CAIE A Level 9701

Chemistry 2025 · Oct/Nov

92 Questions Q&A Available

Questions & Model Answers 92 Qs

Q1
Short Answer 1 mark
Magnesium nitrate, Mg(NO , and strontium nitrate, Sr(NO , both decompose when heated to form the metal oxide and a mixture of gases.
Q1(a)
Short Answer 1 mark
Write an equation for the thermal decomposition of Mg(NO
Q1(b)
Short Answer 2 marks
State which of Mg(NO or Sr(NO decomposes at a lower temperature. Explain your answer . compound that decomposes at a lower temperature explanation
Q1(c)
Short Answer 1 mark
Magnesium oxide, MgO, and strontium oxide, SrO, both react with dilute sulfuric acid. MgO forms a soluble salt, A. SrO forms an insoluble salt, B.
Q1(i)
Short Answer 1 mark
Identify the products formed when MgO reacts with dilute sulfuric acid.
Q1(ii)
Short Answer 1 mark
Explain why A is more soluble than B. [Total: 7] * 0000800000003 * 2 Ethanal, CH CHO, reacts with nitrogen dioxide, NO . The products of the first step of this reaction are a CH =O radical and a molecule of nitrous acid, HNO CHO + NO =O + HNO
Q1(a)
Short Answer 1 mark
(i) Use two words to complete the sentence. This reaction involves of the single covalent bond between a hydrogen atom and a carbon atom in CH CHO.
Q1(ii)
Short Answer 1 mark
The hydrogen atom mentioned in (a)(i) forms a covalent bond with one of the oxygen atoms of an NO molecule. An NO molecule has a single, unpaired electron on the nitrogen atom. All electrons are paired in an HNO molecule. Draw dot-and-cross diagrams of NO and HNO in the boxes. Show outer shell electrons only. HNO
Q1(iii)
Short Answer 1 mark
Use VSEPR theory to predict the bond angle at the nitrogen atom in an HNO molecule. bond angle =
Q1(b)
Short Answer 1 mark
The rate equation for the reaction between CH CHO and NO is shown. rate = k [CH CHO][NO Under certain conditions, when the concentrations of both CH CHO and NO are 0.200 mol dm the rate of the reaction is 1.53 × 10 mol dm Calculate the value of the rate constant, k, under these conditions. Give the units of k. k = units [2] * 0000800000004 *
Q1(c)
Short Answer 1 mark
The reaction mixture described in
Q1(c)(b)
Short Answer 1 mark
is monitored over a period of time. Predict whether the graph of [NO ] against time shows a constant half-life. Explain your answer . prediction explanation
Q1(d)
Short Answer 1 mark
NO also reacts with ozone, O The rate equation is shown. rate = k [NO Under certain conditions, the value of k is 0.0848 s The reaction has a constant half-life under these conditions. Calculate the half-life in seconds. half-life = s
Q1(e)
Short Answer 1 mark
NO is present in the exhaust gases of cars. It can react with carbon monoxide, CO, on the surface of a heterogeneous catalyst in the car's catalytic converter. Describe the mode of action of this heterogeneous catalyst. [Total: 9] * 0000800000005 *
Q3(a)
Short Answer 1 mark
(i) Define conjugate acid-base pair.
Q3(ii)
Short Answer 1 mark
Give the formulas of the conjugate acid and the conjugate base of the hydrogen phosphate ion, HPO conjugate acid of HPO conjugate base of HPO
Q3(b)
Short Answer 1 mark
The K of propanoic acid, CH COOH, is 1.35 × 10 mol dm at 298 K. Solution C is a solution of CH COOH with a pH of 3.60 at 298 K.
