An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V3) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
Paper 23 (Extended MCQ)
Answer all 40 multiple-choice questions. For each question, choose from the four possible options.
1. It is a polyester formed by condensation polymerisation. 2. It is a polyamide formed by addition polymerisation. 3. A molecule of water is eliminated during the formation of each linkage. 4. The monomers are a dicarboxylic acid and a diamine.
A.1 and 3 only
B.1 and 4 only
C.2 and 3 only
D.3 and 4 only
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. The polymer contains ester linkages (\(-O-CO-\)), which makes it a polyester. Polyesters are formed by condensation polymerisation. 2. It is not a polyamide, nor is it formed by addition polymerisation. 3. In condensation polymerisation, a small molecule such as water (\(H_2O\)) is eliminated when each ester linkage is formed. 4. The monomers used to make this polyester are a dicarboxylic acid and a diol (not a diamine).
Therefore, statements 1 and 3 are correct.
Marking scheme
Award 1 mark for the correct option A.
Question 2 · multiple-choice
1 marks
Which statement explains why zinc powder reacts faster with dilute hydrochloric acid than zinc lumps of the same mass?
A.The activation energy of the reaction is lower for zinc powder.
B.The particles in the zinc powder have more kinetic energy.
C.There is a greater frequency of collisions between reactant particles.
D.The concentration of the hydrochloric acid increases.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Zinc powder has a much larger total surface area than the same mass of zinc lumps. This means there are more exposed zinc particles available to collide with the acid, resulting in a greater frequency of successful collisions per unit time. The activation energy remains unchanged as no catalyst is added, and the kinetic energy of the particles is determined by temperature, which is constant here.
Marking scheme
Award 1 mark for the correct option C.
Question 3 · multiple-choice
1 marks
In the extraction of iron in a blast furnace, several chemical reactions occur. Which equation represents the reaction that produces the reducing agent for the iron(III) oxide?
A.\(C + O_2 \rightarrow CO_2\)
B.\(CO_2 + C \rightarrow 2CO\)
C.\(Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\)
D.\(CaCO_3 \rightarrow CaO + CO_2\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
The main reducing agent in the blast furnace is carbon monoxide (\(CO\)). It is produced when carbon dioxide reacts with excess hot carbon (coke): \[CO_2(g) + C(s) \rightarrow 2CO(g)\] Reaction A produces carbon dioxide, reaction C is the reduction of iron(III) oxide itself, and reaction D is the thermal decomposition of limestone.
Marking scheme
Award 1 mark for the correct option B.
Question 4 · multiple-choice
1 marks
An aqueous solution of copper(II) sulfate is electrolysed using inert carbon electrodes.
Which row describes the observations at each electrode?
A.Cathode: pink solid | Anode: bubbles of a colourless gas
B.Cathode: pink solid | Anode: bubbles of a green gas
C.Cathode: bubbles of a colourless gas | Anode: pink solid
D.Cathode: bubbles of a green gas | Anode: bubbles of a colourless gas
Show answer & marking schemeHide answer & marking scheme
Worked solution
At the cathode (negative electrode), copper ions (\(Cu^{2+}\)) are reduced to copper metal: \[Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\] This is observed as a pink/brown solid coating on the electrode.
At the anode (positive electrode), hydroxide ions (\(OH^-\)) from water are oxidised to oxygen gas: \[4OH^-(aq) \rightarrow O_2(g) + 2H_2O(l) + 4e^-\] This is observed as bubbles of a colourless gas.
Marking scheme
Award 1 mark for the correct option A.
Question 5 · multiple-choice
1 marks
The reaction shown is at equilibrium in a sealed container.
\[2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\]
\(\Delta H = -197\text{ kJ/mol}\)
Which changes will both increase the equilibrium yield of \(SO_3(g)\)?
A.decreasing the temperature and decreasing the pressure
B.decreasing the temperature and increasing the pressure
C.increasing the temperature and decreasing the pressure
D.increasing the temperature and increasing the pressure
Show answer & marking schemeHide answer & marking scheme
Worked solution
1. Temperature: The forward reaction is exothermic (\(\Delta H < 0\)). Decreasing the temperature shifts the equilibrium position to the right (in the exothermic direction) to release heat, increasing the yield of \(SO_3(g)\). 2. Pressure: There are 3 moles of gaseous reactants on the left and 2 moles of gaseous products on the right. Increasing the pressure shifts the equilibrium position to the side with fewer moles of gas (the right side), increasing the yield of \(SO_3(g)\).
Therefore, decreasing the temperature and increasing the pressure both increase the equilibrium yield of \(SO_3(g)\).
Marking scheme
Award 1 mark for the correct option B.
Question 6 · multiple-choice
1 marks
Gallium exists naturally as two isotopes, \(^{69}\text{Ga}\) and \(^{71}\text{Ga}\). The relative atomic mass of gallium is 69.8.
What is the percentage abundance of each isotope in a natural sample of gallium?
A.40% \(^{69}\text{Ga}\) and 60% \(^{71}\text{Ga}\)
B.50% \(^{69}\text{Ga}\) and 50% \(^{71}\text{Ga}\)
C.60% \(^{69}\text{Ga}\) and 40% \(^{71}\text{Ga}\)
D.80% \(^{69}\text{Ga}\) and 20% \(^{71}\text{Ga}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
Let the fractional abundance of \(^{69}\text{Ga}\) be \(x\). The fractional abundance of \(^{71}\text{Ga}\) is therefore \(1-x\).
Thus, the abundance of \(^{69}\text{Ga}\) is 60% and the abundance of \(^{71}\text{Ga}\) is 40%.
Marking scheme
Award 1 mark for the correct option C.
Question 7 · multiple-choice
1 marks
Methane reacts with chlorine to form chloromethane and hydrogen chloride:
\[CH_4 + Cl_2 \rightarrow CH_3Cl + HCl\]
Some bond energies are shown in the table.
| Bond | Bond energy in kJ/mol | | :---: | :---: | | C-H | 410 | | Cl-Cl | 240 | | C-Cl | 340 | | H-Cl | 430 |
What is the enthalpy change, in kJ/mol, for this reaction?
A.-120 kJ/mol
B.-20 kJ/mol
C.+20 kJ/mol
D.+120 kJ/mol
Show answer & marking schemeHide answer & marking scheme
Worked solution
To calculate the enthalpy change (\(\Delta H\)):
1. Energy required to break bonds (reactants side): - Break 1 mol of C-H bonds: \(410 \text{ kJ/mol}\) - Break 1 mol of Cl-Cl bonds: \(240 \text{ kJ/mol}\) - Total energy input = \(410 + 240 = 650 \text{ kJ/mol}\)
2. Energy released during bond formation (products side): - Form 1 mol of C-Cl bonds: \(340 \text{ kJ/mol}\) - Form 1 mol of H-Cl bonds: \(430 \text{ kJ/mol}\) - Total energy output = \(340 + 430 = 770 \text{ kJ/mol}\)
An aqueous solution of a salt contains a single cation.
- Addition of aqueous sodium hydroxide produces a green precipitate which is insoluble in excess. - Addition of aqueous ammonia also produces a green precipitate which is insoluble in excess.
Which cation is present in the solution?
A.\(Cr^{3+}\)
B.\(Fe^{2+}\)
C.\(Fe^{3+}\)
D.\(Cu^{2+}\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
According to qualitative analysis tests: - \(Fe^{2+}\) forms a green precipitate with both aqueous sodium hydroxide and aqueous ammonia, and this precipitate is insoluble in excess of either reagent. - \(Cr^{3+}\) also forms a green precipitate with aqueous sodium hydroxide, but it is soluble in excess to give a green solution. - \(Fe^{3+}\) forms a red-brown precipitate. - \(Cu^{2+}\) forms a light blue precipitate.
