A-Level Chemistry Practice Test 6

Correct: (B)

There is a very large jump between the first and second ionisation energies, so the second electron is removed from an inner shell; X has only one outer electron and is in Group 1.

Correct: (D)

Ar = (0.790 × 24) + (0.100 × 25) + (0.110 × 26) = 18.96 + 2.50 + 2.86 = 24.3

Correct: (A)

Copper is an exception to the usual filling order: the arrangement [Ar] 4s^1 3d^10, with a full 3d subshell, is more stable than [Ar] 4s^2 3d^9.

Correct: (C)

Silicon has a giant covalent structure with strong covalent bonds throughout, giving it a very high melting point; sodium and aluminium are metals and sulfur is a simple molecular solid.

Correct: (D)

First ionisation energy is defined for gaseous atoms forming gaseous 1+ ions, with one electron removed from each atom.

Correct: (A)

n = pV/RT = (100000 × 500 × 10^-6)/(8.31 × 300) = 50.0/2493 = 0.0201 mol

Correct: (C)

Percentage yield = (4.55/7.00) × 100 = 65.0%

Correct: (B)

Moles of HCl = 0.0250 × 0.100 = 2.50 × 10^-3; NaOH reacts 1:1 with HCl, so volume = 2.50 × 10^-3/0.150 = 0.0167 dm^3 = 16.7 cm^3

Correct: (D)

Atom economy = 120.0/(24.0 + 98.0) × 100 = 120.0/122.0 × 100 = 98.4%

Correct: (A)

Moles: C = 52.2/12.0 = 4.35; H = 13.0/1.00 = 13.0; O = 34.8/16.0 = 2.18; dividing by 2.18 gives the ratio 2 : 6 : 1, so the empirical formula is C2H6O.

Correct: (C)

Moles of Na2CO3 = 1.06/106 = 0.0100; concentration = 0.0100/0.250 = 0.0400 mol dm^-3

Correct: (B)

BF3 has three bonding pairs and no lone pair on the boron atom, so the electron pairs spread out in a trigonal planar arrangement with bond angles of 120°.

Correct: (D)

In NH4^+, the lone pair on the nitrogen atom of NH3 is donated to an H^+ ion, forming a coordinate bond; once formed it is identical to the other N-H bonds.

Correct: (A)

Li^+ and F^- are the smallest pair of ions here, so the electrostatic attractions in the lattice are the strongest and the lattice enthalpy is the most exothermic.

Correct: (C)

Fluorine is the most electronegative element, so the difference in electronegativity between H and F is the largest of these, making the H-F bond the most polar.

Correct: (B)

Melting diamond requires breaking many strong covalent bonds in the giant lattice, which takes a great deal of energy.

Correct: (D)

The larger number of electrons in Br2 gives stronger instantaneous dipole-induced dipole forces, so more energy is needed to separate the molecules; the covalent bonds are not broken on boiling.

Correct: (A)

ΔHf = ΔHc(C) + 2ΔHc(H2) − ΔHc(CH4) = −394 + 2(−286) − (−891) = −966 + 891 = −75 kJ mol^-1

Correct: (C)

q = mcΔT = 100 × 4.18 × 20.0 = 8360 J = 8.36 kJ; ΔH = −8.36/0.100 = −83.6 kJ mol^-1

Correct: (B)

ΔG = ΔH − TΔS = −200 − 400 × (−0.100) = −200 + 40 = −160 kJ mol^-1

Correct: (D)

When water freezes, the particles become arranged in an ordered solid lattice, so the disorder (entropy) decreases.

Correct: (A)

Energy absorbed breaking bonds = 945 + 3(436) = 2253; energy released forming bonds = 6 × 391 = 2346; ΔH = 2253 − 2346 = −93 kJ mol^-1

Correct: (C)

The enthalpy of formation is for making one mole of the compound from its elements in their standard states, which is H2(g) + 1/2 O2(g) → H2O(l).

Correct: (B)

Falling to 1/8 takes three half-lives (1 → 1/2 → 1/4 → 1/8), so the time taken is 3 × 20 = 60 s.

Correct: (D)

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

Correct: (A)

The exponential term gives the fraction of particles with energy at least equal to Ea; the pre-exponential factor A covers the collision frequency and orientation.

Correct: (C)

k = rate/[A] = 4.00 × 10^-3/0.200 = 0.0200 s^-1

Correct: (B)

A catalyst works by providing an alternative reaction pathway with a lower activation energy; it does not change the energy of the particles or the overall enthalpy change.

Correct: (D)

0.20 mol dm^-3 of HI reacts, forming 0.10 mol dm^-3 each of H2 and I2; Kc = [H2][I2]/[HI]^2 = (0.10 × 0.10)/(0.80)^2 = 0.0100/0.64 = 0.0156 (no units, as there are equal numbers of moles on each side).

Correct: (A)

Kp = p(NO2)^2/p(N2O4), so the units are kPa^2/kPa = kPa.

