IGCSE Chemistry 0620/41 May/June 2025 Past Paper with Answers

Question 1

From the list bauxite, calcium oxide, ethanol, graphite, oxygen and sulfur dioxide, which substance is a basic oxide?

1. calcium oxide, a metal oxide that reacts with acids
2. sulfur dioxide, a non-metal oxide that dissolves to form an acid
3. bauxite, an ore of aluminium
4. graphite, a form of the element carbon

Question 2

From the list ethanol, graphite, methane, propane and propene, which substance contains carbon atoms only?

1. graphite, a giant covalent form of the element carbon
2. methane, which contains carbon and hydrogen atoms
3. ethanol, which contains carbon, hydrogen and oxygen atoms
4. propene, which contains carbon and hydrogen atoms

Question 3

From the list calcium oxide, ethanol, methane, propene and sulfur dioxide, which substance is manufactured by fermentation?

1. ethanol, made by the fermentation of sugars by yeast
2. methane, made by the decomposition of vegetation
3. sulfur dioxide, made by burning sulfur
4. propene, made by cracking hydrocarbons

Question 4 · Answer A

Step 1: When vegetation (plant matter) decomposes without much air, methane gas is produced. Step 2: So the gas produced by the decomposition of vegetation is methane. Step 3: Ethanol comes from fermentation, propane from petroleum and nitrogen from the air, so the answer is methane.

Question 5 · Answer A

Step 1: Aluminium is extracted from its ore, which contains aluminium oxide. Step 2: Bauxite is the ore of aluminium and contains aluminium oxide. Step 3: Calcium oxide, graphite and sulfur dioxide do not contain aluminium oxide, so the answer is bauxite.

Question 6 · Answer A

Step 1: Acid rain is caused by acidic gases that dissolve in rain water. Step 2: Sulfur dioxide dissolves in rain to form an acidic solution, so it causes acid rain. Step 3: Nitrogen and oxygen do not cause acid rain, and methane is a greenhouse gas, so the answer is sulfur dioxide.

Question 7 · Answer A

Step 1: Count the atoms in each molecule using its formula. Step 2: Propane is $\text{C}_3\text{H}_8$, which has 3 carbon atoms + 8 hydrogen atoms = 11 atoms. Step 3: Propene has 9 atoms, methane 5 and ethanol 9, so the molecule with 11 atoms is propane.

Question 8 · Answer A

Step 1: In the blast furnace, limestone (calcium carbonate) is thermally decomposed. Step 2: This produces calcium oxide and carbon dioxide; the calcium oxide then removes acidic impurities as slag. Step 3: Bauxite, graphite and sulfur dioxide are not made by this decomposition, so the answer is calcium oxide.

Question 9 · Answer A

Step 1: Clean dry air is about 78% nitrogen and 21% oxygen, with small amounts of other gases. Step 2: The gas making up roughly 21% is oxygen. Step 3: Nitrogen is about 78%, while sulfur dioxide and methane are only present in small or trace amounts, so the answer is oxygen.

Question 10 · Answer A

Step 1: Addition polymerisation needs monomers that contain a carbon-carbon double bond (C=C). Step 2: Propene is an alkene with a C=C double bond, so it can act as a monomer. Step 3: Propane and methane are saturated (no C=C) and ethanol has no C=C, so the monomer is propene.

Question 11 · Answer A

Step 1: An electron carries a relative charge of −1 and has a very small relative mass (1/1840). Step 2: A proton carries a relative charge of +1 and has a relative mass of 1. Step 3: So the electron is −1, and the proton is +1 with a relative mass of 1.

Question 12 · Answer A

Step 1: Protons = proton number = 19 for both. Neutrons = nucleon number − proton number, so $^{39}\text{K}$ has 39 − 19 = 20 and $^{41}\text{K}$ has 41 − 19 = 22. Step 2: A neutral atom has electrons = protons, so $^{39}\text{K}$ has 19 electrons. Step 3: $^{41}\text{K}^+$ has lost one electron, leaving 18 electrons (protons and neutrons unchanged), giving 18 electrons, 22 neutrons, 19 protons.