Q3(i)
Short Answer 2 marks
Calculate the concentration of CH COOH in solution C. [CH COOH] = mol dm
Q3(ii)
Short Answer 1 mark
Calculate the concentration of hydroxide ions in solution C. [OH ] = mol dm
Q3(iii)
Short Answer 1 mark
Calculate the concentration of a solution of hydrochloric acid with the same pH as solution C. concentration = m ol dm [1] * 0000800000006 *
Q3(iv)
Short Answer 1 mark
Table 3.1 shows three possible values of the K of dimethylpropanoic acid, (CH CCOOH. Place a tick in Table 3.1 to show the correct value. Explain your answer. Table 3.1 value of K / mol dm in this column 9.33 × 10 1.35 × 10 3.35 × 10 explanation
Q3(c)
Short Answer 1 mark
Solution D is made by mixing 100 cm of 0.100 mol dm COOH and 100 cm 0.100 mol dm NaCl. The pH of solution D is measured as small amounts of H
Q3(aq)
Short Answer 1 mark
are added to it, and when small amounts of NaOH(aq) are added to it. Solution D only acts as a buf fer solution when one of these solutions is added to it.
Q3(i)
Short Answer 1 mark
Complete the sentence and write an equation for the reaction that occurs. Solution D acts as a buf fer when is added to it. equation
Q3(ii)
Short Answer 1 mark
Complete the sentence and explain why solution D does not act as a buf fer when the other solution is added. Solution D does not act as a buf fer when is added to it. explanation [1] * 0000800000007 *
Q3(d)
Short Answer 1 mark
Manganese(II) hydroxide, Mn(OH) , is only slightly soluble in water. The solubility of Mn(OH) in water is 3.28 × 10 g dm at 298 K.
Q3(i)
Short Answer 1 mark
Calculate the concentration of a saturated solution of Mn(OH) at 298 K. [Mn(OH) ] = mol dm
Q3(ii)
Short Answer 2 marks
Write an expression for the K of Mn(OH) . Give the units of K units =
Q3(iii)
Short Answer 1 mark
Use your answers to
Q3(iii)(ii)
Short Answer 1 mark
to calculate the value of K of Mn(OH) at 298 K. = [1] [Total: 15] * 0000800000008 *
Q4(a)
Short Answer 1 mark
Define enthalpy change of atomisation, ΔH
Q4(b)
Short Answer 1 mark
Define first electron affinity, EA.
Q4(c)
Short Answer 1 mark
Explain why the first electron af finity of chlorine is more exothermic than the first electron affinity of iodine.
Q4(d)
Short Answer 1 mark
The enthalpy change for the reaction Cl
Q4(g)
Short Answer 1 mark
+ 2e 2Cl
Q4(g)
Short Answer 1 mark
is - 486 kJ mol The first electron af finity of chlorine is -364 kJ mol Calculate the enthalpy change of atomisation of chlorine. of chlorine = kJ mol [2] [Total: 6] * 0000800000009 *
Q5
Short Answer 1 mark
Cobalt is a transition element which forms compounds containing Co and Co ions. Cobalt(II) sulfate dissolves in water to form a solution containing the [Co(H complex ion.
Q5(a)
Short Answer 1 mark
(i) Complete the electronic configurations of a Co ion and a Co ion. = [Ar] = [Ar]
Q5(ii)
Short Answer 1 mark
Explain why transition elements can form complex ions.
Q5(iii)
Short Answer 2 marks
An excess of concentrated HCl is added to a solution containing [Co(H Describe the colour change observed and the state of the cobalt-containing product. The colour changes from to . The state of the cobalt-containing product is .
Q5(iv)
Short Answer 1 mark
Write an equation for the reaction occurring in (a)(iii).
Q5(v)
Short Answer 1 mark
Name the type of reaction occurring in (a)(iii).
Q5(vi)
Short Answer 1 mark
Write an equation for the reaction that occurs when an excess of NaOH(aq) is added to a solution containing [Co(H [ i
Q5(b)
Short Answer 1 mark
Cobalt metal can be oxidised by acidified K . The relevant half-equations, and their values, are shown. + 2e Co E = -0.28 V + 14H + 6e 2Cr + 7HO E = +1.33 V
Q5(i)
Short Answer 2 marks
A Co /Co electrode is constructed in which [Co ] is 0.020 mol dm at 298 K. Use the Nernst equation to show that the E value for this Co / Co electrode is -0.33 V .
