Chemistry 9701/42 October/November 2024: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) The equation for reaction 1 is shown.
reaction 1 X 2Y
Reaction 1 is first order with respect to the concentration of X. The half-life of the reaction, t1 ,
2 is 900 s at 20 °C.
(i) A solution of X with a concentration of 0.180 mol dm–3 is prepared at 20 °C. Calculate the average rate of reaction 1 over the first 1800 s.
average rate of reaction 1 = …… [2]
(ii) Complete the rate equation for reaction 1.
rate = …… [1]
(iii) Show that the rate constant, k, is 7.70 × 10–4 s–1 at 20 °C.
[1]
(iv) Calculate the initial rate of reaction 1 when the concentration of X is 0.150 mol dm–3.
Include units.
rate = …… units …… [2]
(b) Catalysts may be homogeneous or heterogeneous.
(i) Platinum is a transition element. Explain why transition elements behave as catalysts.
…… [1]
(ii) Name the metal catalyst in the Haber process and explain why it is a heterogeneous catalyst.
metal ……
…… [1]
(iii) Platinum acts as a heterogeneous catalyst in the removal of nitrogen dioxide, NO2, from the exhaust gases of car engines.
Describe the mode of action of a platinum catalyst in this process.
…… [2]
(iv) NO2 acts as a homogeneous catalyst in the oxidation of atmospheric sulfur dioxide, SO2.
Write equations for the two reactions that occur.
equation 1 ……
equation 2 ……
[1]
(c) SO2 dissolves in water, forming H2SO3.
H2SO3 can be oxidised under acidic conditions.
The relevant electrode reaction and its E o– value are shown.
SO42– + 4H+ + 2e– H2SO3 + H2O E o– = +0.17 V
Four more half-equations for reactions occurring under acidic conditions, and their E o– values, are shown.
H3BO3 + 3H+ + 3e– B + 3H2O E o– = –0.73 V
+ +
BiO + 2H + 3e – Bi + H2O E o– = +0.28 V
+
S + 2H + 2e – H2S E o– = +0.14 V
+
Sb + 3H + 3e – SbH3 E o– = –0.51 V
Select the oxidising agent that could oxidise H2SO3 to SO42– ions under acidic conditions.
Write an equation, and give the E o–cell value, for the reaction that occurs.
oxidising agent ……
equation ……
E o–cell = …… V
[3]
[Total: 14]
Mark scheme
1(a)(i) Change in concentration = 0.135 [1] 2
0.135 / 1800 = 7.5 × 10–5 [1]
1(a)(ii) Rate = k[X] [1] 1
1(a)(iii) k = 0.693 / t½ / k = 0.693 / 900 / k = ln2 / t½ / k = ln2 / 900 / t½ = ln2 / k [1] 1
1(a)(iv) 1.16 × 10–4 [1] 2
mol dm–3 s–1 [1]
1(b)(i) Any two from: 1
• variable oxidation state
• vacant / empty / unfilled d orbitals [1]
• can form dative bonds / can accept electrons
1(b)(ii) Iron AND 1
iron is solid, reactants are gases OR
catalyst and reactants are in different phases / states [1]
1(b)(iii) Two for one mark, three for two marks: 2
• adsorption of reactants by Pt
• bonds of reactants weaken [1]
• desorption of products [1]
1(b)(iv) NO2 + SO2 → NO + SO3 AND 1
2NO + O2 → 2NO2 [1]
1(c) BiO+ [1] 3
3H2SO3 + 2BiO+ + H2O → 3SO42- + 8H+ + 2Bi [1]
0.11 V [1]
(a) Predict and explain the variation in enthalpy change of hydration for the ions Na+, Mg2+ and Al 3+.
…… [3]
(b) Fig. 2.1 shows an incomplete energy cycle.
line B: Mg2+(g) + 2Cl –(g)
change 2: change 1: lattice energy of magnesium chloride,
ΔHlatt MgCl 2(s)
line A: MgCl 2(s)
change 3: enthalpy change of solution of magnesium chloride, line C:
ΔHsol MgCl 2(s)
Fig. 2.1
(i) Complete line C on Fig. 2.1. Include state symbols. [1]
(ii) Use both words and symbols to identify change 2 on Fig. 2.1.
Use changes 1 and 3 as examples of how this should be done.
[2]
(iii) Calculate a value for the lattice energy of magnesium chloride, ΔHlatt MgCl 2(s), by selecting and using appropriate data from Table 2.1.
