IGCSE Chemistry 0620 Paper 41, Oct/Nov 2025

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0620/41 · Oct/Nov 2025

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Protons, neutrons and electrons are particles found in atoms.

Complete Table 1.1 to show the relative mass and relative charge of a proton, a neutron and an electron. Table 1.1

particle	relative mass	relative charge
proton		+1
neutron
electron	$\frac{1}{1840}$

[2]

b(i)

Some elements have many isotopes. Define the term isotopes.

[2]

b(ii)

Explain why all isotopes of the same element have the same chemical properties.

[1]

Complete Table 1.2. Table 1.2

atom or ion	number of protons	number of neutrons	number of electrons
$_{18}^{40} Ar$	18		18
$_{16}^{32} S^{2 -}$		16
	22	28	20

[5]

The term mass number is defined as the total number of protons and neutrons in the nucleus of an atom. State the name of one other term which is defined as the total number of protons and neutrons in the nucleus of an atom.

[1]

Calculate the number of atoms in $2.00 g$ of argon. Give your answer in standard form. number of atoms = ..............................

[2]

Calcium is an element in Group II of the Periodic Table.

Identify the element in Group II which has only five occupied electron shells.

[1]

Name and describe the type of bonding found in elements in Group II. name ......................................................................................................................................... description ................................................................................................................................

[4]

c(i)

When a piece of calcium is added to some cold water containing universal indicator a reaction takes place. Give three observations when this reaction takes place. 1 ........................................................................................................................................ 2 ........................................................................................................................................ 3 ........................................................................................................................................

[3]

c(ii)

Name the two products of this reaction. ................................................................. and ............................................................

[2]

d(i)

Calcium burns in oxygen. State the colour of the flame.

[1]

d(ii)

Write the symbol equation for this reaction.

[2]

e(i)

Crystals of hydrated calcium nitrate contain water molecules. The formula of hydrated calcium nitrate is $C a (N O_{3})_{2} \cdot x H_{2} O$ . $x$ is a whole number. State the term given to the water molecules present in hydrated crystals.

[1]

e(ii)

When hydrated calcium nitrate is heated gently, the following reaction occurs. $C a (N O_{3})_{2} \cdot x H_{2} O \to C a (N O_{3})_{2} + x H_{2} O$ A sample of hydrated calcium nitrate is heated gently. $3.28 g$ of $C a (N O_{3})_{2}$ forms and the mass of the crystals decreases by $1.44 g$ . [ $M_{r}$ : $C a (N O_{3})_{2}$ , 164; $H_{2} O$ , 18] Determine the value of $x$ in $C a (N O_{3})_{2} \cdot x H_{2} O$ using the following steps. • Calculate the number of moles of $C a (N O_{3})_{2}$ that remain. .............................. mol • Calculate the number of moles of $H_{2} O$ given off. .............................. mol • Determine the value of $x$ . x = ..............................

[3]

A student makes crystals of the salt sodium sulfate, $N a_{2} S O_{4}$ . The student reacts $0.200 {mol/dm}^{3}$ dilute sulfuric acid, $H_{2} S O_{4} (a q)$ , with aqueous sodium hydroxide, $N a O H (a q)$ . The student uses the following steps. step 1 The student places $40.0 {cm}^{3}$ of $N a O H (a q)$ into a conical flask. This volume contains $0.0100 moles$ of $N a O H$ . step 2 The student adds a few drops of methyl orange indicator to the $N a O H (a q)$ in the conical flask. step 3 The student adds $0.200 {mol/dm}^{3}$ $H_{2} S O_{4} (a q)$ to the flask until the end-point is reached. step 4 The student transfers the mixture from the conical flask to an evaporating basin and obtains dry crystals.

Complete the symbol equation for the reaction. Include state symbols. $H_{2} S O_{4} (a q) + 2 N a O H (a q) \to N a_{2} S O_{4} (\dots \dots) + \dots \dots \dots \dots \dots \dots (\dots \dots)$

[2]

State the type of exothermic reaction taking place.

[1]

Calculate the concentration of $N a O H (a q)$ used in step 1. concentration of $N a O H (a q) =$ .............................. ${mol/dm}^{3}$

[1]

Name the item of apparatus the student uses to add $H_{2} S O_{4} (a q)$ in step 3.

[1]

Calculate the volume of $H_{2} S O_{4} (a q)$ , in ${cm}^{3}$ , added in step 3. volume of $H_{2} S O_{4} (a q) =$ .............................. ${cm}^{3}$

[2]

State the colour change observed in step 3. from ................................................................. to ...............................................................