Question 13 · Answer A

Step 1: Relative atomic mass = [(39 × 90) + (41 × 10)] ÷ 100. Step 2: This is (3510 + 410) ÷ 100 = 3920 ÷ 100. Step 3: So the relative atomic mass is 39.2.

Question 14 · Answer A

Step 1: The relative atomic mass is the weighted average of all the isotopes of an element. Step 2: If the relative atomic mass (27) equals the nucleon number of the only quoted isotope (27), there can be no other isotope pulling the average away from 27. Step 3: So aluminium must have only one isotope, of nucleon number 27.

Question 15 · Answer A

Step 1: $\text{Ca}^{2+}$ has 18 electrons arranged 2,8,8 (the argon configuration). Step 2: A neutral atom with 18 electrons is argon (Ar). Step 3: A negative ion with 18 electrons is $\text{Cl}^-$ (17 + 1 = 18 electrons), so the pair is Ar and $\text{Cl}^-$.

Question 16 · Answer A

Step 1: In a metal, the atoms lose their outer electrons to form positive metal ions (cations). Step 2: These released electrons become delocalised, forming a sea of mobile electrons around the positive ions. Step 3: There is strong electrostatic attraction between the positive ions and the delocalised electrons, which holds the metal together.

Question 17 · Answer A

Step 1: Solid copper contains a sea of delocalised electrons that are free to move. Step 2: When a voltage is applied, these mobile electrons drift through the metal, carrying the electric charge. Step 3: So copper conducts electricity because of its free-moving (delocalised) electrons.

Question 18 · Answer A

Step 1: An alloy contains atoms of different sizes, which disrupt the regular layers of metal atoms. Step 2: This makes it harder for the layers to slide over each other, so the alloy is harder and stronger than the pure metal. Step 3: So a key reason alloys are more useful is that they are harder and stronger.

Question 19 · Answer A

Step 1: Brass is an alloy made from two metals. Step 2: Besides copper, brass contains zinc. Step 3: Tin, lead and iron are not the metal in brass, so the answer is zinc.

Question 20 · Answer A

Step 1: The acid reacts with the copper(II) carbonate, so the solid is used up and dissolves (disappears). Step 2: The carbonate reacts with the acid to release carbon dioxide gas, seen as fizzing/bubbling (effervescence). Step 3: So the two observations are the solid dissolving and effervescence.

Question 21 · Answer A

Step 1: Copper(II) carbonate is added until no more reacts, so it is added in excess. Step 2: The reaction stops because all of the acid has been used up (there is no more acid to react with the carbonate). Step 3: So the reaction stops because the sulfuric acid has all reacted.

Question 22 · Answer A

Step 1: Excess copper(II) carbonate was added, so some did not react. Step 2: Filtration removes this unreacted solid as the residue, while the aqueous copper(II) sulfate passes through as the filtrate. Step 3: So the residue is the unreacted copper(II) carbonate.

Question 23 · Answer A

Step 1: Copper(II) sulfate forms when a copper base or carbonate reacts with sulfuric acid. Step 2: Copper(II) oxide is a base that reacts with sulfuric acid to give copper(II) sulfate and water. Step 3: Copper metal is too unreactive to react with dilute sulfuric acid, so the best answer is copper(II) oxide (copper(II) hydroxide would also work).

Question 24 · Answer A

Step 1: A solution that has dissolved the maximum possible solute at a given temperature cannot dissolve any more. Step 2: The term for this is a saturated solution. Step 3: Dilute and concentrated describe relative amounts but can still dissolve more, so the correct term is saturated.

Question 25 · Answer A

Step 1: Increasing the concentration packs more acid particles into the same volume. Step 2: With more particles per unit volume, the particles collide more often, so there are more collisions per unit time. Step 3: More frequent collisions mean a faster rate of reaction.

Question 26 · Answer A

Step 1: The dot formula $\text{CuSO}_4{\cdot}5\text{H}_2\text{O}$ shows water chemically combined in the crystal (water of crystallisation). Step 2: A substance chemically combined with water in this way is described as hydrated. Step 3: Anhydrous means without water, so the correct term is hydrated.

Question 27 · Answer A

Step 1: Each nitrogen atom has 5 outer electrons and needs 3 more to reach a full octet. Step 2: The two atoms share three pairs of electrons (a triple bond), giving each atom a share of 8 outer electrons. Step 3: After the three shared pairs, each nitrogen still has one non-bonding (lone) pair, so the diagram shows a triple bond plus one lone pair on each atom.