Correct: (C)

Raising the temperature favours the endothermic (reverse) reaction, so the yield of SO3 falls; Kc also decreases, because equilibrium constants for exothermic reactions become smaller as the temperature rises.

Correct: (B)

Kc is the concentrations of the products (ester and water) divided by the concentrations of the reactants (acid and alcohol), each raised to its stoichiometric power.

Correct: (D)

Kp = p(HI)^2/(p(H2) × p(I2)) = 80^2/(10 × 10) = 6400/100 = 64 (no units, as there are equal numbers of moles of gas on each side).

Correct: (A)

2(+1) + x + 4(−2) = 0, so x = +6.

Correct: (C)

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

Correct: (B)

EMF = 0.34 − (−2.37) = +2.71 V

Correct: (D)

Each Al^3+ ion needs three electrons, so moles of electrons = 3 × 0.0100 = 0.0300; Q = 0.0300 × 96500 = 2895 C.

Correct: (A)

In a hydrogen fuel cell, hydrogen and oxygen react to form water only, generating an electric current; the cell keeps working for as long as fuel is supplied.

Correct: (C)

Moles of H^+ = 2.50 × 10^-3 and moles of OH^- = 1.25 × 10^-3, so excess H^+ = 1.25 × 10^-3 mol in 50.0 cm^3; [H^+] = 0.0250 mol dm^-3 and pH = −log(0.0250) = 1.60

Correct: (B)

pKa = −log(3.2 × 10^-4) = 3.49

Correct: (D)

Added OH^- is removed by reaction with the large reserve of undissociated acid: CH3COOH + OH^- → CH3COO^- + H2O, so the pH barely changes.

Correct: (A)

[H^+] = 10^-pH = 10^-2.50 = 3.16 × 10^-3 mol dm^-3

Correct: (C)

The equivalence point of a weak acid-strong base titration lies above pH 7 because the conjugate base hydrolyses, so an indicator that changes colour in alkaline solution, such as phenolphthalein, must be used.

Correct: (B)

The solubility of the Group 2 hydroxides increases down the group, so barium hydroxide is the most soluble; the sulfates show the opposite trend.

Correct: (D)

Silver bromide is a cream precipitate; silver chloride is white, silver iodide is yellow, and silver fluoride is soluble so no precipitate forms with fluoride.

Correct: (A)

Vanadium is in oxidation state +5 in V2O5; it is reduced by SO2 to VO2 (vanadium +4) and is then re-oxidised by oxygen back to V2O5.

Correct: (C)

The initial pale blue precipitate of Cu(OH)2 redissolves in excess ammonia to form the deep blue complex ion [Cu(NH3)4(H2O)2]^2+.

Correct: (B)

3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O; with hot concentrated alkali, chlorine disproportionates to chloride ions and chlorate(V) ions.

Correct: (D)

The hydroxide ion is a nucleophile and substitutes the bromine atom, giving ethanol by nucleophilic substitution.

Correct: (A)

Electrophilic addition of sulfuric acid follows Markovnikov's rule, placing the sulfate group on the middle carbon, and hydrolysis of the intermediate gives propan-2-ol.

Correct: (C)

C10H22 → C2H4 + C8H18; the atoms balance (C: 2 + 8 = 10; H: 4 + 18 = 22), and the other product is the alkane octane.

Correct: (B)

Secondary alcohols are oxidised to ketones; propan-1-ol gives a carboxylic acid under reflux, and the tertiary alcohol 2-methylpropan-2-ol resists oxidation.

Correct: (D)

Heated aluminium oxide dehydrates ethanol to ethene (concentrated sulfuric acid at 170°C can also be used).

Correct: (A)

A nucleophile is an electron pair donor; OH^- has lone pairs that it can donate to form a new covalent bond, whereas H^+, NO2^+ and Br^+ are electron pair acceptors (electrophiles).

Correct: (C)

2,4-dinitrophenylhydrazine reacts with carbonyl compounds (aldehydes and ketones) to form an orange or yellow crystalline precipitate, which can be used to identify the carbonyl compound from its melting point.

Correct: (B)

Alkaline hydrolysis breaks the peptide links in the protein, giving the amino acids as their carboxylate (sodium) salts.

Correct: (D)

AlCl3 accepts a chloride ion from CH3Cl, generating the carbocation CH3^+, which is the electrophile that attacks the benzene ring.

Correct: (A)

The very broad 2500-3300 cm^-1 absorption is the O-H bond of a carboxylic acid and the peak near 1720 cm^-1 is its C=O bond; an alcohol shows a broad O-H peak at 3200-3600 cm^-1 but has no C=O peak.

Correct: (C)

Rf = distance moved by the spot/distance moved by the solvent front = 4.5/9.0 = 0.50

Correct: (B)

Nylon-6,6 is a polyamide formed when monomers join together with the loss of small molecules; the others are addition polymers formed from alkene monomers with no by-product.