[2]

The dry crystals formed in step 4 are coloured and not white. This is because the student should do an additional step between step 3 and step 4. Suggest what the student should do in this additional step to produce white crystals.

[1]

h(i)

In step 4, the student gently heats the solution in the evaporating basin until the solution is saturated. The student then stops heating and leaves the hot solution to cool. Crystals start to appear. Explain the term saturated solution.

[2]

h(ii)

Explain why crystals start to appear as the hot solution cools.

[1]

h(iii)

Suggest the effect, if any, on the mass of crystals collected in step 4 if the solution in the evaporating basin is allowed to dry without gentle heating.

[1]

Gaseous sulfur tetrafluoride, $S F_{4}$ , reacts with steam in a reversible reaction. $S F_{4} (g) + 2 H_{2} O (g) ⇌ S O_{2} (g) + 4 H F (g) Δ H = - 54 kJ/mol$

Complete the reaction pathway diagram in Fig. 4.1 for this reaction. Include in your diagram: • the position and the formulae of the products • an arrow, labelled $E_{a}$ , to show the activation energy • an arrow, labelled $Δ H$ , to show the enthalpy change of the reaction.

[4]

The equation for the reaction can be represented as shown in Fig. 4.2. Table 4.1 shows some bond energies. Table 4.1

bond	S–F	O–H	H–F
bond energy in kJ/mol	330	460	570

Use the bond energies in Table 4.1 and the value of $Δ H$ of the reaction to calculate the S=O bond energy in kJ/mol. Use the following steps. • Calculate the energy needed to break the bonds in the reactants. .............................. kJ • Calculate the energy released when the bonds in the products form. .............................. kJ • Calculate the S=O bond energy. .............................. kJ/mol

[4]

The equation for the reaction is shown. $S F_{4} (g) + 2 H_{2} O (g) ⇌ S O_{2} (g) + 4 H F (g) Δ H = - 54 kJ/mol$ State the effect, if any, on the position of equilibrium when the following changes are made. Give a reason for each of your answers. • The temperature is increased. ................................................................................................................................................... ................................................................................................................................................... • The pressure is increased. ................................................................................................................................................... ................................................................................................................................................... • A catalyst is added. ................................................................................................................................................... ...................................................................................................................................................

[5]

Explain, in terms of collision theory, why reducing the temperature decreases the rate of the reverse reaction.

[3]

This question is about the homologous series of alcohols.

a(i)

A homologous series is a family of organic compounds whose members have the same general formula. State the general formula for alcohols.

[1]

a(ii)

Give one other characteristic that is the same for all members of a homologous series.

[1]

b(i)

Ethanol can be manufactured by two methods: • method 1 uses glucose as the starting material • method 2 uses ethene as the starting material. Complete Table 5.1. Table 5.1

	method 1 glucose as starting material	method 2 ethene as starting material
typical temperature used / °C
two other essential conditions	1	1
	2	2

[6]

b(ii)

Write the symbol equation for the reaction in method 1.

[2]

b(iii)

Write the symbol equation for the reaction in method 2.

[2]

c(i)

Butane-1,4-diol has the structural formula $H O - C H_{2} - C H_{2} - C H_{2} - C H_{2} - O H$ . Deduce the molecular formula of butane-1,4-diol.

[1]

c(ii)

Butane-1,4-diol reacts with ethanoic acid. Determine the number of moles of ethanoic acid which react fully with one mole of butane-1,4-diol.

[1]

d(i)

Butanedioic acid has the structural formula $H O O C - C H_{2} - C H_{2} - C O O H$ . Deduce the empirical formula of butanedioic acid.

[1]

d(ii)

Name the gas formed when butanedioic acid reacts with sodium.

[1]

e(i)

Butane-1,4-diol can be represented as shown. $H O - □ - O H$ Butanedioic acid can be represented as shown. $H O O C - □ - C O O H$ Butane-1,4-diol reacts with butanedioic acid to form a polymer. Draw two repeat units of the polymer formed from the reaction of butane-1,4-diol with butanedioic acid. Show all the atoms and all the bonds in the ester linkages.

[3]

e(ii)

State the type of polymerisation when butane-1,4-diol reacts with butanedioic acid.