Table 2.1
energy change value / kJ mol–1 enthalpy change of solution of magnesium chloride –155 enthalpy change of formation of magnesium chloride –642 first ionisation energy of magnesium +736
second ionisation energy of magnesium +1450 electron affinity of chlorine –349 enthalpy change of hydration of Mg2+ –1920 enthalpy change of hydration of Cl – –364
ΔHlatt MgCl 2(s) = …… kJ mol–1 [3]
(c) Define entropy.
…… [1]
(d) At 25 °C the enthalpy change of solution of compound Z is +26 kJ mol–1. The entropy change
of solution of Z at the same temperature is +52 J K–1 mol–1.
Calculate the value of the Gibbs free energy change, ΔG, for the solution of Z at 25 °C.
ΔG = …… kJ mol–1 [2]
(e) (i) Use your answer to (d) to predict whether or not Z is soluble in water at 25 °C. Explain your answer.
…… [1]
(ii) Predict whether Z becomes more or less soluble as the water is heated from 25 °C to
95 °C. Explain your answer.
…… [1]
[Total: 14]
Mark scheme
2(a) ΔHhyd increases from left to right due to increase in charge [1] 3
ionic radius decreases from left to right [1]
causing increased attractive force to water molecules [1]
2(b)(i) Mg2+ (aq) + 2Cl– (aq) [1] 1
2(b)(ii) Enthalpy change of hydration of magnesium ions and chloride ions [1] 2
ΔHhydMg2+ + 2ΔHhydCl– [1]
2(b)(iii) Selects 155, 1920 and 364 only [1] 3
2 × 364 [1]
answer –2493 [1]
2(c) The number of arrangements of the particles and of the energy in the system [1] 1
2(d) ΔG = ΔH – TΔS [1] 2
answer +10.5 [1]
2(e)(i) No, ΔG is positive [1] 1
2(e)(ii) Becomes more / soluble because ΔG becomes more negative / less positive / smaller / closer to zero [1] 1
(a) The pH of a saturated solution of calcium hydroxide is 12.35 at 298 K.
(i) Show that the concentration of hydroxide ions in a saturated solution of calcium hydroxide is 0.0224 mol dm–3 at 298 K.
[2]
(ii) Use data given in (i) to calculate the solubility product, Ksp, of calcium hydroxide at 298 K.
Include the units of Ksp in your answer.
Ksp = …… units …… [3]
(iii) A spatula measure of solid calcium chloride is stirred into a sample of saturated calcium hydroxide solution. All of the calcium chloride dissolves.
Describe one other observation that would be made and give an estimated value of the
pH of the solution obtained.
Explain both your answers.
observation ……
pH of solution ……
explanation ……
[3]
(iv) Calcium hydroxide reacts with dilute sulfuric acid to form calcium sulfate. Barium hydroxide behaves in a similar way, forming barium sulfate.
Explain why calcium sulfate is more soluble in water than barium sulfate.
…… [3]
(b) Some solid calcium is added to an excess of aqueous ethanoic acid, CH3COOH, and left until all the calcium has reacted. The resulting mixture, mixture D, contains no undissolved solids.
(i) Write an equation for the reaction of calcium with CH3COOH.
…… [1]
(ii) Use formulae of molecules and ions to identify two conjugate acid–base pairs present in mixture D.
Pair 1 should consist of organic species.
pair 1: …… ……
conjugate acid conjugate base
pair 2: …… ……
conjugate acid conjugate base
[2]
(iii) Write the expression for the Ka of CH3COOH.
Ka =
[1]
(iv) The concentration of calcium ethanoate, (CH3COO)2Ca, in mixture D is 0.394 mol dm–3.
The concentration of CH3COOH in mixture D is 0.270 mol dm–3.
The Ka of CH3COOH is 1.74 × 10–5 mol dm–3 at 298 K.
Calculate the pH of mixture D.
pH = …… [2]
(v) Write two equations to show how mixture D can act as a buffer solution.
