A-Level Chemistry Practice Test 19

Correct: (B)

Ar = (69 × 0.600) + (71 × 0.400) = 41.4 + 28.4 = 69.8.

Correct: (A)

After one electron is removed, Na^+ has the stable electronic configuration of neon; removing a second electron means breaking into this stable inner shell, which requires a very large amount of energy.

Correct: (C)

A chlorine atom has 17 electrons; Cl^- has gained one extra electron, giving 18 electrons, the same number as argon.

Correct: (D)

K is in Period 4 and Group 1: it has one more electron shell than Na and Mg, and in Period 4 it has a smaller effective nuclear charge than Ca, so its outer electron is held least tightly and its radius is largest.

Correct: (A)

In phosphorus the three 3p electrons are unpaired in separate orbitals; in sulfur one 3p orbital contains a pair of electrons, and the repulsion between them makes one electron easier to remove.

Correct: (C)

V = nRT/p = (0.250 × 8.31 × 300) / 100000 = 6.23 × 10^-3 m^3 = 6.23 dm^3.

Correct: (D)

moles = 0.200 × 0.250 = 0.0500 mol; mass = 0.0500 × 58.5 = 2.93 g.

Correct: (B)

Percentage yield = (6.40 / 8.00) × 100 = 80.0%.

Correct: (C)

Atom economy = (Mr of desired product / total Mr of reactants) × 100 = (56 / 100) × 100 = 56%.

Correct: (A)

moles of NaOH = 0.100 × 0.0250 = 0.00250 mol = moles of HCl; concentration = 0.00250 / 0.0200 = 0.125 mol dm^-3.

Correct: (D)

C: 40.0/12 = 3.33; H: 6.7/1 = 6.7; O: 53.3/16 = 3.33. Dividing by 3.33 gives a ratio of 1 : 2 : 1, so the empirical formula is CH2O.

Correct: (C)

NH3 has three bonding pairs and one lone pair of electrons. The lone pair repels the bonding pairs more strongly, giving a trigonal pyramidal shape and reducing the bond angle to 107°.

Correct: (B)

In NH3 the bond dipoles do not cancel because the molecule is pyramidal. CO2, BF3 and CH4 are symmetrical, so their bond dipoles cancel out.

Correct: (D)

Lattice enthalpy (of formation) is the exothermic enthalpy change when 1 mol of an ionic lattice is formed from its separate gaseous ions.

Correct: (A)

CH3OH contains an O-H group. The hydrogen attached to the highly electronegative oxygen atom forms hydrogen bonds with lone pairs on oxygen atoms of neighbouring molecules.

Correct: (C)

CsF has the greatest difference in electronegativity between its elements (caesium is the least electronegative metal and fluorine the most electronegative element), so its bonding has the most ionic character.

Correct: (B)

XeF4 has four bonding pairs and two lone pairs arranged in an octahedral arrangement of electron pairs, with the lone pairs opposite each other, giving a square planar shape.

Correct: (D)

q = mcΔT = 50.0 × 4.18 × 8.00 = 1672 J = 1.67 kJ.

Correct: (A)

By Hess's law: ΔHf = ΔHc(C) + 2ΔHc(H2) − ΔHc(CH4) = (−393) + 2(−286) − (−890) = −393 − 572 + 890 = −75 kJ mol^-1.

Correct: (C)

A gas has far more disorder (entropy) than solids, so forming CO2 gas from a solid carbonate gives a positive entropy change.

Correct: (D)

ΔG = ΔH − TΔS = −100 − (400 × −0.200) = −100 + 80 = −20 kJ mol^-1, so the reaction is feasible at 400 K.

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)

The reaction becomes feasible when ΔG = 0, so T = ΔH/ΔS = 50000 / 100 = 500 K. Above 500 K, ΔG is negative.

Correct: (C)

The change in rate = 2^2 × 0.5 = 2, so the rate doubles.

Correct: (B)

Since rate has units mol dm^-3 s^-1 and rate = k[A]^2, k = rate/[A]^2, giving units mol^-1 dm^3 s^-1 for a second-order reaction.

Correct: (D)

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

Correct: (A)

As T increases, the exponential term e^(-Ea/RT) increases, so a much larger fraction of collisions have enough energy to react.

Correct: (C)

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

Correct: (D)

Kc = [products]/[reactants] with each concentration raised to the power of its stoichiometric coefficient: Kc = [NH3]^2 / ([N2][H2]^3).

Correct: (A)

A reacted = 0.800 − 0.600 = 0.200 mol, so B formed = 0.400 mol. [A] = 0.600 and [B] = 0.400 mol dm^-3. Kc = [B]^2/[A] = (0.400)^2 / 0.600 = 0.267 mol dm^-3.

Correct: (B)

There are 2 moles of gas on each side, so the change in the number of moles of gas is zero and Kp = Kc(RT)^0 = Kc.