[1]

Mark scheme

0620/41 · Oct/Nov 2025

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1a · 2 marks

1 mark
relative mass: neutron 1
1 mark
relative charge: neutron 0 / nil AND electron – 1

1b(i) · 2 marks

M1
atoms of the same element that have the same number of protons
M2
but different numbers of neutrons

1b(ii) · 1 mark

1 mark
they have the same number of electrons and therefore the same electronic configuration

1c · 5 marks

M1
22
M2
16 and 18
M3
50 and 22 and to the left of the symbol
M4
Ti
M5
2+

1d · 1 mark

1 mark
nucleon number

1e · 2 marks

M1
mol Ar = 2.0 ÷ 40 = 0.05 (mol)
M2
number of atoms = 0.05 × 6.02 × 10^23 = 3.01 × 10^22

2a · 1 mark

1 mark
strontium / Sr

2b · 4 marks

M1
metallic
M2
positive ions / cations
M3
sea of electrons / mobile electrons / delocalised electrons
M4
attraction between positive ions and electrons

2c(i) · 3 marks

M1
effervescence
M2
solid dissolves
M3
solution turns blue

2c(ii) · 2 marks

1 mark
calcium hydroxide
1 mark
hydrogen

2d(i) · 1 mark

1 mark
orange-red

2d(ii) · 2 marks

M1
CaO as only product
M2
correct equation

2e(i) · 1 mark

1 mark
water(s) of crystallisation

2e(ii) · 3 marks

M1
3.28 ÷ 164 = 0.02(00)
M2
1.44 ÷ 18 = 0.08(00)
M3
= M2 ÷ M1 = 4

3a · 2 marks

M1
2H2O
M2
state symbols (aq) and (l)

3b · 1 mark

1 mark
neutralisation

3c · 1 mark

1 mark
0.0100 × 1000 ÷ 40 = 0.25(0)

3d · 1 mark

1 mark
burette

3e · 2 marks

M1
mol H2SO4 needed = 0.0100 ÷ 2 = 0.005(00)
M2
vol of H2SO4 needed = 0.005(00) ÷ 0.200 × 1000 = 25 (cm3)

3f · 2 marks

M1
yellow
M2
orange

3g · 1 mark

1 mark
repeat steps 1 and 3 without indicator

3h(i) · 2 marks

M1
(a solution containing the) maximum concentration of a solute dissolved in the solvent
M2
at a specified temperature

3h(ii) · 1 mark

1 mark
solubility decreases as temperature decreases

3h(iii) · 1 mark

1 mark
none

4a · 4 marks

M1
energy level of products below energy level of reactants
M2
label of SO2(g) + 4HF(g) on product line
M3
upward arrow labelled Ea from energy level of reactants to the highest point of the curve
M4
one downward arrow labelled ΔH AND energy change starting from energy level of reactants and finishing at energy level of products

4b · 4 marks

M1
= (4 × S–F) + (4 × O–H) = (4 × 330) + (4 × 460) = 3160
M2
bond energy of products – 3160 = –54 = 3160 – (–54) = 3214
M3
(2 × S=O) + (4 × H–F) = 3214 therefore 2 × S=O = 3214 – (4 × 570) = 934
M4
S=O = M3 ÷ 2 = 934 ÷ 2 = 467

4c · 5 marks

M1
increased temperature: (position of) equilibrium moves to left hand side
M2
increased temperature: reaction is exothermic
M3
increased pressure: (position of) equilibrium moves to left hand side
M4
increased pressure: less (gaseous) moles on left hand side
M5
no change / catalyst does not affect position of equilibrium

4d · 3 marks

M1
kinetic energy of particles decreases
M2
frequency of collisions between particles decreases
M3
lower percentage/proportion/fraction of collisions / particles have energy greater than / equal to activation energy

5a(i) · 1 mark

1 mark
CnH2n+1OH

5a(ii) · 1 mark

1 mark
(same) functional group

5b(i) · 6 marks

M1
25–35
M2
300
M3
yeast
M4
acid
M5
absence of oxygen
M6
60 atm

5b(ii) · 2 marks

M1
C6H12O6 as the only reactant on left
M2
correct equation

5b(iii) · 2 marks

M1
both C2H4 + H2O only on left
M2
correct equation

5c(i) · 1 mark

1 mark
C4H10O2

5c(ii) · 1 mark

1 mark
2

5d(i) · 1 mark

1 mark
C2H3O2

5d(ii) · 1 mark

1 mark
hydrogen

5e(i) · 3 marks

M1
any one correct fully displayed ester link between any two blocks
M2
three fully displayed inter-block ester links with correct orientation
M3
fully correct diagram showing two complete repeat units of polyester including continuation bonds

5e(ii) · 1 mark

1 mark
condensation (polymerisation)

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