Chemistry Paper 3 November 2018

1. A$24.00$
2. B$24.32$
3. C$24.79$
4. D$25.00$

1. A$1s^2 2s^2 2p^6 3s^2$; the peaks are the three isotopes, with heights proportional to abundance
2. B$1s^2 2s^2 2p^6$; the peaks are doubly and triply charged ions
3. C$1s^2 2s^2 2p^6 3s^1$; the peaks are from three different ionisation energies
4. D$1s^2 2s^2 2p^6 3s^2 3p^2$; the peaks are fragmentation products

1. A$4.20$
2. B$4.23$
3. C$4.54$
4. D$4.85$

1. Athe mass in grams of exactly one mole of atoms of the element, measured directly and precisely on an analytical balance
2. Bthe total number of protons and neutrons found together in the nucleus of a single atom of the element
3. Cthe mass of one atom of an element compared directly with the mass of one atom of hydrogen-1, atom for atom
4. Dthe weighted average mass of the isotopes of an element relative to 1/12 the mass of a carbon-12 atom

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1. ABecause Kw = [H+][OH-] = 1x10^-14 and [H+] = [OH-] in pure water, [H+]^2 = 1x10^-14, so [H+] = 1x10^-7 mol dm-3 and pH = -log(1x10^-7) = 7
2. BBecause the concentration of water itself is defined as 7 mol dm-3 at 25 degrees C, fixing the pH at that value
3. CBecause Kw doubles the concentration of H+ relative to OH-, so pH is fixed at half of 14 by definition alone
4. DBecause water has exactly 7 hydrogen ions present for every 7 hydroxide ions in any sample, regardless of the temperature, concentration or degree of dilution involved

1. Athe equilibrium constant for the complete ionisation of a strong acid dissolved in water
2. Bthe concentration of hydrogen ions alone in pure water at a stated temperature
3. Cthe sum of the concentrations of hydrogen and hydroxide ions present in any aqueous solution
4. Dthe product of the concentrations of hydrogen ions and hydroxide ions in water, [H+][OH-]

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1. Amol dm-3
2. Bmol dm-6
3. Cno units, since Kw is a pure equilibrium ratio
4. Dmol2 dm-6

1. A$+844$ kJ mol$^{-1}$
2. B$-1446$ kJ mol$^{-1}$
3. C$-844$ kJ mol$^{-1}$
4. D$+142$ kJ mol$^{-1}$

1. AThe oxide ion has a smaller radius than the oxygen atom
2. BThe O$^-$ ion is unstable and decomposes
3. CEnergy must be supplied to overcome the repulsion between the incoming electron and the already negative O$^-$ ion
4. DOxygen has a very high electronegativity

1. ARate $= k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]^2$; $k = 4.76$
2. BRate $= k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]$; $k = 0.081$ mol$^{-1}$ dm$^3$ s$^{-1}$
3. CRate $= k[\text{S}_2\text{O}_8^{2-}]$; $k = 1.4 \times 10^{-3}$ s$^{-1}$
4. DRate $= k[\text{S}_2\text{O}_8^{2-}]^2[\text{I}^-]$; $k = 1.01$

1. AMgO reacts readily with the hot furnace gases, forming a fresh protective coating on the lining that regenerates itself continuously as it wears
2. BMgO has a very high melting point, because it is an ionic lattice of small, doubly-charged ions held together by strong electrostatic forces
3. CMgO is a soft, low-melting solid that can be easily reshaped to fit the internal contours of a furnace
4. DMgO conducts heat away extremely quickly, in the same way a metal does, cooling the furnace walls

1. A2S2O8^2- + I- -> 2SO4^2- + I+
2. BS2O8^2- + 2I- -> 2SO3^2- + I2 + O2
3. CS2O8^2- + 2I- -> 2SO4^2- + I2
4. DS2O8^2- + I- -> SO4^2- + IO-