Marking scheme
Award 1 mark for the correct option B.
Question 9 · multiple-choice
1 marks
A polymer is formed by reacting ethane-1,2-diol, \(\text{HO-CH}_2\text{-CH}_2\text{-OH}\), with benzene-1,4-dicarboxylic acid, \(\text{HOOC-C}_6\text{H}_4\text{-COOH}\).
Which option correctly describes the type of polymerisation and the small molecule released during the reaction?
A.type of polymerisation: addition; small molecule released: water
B.type of polymerisation: condensation; small molecule released: water
C.type of polymerisation: addition; small molecule released: hydrogen chloride
D.type of polymerisation: condensation; small molecule released: hydrogen chloride
Show answer & marking schemeHide answer & marking scheme
Worked solution
Ethane-1,2-diol (a diol) and benzene-1,4-dicarboxylic acid (a dicarboxylic acid) react by condensation polymerisation to form a polyester. Each ester linkage formed releases a water molecule (\(\text{H}_2\text{O}\)).
Marking scheme
1 mark for identifying condensation polymerisation and water as the correct option (B).
Question 10 · multiple-choice
1 marks
A student reacts excess calcium carbonate chips with dilute hydrochloric acid. The volume of carbon dioxide gas collected over time is plotted. In a second experiment, the student repeats the reaction but modifies one variable. The resulting curve has a steeper initial gradient but levels off at the same final volume of gas.
Which modification explains this result?
A.using a higher concentration of hydrochloric acid of the same volume
B.using a larger volume of the same concentration of hydrochloric acid
C.using calcium carbonate powder instead of chips, with all other conditions kept constant
D.conducting the reaction at a lower temperature
Show answer & marking schemeHide answer & marking scheme
Worked solution
Using calcium carbonate powder increases the surface area, which increases the rate of reaction, resulting in a steeper initial gradient. Since the acid is the limiting reactant and its volume and concentration are kept constant, the same number of moles of acid react, producing the same final volume of carbon dioxide.
Marking scheme
1 mark for selecting option C.
Question 11 · multiple-choice
1 marks
A student wants to obtain a pure sample of hydrated copper(II) sulfate crystals from a mixture of solid copper(II) oxide and copper(II) sulfate solution.
Which sequence of steps should the student carry out?
A.filter \(\rightarrow\) evaporate to dryness
B.filter \(\rightarrow\) heat until saturated \(\rightarrow\) cool to crystallise \(\rightarrow\) filter and dry the crystals
C.heat until saturated \(\rightarrow\) cool to crystallise \(\rightarrow\) filter
D.recrystallise \(\rightarrow\) filter \(\rightarrow\) dry in an oven
Show answer & marking schemeHide answer & marking scheme
Worked solution
First, the insoluble copper(II) oxide is removed by filtration. The copper(II) sulfate solution (filtrate) is then heated to its crystallisation point (saturated solution) and allowed to cool so that hydrated copper(II) sulfate crystals form. Finally, the crystals are separated by filtration and dried between sheets of filter paper.
Marking scheme
1 mark for selecting option B.
Question 12 · multiple-choice
1 marks
Iron is extracted from hematite, \(\text{Fe}_2\text{O}_3\), in a blast furnace.
Which reaction represents a redox process where carbon monoxide acts as the reducing agent?
Show answer & marking schemeHide answer & marking scheme
Worked solution
In the reaction \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\), iron(III) oxide loses oxygen (is reduced) and carbon monoxide gains oxygen (is oxidised). Thus, carbon monoxide (\(\text{CO}\)) acts as the reducing agent.
Marking scheme
1 mark for selecting option C.
Question 13 · multiple-choice
1 marks
Two separate solutions, solution X (hydrochloric acid) and solution Y (ethanoic acid), both have a concentration of \(0.1\text{ mol/dm}^3\).
Which statement about these solutions is correct?
A.Solution X has a higher pH than solution Y.
B.Solution Y reacts more vigorously with magnesium ribbon than solution X.
C.Solution X is a better electrical conductor than solution Y.
D.Both solutions require different volumes of \(0.1\text{ mol/dm}^3\) sodium hydroxide for complete neutralisation of equal volumes.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Hydrochloric acid (solution X) is a strong acid that fully dissociates in water, whereas ethanoic acid (solution Y) is a weak acid that only partially dissociates. As a result, solution X has a higher concentration of mobile ions than solution Y, making it a better conductor of electricity.
Marking scheme
1 mark for selecting option C.
Question 14 · multiple-choice
1 marks
An element E has two naturally occurring isotopes: \(^{69}\text{E}\) and \(^{71}\text{E}\).
The relative atomic mass of E is 69.8.
What is the percentage abundance of the heavier isotope, \(^{71}\text{E}\)?
A.30%
B.40%
C.60%
D.70%
Show answer & marking schemeHide answer & marking scheme
Worked solution
Let \(x\) be the abundance fraction of \(^{71}\text{E}\). The abundance fraction of \(^{69}\text{E}\) is \(1 - x\).
Converting to a percentage, the abundance of \(^{71}\text{E}\) is 40%.
Marking scheme
1 mark for selecting option B.
Question 15 · multiple-choice
1 marks
An unknown salt solution is tested.
- Addition of dilute nitric acid followed by aqueous silver nitrate produces a cream precipitate. - Addition of aqueous sodium hydroxide dropwise produces a green precipitate, which is insoluble in excess sodium hydroxide.
What is the identity of the salt?
A.chromium(III) bromide
B.iron(II) bromide
C.iron(II) chloride
D.chromium(III) chloride
Show answer & marking schemeHide answer & marking scheme
Worked solution
The reaction with nitric acid and silver nitrate producing a cream precipitate indicates the presence of bromide ions (\(\text{Br}^-\)). The reaction with sodium hydroxide producing a green precipitate that is insoluble in excess indicates the presence of iron(II) ions (\(\text{Fe}^{2+}\)). Chromium(III) also produces a green precipitate, but it is soluble in excess sodium hydroxide to form a green solution. Therefore, the salt is iron(II) bromide.
Marking scheme
1 mark for selecting option B.
Question 16 · multiple-choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert platinum electrodes.
Which option correctly identifies the products formed at each electrode and the change in pH of the remaining electrolyte?
A.product at cathode: hydrogen; product at anode: chlorine; change in pH: increases
B.product at cathode: hydrogen; product at anode: oxygen; change in pH: decreases
C.product at cathode: sodium; product at anode: chlorine; change in pH: no change
D.product at cathode: sodium; product at anode: oxygen; change in pH: increases
Show answer & marking schemeHide answer & marking scheme
Worked solution
During the electrolysis of concentrated aqueous sodium chloride, hydrogen ions (\(\text{H}^+\)) are selectively discharged at the cathode to form hydrogen gas, and chloride ions (\(\text{Cl}^-\)) are discharged at the anode to form chlorine gas. The remaining sodium (\(\text{Na}^+\)) and hydroxide (\(\text{OH}^-\)) ions form alkaline sodium hydroxide solution, causing the pH of the electrolyte to increase.
Marking scheme
1 mark for selecting option A.