equation 1 ……
equation 2 ……
[2]
[Total: 19]
Mark scheme
3(a)(i) [H+] = 10–12.35 OR [H+] = 4.47 × 10–13 [1] 2
Kw / 4.47 × 10–13 [1]
OR
pOH = 1.65 [1]
[OH–] = 10–1.65 [1]
3(a)(ii) Ksp = (0.0112)(0.0224)2 OR Ksp = [Ca2+][OH–]2 [1] 3
answer 5.62 × 10–6 [1]
mol3 dm–9 [1]
3(a)(iii) white solid / white ppt AND pH between 7.01 and 12.34 [1] 3
common ion effect [1]
hydroxide removed by precipitation OR ppt is Ca(OH)2 [1]
3(a)(iv) lattice energy and hydration energy greater for CaSO4 OR 3
lattice energy and hydration energy decrease down group [1]
hydration energy decreases more / is dominant factor [1]
enthalpy of solution is more endothermic for BaSO4 OR is more endothermic down grp [1]
3(b)(i) Ca + 2CH3COOH → Ca(CH3COO)2 + H2 [1] 1
3(b)(ii) Pair 1: CH3COOH and CH3COO– [1] 2
Pair 2: H3O+ and H2O OR H2O and OH– [1]
3(b)(iii) [H+ ][CH3COO− ] 1
[1]
[CH3COOH]
3(b)(iv) [CH3COO–] = 0.788, [CH3COOH] = 0.270 2
[H+] = 5.96 × 10–6 [1]
pH = 5.22 [1]
3(b)(v) CH3COO– + H+ → CH3COOH 2
OR Ca(CH3COO)2 + 2H+ → 2CH3COOH + Ca2+ [1]
CH3COOH + OH– → CH3COO– + H2O [1]
Transition metal atoms and transition metal ions form complexes by combining with species called ligands.
(a) When NaOH(aq) is added to an aqueous solution containing [Co(H2O)6]2+ a precipitation reaction occurs accompanied by a colour change.
In this reaction two of the water ligands each lose one H+ ion. The H+ ions are gained by OH– ions from the NaOH(aq).
(i) State the colour change seen in this precipitation reaction.
from …… to …… [1]
(ii) Complete the ionic equation for this precipitation reaction.
[Co(H2O)6]2+ + …… …… + ……
[1]
(iii) This precipitation reaction can also be described as a different type of reaction.
Name this type of reaction.
…… [1]
(b) L is an uncharged tridentate ligand. L donates three lone pairs to a metal atom or ion.
Cobalt forms an octahedral complex ion, E, with L. Complex ion E has a 2+ charge.
(i) Give the formula of E.
…… [1]
(ii) Identify the oxidation state of cobalt in E.
…… [1]
(iii) The d-orbitals of the cobalt atom or ion present in E are split in energy.
State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level.
number of d-orbitals at a higher energy level number of d-orbitals at a lower energy level
[1]
(iv) Define the term non-degenerate d-orbitals.
…… [1]
(c) The mineral chromite contains a compound which has the formula FeCrnO4. The oxidation state of iron in FeCrnO4 is +2.
A sample of 4.18 g of FeCrnO4 is dissolved in an excess of sulfuric acid. The resulting solution is made up to 250 cm3. This is solution F.
All the Fe2+ ions in 25.0 cm3 of solution F are oxidised to Fe3+ ions by exactly 18.7 cm3 of
0.0200 mol dm–3 KMnO4.
One MnO4– ion reacts with five Fe2+ ions. Assume no other oxidation reaction occurs.
(i) Write an equation for the reaction of Fe2+ ions with MnO4– ions in acid solution.
…… [1]
(ii) Calculate the number of moles of Fe2+ ions in 25.0 cm3 of solution F.
number of moles of Fe2+ ions = …… [2]
(iii) Calculate the Mr of FeCrnO4 and use your answer to deduce the value of n.
Mr of FeCrnO4 = ……
value of n = ……
[2]
[Total: 12]
Ni2+ ions form a number of different complex ions, including [Ni(H2O)6]2+, [Ni(NH3)6]2+ and
[Ni(en)3]2+.
The abbreviation en represents 1,2-diaminoethane. The numerical values of two stability constants, Kstab, are given in Table 5.1.
Table 5.1
complex Kstab
[Ni(NH3)6 ]2+ 4.8 × 107
[Ni(en)3]2+ 2.0 × 1018
(a) Complete the expression for the Kstab of [Ni(en)3]2+.
Kstab =
[1]
(b) A solution of [Ni(H2O)6]2+ is added to a solution that contains 0.10 mol dm–3 NH3 and
0.10 mol dm–3 en.
(i) Predict which complex ion, [Ni(NH3)6 ]2+ or [Ni(en)3]2+, is present in the resulting mixture in the highest concentration. Explain your answer.
complex ion present in largest concentration = ……
explanation ……
[1]
(ii) Complete the equation for the ligand exchange reaction occurring in (i).
[Ni(H2O)6]2+ + …… …… + ……
[1]
(c) Complete Fig. 5.1 to show the three-dimensional structures of the two isomers of [Ni(en)3]2+.
Use N N to represent the en ligand.
Name the type of isomerism shown.