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1. AWithdraw a series of samples at fixed time intervals, quench each one immediately with excess sodium carbonate solution to stop the reaction, then measure the mass of carbon dioxide gas lost from each flask over the whole experiment
2. BWithdraw samples at intervals and titrate the iodine formed against standard sodium thiosulphate solution (an 'iodine clock' variant with starch as indicator), since the volume needed at each time gives the amount of iodine present
3. CMeasure the temperature of the reaction mixture continuously throughout the experiment, since the reaction rate is assumed to be directly proportional to the temperature rise recorded on an ordinary laboratory thermometer
4. DWeigh the sealed reaction flask continuously on an electronic balance throughout the experiment, since the total mass of the mixture is assumed to decrease steadily and measurably as the reaction proceeds towards completion

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1. AFirst order overall only
2. BSecond order overall
3. CThird order overall
4. DZero order overall

1. AIt would stay the same, since the two changes cancel each other out completely
2. BIt would double, since only one concentration change matters for a second-order reaction
3. CIt would increase eight-fold overall, since three separate reacting species are counted in the balanced overall equation
4. DIt would increase four-fold, since the rate depends on the product of both concentrations, each doubled

1. A$+0.83$ V
2. B$+1.23$ V
3. C$-1.23$ V
4. D$+0.43$ V

1. A$0.67$ g
2. B$1.34$ g
3. C$2.69$ g
4. D$129.6$ g

1. A$0.0095$ g
2. B$0.0148$ g
3. C$0.0237$ g
4. D$0.0592$ g

1. AH2 + O2 -> H2O2
2. B2H2 + O2 -> 2H2O2
3. CH2 + 2O2 -> 2H2O
4. D2H2 + O2 -> 2H2O

1. AFuel cells are able to operate safely at much higher temperatures than a lead-acid accumulator could reasonably be expected to tolerate
2. BFuel cells keep producing current for as long as fuel is supplied, without needing to be recharged, and their only product is water rather than toxic lead compounds
3. CFuel cells tend to be physically smaller and considerably lighter than a lead-acid accumulator of broadly similar electrical capacity, voltage and overall power output
4. DFuel cells can typically be recharged far more times than a lead-acid accumulator before either system eventually wears out and needs full replacement

1. AThe two electrolytes somehow react directly with each other in solution to cancel out any potential difference that might otherwise exist between the two types of cell entirely
2. BThe acidic and alkaline electrolytes simply happen to have numerically equal standard electrode potentials by coincidence, as listed in the data booklet
3. CThe overall cell reaction, 2H2 + O2 -> 2H2O, and hence the free energy change and e.m.f., is identical in both cases, regardless of which electrolyte is used
4. DFuel cells do not actually depend on electrode potentials at all in practice, so the choice of electrolyte has no effect on any measurable cell property

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1. A5SO2 + 2MnO4- + 2H2O -> 5SO4^2- + 2Mn2+ + 4H+
2. BSO2 + MnO4- + H2O -> SO4^2- + MnO2 + 2H+
3. C5SO2 + 2MnO4- -> 5SO3 + 2MnO + O2
4. D2SO2 + MnO4- + 2H2O -> 2SO4^2- + Mn2+ + 4H+ + O2

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1. ADecomposition temperature falls: cation charge increases down the group
2. BDecomposition temperature rises: the nitrate ion becomes smaller down the group
3. CThere is no trend; all Group II nitrates decompose at the same temperature
4. DDecomposition temperature rises: cation size increases, charge density falls, so the nitrate ion is polarised less

1. AIodide, because the outer electron is further from the nucleus and more shielded, so it is lost more readily
2. BBromide, because Br$^-$ is smaller and more polarising
3. CBromide, because HBr is a stronger acid than HI
4. DIodide, because HI has the strongest H–X bond

1. A$\text{SO}_4^{2-} + 8\text{H}^+ + 6e^- \rightarrow \text{S} + 4\text{H}_2\text{O}$
2. B$\text{SO}_4^{2-} + 4\text{H}^+ + 2e^- \rightarrow \text{SO}_2 + 2\text{H}_2\text{O}$
3. C$\text{S} + 2e^- \rightarrow \text{S}^{2-}$
4. D$\text{SO}_4^{2-} + 10\text{H}^+ + 8e^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}$