Question 17 · multiple-choice
1 marks
A student investigates the reaction between excess marble chips and dilute hydrochloric acid. Which changes would increase the rate of reaction by increasing the collision frequency of the reacting particles? 1. using more concentrated hydrochloric acid 2. heating the reaction mixture 3. using larger marble chips
A.1 only
B.1 and 2 only
C.2 and 3 only
D.1, 2 and 3
Show answer & marking schemeHide answer & marking scheme
Worked solution
Concentration increases the number of reacting particles per unit volume, which increases collision frequency. Heating increases the kinetic energy of particles, making them move faster and collide more frequently. Using larger marble chips decreases the surface area, which decreases the collision frequency.
Marking scheme
Award 1 mark for the correct option (B).
Question 18 · multiple-choice
1 marks
Two monomers, X and Y, react together to form a condensation polymer. Monomer X is a dicarboxylic acid with the formula HOOC-C4H8-COOH. Monomer Y is a diol with the formula HO-C3H6-OH. Which type of linkage is formed in the polymer, and which small molecule is released during the reaction?
A.amide linkage, ammonia
B.amide linkage, water
C.ester linkage, hydrogen chloride
D.ester linkage, water
Show answer & marking schemeHide answer & marking scheme
Worked solution
A dicarboxylic acid reacts with a diol to form a polyester, which contains ester linkages. Since this is a condensation polymerisation, a small molecule of water (H2O) is released as each linkage forms.
Marking scheme
Award 1 mark for the correct option (D).
Question 19 · multiple-choice
1 marks
Aqueous copper(II) sulfate is electrolysed using copper electrodes. What happens to the mass of each electrode during this electrolysis?
A.Anode mass decreases, cathode mass increases.
B.Anode mass decreases, cathode mass remains unchanged.
C.Anode mass remains unchanged, cathode mass increases.
D.Anode mass increases, cathode mass decreases.
Show answer & marking schemeHide answer & marking scheme
Worked solution
At the anode, copper atoms lose electrons and dissolve into the solution as copper ions: Cu(s) -> Cu2+(aq) + 2e-, so its mass decreases. At the cathode, copper ions gain electrons and deposit as copper metal: Cu2+(aq) + 2e- -> Cu(s), so its mass increases.
Marking scheme
Award 1 mark for the correct option (A).
Question 20 · multiple-choice
1 marks
What is the volume occupied by 3.01 * 10^(23) molecules of oxygen gas, O2, at room temperature and pressure (r.t.p.)? (Assume Avogadro's constant is 6.02 * 10^(23) mol^(-1) and the volume of 1 mol of gas at r.t.p. is 24 dm^3)
A.6.0 dm^3
B.12.0 dm^3
C.24.0 dm^3
D.48.0 dm^3
Show answer & marking schemeHide answer & marking scheme
Worked solution
First, calculate the number of moles of oxygen gas: moles = (3.01 * 10^(23)) / (6.02 * 10^(23)) = 0.5 mol. Then calculate the volume: volume = 0.5 mol * 24 dm^3/mol = 12.0 dm^3.
Marking scheme
Award 1 mark for the correct option (B).
Question 21 · multiple-choice
1 marks
The equation shows the synthesis of ammonia from nitrogen and hydrogen: N2(g) + 3H2(g) -> 2NH3(g). The relevant bond energies are: N#N: 945 kJ/mol; H-H: 436 kJ/mol; N-H: 391 kJ/mol. What is the enthalpy change, delta H, for this reaction?
A.-93 kJ/mol
B.-57 kJ/mol
C.+57 kJ/mol
D.+93 kJ/mol
Show answer & marking schemeHide answer & marking scheme
Worked solution
Energy needed to break bonds: (1 * 945) + (3 * 436) = 2253 kJ/mol. Energy released in making bonds: 6 * 391 = 2346 kJ/mol. Enthalpy change (delta H) = energy input - energy released = 2253 - 2346 = -93 kJ/mol.
Marking scheme
Award 1 mark for the correct option (A).
Question 22 · multiple-choice
1 marks
An oxide of metal X reacts with metal Y to produce metal X and an oxide of metal Y. Metal Y does not react with cold water but reacts with steam. Which statement about the metals X and Y is correct?
A.Metal X is more reactive than metal Y.
B.Metal Y is below hydrogen in the reactivity series.
C.Metal Y is more reactive than sodium.
D.Metal X could be copper.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Since metal Y displaces X from its oxide, Y is more reactive than X (Y > X). Metal Y reacts with steam but not cold water, placing it in the middle of the reactivity series (e.g., magnesium, zinc, or iron). This means Y is less reactive than sodium. Because X is less reactive than Y, X is low in the reactivity series and could be copper.
Marking scheme
Award 1 mark for the correct option (D).
Question 23 · multiple-choice
1 marks
In which reaction does the underlined substance act as a Brønsted-Lowry base?
A.NH3 in the reaction: NH3 + HCl -> NH4+ + Cl-
B.HCl in the reaction: HCl + H2O -> H3O+ + Cl-
C.H2SO4 in the reaction: H2SO4 + HNO3 -> HSO4- + H2NO3+
D.CH3COOH in the reaction: CH3COOH + H2O -> CH3COO- + H3O+
Show answer & marking schemeHide answer & marking scheme
Worked solution
A Brønsted-Lowry base is defined as a proton (H+) acceptor. In reaction A, NH3 accepts a proton from HCl to become NH4+, so it acts as a base. In all other reactions, the underlined species donate protons, acting as acids.
Marking scheme
Award 1 mark for the correct option (A).
Question 24 · multiple-choice
1 marks
An element E exists as two isotopes, 69E and 71E. The relative atomic mass of E is 69.8. What is the percentage abundance of the 71E isotope?
A.30%
B.40%
C.50%
D.60%
Show answer & marking schemeHide answer & marking scheme
Worked solution
Let the percentage abundance of 71E be x%. Therefore, the abundance of 69E is (100 - x)%. Relative atomic mass = [69 * (100 - x) + 71 * x] / 100 = 69.8. This simplifies to 6900 - 69x + 71x = 6980, so 2x = 80, giving x = 40%.
Marking scheme
Award 1 mark for the correct option (B).
Question 25 · multiple-choice
1 marks
A sample of element X contains two isotopes, \(^{69}\text{X}\) and \(^{71}\text{X}\). The relative atomic mass of this sample of X is 69.8. What is the percentage of the heavier isotope, \(^{71}\text{X}\), in this sample?
A.20%
B.40%
C.60%
D.80%
Show answer & marking schemeHide answer & marking scheme
Worked solution
Let \(p\) be the fractional abundance of \(^{71}\text{X}\). The abundance of \(^{69}\text{X}\) is therefore \(1 - p\). The relative atomic mass is the weighted average: \(69(1 - p) + 71p = 69.8\). Solving for \(p\): \(69 - 69p + 71p = 69.8 \Rightarrow 2p = 0.8 \Rightarrow p = 0.4\). This corresponds to 40%.
Marking scheme
1 mark for the correct option B.
Question 26 · multiple-choice
1 marks
Concentrated aqueous potassium bromide is electrolysed using inert platinum electrodes. Which row correctly identifies the products formed at the cathode and the anode?
A.cathode: hydrogen; anode: bromine
B.cathode: potassium; anode: bromine
C.cathode: hydrogen; anode: oxygen
D.cathode: potassium; anode: oxygen
Show answer & marking schemeHide answer & marking scheme
Worked solution
At the cathode, hydrogen ions (\(\text{H}^+\)) are discharged in preference to potassium ions (\(\text{K}^+\)) because hydrogen is lower in the reactivity series, producing hydrogen gas. At the anode, bromide ions (\(\text{Br}^-\)) are discharged in preference to hydroxide ions (\(\text{OH}^-\)) because it is a concentrated halide solution, producing bromine.