Q5(ii)
Short Answer 1 mark
An electrochemical cell is constructed using the Co / Co electrode described in (b)(i) and a Cr / Cr electrode in which all conditions are standard. Calculate the value of E cell cell =
Q5(iii)
Short Answer 1 mark
A current is drawn from the electrochemical cell described in (b)(ii). Write an equation for the reaction taking place in the cell.
Q5(iv)
Short Answer 1 mark
Complete the sentences to identify the negative electrode and the direction of electron flow when a current is drawn from the cell described in (b)(ii). The electrode is the negative electrode. Electrons flow from the electrode to the electrode.
Q5(c)
Short Answer 1 mark
A molten Co salt is electrolysed using a current of 0.500 A. 0.547 g of cobalt metal forms at the cathode. Under the conditions used no other reduction reaction occurs at the cathode. Calculate the time in minutes for which the current flows to produce this mass of cobalt. Give your answer to three significant figures. time = min [3] [Total: 15] * 0000800000011 * [[i
Q6(a)
Short Answer 1 mark
Nickel forms complexes.
Q6(i)
Short Answer 1 mark
Give the formula and charge of the tetrahedral complex formed by Ni atoms with carbon monoxide molecules. Carbon monoxide is a monodentate ligand. This is complex E. E =
Q6(ii)
Short Answer 2 marks
Give the formula and charge of the octahedral complex formed by Ni ions with ethanedioate ions. This is complex F. F =
Q6(iii)
Short Answer 1 mark
Identify which complex, E or F, ex ists as a mixture of two stereoisomers and the type of stereoisomerism involved. The complex which exists as a mixture of two stereoisomers is . The type of stereoisomerism involved is .
Q6(b)
Short Answer 1 mark
Cadmium forms complexes with methylamine, CH , and 1,2-diaminoethane, en. The values of the stability constants, K stab , of these complex ions are given in Table 6.1. Table 6.1 complex K stab [Cd(CH 3.5 × 10 [Cd(en) 4.0 × 10
Q6(i)
Short Answer 1 mark
Explain, by reference to its structure, why CH acts as a monodentate ligand.
Q6(ii)
Short Answer 1 mark
Some Cd
Q6(aq)
Short Answer 1 mark
is added to a solution containing equal concentrations of CH and en. Predict which of the two complexes in Table 6.1 forms at the higher concentration. Explain your answer. complex that forms at the higher concentration explanation
Q6(iii)
Short Answer 1 mark
Complete the expression for the K stab of [Cd(CH stab [1] [Total: 7] * 0000800000012 * [Ti
Q7
Short Answer 1 mark
P, Q, R, S, T, U, and V are the seven structural isomers with molecular form ula C O that have a carbonyl group. P CH (CH CHO Q CH CH(CH )CHO R (CH CHCHCHO S (CH CCHO T CH COCH U CH COCH V (CH CHCOCH
Q7(a)
Short Answer 2 marks
Only one of these seven compounds has stereoisomers. Draw three-dimensional diagrams of the two stereoisomers of this compound.
Q7(b)
Short Answer 1 mark
P, Q, R, S, T, U, and V are treated separately with alkaline I
Q7(aq)
Short Answer 1 mark
and the product mixture is acidified.
Q7(i)
Short Answer 1 mark
Identify the two compounds that give a positive result with alkaline I
Q7(ii)
Short Answer 1 mark
Describe the observations when one of the compounds you have identified in (b)(i) is treated with alkaline I
Q7(aq)
Short Answer 1 mark
and give the structural formulae of the two carbon-containing products of this reaction. observations two carbon-containing products and [2] * 0000800000013 * [ui P CH (CH CHO Q CH CH(CH )CHO R (CH CHCHCHO S (CH CCHO T CH COCH U CH COCH V (CH CHCOCH
Q7(c)
Short Answer 1 mark
The proton ( H) NMR spectra of P, Q, R, S, T, U, and V are compared.