1. AMelting points increase then decrease down the group in a regular zig-zag pattern, caused by alternating crystal structures at each successive element in turn
2. BMelting points generally increase steadily all the way down the group, because the number of delocalised electrons contributed by each atom increases considerably from Be all the way to Ba
3. CMelting points stay essentially constant all the way down the entire group, because metallic bonding strength does not depend on atomic radius at all
4. DMelting points generally decrease down the group, because the metallic radius increases so the delocalised electrons are further from the nuclei and metallic bonding weakens

1. A2NaBr + H2SO4 -> Na2SO4 + Br2 + H2
2. B2NaBr + 3H2SO4 -> 2Na2SO4 + Br2 + SO2 + 3H2O
3. C2NaBr + 3H2SO4 -> 2NaHSO4 + Br2 + SO2 + 2H2O
4. DNaBr + H2SO4 -> NaHSO4 + HBr

1. ANaI + H2SO4 -> NaHSO4 + HI
2. B8NaI + 5H2SO4 -> 8NaHSO4 + 4I2 + H2S + 2 H2O (g)
3. C8NaI + 9H2SO4 -> 8NaHSO4 + 4I2 + H2S + 4H2O
4. D2NaI + H2SO4 -> Na2SO4 + I2 + H2S

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1. AA brown/black solid forms immediately, showing that chlorine has been reduced to elemental chlorine gas
2. BOnly steamy fumes of hydrogen chloride gas are produced, with no colour change and no redox reaction taking place
3. CA colourless gas with a distinct rotten-egg smell is produced, showing that sulphur has been reduced all the way to hydrogen sulphide
4. DNo reaction occurs at all, since NaCl does not react with any acid under any conditions

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1. AVolatility increases; the molecules become lighter down the group
2. BVolatility decreases; larger molecules with more electrons have stronger van der Waals forces
3. CVolatility decreases; the covalent bonds within the molecules get stronger
4. DVolatility is unchanged; all halogens are diatomic

1. AStability decreases down the group because the H–X bond gets longer and weaker
2. BStability decreases down the group because the halogens become more electronegative
3. CStability increases down the group because the hydrides become more polar
4. DStability is constant because all have one H–X bond

1. AHeat the sample and weigh the residue
2. BTitrate a solution of a weighed sample against standard AgNO$_3$ with potassium chromate indicator
3. CTitrate a weighed sample against standard NaOH using phenolphthalein
4. DMeasure the pH of a solution of the sample

1. ANeither MgO nor CaO dissolves in water to any measurable extent whatsoever, since both are extremely stable, essentially insoluble ionic oxides of identical overall lattice type and structure
2. BMgO is actually more soluble in water than CaO, because the smaller Mg2+ ion hydrates considerably more strongly than the larger Ca2+ ion does in solution
3. CMgO and CaO are equally soluble in water, since both are Group (II) oxides built from exactly the same overall ionic lattice type and ionic charge
4. DCaO is more soluble in water than MgO, because the fall in hydration enthalpy down the group is smaller than the fall in lattice enthalpy, making dissolving more favourable

1. Athey conduct electricity extremely well when heated, generating light in the same way as a filament bulb
2. Bthey react rapidly with atmospheric oxygen to form a stable, brightly coloured solid residue that continues glowing for several minutes afterwards
3. Cheating excites their electrons to higher energy levels, and visible light of a characteristic colour is emitted as the electrons fall back down
4. Dtheir compounds explode violently on contact with air, producing a bright flash of white light

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1. AAg+ + Cl- -> AgCl
2. BAg+ + 2Cl- -> AgCl2-
3. C2Ag+ + Cl- -> Ag2Cl+
4. DAg2+ + Cl- -> AgCl+

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1. AHF
2. BHCl
3. CHBr
4. DHI