Marking scheme
1 mark for the correct option A.
Question 27 · multiple-choice
1 marks
Excess calcium carbonate chips are reacted with dilute hydrochloric acid, and the volume of carbon dioxide gas produced is measured over time. The experiment is repeated under the same conditions but using the same mass of calcium carbonate powder instead of chips. Which statement about the second experiment is correct?
A.The activation energy of the reaction decreases, increasing the rate of reaction.
B.The total volume of carbon dioxide produced increases because of the larger surface area.
C.The collision rate increases because more particles are exposed, but the total volume of gas remains the same.
D.The average kinetic energy of the reactant particles increases, leading to more successful collisions.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Using powder instead of chips increases the surface area, exposing more particles to collisions and thus increasing the collision rate and the rate of reaction. The total volume of gas produced depends on the limiting reactant (hydrochloric acid, as calcium carbonate is in excess), which does not change, so the total volume remains the same. Surface area changes do not affect activation energy or kinetic energy.
Marking scheme
1 mark for the correct option C.
Question 28 · multiple-choice
1 marks
The following gas-phase reaction has reached dynamic equilibrium in a sealed container: \(\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)} \quad \Delta H = +57\text{ kJ/mol}\). \(\text{N}_2\text{O}_4\) is colorless and \(\text{NO}_2\) is brown. Which change would cause the mixture to become darker brown?
A.decreasing the temperature at constant volume
B.increasing the pressure by reducing the volume
C.increasing the temperature at constant volume
D.adding a catalyst to the mixture
Show answer & marking schemeHide answer & marking scheme
Worked solution
The forward reaction is endothermic (\(\Delta H > 0\)). According to Le Chatelier's principle, increasing the temperature shifts the equilibrium in the endothermic direction (to the right) to absorb the added heat. This increases the concentration of brown \(\text{NO}_2\) gas, making the mixture darker brown.
Marking scheme
1 mark for the correct option C.
Question 29 · multiple-choice
1 marks
A section of an addition polymer chain is shown: \(-\text{CH}_2-\text{CH(CH}_3)-\text{CH}_2-\text{CH(CH}_3)-\text{CH}_2-\text{CH(CH}_3)-\). Which monomer is used to make this polymer?
A.ethane
B.ethene
C.propene
D.but-1-ene
Show answer & marking schemeHide answer & marking scheme
Worked solution
The repeat unit of this addition polymer is \(-\text{CH}_2-\text{CH(CH}_3)-\), which contains three carbon atoms in the backbone/side-chain group. The monomer must be an alkene with the same number of carbon atoms, which is propene (\(\text{CH}_2=\text{CH-CH}_3\)).
Marking scheme
1 mark for the correct option C.
Question 30 · multiple-choice
1 marks
In which reaction is water acting as a proton acceptor (Brønsted-Lowry base)?
Show answer & marking schemeHide answer & marking scheme
Worked solution
A Brønsted-Lowry base is defined as a proton acceptor. In reaction B, hydrochloric acid (\(\text{HCl}\)) donates a proton to water (\(\text{H}_2\text{O}\)), converting it into the hydronium ion (\(\text{H}_3\text{O}^+\)). Therefore, water acts as a proton acceptor in this reaction.
Marking scheme
1 mark for the correct option B.
Question 31 · multiple-choice
1 marks
Which reaction occurring in the blast furnace is highly exothermic and provides most of the heat required to maintain the high temperature in the furnace?
Show answer & marking schemeHide answer & marking scheme
Worked solution
The reaction between carbon (coke) and oxygen to form carbon dioxide is highly exothermic. This combustion reaction produces the heat necessary to keep the furnace at the very high temperatures required for iron extraction.
Marking scheme
1 mark for the correct option A.
Question 32 · multiple-choice
1 marks
A student wants to investigate the rate of reaction between zinc and dilute sulfuric acid by measuring the volume of hydrogen gas produced over time. Which pieces of apparatus are essential for this investigation?
A.balance, thermometer, gas syringe
B.gas syringe, stop-watch, conical flask with delivery tube
C.pipette, beaker, gas syringe
D.burette, stop-watch, thermometer
Show answer & marking schemeHide answer & marking scheme
Worked solution
To measure the volume of a gas produced over time, we need a container for the reaction to take place (conical flask), a way to direct and collect the gas (delivery tube and gas syringe), and a way to measure elapsed time (stop-watch).
Marking scheme
1 mark for the correct option B.
Question 33 · multiple-choice
1 marks
A synthetic polyamide has the repeat unit shown below: \([-NH-CH_{2}-CH_{2}-NH-CO-CH_{2}-CH_{2}-CO-]\) Which pair of monomers can react together to form this polymer?
A.\(H_{2}N-CH_{2}-CH_{2}-NH_{2}\) and \(HOOC-CH_{2}-CH_{2}-COOH\)
B.\(H_{2}N-CH_{2}-CH_{2}-COOH\) and \(H_{2}N-CH_{2}-CH_{2}-COOH\)
C.\(HO-CH_{2}-CH_{2}-OH\) and \(HOOC-CH_{2}-CH_{2}-COOH\)
D.\(H_{2}N-CH_{2}-CH_{2}-NH_{2}\) and \(HO-CH_{2}-CH_{2}-OH\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
Polyamides are formed by the condensation reaction between a diamine and a dicarboxylic acid. The monomer with the amine groups is \(H_{2}N-CH_{2}-CH_{2}-NH_{2}\) and the monomer with the carboxylic acid groups is \(HOOC-CH_{2}-CH_{2}-COOH\).
Marking scheme
Award 1 mark for the correct option A.
Question 34 · multiple-choice
1 marks
A reaction between calcium carbonate and excess dilute hydrochloric acid is carried out at two different temperatures, \(20^\circ\text{C}\) and \(40^\circ\text{C}\), with all other variables kept constant. Which statement explains why the rate of reaction increases when the temperature is increased?
A.The activation energy of the reaction is lowered.
B.The reactant particles collide more frequently, and a higher proportion of collisions have energy greater than the activation energy.
C.The particles move slower, increasing the probability of successful collisions.
D.The acid becomes more concentrated as water evaporates rapidly.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Increasing the temperature increases the kinetic energy of the particles. Consequently, they move faster and collide more frequently. Additionally, a much larger fraction of the colliding particles possess energy equal to or greater than the activation energy, leading to a higher frequency of successful collisions.
Marking scheme
Award 1 mark for the correct option B.
Question 35 · multiple-choice
1 marks
A gas-phase reversible reaction is represented by the equation shown: \(X(g) + 3Y(g) \rightleftharpoons 2Z(g)\) \(\Delta H = -92\text{ kJ/mol}\) Which combination of temperature and pressure changes will shift the position of equilibrium to the right to produce the highest yield of \(Z(g)\)?
A.increasing temperature and decreasing pressure
B.decreasing temperature and increasing pressure
C.increasing temperature and increasing pressure
D.decreasing temperature and decreasing pressure
Show answer & marking schemeHide answer & marking scheme
Worked solution
To shift the position of equilibrium to the right (exothermic direction), the temperature should be decreased. To shift it towards the side with fewer gas molecules (4 moles on the left, 2 moles on the right), the pressure should be increased.
Marking scheme
Award 1 mark for the correct option B.
Question 36 · multiple-choice
1 marks
Concentrated aqueous potassium bromide is electrolysed using inert platinum electrodes. Which products are obtained at the anode and the cathode?