Question 28 · Answer A

Step 1: The Haber process uses a moderately high temperature of about 450 °C to give a reasonable rate and yield. Step 2: It uses a high pressure of about 200 atmospheres to favour the side with fewer gas molecules (ammonia). Step 3: An iron catalyst speeds up the reaction, so the conditions are about 450 °C, about 200 atmospheres and an iron catalyst.

Question 29 · Answer A

Step 1: Nitrogen and hydrogen combine to form ammonia, $\text{NH}_3$. Step 2: Balancing gives 1 nitrogen molecule and 3 hydrogen molecules forming 2 ammonia molecules: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$. Step 3: This balances 2 N and 6 H on each side.

Question 30 · Answer A

Step 1: With 4 $\text{NH}_3$, there are 4 N and 12 H on the left, so 4 NO and 6 $\text{H}_2\text{O}$ balance N and H. Step 2: The right now has 4 O (in NO) + 6 O (in water) = 10 O atoms, which needs 5 $\text{O}_2$. Step 3: So the coefficients are 5 $\text{O}_2$, 4 NO and 6 $\text{H}_2\text{O}$.

Question 31 · Answer A

Step 1: In $\text{NH}_3$, each hydrogen is +1, and the molecule is neutral, so nitrogen must be −3 (since −3 + 3(+1) = 0). Step 2: In NO, oxygen is −2 and the molecule is neutral, so nitrogen must be +2 (since +2 + (−2) = 0). Step 3: So nitrogen is −3 in $\text{NH}_3$ and +2 in NO.

Question 32 · Answer A

Step 1: Oxidation number tracks how oxidised or reduced an element is. Step 2: In terms of oxidation number, oxidation is defined as an increase in oxidation number. Step 3: A decrease in oxidation number is reduction, so oxidation is an increase.

Question 33 · Answer A

Step 1: The product is $\text{HNO}_3$, which contains H, N and O. Step 2: Balancing N gives 4 NO making 4 $\text{HNO}_3$; the 4 H need 2 $\text{H}_2\text{O}$. Step 3: Counting oxygen: the right has 12 O; NO gives 4 O and water gives 2 O, so 6 more O atoms (3 $\text{O}_2$) are needed, giving $4\text{NO} + 3\text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{HNO}_3$.

Question 34 · Answer A

Step 1: Moles of $\text{NH}_3$ = 360 ÷ 24 = 15 mol. Step 2: From the 2:1 ratio, moles of $(\text{NH}_4)_2\text{SO}_4$ = 15 ÷ 2 = 7.5 mol. Step 3: Mass = 7.5 × 132 = 990 g.

Question 35 · Answer A

Step 1: At equilibrium in a closed system, the forward and reverse reactions occur at the same rate. Step 2: Because the rates are equal, $\text{CH}_3\text{OH}$ is being made as fast as it is broken down, so its concentration no longer changes. Step 3: The reactions have not stopped; they are still going at equal rates, so the concentration stays constant.

Question 36 · Answer A

Step 1: Lowering the pressure means the particles are further apart, so they collide less often and the rate of reaction decreases. Step 2: The forward reaction makes fewer gas molecules, so a lower pressure shifts the equilibrium towards the reactants, decreasing the yield of methanol. Step 3: So both the rate and the yield of methanol decrease at lower pressure.

Question 37 · Answer A

Step 1: A catalyst speeds up the forward and reverse reactions equally, so the position of equilibrium is unchanged. Step 2: Therefore the concentration of $\text{CH}_3\text{OH}$ at equilibrium does not change. Step 3: However, the catalyst does increase the rate of the reverse reaction (it increases both rates), so the answer is no change to the concentration but an increase in the reverse rate.

Question 38 · Answer A

Step 1: Raising the temperature shifts an equilibrium in the endothermic direction. Step 2: Since a higher temperature decreases the yield of methanol, the reverse reaction must be endothermic, so the forward reaction is exothermic. Step 3: So the forward reaction is exothermic.