Q7(i)
Short Answer 1 mark
Identify the only compound that gives a spectrum with two singlets and no other peaks. Fig. 7.1 shows the spectrum obtained from one of the compounds. 11 10 9 8 7 6 5 4 3 2 1 0 δ / ppm
Q7(ii)
Short Answer 1 mark
Identify the compound that gives this spectrum.
Q7(iii)
Short Answer 1 mark
Name the splitting pattern of the peak at δ = 1.1 in Fig. 7.1. Give the reason for this splitting. name reason
Q7(iv)
Short Answer 1 mark
Identify the substance that gives the small peak at δ = 0 in Fig. 7.1. [ri
Q7(d)
Short Answer 1 mark
The carbon-13 NMR spectra of R, S, T and U are compared. Complete Table 7.1 to state the number of peaks in the spectrum of each compound. Table 7.1 compound number of peaks R (CH CHCHCHO S (CH CCHO T CH COCH U CH COCH [2] [Total: 11] * 0000800000015 * [ni
Q8
Short Answer 1 mark
Asparagine and aspartic acid are two naturally occurring amino acids. Their structures and isoelectric points are shown in Table 8.1. Table 8.1 amino acid structure isoelectric point)CH CONH 5.41)CH COOH 2.77
Q8(a)
Short Answer 1 mark
Define isoelectric point.
Q8(b)
Short Answer 2 marks
Draw the structures of asparagine and aspartic acid at pH 2.
Q8(c)
Short Answer 1 mark
Asparagine and aspartic acid are treated separately with an ex cess of LiAl H Draw the structures of the organic products of these reactions. product of asparagine treated with an excess of LiAl H product of aspartic acid treated with an excess of LiAl H [2] * 0000800000016 * [oi
Q8(d)
Short Answer 1 mark
Propanedioic acid, HOOCCH COOH, is treated with an excess of thionyl chloride, SOCl Propanedioyl chloride, Cl OCCH COCl , is formed.
Q8(i)
Short Answer 1 mark
Write an equation for this reaction.
Q8(ii)
Short Answer 2 marks
Propanedioyl chloride reacts with an ex cess of asparagine to form compound G with molecular formula C Each molecule of compound G has four amide groups. Draw the structure of compound G. Compound G, C
Q8(e)
Short Answer 2 marks
Asparagine is hydrolysed with an excess of hot NaOH(aq). Draw the structure of the organic product of this reaction.
Q8(f)
Short Answer 1 mark
A polymer can form from asparagine, HOOCCH(NH)CH CONH , as the only monomer. Draw a length of the polymer chain containing three monomer residues. Clearly label the repeat unit of the polymer on your diagram. [3] * 0000800000017 * [vi
Q8(g)
Short Answer 1 mark
Aspartic acid exists in two optically active forms.
Q8(i)
Short Answer 2 marks
Plane polarised light is passed through pure samples of these two optically active forms in solutions of the same concentration. Describe two similarities and one dif ference in their effect on the plane polarised light. similarities difference
Q8(ii)
Short Answer 1 mark
Give the term used to describe a mixture of equal amounts of the two optically active forms. [Total: 16] * 0000800000018 * [ei Question 9 starts on page 20. * 0000800000019 * [Ci
Q9
Short Answer 1 mark
Compound X is made from benzene by the route shown in Fig. 9.1. step 1 step 2 step 3 COOH step 4 COOH benzene
Q9(a)
Short Answer 1 mark
Describe the bonding in benzene, C Y our answer should include: hydrogen atoms.