A.Anode: oxygen; Cathode: potassium
B.Anode: bromine; Cathode: hydrogen
C.Anode: hydrogen; Cathode: bromine
D.Anode: bromine; Cathode: potassium
Show answer & marking schemeHide answer & marking scheme
Worked solution
At the cathode, hydrogen ions (from water) are preferentially discharged over potassium ions because hydrogen is less reactive than potassium, producing hydrogen gas. At the anode, bromide ions are preferentially discharged because halide ions are in high concentration in a concentrated solution, producing bromine gas.
Marking scheme
Award 1 mark for the correct option B.
Question 37 · multiple-choice
1 marks
What is the maximum volume of carbon dioxide gas, measured at r.t.p., produced when \(5.0\text{ g}\) of calcium carbonate (\(M_r = 100\)) reacts completely with excess dilute hydrochloric acid?
A.\(0.6\text{ dm}^3\)
B.\(1.2\text{ dm}^3\)
C.\(2.4\text{ dm}^3\)
D.\(12.0\text{ dm}^3\)
Show answer & marking schemeHide answer & marking scheme
Worked solution
The reaction is: \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\). Moles of \(\text{CaCO}_3 = \frac{5.0}{100} = 0.05\text{ mol}\). Since \(1\text{ mol}\) of \(\text{CaCO}_3\) produces \(1\text{ mol}\) of \(\text{CO}_2\), the moles of \(\text{CO}_2\) produced = \(0.05\text{ mol}\). Volume of \(\text{CO}_2\) at r.t.p. = \(0.05\text{ mol} \times 24\text{ dm}^3/\text{mol} = 1.2\text{ dm}^3\).
Marking scheme
Award 1 mark for the correct option B.
Question 38 · multiple-choice
1 marks
Which substance reacts with dilute sulfuric acid to produce a gas that turns limewater cloudy?
A.copper(II) oxide
B.sodium carbonate
C.magnesium ribbon
D.zinc hydroxide
Show answer & marking schemeHide answer & marking scheme
Worked solution
Dilute acids react with metal carbonates to produce carbon dioxide gas, which turns limewater cloudy. Therefore, sodium carbonate is the correct substance.
Marking scheme
Award 1 mark for the correct option B.
Question 39 · multiple-choice
1 marks
Two atoms are isotopes of the same element. Which statement about these two atoms is correct?
A.They have different chemical properties because they have different numbers of neutrons.
B.They have the same physical properties because they have the same nucleon number.
C.They have the same chemical properties because they have the same number of outer-shell electrons.
D.They have different numbers of protons but the same number of neutrons.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Isotopes have the same number of protons and outer-shell electrons, which means they display identical chemical properties. They differ only in their number of neutrons, which leads to different physical properties.
Marking scheme
Award 1 mark for the correct option C.
Question 40 · multiple-choice
1 marks
Which chemical equation represents a reaction in the extraction of iron in a blast furnace where carbon dioxide is reduced?
Show answer & marking schemeHide answer & marking scheme
Worked solution
In the reaction \(\text{CO}_2 + \text{C} \rightarrow 2\text{CO}\), the carbon atom in carbon dioxide goes from an oxidation state of +4 to +2, meaning it is reduced to carbon monoxide.
Marking scheme
Award 1 mark for the correct option B.
Ready to test yourself?
Turn these notes into exam-style practice. Get unlimited AI questions on this topic with instant marking and explanations.
Answer all structured and short-answer questions. Show all calculations and draw displayed structures where specified.
6 Question · 79.98 marks
Question 1 · structured-theory
13.33 marks
Germanium is an element in Group IV of the Periodic Table.
(a) Complete the table below to show the number of protons, neutrons and electrons in the specified atoms and ions of germanium.
| Particle | Number of protons | Number of neutrons | Number of electrons | | :--- | :---: | :---: | :---: | | \(^{74}_{32}\text{Ge}\) | | | | | \(^{72}_{32}\text{Ge}^{2+}\) | | | |
(b) A naturally occurring sample of germanium contains three isotopes: - \(^{70}\text{Ge}\) with an abundance of \(20.0\%\) - \(^{72}\text{Ge}\) with an abundance of \(27.0\%\) - \(^{74}\text{Ge}\) with an abundance of \(53.0\%\)
Calculate the relative atomic mass (\(A_r\)) of this sample of germanium. Give your answer to 1 decimal place.
(c) Define the term *isotopes*.
(d) Germanium reacts with chlorine to form germanium(IV) chloride, \(\text{GeCl}_4\).
(i) Draw a dot-and-cross diagram of a molecule of \(\text{GeCl}_4\). Show outer shell electrons only.
(ii) State the physical state of \(\text{GeCl}_4\) at room temperature and pressure, given that it has a simple molecular structure. Explain your answer.
Show answer & marking schemeHide answer & marking scheme
(c) Atoms of the same element with the same proton (atomic) number but different nucleon (mass) numbers / different number of neutrons.
(d) (i) Central Ge atom with 4 single covalent bonds sharing electrons with 4 Cl atoms. Ge has 8 electrons in its outer shell (4 shared pairs). Each Cl has 8 electrons in its outer shell (1 shared pair and 3 lone pairs). (ii) Liquid (or gas) because simple molecular substances have weak intermolecular forces which require little energy to overcome.
Marking scheme
(a) [4 marks] - 1 mark for each correct column of the table (4 columns total represented by protons/neutrons/electrons across both particles). - Protons: 32 and 32 [1] - Neutrons in Ge-74: 42 [1] - Neutrons in Ge-72: 40 [1] - Electrons: 32 and 30 [1]
(b) [3 marks] - \(70 \times 20\) and \(72 \times 27\) and \(74 \times 53\) shown [1] - Division by 100 [1] - Correct calculation to 1 d.p. (72.7) [1]
(c) [2 marks] - Atoms of the same element / same proton number [1] - Different number of neutrons / different nucleon number [1]
(d) [4 marks] - (i) Four shared pairs of electrons between Ge and Cl [1] - (i) Six non-bonding electrons on each of the four chlorine atoms [1] - (ii) Liquid / gas [1] - (ii) Weak intermolecular forces (require little energy to break) [1]
Question 2 · structured-theory
13.33 marks
Lead is extracted from its principal ore, galena (lead(II) sulfide, \(\text{PbS}\)), in a two-stage industrial process.
(a) In the first stage, the galena is roasted in air to produce lead(II) oxide, \(\text{PbO}\), and sulfur dioxide, \(\text{SO}_2\).
(i) Write a balanced chemical equation for this reaction. State symbols are not required.
(ii) State one major environmental consequence of releasing sulfur dioxide gas directly into the atmosphere.
(b) In the second stage, the lead(II) oxide is reduced by heating it with carbon in a furnace.
(i) State which substance is reduced in this reaction. Explain your answer in terms of oxygen transfer.
(ii) Identify the oxidizing agent in this reaction.
(iii) Carbon monoxide, \(\text{CO}\), can also reduce lead(II) oxide in the furnace. Write a balanced chemical equation for the reduction of lead(II) oxide by carbon monoxide.
(c) Pure lead can also be obtained in the laboratory by the electrolysis of molten lead(II) bromide using inert carbon electrodes.
(i) Write the ionic half-equation, including state symbols, for the reaction occurring at the negative electrode (cathode).
(ii) Describe what is observed at the positive electrode (anode) during this electrolysis.
(iii) Explain why solid lead(II) bromide does not conduct electricity, whereas molten lead(II) bromide does.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) \(2\text{PbS} + 3\text{O}_2 \rightarrow 2\text{PbO} + 2\text{SO}_2\) (ii) It causes acid rain (which damages buildings and acidifies lakes/soils).