Correct: (C)

There are 3 moles of gas on the left and 2 moles on the right, so increasing the pressure shifts the equilibrium to the right, towards fewer moles of gas, increasing the yield of SO3.

Correct: (D)

A catalyst increases the rates of the forward and reverse reactions equally, so the position of equilibrium and the value of Kc are unchanged. Kc depends only on temperature.

Correct: (A)

Using oxidation states: 2(+1) + 2(Cr) + 7(−2) = 0, so 2 + 2Cr − 14 = 0 and Cr = +6.

Correct: (B)

The most negative electrode potential corresponds to the metal that loses electrons most readily, so zinc is the strongest reducing agent of the three.

Correct: (C)

E(cell) = E(more positive) − E(more negative) = +0.34 − (−0.76) = +1.10 V.

Correct: (D)

Q = It = 0.500 × 1930 = 965 C, which is 965/96500 = 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)

EMF = +0.77 − (+0.54) = +0.23 V. The EMF is positive, so Fe^3+ can oxidise I^- to I2 and the reaction is feasible.

Correct: (B)

HCl is a strong acid and is fully ionised, so [H^+] = 0.0100 mol dm^-3. pH = −log(0.0100) = 2.00.

Correct: (C)

For a weak acid, [H^+] = √(Ka × c) = √(1.00 × 10^-5 × 0.100) = √(1.00 × 10^-6) = 1.00 × 10^-3 mol dm^-3. pH = 3.00.

Correct: (D)

pOH = −log(0.0500) = 1.30. pH = 14.00 − 1.30 = 12.70.

Correct: (A)

pH = pKa + log([salt]/[acid]) = 4.74 + log(0.100/0.100) = 4.74 + 0 = 4.74.

Correct: (B)

At the half-neutralisation point, half the acid has been converted to its salt, so [acid] = [salt] and pH = pKa + log(1) = pKa.

Correct: (B)

The solubility of Group 2 hydroxides increases down the group: Mg(OH)2 is sparingly soluble while Ba(OH)2 is appreciably soluble. Reactivity of the metals also increases down the group.

Correct: (D)

Cl2 + 2NaOH → NaCl + NaClO + H2O. Chlorine is both oxidised (0 to +1 in NaClO) and reduced (0 to −1 in NaCl), so this is a disproportionation reaction.

Correct: (A)

A pale blue precipitate of Cu(OH)2 forms first and then dissolves in excess ammonia as ligand substitution forms the [Cu(NH3)4(H2O)2]^2+ complex, which is deep blue.

Correct: (C)

Ba^2+ ions form a white precipitate of BaSO4 with sulfate ions. The acid is added first to remove carbonate ions, which would also give a white precipitate with Ba^2+.

Correct: (B)

Fe^3+ ions form a red-brown (rust-coloured) precipitate of iron(III) hydroxide, Fe(OH)3.

Correct: (D)

Ultraviolet light provides the energy for homolytic fission of the Cl-Cl bond, producing two chlorine radicals, which start the chain reaction.

Correct: (A)

In nucleophilic substitution the carbon-halogen bond must break. The C-I bond is the weakest of the carbon-halogen bonds, so 1-iodobutane reacts fastest.

Correct: (C)

By Markovnikov's rule, the hydrogen adds to the carbon of the double bond that already has the most hydrogen atoms, because this gives the more stable secondary carbocation intermediate. The major product is 2-bromopropane.

Correct: (B)

A primary alcohol is oxidised to an aldehyde and then, when heated under reflux with excess oxidising agent, further oxidised to the carboxylic acid, propanoic acid.

Correct: (D)

Hot ethanolic KOH gives elimination (dehydrohalogenation) rather than substitution, removing HBr from the halogenoalkane to form the alkene, propene.

Correct: (A)

Cracking breaks a long-chain alkane into a shorter-chain alkane and an alkene. The equation is balanced: C, 10 = 8 + 2; H, 22 = 18 + 4.

Correct: (C)

Tollens' reagent (ammoniacal silver nitrate) is a mild oxidising agent. Aldehydes reduce Ag^+ to silver metal, forming a silver mirror, but ketones cannot be oxidised further and do not react.

Correct: (B)

Alkaline hydrolysis of an ester gives the sodium salt of the carboxylic acid (sodium ethanoate) and the alcohol (ethanol).

Correct: (D)

Concentrated sulfuric acid protonates nitric acid and removes water, generating the nitronium ion, NO2^+, which is the electrophile that attacks the benzene ring.

Correct: (A)

In phenylamine the nitrogen lone pair overlaps with the pi system of the benzene ring and becomes delocalised, so it is less available to accept a proton. In ethylamine the lone pair is localised on nitrogen.

Correct: (C)

At very low pH there is a high concentration of H^+ ions, so the basic -NH2 group is protonated to -NH3^+ while the -COOH group stays protonated. The predominant species is ^+NH3CH2COOH.

Correct: (B)

Rf = distance moved by the spot / distance moved by the solvent front = 4.2 / 6.0 = 0.70.