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1. AMgO and CaO react with water at exactly the same rate, since both contain the same O2- ion
2. BCaO is a covalent oxide while MgO is ionic, and covalent oxides generally react faster with water than ionic ones do
3. CCaO has a lower lattice enthalpy than MgO (larger Ca2+ ion), so its lattice is broken up more readily, allowing faster reaction with water
4. DCaO has a much smaller ionic radius than MgO, giving it a much higher charge density that speeds up the reaction considerably more than expected

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1. AEach carbon is sp$^3$ hybridised with four sigma bonds and no $\pi$ bonds
2. BThree localised C=C double bonds alternate with three single bonds
3. CEach carbon forms two sigma bonds and two localised $\pi$ bonds
4. DEach carbon is sp$^2$ hybridised with three sigma bonds; the remaining p orbitals overlap sideways to give a delocalised $\pi$ system

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1. AFree-radical substitution initiated by UV light
2. BElectrophilic addition across one of the C=C bonds
3. CElectrophilic substitution: FeCl$_3$ generates Cl$^+$, which attacks the ring to give an arenium ion, then H$^+$ is lost
4. DNucleophilic addition: Cl$^-$ attacks the ring directly

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1. AAn alkene; isolated by fractional distillation of the residue
2. BA ketone; isolated by recrystallisation
3. CAn aldehyde; isolated by distilling it off as it forms
4. DAn ester; isolated by solvent extraction

1. ACH3CH2CH=CHCH2OH remains completely unchanged, since bromine does not react at all with alkenols of this type
2. BCH3CH2CH=CHCH2Br, from simple substitution of the -OH group by bromine
3. CCH3CH2CBr2-CH2-CH2OH, from addition of two bromine atoms onto a single carbon
4. DCH3CH2CHBr-CHBr-CH2OH, from electrophilic addition of Br2 across the C=C double bond

1. ANucleophilic substitution, replacing the double bond with two single bonds
2. BOxidation, converting the alcohol directly into a carboxylic acid
3. CEsterification, forming an ester at the -OH group and releasing water
4. DElectrophilic addition, forming a saturated diol

1. AThey are optical isomers (enantiomers), non-superimposable mirror images of each other
2. BThey are structural (positional) isomers, since they have the same molecular formula but the Br atom is on a different carbon in each
3. CThey are the same compound, simply drawn in two different orientations on the page
4. DThey are geometric (cis-trans) isomers, differing only in the spatial arrangement of groups across a shared double bond in each molecule

1. AConcentrated sulphuric acid alone, at room temperature
2. BCold, dilute potassium manganate(VII) solution, with no heating required at any stage
3. CExcess acidified potassium manganate(VII) or dichromate(VI), heated under reflux
4. DLithium aluminium hydride in dry ether

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1. AFeCl4- is inherently unstable and forces the arenium ion to lose H+ immediately after it forms, regardless of any relative stability difference between the two species
2. BLosing H+ restores the fully delocalised pi system of benzene, which is significantly more energetically stable than the localised arenium intermediate
3. CThe arenium intermediate cannot exist for more than a brief instant because chlorine itself is far too reactive a species to allow it to persist
4. DThe arenium intermediate itself behaves as a strong base, and H+ is simply too weak an acid to remain attached to it for any meaningful length of time

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1. ACH$_3$CH$_2$OH
2. BCH$_3$COOH
3. CCH$_2$=CH$_2$
4. DCH$_2$=CH–OOCCH$_3$

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1. A$-\text{CH}=\text{CH}-$
2. B$-\text{CH}_2-\text{CH(OH)}-$
3. C$-\text{CH}_2-\text{CH(ONa)}-$
4. D$-\text{CH}_2-\text{CH(COOH)}-$

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1. AH$_2$N–CH(CH$_3$)–COO$^-$; only the carboxyl group reacts
2. BH$_2$N–CH(CH$_3$)–COOH; it is amphoteric because it contains carbon
3. C$^+$H$_3$N–CH(CH$_3$)–COO$^-$; the –NH$_2$ accepts H$^+$ and the –COOH donates H$^+$
4. D$^+$H$_3$N–CH(CH$_3$)–COOH; only the amine group reacts