(b) (i) Lead(II) oxide (\(\text{PbO}\)) is reduced because it loses oxygen to form lead. (ii) Lead(II) oxide (\(\text{PbO}\)). (iii) \(\text{PbO} + \text{CO} \rightarrow \text{Pb} + \text{CO}_2\)
(c) (i) \(\text{Pb}^{2+}\text{(l)} + 2\text{e}^- \rightarrow \text{Pb(l)}\) (ii) Red-brown vapour / gas is evolved. (iii) In the solid state, ions are in fixed positions in a giant lattice and cannot move; when molten, the ions are free to move and carry the charge.
Marking scheme
(a) [3 marks] - (i) Correct formulae of reactants and products: \(\text{PbS}\), \(\text{O}_2\), \(\text{PbO}\), \(\text{SO}_2\) [1] - (i) Correct balancing: 2, 3, 2, 2 [1] - (ii) Acid rain / kills aquatic life / damages limestone structures [1]
(a) (i) Draw a fully labeled diagram of the apparatus that the student could use to carry out this experiment and collect and measure the volume of carbon dioxide gas produced.
(ii) Describe the shape of the graph of volume of gas plotted against time from the start of the reaction until the reaction is complete.
(b) The experiment is repeated using the same mass of calcium carbonate chips and the same volume of nitric acid of the same concentration, but at a higher temperature.
(i) Explain, in terms of collision theory, why the rate of reaction increases at a higher temperature.
(ii) Describe how the curve for this second experiment would differ from the original curve if plotted on the same axes.
(c) The reaction is repeated a third time at the original temperature, but using calcium carbonate powder instead of calcium carbonate chips. All other variables are kept constant.
State and explain the effect of this change on the rate of reaction.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) The diagram should show a conical flask containing calcium carbonate chips and dilute nitric acid, sealed with a rubber bung, with a delivery tube leading to a gas syringe (or an inverted measuring cylinder filled with water over a trough). (ii) The curve starts steep at the beginning (high rate), then its gradient decreases (rate slows down), and finally it goes completely flat/horizontal (reaction complete/rate is zero).
(b) (i) At a higher temperature, particles have more kinetic energy and move faster, so there are more frequent collisions. Also, a greater proportion of colliding particles have energy greater than or equal to the activation energy, leading to a higher percentage of successful collisions. (ii) The curve is steeper at the start but levels off at the exact same final volume of carbon dioxide.
(c) The rate of reaction increases. Powder has a much larger surface area than large chips, meaning more particles are exposed, leading to a higher frequency of collisions per unit time.
Marking scheme
(a) [5 marks] - (i) Conical flask with reacting mixture [1] - (i) Delivery tube with gas-tight connection [1] - (i) Gas syringe or inverted measuring cylinder in water trough [1] - (ii) Steepest at start and curve decreasing in gradient [1] - (ii) Levels off / becomes horizontal at the end [1]
(b) [5 marks] - (i) Particles have more kinetic energy / move faster [1] - (i) More frequent collisions [1] - (i) More particles have energy \(\ge\) activation energy / more successful collisions [1] - (ii) Shorter time to finish / steeper curve [1] - (ii) Same final volume of gas [1]
(c) [3 marks] - Rate increases [1] - Larger surface area [1] - More frequent collisions / more collisions per second [1]
Question 4 · structured-theory
13.33 marks
This question is about preparing and analyzing salts.
(a) Describe the experimental steps required to prepare a pure, dry sample of hydrated copper(II) sulfate crystals, starting from solid copper(II) oxide (an insoluble base) and dilute sulfuric acid.
(b) In a separate experiment, a student performs an acid-base titration to determine the concentration of a solution of sodium hydroxide, \(\text{NaOH(aq)}\).
\(25.0\text{ cm}^3\) of the \(\text{NaOH(aq)}\) is neutralized by exactly \(20.0\text{ cm}^3\) of \(0.150\text{ mol/dm}^3\) sulfuric acid, \(\text{H}_2\text{SO}_4\text{(aq)}\).
(i) Calculate the number of moles of \(\text{H}_2\text{SO}_4\) used in the titration.
(ii) Deduce the number of moles of \(\text{NaOH}\) that reacted with this amount of acid.
(iii) Calculate the concentration of the sodium hydroxide solution in \(\text{mol/dm}^3\).
(c) (i) Name a suitable indicator for this titration.
(ii) State the colour change of this indicator at the end-point when acid is added from the burette to the alkali in the conical flask.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) 1. Add excess copper(II) oxide to a fixed volume of warm dilute sulfuric acid. 2. Stir to ensure complete reaction (until no more oxide dissolves). 3. Filter the mixture to remove the unreacted excess copper(II) oxide. 4. Heat the filtrate (copper(II) sulfate solution) to the point of crystallization. 5. Leave the saturated solution to cool slowly so crystals form. 6. Filter the crystals and dry them with filter paper.
(b) (i) \(\text{Moles of } \text{H}_2\text{SO}_4 = \text{concentration} \times \text{volume in dm}^3 = 0.150 \times \frac{20.0}{1000} = 0.00300\text{ mol}\). (ii) From the equation, the mole ratio of \(\text{NaOH} : \text{H}_2\text{SO}_4\) is \(2 : 1\). \(\text{Moles of } \text{NaOH} = 2 \times 0.00300 = 0.00600\text{ mol}\). (iii) \(\text{Concentration of } \text{NaOH} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00600}{25.0 / 1000} = 0.240\text{ mol/dm}^3\).
(c) (i) Methyl orange (or phenolphthalein). (ii) For methyl orange: Yellow to orange/red. (For phenolphthalein: Pink to colourless).
Marking scheme
(a) [6 marks] - Add excess copper(II) oxide [1] - Heat / warm acid [1] - Filter to remove excess copper(II) oxide [1] - Heat filtrate to point of crystallization / saturate [1] - Leave to cool (to crystallize) [1] - Filter and dry crystals with filter paper [1]
(c) [3 marks] - (i) Methyl orange / phenolphthalein [1] - (ii) Yellow to red/orange (if methyl orange used) OR pink to colourless (if phenolphthalein used) [2] - 1 mark for starting colour - 1 mark for final colour
Question 5 · structured-theory
13.33 marks
Polymers are long-chain molecules formed from many smaller monomer units.
(a) Poly(phenylethene), commonly known as polystyrene, is an addition polymer. The monomer used to make it is phenylethene, \(\text{C}_6\text{H}_5\text{CH}=\text{CH}_2\).
(i) Draw the displayed formula of phenylethene. Show all atoms and all bonds.
(ii) Draw the structure of poly(phenylethene) showing two repeat units.
(iii) State the fundamental difference between addition polymerization and condensation polymerization.
(b) Terylene is a synthetic condensation polyester. It is formed by reacting a dicarboxylic acid and a diol together.
(i) The dicarboxylic acid can be represented as \(\text{HOOC}-\square-\text{COOH}\) and the diol can be represented as \(\text{HO}-\bigcirc-\text{OH}\).
Draw the structure of the polyester formed from these two monomers, showing one complete repeat unit and all the bonds in the ester linkage.
(ii) Name the small molecule that is eliminated during the formation of Terylene.
(c) Nylon-6,6 is a polyamide.
(i) Name the two different functional groups that react together to form a polyamide linkage.
