A-Level Chemistry Practice Test 7

Correct: (B)

Ar = (10 × 19.9 + 11 × 80.1) / 100 = (199 + 881.1) / 100 = 10.80

Correct: (C)

The large jump after four electrons are removed shows the element has four electrons in its outer shell, so it is in Group 4.

Correct: (A)

A neutral Fe atom has 26 electrons; forming Fe^3+ removes 3 electrons, leaving 26 − 3 = 23.

Correct: (D)

The configuration contains 2 + 2 + 6 + 2 + 3 = 15 electrons, so the atomic number is 15, which is phosphorus.

Correct: (B)

The proton number rises while electrons enter the same shell at a similar distance, so the attraction on the outer electrons increases.

Correct: (C)

V = nRT / p = (0.250 × 8.31 × 300) / 100 000 = 6.23 × 10^-3 m^3 = 6.2 dm^3.

Correct: (A)

Percentage yield = (19.5 / 25.0) × 100 = 78%.

Correct: (D)

Moles NaOH = 0.100 × 0.0250 = 0.00250 mol; moles H2SO4 = 0.00250 / 2 = 0.00125 mol; volume = 0.00125 / 0.0400 = 0.0313 dm^3 = 31.3 cm^3.

Correct: (B)

Moles: C = 40.0 / 12.0 = 3.33; H = 6.7 / 1.0 = 6.7; O = 53.3 / 16.0 = 3.33. Ratio 1 : 2 : 1 gives CH2O.

Correct: (C)

Atom economy = Mr of desired product / total Mr of all products × 100 = 56 / (56 + 44) × 100 = 56%.

Correct: (A)

Moles = 8.0 / 16 = 0.50 mol; molecules = 0.50 × 6.02 × 10^23 = 3.01 × 10^23.

Correct: (D)

BF3 has three bonding pairs and no lone pair on boron, so the pairs repel to give a trigonal planar shape with 120° angles.

Correct: (B)

Ammonia has N−H bonds and a lone pair on nitrogen, so hydrogen bonds form between NH3 molecules.

Correct: (C)

Lattice enthalpy (of formation) is the enthalpy change when one mole of an ionic lattice is formed from its gaseous ions.

Correct: (A)

Fluorine has the greatest electronegativity, so the electronegativity difference (about 1.8) is largest for the H−F bond.

Correct: (D)

Ethene has four C−H single bonds and one C−C single bond (5 sigma bonds), plus one pi bond in the C=C double bond.

Correct: (B)

Melting diamond requires breaking very many strong C−C covalent bonds in the three-dimensional giant structure, which needs a large amount of energy.

Correct: (A)

q = mcΔT = 500 × 4.18 × 11.5 = 24 035 J = 24.0 kJ. ΔH = −24.0 / 0.0200 = −1200 kJ mol^-1.

Correct: (C)

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

Correct: (D)

A solid decomposes to give a solid and a gas, so the number of moles of gas increases and the system becomes more disordered.

Correct: (B)

At the limit ΔG = 0, so T = ΔH / ΔS = 100 000 / 200 = 500 K. Above 500 K, ΔG becomes positive.

Correct: (A)

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

Correct: (C)

Standard enthalpy of formation is for forming 1 mol of a compound from its elements in their standard states, which is H2(g) + 0.5O2(g) → H2O(l).

Correct: (D)

Rate quadrupling when [A] doubles means second order in A; doubling [B] doubles the rate, so first order in B. Overall order = 2 + 1 = 3.

Correct: (A)

For rate = k[A]^2, k = rate / [A]^2 = (mol dm^-3 s^-1) / (mol^2 dm^-6) = mol^-1 dm^3 s^-1.

Correct: (B)

At higher temperature the Maxwell−Boltzmann distribution shifts so that many more molecules possess at least the activation energy, so successful collisions are much more frequent.

Correct: (C)

A constant half-life, independent of concentration, is the characteristic behaviour of a first-order reaction.

Correct: (D)

k = rate / ([A][B]) = 4.0 × 10^-5 / (0.10 × 0.20) = 4.0 × 10^-5 / 0.020 = 2.0 × 10^-3 mol^-1 dm^3 s^-1.

Correct: (B)

At equilibrium [A] = [B] = 1.00 − 0.60 = 0.40 and [C] = [D] = 0.60. Kc = (0.60 × 0.60) / (0.40 × 0.40) = 0.36 / 0.16 = 2.25.