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1. AA carboxylic acid group and a separate alcohol group
2. BAn amide (peptide-type) group and a carbon-carbon double bond (alkene)
3. CAn ester group and a separate primary amine (-NH2) group
4. DAn ester group and a carbon-carbon double bond (alkene)

1. ALithium aluminium hydride, $LiAlH_4$, in dry ether
2. BWarm aqueous sodium hydroxide solution, with no separate catalyst needed at all
3. CConcentrated sulphuric acid, heated under reflux with the mixture for several hours
4. DAqueous bromine, added dropwise at room temperature until the colour persists

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1. ANeither W nor Z forms enantiomers, since neither has four different groups on the same carbon in this case
2. BOnly Z forms enantiomers, because W's methyl side chain is not different enough from a hydrogen atom to count
3. COnly W forms enantiomers, since Z's extra -CH2OH side chain makes its central carbon bonded to two effectively identical groups instead of four different ones
4. DBoth W and Z form enantiomers, since each has a carbon bonded to four different groups: an amine, a carboxyl group, a hydrogen, and a distinct R group

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1. AAddition polymerisation
2. BCo-polymerisation with a diol
3. CCondensation polymerisation
4. DRing-opening polymerisation

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1. AThe -COOH group is too weak an acid to lose a proton under any circumstances in aqueous solution
2. BThe -NH2 group is not basic enough to accept a proton from water at neutral pH
3. CZwitterions are simply more soluble in water than any comparable neutral molecule, so the equilibrium favours them regardless of the relative acid-base strength involved
4. DIntramolecular proton transfer occurs from the more acidic -COOH group to the more basic -NH2 group, giving -COO- and -NH3+ on the same molecule

1. AX is acetanilide, C$_6$H$_5$NHCOCH$_3$; the –NH$_2$ group is protected as an amide
2. BX is chlorobenzene; to activate the ring
3. CX is phenyl ethanoate; to make the ring more soluble
4. DX is nitrobenzene; to introduce the nitro group

1. AHydrolysis; dilute HCl (or NaOH), heat under reflux
2. BReduction; Sn and concentrated HCl
3. COxidation; acidified K$_2$Cr$_2$O$_7$
4. DElimination; ethanolic KOH

1. AKetone; butanone, CH$_3$COCH$_2$CH$_3$
2. BEster; methyl propanoate, CH$_3$CH$_2$COOCH$_3$
3. CAldehyde; butanal, CH$_3$CH$_2$CH$_2$CHO
4. DAlcohol; butan-2-ol, CH$_3$CH(OH)CH$_2$CH$_3$

1. ASulphonation, using concentrated sulphuric acid alone, heated gently for several minutes at moderate temperature
2. BNitration (electrophilic substitution), using concentrated nitric acid and concentrated sulphuric acid
3. CBromination (electrophilic substitution), using bromine water alone, with no other reagent required at any stage
4. DReduction, using tin metal and concentrated hydrochloric acid, followed by prolonged warming under reflux conditions

1. AC6H6 + HNO3 -> C6H5NO2 + H2O (with concentrated H2SO4 as catalyst, 55 degrees C)
2. BC6H6 + 3HNO3 -> C6H3(NO2)3 + 3H2O, introducing all three nitro groups in a single step
3. CC6H6 + NaNO2 + HCl -> C6H5N2+ + NaCl + H2O
4. DC6H6 + HNO3 -> C6H5NH2 + O2

1. AConcentrated sulphuric acid alone, heated under reflux
2. BAqueous bromine solution alone, used at room temperature, with no other reagent needed at all
3. CLithium aluminium hydride in dry ether, then dilute acid
4. DTin and concentrated hydrochloric acid, then aqueous sodium hydroxide to free the amine

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1. ACarbon-14, because it has a very long half-life
2. BIodine-131, because it is a gamma emitter
3. CUranium-235, because it is highly radioactive
4. DPhosphorus-32, because it is a beta emitter that exposes X-ray film and is readily incorporated into DNA

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1. AIt has the largest relative molecular mass in the mixture
2. BIt carries a net positive charge — the pH is below its isoelectric point (e.g. a basic amino acid such as lysine)
3. CIt carries a net negative charge — the pH is above its isoelectric point
4. DIt is uncharged and at its isoelectric point