(ii) Draw the displayed structure of the amide linkage (peptide link) formed, showing all atoms and bonds.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) A carbon-to-carbon double bond (\(\text{C}=\text{C}\)) with three \(\text{H}\) atoms and one phenyl group (\(-\text{C}_6\text{H}_5\)) attached to them. (ii) A chain of four single-bonded carbon atoms: \(-(\text{CH}(\text{C}_6\text{H}_5)-\text{CH}_2-\text{CH}(\text{C}_6\text{H}_5)-\text{CH}_2)-\), with continuation bonds on both ends. (iii) Addition polymerization involves monomers with \(\text{C}=\text{C}\) double bonds reacting to form a polymer as the only product; condensation polymerization involves monomers with two functional groups reacting, eliminating a small molecule (like water) at each linkage.
(b) (i) \(-\text{O}-\bigcirc-\text{O}-\text{CO}-\square-\text{CO}-\) with continuation bonds on the outer oxygens/carbonyl carbons. (ii) Water (\(\text{H}_2\text{O}\)).
(c) (i) Amine (\(-\text{NH}_2\)) and carboxylic acid (\(-\text{COOH}\)). (ii) \(-\text{CO}-\text{NH}-\) displayed showing all bonds: a carbon double-bonded to oxygen, single-bonded to a nitrogen, which is single-bonded to a hydrogen.
Marking scheme
(a) [6 marks] - (i) Correctly drawn \(\text{C}=\text{C}\) double bond with three hydrogens [1] - (i) Correctly drawn phenyl group (\(-\text{C}_6\text{H}_5\) or ring) attached to a carbon [1] - (ii) Correctly drawn chain backbone of 4 carbons with single bonds and continuation bonds [1] - (ii) Correct attachment of two phenyl groups on alternate carbons [1] - (iii) Addition: only polymer formed / no side product [1] - (iii) Condensation: polymer and a small molecule (e.g. water) formed [1]
(b) [4 marks] - (i) Correctly drawn ester linkage: \(-\text{CO}-\text{O}-\) with all bonds shown [1] - (i) Correctly positioned blocks: \(\square\) and \(\bigcirc\) in the main chain [1] - (i) Continuation bonds shown at both ends of the repeat unit [1] - (ii) Water [1]
(c) [4 marks] - (i) Amine [1] - (i) Carboxylic acid [1] - (ii) \(-\text{C}=\text{O}\) shown [1] - (ii) \(-\text{N}-\text{H}\) single bond connected to carbonyl carbon [1]
Question 6 · structured-theory
13.33 marks
Carboxylic acids and alcohols are important families of organic compounds.
(a) An organic compound, **Y**, contains carbon, hydrogen and oxygen. Analysis shows that **Y** has the following composition by mass: - Carbon: \(40.00\%\) - Hydrogen: \(6.67\%\) - Oxygen: \(53.33\%\)
(i) Calculate the empirical formula of compound **Y**.
(ii) The relative molecular mass of compound **Y** is 60. Determine the molecular formula of **Y**.
(b) Compound **Y** is a weak carboxylic acid.
(i) Write the IUPAC name of compound **Y**.
(ii) Draw the displayed formula of compound **Y**, showing all atoms and all bonds.
(c) Carboxylic acids react with alcohols to form esters in a reversible reaction.
(i) State the name of a catalyst used to increase the rate of this esterification reaction.
(ii) Propanoic acid reacts with ethanol when heated with the catalyst. Write the chemical equation for this reaction. State symbols are not required.
(iii) Draw the displayed formula of the ester formed in this reaction.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Moles of C = \(\frac{40.00}{12} = 3.33\) - Moles of H = \(\frac{6.67}{1} = 6.67\) - Moles of O = \(\frac{53.33}{16} = 3.33\) Dividing by the smallest value (3.33): - C : H : O = \(1 : 2 : 1\). Empirical formula is \(\text{CH}_2\text{O}\).
(ii) Empirical formula mass of \(\text{CH}_2\text{O} = 12 + (2 \times 1) + 16 = 30\). Multiplier = \(\frac{60}{30} = 2\). Molecular formula is \(\text{C}_2\text{H}_4\text{O}_2\).
(b) (i) Ethanoic acid. (ii) Displayed formula of ethanoic acid: \(\text{CH}_3\text{COOH}\) showing all atoms and bonds, including the \(\text{C}=\text{O}\) and \(\text{O}-\text{H}\) bonds.
(c) (i) Concentrated sulfuric acid. (ii) \(\text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}\) (iii) Displayed formula of ethyl propanoate showing all atoms and bonds: \(\text{CH}_3-\text{CH}_2-\text{CO}-\text{O}-\text{CH}_2-\text{CH}_3\).
Marking scheme
(a) [5 marks] - (i) Dividing percentages by relative atomic masses (C: 3.33, H: 6.67, O: 3.33) [1] - (i) Dividing by the smallest to get 1:2:1 ratio [1] - (i) Empirical formula: \(\text{CH}_2\text{O}\) [1] - (ii) Calculating empirical mass (30) [1] - (ii) Molecular formula: \(\text{C}_2\text{H}_4\text{O}_2\) [1]
(b) [3 marks] - (i) Ethanoic acid [1] - (ii) Correct structure of \(-\text{COOH}\) group with all bonds shown [1] - (ii) Correct methyl group attached to carbonyl carbon [1]
(c) [5 marks] - (i) Concentrated sulfuric acid / acid catalyst [1] - (ii) Correct reactants: propanoic acid and ethanol [1] - (ii) Correct products: ethyl propanoate and water [1] - (iii) Correct displayed ester linkage (\(-\text{COO}-\)) shown [1] - (iii) Rest of ethyl propanoate structure fully correct [1]
Paper 63 (Alternative to Practical)
Answer all practical questions, including one comprehensive 6-mark experimental planning question.
4 Question · 40 marks
Question 1 · practical-structured
10 marks
A student investigates the rate of reaction between marble chips (calcium carbonate) and dilute hydrochloric acid.
The equation for the reaction is: $$\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$$
(a) Identify the piece of apparatus that should be used to: (i) measure exactly $25.0\text{ cm}^3$ of dilute hydrochloric acid, [1] (ii) collect and measure the volume of carbon dioxide gas produced. [1]
(b) The student noticed that the plunger of the gas syringe was tight and did not move smoothly. State how this would affect the measured rate of reaction. [1]
(c) Describe a chemical test to confirm that the gas produced is carbon dioxide, including the expected positive result. test: ........................................................................................................................... result: ......................................................................................................................... [2]
(d) Explain why a gas syringe is more suitable for collecting carbon dioxide than collecting it over water in an inverted measuring cylinder. [2]
(e) The reaction eventually slows down and stops. State why the reaction stops. [1]
(f) Sketch on a graph of volume of gas (y-axis) against time (x-axis) how the curve would change if the student repeated the experiment using the same mass of calcium carbonate but as a fine powder instead of large marble chips. Explain your answer. sketch description: .................................................................................................... explanation: .................................................................................................................. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) Volumetric pipette / burette (ii) Gas syringe (b) The measured rate of reaction would be lower than the actual rate / volume of gas recorded would be less than expected. (c) Test: bubble the gas through limewater. Result: limewater turns cloudy / milky. (d) Carbon dioxide is moderately soluble in water, so some gas would dissolve if collected over water, resulting in an underestimate of the volume of gas produced. (e) One or both of the reactants (calcium carbonate or hydrochloric acid) have been completely used up / reaction goes to completion. (f) Sketch description: The curve would be steeper at the start but level off at the same final volume. Explanation: Powder has a larger surface area, leading to a faster rate of reaction, but the same amount of reactants produces the same final volume of gas.