Ni Ni
Fig. 5.1
type of isomerism shown ……
[3]
[Total: 6]
Mark scheme
5(a) [[Ni(en )3 ]2+ ] [[Ni(en )3 ]2+ ] 1
Kstab = OR [1]
[Ni2+ ][en]3 [Ni(H2O)62+ ][en]3
5(b)(i) [Ni(en)3]2+ AND 1
larger Kstab OR more stable [1]
5(b)(ii) [Ni(H2O)6]2+ + 3en → [Nien3]2+ + 6H2O [1] 1
5(c) 3
octahedral with correct 3D for one [Ni(en)3]2+ [1]
second optical isomer [1]
optical isomerism [1]
Fig. 6.1 shows two reactions of ethanedioic acid, HOOCCOOH.
an excess G of SOCl 2
HOOCCOOH
2CO2 + H2O
Fig. 6.1
(a) (i) Draw the organic product G in the box in Fig. 6.1.
[1]
(ii) In Fig. 6.1, SOCl 2 is given as the reagent that reacts with HOOCCOOH to produce G.
Identify a different reagent that also reacts with HOOCCOOH to produce G.
…… [1]
(b) Identify two different reagents that oxidise HOOCCOOH to form carbon dioxide and water.
…… [2]
(c) HOOCCOOH ionises as shown.
HOOCCOOH HOOCCOO– + H+
HOOCCOOH is a much stronger acid than methanoic acid, HCOOH.
Suggest an explanation for this difference in acidity.
…… [2]
(d) Benzene-1,4-dicarboxylic acid, HOOCC6H4COOH, can be made from benzene, C6H6, in two steps as shown in Fig. 6.2.
J
step 1 step 2
HOOC COOH
benzene-1,4-dicarboxylic acid
Fig. 6.2
(i) Suggest the identity of J by drawing its structure in the box in Fig. 6.2.
[1]
(ii) Identify the reagents and conditions for step 1 and step 2.
step 1 ……
step 2 ……
[2]
(iii) Draw the structure of exactly one repeat unit of the polymer formed when benzene-1,4-dicarboxylic acid reacts with ethane-1,2-diol, HOCH2CH2OH. The linkage formed between the monomers should be shown fully displayed.
[2]
(iv) State the type of polymerisation that occurs when benzene-1,4-dicarboxylic acid reacts
with ethane-1,2-diol and name the linkage formed between the monomers.
type of polymerisation ……
linkage ……
[1]
[Total: 12]
Mark scheme
6(a)(i) ClOCCOCl [1] 1
6(a)(ii) PCl3 / PCl5 [1] 1
6(b) oxygen [1] 2
acidified KMnO4 [1]
6(c) C=O are electron withdrawing / electronegative [1] 2
weakens O–H bond OR stabilises anion [1]
6(d)(i) 1
[1]
6(d)(ii) step 1: CH3Cl + AlCl3 [1] 2
step 2: hot alkaline KMnO4 [1]
6(d)(iii) 2
correctly displayed ester linkage [1]
rest of structure [1]
6(d)(iv) condensation AND ester [1] 1
Benzene reacts with chlorine gas to form chlorobenzene. This reaction can be described as the reaction between benzene molecules and Cl + ions. The Cl + ions are formed by adding a suitable catalyst to the chlorine gas.
(a) Give the name or formula of a catalyst that can be used for this reaction.
…… [1]
(b) The mechanism for this reaction is shown.
Cl +
x y H step 1 step 2 Cl
Cl
+
diagram 1 diagram 2 diagram 3
(i) The movement of a pair of electrons is represented by x in diagram 1.
• State where this pair of electrons is before step 1 takes place.
• State where this pair of electrons is after step 1 has taken place.
[2]
(ii) The movement of another pair of electrons is represented by y in diagram 2.
• State where this pair of electrons is before step 2 takes place.
• State where this pair of electrons is after step 2 has taken place.
[2]
(c) There are six carbon atoms in diagram 2.
State how many of these carbon atoms are sp hybridised, sp2 hybridised, and sp3 hybridised.
sp hybridised ……
sp2 hybridised ……
sp3 hybridised ……
[1]
(d) Complete the equation for this reaction between benzene and chlorine.
C6H6 + …… …… + ……
[1]
(e) The mechanism for this reaction is electrophilic substitution.
Complete the following sentence. Write formulae in the gaps provided.
During this reaction, the electrophile is …… and a …… atom
in benzene is substituted by a …… atom.
[1]
(f) Chloroethane reacts with NaOH(aq). Chlorobenzene does not.
(i) Name the mechanism of the reaction that chloroethane undergoes with NaOH(aq), and identify the major organic product that is formed.
mechanism ……
major organic product ……
[1]
(ii) Explain the difference in reactivity of chloroethane and chlorobenzene when treated with
NaOH(aq).