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1. AThe brand of gel paper used to run the electrophoresis
2. BThe size (mass) of the amino acid ion, and the voltage applied across the gel
3. CThe colour of the amino acid in white light
4. DThe exact time of day at which the electrophoresis experiment happens to be carried out

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1. AAspartic acid and lysine, since both are strongly attracted to the positive electrode by their charged amine side groups
2. BAlanine and glutamic acid, since both happen to have exactly zero net charge at this pH and so simply drift randomly towards either electrode over time
3. CAspartic acid and glutamic acid, since both carry a net negative charge at this pH from their extra -COOH side chain, which is deprotonated
4. DLysine and alanine, since both carry a net negative charge at this pH because of their basic side-chain amine groups

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1. ALysine reacts with the gel buffer to form a temporary negative ion that then migrates towards the negative electrode
2. BLysine has no charge at all at this pH, but drifts towards the negative electrode purely by diffusion
3. CIts extra side-chain -NH2 group is protonated at this pH, giving lysine an overall positive charge that is attracted to the negative electrode
4. DIts side-chain -COOH group is deprotonated at this pH, giving lysine an overall negative charge that happens to move towards the negative electrode anyway

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1. AThey give a very large surface-area-to-volume ratio, so less of the expensive Pt/Rh is needed for the same activity
2. BThey allow the catalyst to dissolve in the exhaust gases
3. CThey lower the temperature of the exhaust gases
4. DThey make the converter heavier and more durable

1. AHeterogeneous catalysis; the metal is consumed and regenerated in a redox cycle
2. BHeterogeneous catalysis; gases adsorb onto active sites, bonds weaken and align, lowering the activation energy, then products desorb
3. CHomogeneous catalysis; the metal dissolves in the exhaust stream and forms intermediates
4. DHomogeneous catalysis; the metal raises the activation energy of the reverse reaction

1. A$\text{SO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_3$
2. B$2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$
3. C$\text{CaCO}_3 + \text{SO}_2 \rightarrow \text{CaSO}_3 + \text{CO}_2$
4. D$\text{S} + \text{O}_2 \rightarrow \text{SO}_2$

1. A2CO(g) + 2NO(g) -> 2CO2(g) + N2(g)
2. B2CO(g) + N2(g) -> 2CNO(g), formed when carbon monoxide reacts directly with atmospheric nitrogen gas
3. C2CO(g) + O2(g) -> 2CO3(g), the carbon monoxide reacting directly with oxygen at the catalyst's active sites
4. DCO(g) + NO(g) -> CNO2(g), formed when carbon monoxide reacts directly with nitrogen monoxide in a 1:1 ratio

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1. AS + O2 -> SO3
2. B2S + O2 -> S2O
3. CS + 2O2 -> SO4
4. DS + O2 -> SO2

1. AIt has no lasting environmental effect, since it is rapidly broken down into oxygen and sulphur within minutes of release
2. BIt dissolves in rainwater and is oxidised further to form acid rain, which lowers soil and water pH, damages vegetation and aquatic life, and corrodes buildings
3. CIt reacts with atmospheric nitrogen gas to form a harmless, chemically stable salt that simply falls to the ground over time as fine dust
4. DIt rises essentially unchanged into the stratosphere, where it is thought to directly destroy the ozone layer in exactly the same way that CFC gases are known to do

1. ACa(OH)2 + SO2 -> CaSO3 + H2O, spraying a slurry of calcium hydroxide into the flue gas
2. BCaSO3 + SO2 -> CaSO4 + S, mixing calcium sulphite directly with more sulphur dioxide
3. CCa(OH)2 + 2SO2 -> Ca(HSO3)2, forming a soluble salt that is simply released with the flue gas
4. DCaO + SO2 -> Ca + SO3, reducing sulphur dioxide directly to sulphur trioxide

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1. AWater vapour
2. BCarbon dioxide itself
3. CUnburnt hydrocarbons
4. DNitrogen gas

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