enthalpy of hydrogenation is less exothermic than expected). Addition reactions would destroy this delocalisation, so benzene reacts by substitution, which preserves it.

Correct: (B)

Nitrobenzene is reduced to phenylamine using tin and concentrated hydrochloric acid (followed by alkali to release the free amine). Nitration of benzene is the earlier step used to make the nitrobenzene.

Correct: (A)

Nylon-6,6 forms when hexanedioic acid (a dicarboxylic acid) condenses with hexane-1,6-diamine (a diamine), forming amide (peptide) links and eliminating water. Each monomer contributes 6 carbon atoms, hence the name.

Correct: (C)

The two CH3 groups in propanone are in identical environments, so all six hydrogen atoms are chemically equivalent and give a single signal in the 1H NMR spectrum.