Question 39 · Answer A

Step 1: Transition elements and their compounds are widely used as catalysts. Step 2: From the list, copper is the only transition element, so it is a suitable catalyst. Step 3: Barium, potassium, carbon and sulfur are not transition elements, so the answer is copper because it is a transition element.

Question 40 · Answer A

Step 1: Members of a homologous series share the same general formula and the same functional group. Step 2: They therefore have similar chemical properties, and consecutive members differ by $\text{CH}_2$. Step 3: Two correct characteristics are the same general formula and the same functional group.

Question 41 · Answer A

Step 1: Alcohols have the general formula $\text{C}_n\text{H}_{2n+1}\text{OH}$. Step 2: For n = 5 this is $\text{C}_5\text{H}_{11}\text{OH}$, which as a molecular formula is $\text{C}_5\text{H}_{12}\text{O}$. Step 3: So a five-carbon alcohol has the molecular formula $\text{C}_5\text{H}_{12}\text{O}$.

Question 42 · Answer A

Step 1: An ester contains the ester linkage, a carbon double-bonded to one oxygen and single-bonded to a second oxygen (-C(=O)-O-). Step 2: Methyl ethanoate has a CH3 group joined to this ester carbon, whose second oxygen is joined to another CH3 group, giving three carbon atoms in total. Step 3: So the correct displayed formula shows the CH3-C(=O)-O-CH3 arrangement with all bonds drawn.

Question 43 · Answer A

Step 1: In an ester name, the first part comes from the alcohol and the second part from the acid. Step 2: "Butyl" comes from butan-1-ol, and "ethanoate" comes from ethanoic acid. Step 3: So butyl ethanoate is made from butan-1-ol and ethanoic acid.

Question 44 · Answer A

Step 1: Divide each percentage by the $A_r$: C = 64.87 ÷ 12 = 5.41, H = 13.51 ÷ 1 = 13.51, O = 21.62 ÷ 16 = 1.35. Step 2: Divide by the smallest (1.35): C = 4.0, H = 10.0, O = 1.0. Step 3: So the empirical formula is $\text{C}_4\text{H}_{10}\text{O}$.

Question 45 · Answer A

Step 1: Group VII is the second-to-last group of the Periodic Table. Step 2: These elements (fluorine, chlorine, bromine, iodine) are called the halogens. Step 3: Noble gases are Group VIII/0, alkali metals are Group I, so Group VII are the halogens.

Question 46 · Answer A

Step 1: In Group VII, reactivity decreases going down the group. Step 2: Fluorine is at the top of the group, so it is the most reactive halogen. Step 3: Iodine and astatine are lower down and less reactive, so the most reactive is fluorine.

Question 47 · Answer A

Step 1: Down Group VII the elements change from gases to liquid to solid. Step 2: Iodine is near the bottom, so it is a solid at room temperature and pressure. Step 3: Its colour is grey-black, so iodine is a grey-black solid.

Question 48 · Answer A

Step 1: The potassium ions ($\text{K}^+$) are spectator ions and are left out of the ionic equation. Step 2: Bromine displaces iodine: $\text{Br}_2$ gains electrons to form $\text{Br}^-$, and $\text{I}^-$ loses electrons to form $\text{I}_2$. Step 3: Balancing gives $\text{Br}_2 + 2\text{I}^- \rightarrow 2\text{Br}^- + \text{I}_2$.

Question 49 · Answer A

Step 1: Energy to break bonds (1 I-I + 1 Br-Br) = 150 + 193 = 343 kJ. Step 2: Energy released forming bonds (2 I-Br) = 2 × 175 = 350 kJ. Step 3: $\Delta H$ = energy in − energy out = 343 − 350 = −7 kJ/mol (exothermic).

Question 50 · Answer A

Step 1: Sodium is less dense than water, so it floats on the surface. Step 2: The reaction releases heat and hydrogen gas, so the sodium melts into a ball, fizzes (effervescence) and moves around the surface. Step 3: So two valid observations are that the sodium floats and moves around on the surface.

Question 51 · Answer A

Step 1: Sodium reacts with water to form sodium hydroxide, which is an alkaline solution. Step 2: Thymolphthalein is blue in alkaline solutions and colourless in neutral or acidic ones. Step 3: Since the solution is alkaline (sodium hydroxide), the thymolphthalein turns blue.