Q9(b)
Short Answer 1 mark
Describe the reagents and conditions required for step 1 in Fig. 9.1. T i
Q9(c)
Short Answer 3 marks
In step 1 of Fig. 9.1 benzene reacts with Complete Fig. 9.2 to show the mechanism for this reaction, including: 3 CH intermediate +
Q9(d)
Short Answer 1 mark
Describe the reagents and conditions required for step 2 of Fig. 9.1.
Q9(e)
Short Answer 1 mark
Identify the reagents required for step 3 of Fig. 9.1. Compound W is the product of this step.
Q9(f)
Short Answer 1 mark
Name compound W.
Q9(g)
Short Answer 1 mark
The reagents commonly used for step 4 will not reduce the -COOH group. Identify the reagents and conditions required for step 4 of Fig. 9.1. T[i
Q9(h)
Short Answer 1 mark
Benzene can also be used as a starting material to make compound Y. COOH compound Y Describe how the route described in Fig. 9.1 (repeated below) can be changed to give compound Y instead of compound X. Explain your answer . step 1 step 2 step 3 COOH step 4 COOH benzene [Total: 14] * 0000800000022 * TTi Important values, constants and standards molar gas constant R = 8.31 J Kmol Faraday constant F = 9.65 × 10 C mol Avogadro constant L = 6.02 × 10mol electronic charge e = -1.60 × 10-19 molar volume of gas V = 22.4 dmmol at s.t.p. (101 kPa and 273 K) = 24.0 dmmol at room conditions ionic product of water K = 1.00 × 10-14mol (at 298 K (25 °C)) (4.18 J g ) * 0000800000023 * Tui To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Group The Periodic Ta ble of Elements hydrogen 1.0 helium 4.0 12 13 14 15 16 17 18 3456789 10 11 12 lithium 6.9 beryllium 9.0 atomic number atomic symbolKey name relative atomic mass sodium 23.0 magnesium 24.3 potassium 39.1 calcium 40.1 rubidium 85.5 strontium 87.6 caesium 132.9 barium 137.3 francium radium boron 10.8 aluminium 27.0 gallium 69.7 indium 114.8 thallium 204.4 carbon 12.0 silicon 28.1 germanium 72.6tin 118.7 lead 207.2 titanium 47.9 zirconium 91.2 hafnium 178.5 rutherfordium vanadium 50.9 niobium 92.9 tantalum 180.9 dubnium chromium 52.0 molybdenum 95.9 tungsten 183.8 seaborgium manganese 54.9 technetium rhenium 186.2 bohrium iron 55.8 ruthenium 101.1 osmium 190.2 hassium cobalt 58.9 rhodium 102.9 iridium 192.2 meitnerium nickel 58.7 palladium 106.4 platinum 195.1 darmstadtium copper 63.5 silver 107.9 gold 197.0 roentgenium zinc 65.4 cadmium 112.4 mercury 200.6 copernicium flerovium livermorium nitrogen 14.0 phosphorus 31.0 arsenic 74.9 antimony 121.8 bismuth 209.0 oxygen 16.0 sulfur 32.1 selenium 79.0 tellurium 127.6 polonium fluorine 19.0 chlorine 35.5 bromine 79.9 iodine 126.9 neon 20.2 argon 39.9 krypton 83.8 xenon 131.3 radon nihonium moscovium tennessine oganesson scandium 45.0 yttrium 88.9 57-71 lanthanoids 89-103 actinoids lanthanum 138.9 lanthanoids actinoids actinium cerium 140.1 thorium 232.0 praseodymium 140.9 protactinium 231.0 neodymium 144.2 uranium 238.0 promethium neptunium samarium 150.4 plutonium europium 152.0 americium gadolinium 157.3 curium terbium 158.9 berkelium dysprosium 162.5 californium holmium 164.9 einsteinium erbium 167.3 fermium thulium 168.9 mendelevium ytterbium 173.1 nobelium lutetium 175.0 lawrencium - * 0000800000024 * Tri
Paper Details
Board CAIE
Level A Level
Subject Chemistry
Code 9701
Year 2025
Session Oct/Nov
Q&A 92 questions
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