Correct: (D)

Increasing the pressure shifts the equilibrium towards the side with fewer moles of gas, which is N2O4 (1 mol compared with 2 mol of NO2), so less NO2 is formed.

Correct: (A)

Kp is the partial pressures of the products raised to their stoichiometric powers divided by those of the reactants: p(NH3)^2 divided by p(N2) × p(H2)^3.

Correct: (C)

A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached faster but Kc and the equilibrium composition do not change.

Correct: (B)

Kp = p(HI)^2 / [ p(H2) × p(I2) ] = 60^2 / (20 × 20) = 3600 / 400 = 9.0.

Correct: (C)

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

Correct: (A)

The metal with the most negative electrode potential loses electrons most readily, so magnesium is the strongest reducing agent.

Correct: (D)

EMF = E(positive electrode) − E(negative electrode) = +0.34 − (−0.76) = 1.10 V.

Correct: (B)

Q = It = 2.00 × 1930 = 3860 C; moles of electrons = 3860 / 96 500 = 0.0400 mol; Cu^2+ + 2e- → Cu, so moles Cu = 0.0200; mass = 0.0200 × 63.5 = 1.27 g.

Correct: (C)

Chlorine has the more positive electrode potential, so Cl2 gains electrons and oxidises Br- to Br2; the cell EMF = 1.36 − 1.09 = +0.27 V, so the reaction is feasible.

Correct: (A)

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

Correct: (D)

[H+] = sqrt(Ka × c) = 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: (B)

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

Correct: (C)

[H+] = Ka × [acid] / [salt] = 1.8 × 10^-5 × 0.10 / 0.20 = 9.0 × 10^-6 mol dm^-3; pH = 5.04 (equivalently, pH = pKa + log([salt]/[acid]) = 4.74 + log 2 = 4.74 + 0.30).

Correct: (A)

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

Correct: (D)

The solubility of Group 2 sulfates decreases down the group, so barium sulfate is the least soluble and is effectively insoluble.

Correct: (B)

Reducing power of the halide ions increases down Group 7 because the ions get larger and lose electrons more easily; I- is the strongest reducing agent.

Correct: (C)

Ammonia first forms pale blue Cu(OH)2, which dissolves in excess ammonia to give the deep blue complex [Cu(NH3)4(H2O)2]^2+.

Correct: (A)

Ba^2+ ions give a white precipitate of BaSO4 with sulfate ions; the solution is acidified first to remove carbonate ions that would also give a white precipitate.

Correct: (D)

Down Group 2 the metal ion gets larger and polarises the carbonate ion less, so the carbonates become more stable to heat; BaCO3 needs the highest temperature to decompose.

Correct: (B)

In the initiation step, UV light provides the energy for homolytic fission of the Cl−Cl bond, forming two chlorine free radicals, Cl•.

Correct: (C)

The C−I bond is the weakest carbon−halogen bond, so it breaks most easily and the iodoalkane hydrolyses fastest.

Correct: (A)

H+ adds to the double bond to form the more stable secondary carbocation, so Br ends up on the middle carbon, giving 2-bromopropane as the major (Markownikoff) product.

Correct: (D)

Under reflux a primary alcohol is oxidised fully to the carboxylic acid; propanal would only be obtained if it were distilled off as it formed.

Correct: (B)

Concentrated sulfuric acid acts as a dehydrating agent and an elimination reaction removes water to form ethene.

Correct: (C)

CN- substitutes for Br in a nucleophilic substitution, extending the carbon chain by one carbon to give CH3CH2CN, propanenitrile.

Correct: (A)

Aldehydes reduce Tollens' reagent to give a silver mirror, whereas ketones do not react; 2,4-DNP reacts with both aldehydes and ketones.

Correct: (D)

The ester takes its second name from the acid (propanoate) and its first name from the alcohol (methyl), giving methyl propanoate.

Correct: (B)

Concentrated nitric acid and concentrated sulfuric acid generate the nitronium ion, NO2+, which is the electrophile in electrophilic substitution.

Correct: (C)

The ethyl group pushes electron density onto the nitrogen, making the lone pair more available to accept a proton; in phenylamine the lone pair is delocalised into the benzene ring, making it a much weaker base.

Correct: (D)

A strong absorption in the range 1680−1750 cm^-1 is characteristic of a carbonyl (C=O) bond, as found in esters.

Correct: (A)

Ninhydrin reacts with amino acids to give a purple colour, revealing the positions of the otherwise colourless spots.