…… [2]
[Total: 11]
Mark scheme
7(a) aluminium chloride OR AlCl3 [1] 1
7(b)(i) delocalised system / delocalised ring / pi system / pi ring [1] 2
C–Cl bond [1]
7(b)(ii) C–H bond [1] 2
delocalised system / delocalised ring / pi system / pi ring [1]
7(c) 0, 5, 1 [1] 1
7(d) C6H6 + Cl2 → C6H5Cl + HCl [1] 1
7(e) Cl + 1
hydrogen / H
chlorine / Cl [1]
7(f)(i) nucleophilic substitution AND ethanol [1] 1
7(f)(ii) delocalisation of LP of Cl with π system [1] 2
C–Cl bond is stronger in chlorobenzene / partly double [1]
The amino acid serine, HOCH2CH(NH2)COOH, exists in two optically active forms. These optical isomers, isomer P and isomer Q, are shown in Fig. 8.1.
isomer P isomer Q
CH2OH CH2OH
C C
H H
HOOC COOH
NH2 H2N
Fig. 8.1
(a) Isomer P and isomer Q have identical physical and chemical properties, with the exception of two specific properties. One of these two properties is their differing effect on plane polarised light.
State the other property by which they differ.
…… [1]
(b) A solution of pure isomer P of a particular concentration rotates plane polarised light by 5.0° in a clockwise direction.
Describe how a solution of pure isomer Q of the same concentration affects plane polarised light.
…… [1]
(c) State another term, in addition to stereoisomers, optical isomers and non-superimposable mirror images, which can be used to describe this pair of chiral compounds, isomer P and
isomer Q.
…… [1]
(d) Give the term used to describe a mixture containing equal amounts of isomer P and isomer Q.
…… [1]
(e) Describe one way in which a single pure optical isomer of serine can be produced, instead of making a mixture of isomer P and isomer Q.
…… [1]
(f) Complete Table 8.1 to describe the peaks seen in the proton (1H) NMR spectrum of
HOCH2CH(NH2)COOH dissolved in D2O.
Use as many rows in Table 8.1 as you need to, leaving the other rows blank.
Table 8.1
group responsible name of splitting pattern explanation for splitting pattern for peak shown by peak
[3]
(g) Proline is a naturally occurring amino acid. The skeletal formula of proline is shown.
proline
O
OH
NH
State the number of peaks in the carbon-13 (13C) NMR spectrum of proline.
…… [1]
(h) Glutamic acid is a naturally occurring amino acid.
The skeletal formula of glutamic acid is shown.
glutamic acid
O O
HO OH
NH2
The isoelectric point of glutamic acid is pH 3.
A sample of glutamic acid is dissolved in a solution of pH 1. A strong alkali is then added until the pH of the mixture reaches pH 14. During this process all possible ionised forms of glutamic acid are present at different times, depending on the pH of the solution.
Complete the boxes below to show four different ionised forms of glutamic acid that are present at the stated pH values.
at pH 1 at pH 3
at pH 9 at pH 14
[3]
[Total: 12]
Important values, constants and standards
molar gas constant R = 8.31 J K–1 mol–1
Faraday constant F = 9.65 × 104 C mol–1
Avogadro constant L = 6.022 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K)
Vm = 24.0 dm3 mol–1 at room conditions
ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1)
The Periodic Table of Elements
Group
1 2 13 14 15 16 17 18
1 2
H He hydrogen helium
Key 1.0 4.0
3 4 atomic number 5 6 7 8 9 10
Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon
6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2
11 12 13 14 15 16 17 18
Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon
23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
at www.cambridgeinternational.org after the live examination series.
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton
39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
24
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon
85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids
Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon
132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – –
87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids
Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson
– – – – – – – – – – – – – – – – –
57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium
138.9 140.1 140.9 144.2 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0
89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium
– 232.0 231.0 238.0 – – – – – – – – – – –
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Mark scheme
8(a) biological activity [1] 1
8(b) rotates plane polarised light 5.0° in anticlockwise direction [1] 1
8(c) enantiomers [1] 1
8(d) racemic [1] 1
8(e) use of chiral catalyst [1] 1
8(f) CH2 and CH only in column one [1] 3
CH2 gives a doublet, CH gives a triplet [1]
doublet due to 1 proton on neighbouring carbons
triplet due to 2 protons on neighbouring carbons [1]
8(g) 5 [1] 1
8(h) 3
at pH 1 [1]
at pH 3 AND
or
at pH 9 [1]
at pH 14 [1]
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