Marking scheme
(a) (i) Volumetric pipette / pipette / burette [1] (ii) Gas syringe [1] (b) Measured rate is lower / gas volume recorded is lower [1] (c) Test: Limewater [1] Result: Turns cloudy / milky [1] (d) Carbon dioxide dissolves in water / is soluble in water [1] Leading to lower volume collected / loss of gas [1] (e) Reactant(s) used up / limiting reactant is fully reacted [1] (f) Curve is steeper / reaches plateau earlier [1] Same final volume because same mass of calcium carbonate used [1]
Question 2 · practical-structured
10 marks
A student uses paper chromatography to analyze the pigments in three different black felt-tip pens (X, Y, and Z) and a reference dye (R).
(a) Explain why the starting line on the chromatography paper is drawn in pencil and not in ink. [2]
(b) Describe how the student should place the chromatography paper in the beaker containing the solvent to ensure a successful separation. [2]
(c) State when the student should remove the chromatography paper from the solvent. [1]
(d) Define the term $R_f$ value. [1]
(e) In the chromatogram, the baseline to the centre of the spot for reference dye R is $3.6\text{ cm}$, and the baseline to the solvent front is $8.0\text{ cm}$. Calculate the $R_f$ value of dye R. Show your working. $R_f$ value = ................................................................. [2]
(f) Pen Z produced only one spot that remained on the starting line. State what this indicates about the pigment in pen Z, and suggest how the experiment could be modified to separate it. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Pencil is insoluble in the solvent and will not run/separate. Ink is soluble and would dissolve, contaminating the chromatogram. (b) The paper should be suspended vertically, ensuring the solvent level is below the pencil starting line, and the paper does not touch the sides of the beaker. (c) Just before the solvent front reaches the top of the paper. (d) The ratio of the distance travelled by the solute (pigment) to the distance travelled by the solvent front. (e) $R_f = \frac{\text{distance moved by spot}}{\text{distance moved by solvent}} = \frac{3.6}{8.0} = 0.45$ (f) The pigment is insoluble in the chosen solvent. Modification: repeat the experiment using a different solvent (e.g., ethanol or propanone).
Marking scheme
(a) Pencil is insoluble / does not dissolve in solvent [1] Ink is soluble / would run / would interfere with results [1] (b) Solvent level must be below the starting/pencil line [1] Paper suspended vertically / does not touch the beaker walls [1] (c) When the solvent front is near the top of the paper [1] (d) Distance moved by substance / distance moved by solvent [1] (e) Working: $3.6 / 8.0$ [1] Answer: $0.45$ (no units) [1] (f) Insoluble in the solvent used [1] Use a different solvent / mobile phase (e.g. ethanol / organic solvent) [1]
Question 3 · practical-structured
10 marks
A student is provided with a green crystalline solid, salt T, which contains one cation and one anion. The student performs several tests to identify the ions.
(a) A flame test is performed on solid T. State the flame colour that would confirm the presence of copper(II) ions, $\text{Cu}^{2+}$. [1]
(b) A solution of salt T is prepared in distilled water. (i) To the first portion of this solution, aqueous sodium hydroxide is added dropwise and then in excess. Describe the observations. dropwise: .................................................................................................................... in excess: ................................................................................................................... [2] (ii) To the second portion of the solution, aqueous ammonia is added dropwise and then in excess. Describe the observations. dropwise: .................................................................................................................... in excess: ................................................................................................................... [2]
(c) To the third portion of the solution, dilute nitric acid is added followed by aqueous silver nitrate. A cream precipitate is formed. (i) Identify the anion present in salt T. [1] (ii) State the formula of the cream precipitate. [1] (iii) Deduce the chemical formula of salt T. [1]
(d) Explain why distilled water is used to prepare the solution of salt T rather than tap water. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Blue-green (b) (i) Dropwise: light blue precipitate. In excess: precipitate is insoluble. (ii) Dropwise: light blue precipitate. In excess: precipitate dissolves to form a dark blue solution. (c) (i) Bromide ($\text{Br}^-$) (ii) $\text{AgBr}$ (iii) $\text{CuBr}_2$ (d) Tap water contains dissolved ions (such as chloride or sulfate) which could react with the reagents and produce false positive precipitate results.
Marking scheme
(a) Blue-green [1] (b) (i) Dropwise: light blue precipitate [1] In excess: insoluble / precipitate remains [1] (ii) Dropwise: light blue precipitate [1] In excess: dissolves / soluble AND dark blue solution [1] (c) (i) Bromide [1] (ii) $\text{AgBr}$ / silver bromide [1] (iii) $\text{CuBr}_2$ [1] (d) Tap water contains dissolved impurities/ions [1] These would react with silver nitrate/reagents to form a precipitate / give false results [1]
Question 4 · practical-structured
10 marks
Hydrogen peroxide, $\text{H}_2\text{O}_2$, decomposes slowly at room temperature to produce water and oxygen gas. This reaction is catalysed by transition metal oxides. $$2\text{H}_2\text{O}_2\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)}$$
(a) Plan an investigation to find which of three solid catalysts—manganese(IV) oxide, copper(II) oxide, and iron(III) oxide—is the most effective catalyst for the decomposition of hydrogen peroxide.
You are provided with: - the three solid catalysts as fine powders - aqueous hydrogen peroxide - distilled water - common laboratory apparatus.
Your plan should include: - how you will carry out the experiments - the measurements you will take - how you will ensure a fair test - how you will use your results to determine the most effective catalyst.
You may draw a diagram to help explain your plan. [6]
(b) State two variables, other than the mass of the catalyst, that must be kept constant in this investigation. [2]
(c) Describe the test to confirm that the gas produced is oxygen, including the positive result. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) 1. Measure a fixed volume (e.g. $50\text{ cm}^3$) of hydrogen peroxide of a known concentration using a measuring cylinder and pour it into a conical flask. 2. Weigh out a fixed mass (e.g. $0.5\text{ g}$) of manganese(IV) oxide powder. 3. Connect the conical flask to a gas syringe using a delivery tube and stopper. 4. Add the catalyst to the flask, immediately replace the stopper, and start a stop-watch. 5. Measure the volume of oxygen gas collected in the syringe after a set time (e.g., $1\text{ minute}$), or measure the time taken to collect a fixed volume of gas (e.g., $50\text{ cm}^3$). 6. Repeat steps 1–5 using the same volume and concentration of hydrogen peroxide, and the same mass of copper(II) oxide and iron(III) oxide respectively. 7. The most effective catalyst is the one that produces the largest volume of gas in the set time (or takes the shortest time to collect the fixed volume of gas).
(b) 1. Concentration of hydrogen peroxide. 2. Volume of hydrogen peroxide. 3. Temperature of the mixture. 4. Particle size/surface area of the catalyst (all should be fine powders).
(c) Test: insert a glowing splint into a test-tube of the gas. Result: the splint relights.
Marking scheme
(a) MP1: Measure a fixed volume of hydrogen peroxide (into a conical flask) [1] MP2: Weigh out a fixed mass of one catalyst [1] MP3: Connect flask to gas syringe / collect gas over water [1] MP4: Add catalyst, seal flask, start stopwatch [1] MP5: Record volume of gas in a specified time (or time taken for a fixed volume) [1] MP6: Repeat with the other two catalysts and state that the most effective catalyst gives the fastest rate (largest volume in set time / shortest time) [1]
(b) Any two from: - Concentration of hydrogen peroxide [1] - Volume of hydrogen peroxide [1] - Temperature [1] - Particle size / surface area of catalysts [1]
(c) Test: Glowing splint [1] Result: Relights [1]
Wondering how well you actually know this?
thinka is an AI practice app for IGCSE & IB students: unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.