Question 1
Topic – 6.1
A list of chemical and physical processes, A to H, is shown.
A combustion
B diffusion
C melting
D neutralisation
E photosynthesis
F reversible reaction
G roasting
H thermal decomposition
Answer the following questions about processes A to H.
Each letter may be used once, more than once or not at all.
(a) State which of the processes A to H happens when an acid reacts with an alkali.
(b) State which of the processes A to H reaches a position of equilibrium.
(c) State which of the processes A to H involves particles changing from fixed positions to being mobile, but still touching.
(d) State which two of the processes A to H are physical changes.
(e) State which of the processes A to H is caused by gas particles colliding with each other.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 7.1 — The characteristic properties of acids and bases (Part (a))
• Topic 6.3 — Reversible reactions and equilibrium (Part (b))
• Topic 1.1 — Solids, liquids and gases (Part (c))
• Topic 6.1 — Physical and chemical changes (Part (d))
• Topic 1.2 — Diffusion (Part (e))
▶️ Answer/Explanation
(a) D (neutralisation)
Neutralisation is the specific reaction between an acid and an alkali (base), producing a salt and water. The H⁺ ions from the acid react with OH⁻ ions from the alkali.
(b) F (reversible reaction)
A reversible reaction in a closed system reaches dynamic equilibrium. At this point, the forward and reverse reactions occur at equal rates, so the concentrations of reactants and products remain constant.
(c) C (melting)
Melting is a physical change of state. In a solid, particles vibrate around fixed positions, but when melted, they gain enough energy to slide past each other (become mobile) while still remaining in contact as a liquid.
(d) B (diffusion) and C (melting)
Physical changes alter the form or state of a substance without changing its chemical composition. Diffusion is particle movement, and melting is a state change; neither creates new chemical substances.
(e) B (diffusion)
Diffusion in gases is driven by the random motion and frequent collisions of gas particles. These collisions cause particles to spread out from areas of high concentration to low concentration.
Question 2
This question is about atomic structure and the Periodic Table.
(a) Define the term nucleon number.
(b) State the connection between the number of occupied electron shells in an atom and the period number of that element.
(c) Write the electronic configuration of the following atom and ion.
\(^{28}_{14}\mathrm{Si}\) ………………………………………..
\(^{37}_{17}\mathrm{Cl}^-\) ………………………………………
(d) Complete Table 2.1.
Table 2.1
(e) A sample of thallium, Tl, contains two isotopes, \(^{203}\mathrm{Tl}\) and \(^{205}\mathrm{Tl}\).
(i) Define the term isotopes.
(ii) The relative abundance of \(^{203}\mathrm{Tl}:^{205}\mathrm{Tl}\) is in the ratio 3:7.
Calculate the relative atomic mass of thallium in the sample to one decimal place.
(iii) Suggest why these two isotopes have identical chemical properties.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 2.2 — Atomic structure and the Periodic Table (Parts (a), (b), (c), (d))
• Topic 2.3 — Isotopes (Parts (e)(i), (e)(ii), (e)(iii))
▶️ Answer/Explanation
(a) The nucleon number (mass number) is the total number of protons and neutrons in the nucleus of an atom.
(b) The period number of an element equals the number of occupied electron shells in its atom.
(c) For \(^{28}_{14}\mathrm{Si}\): 2,8,4 (Silicon has 14 electrons; 2 in first shell, 8 in second, 4 in third).
For \(^{37}_{17}\mathrm{Cl}^-\): 2,8,8 (Chlorine atom has 17 electrons; the ion gains 1 electron to make 18, filling the third shell to 8).
(d) Completed table:
| atom or ion | number of protons | number of neutrons | number of electrons |
|---|---|---|---|
| \(^{23}_{11}\mathrm{Na}\) | 11 | 12 | 11 |
| \(^{19}_{9}\mathrm{F}^-\) | 9 | 10 | 10 |
| \(^{69}_{31}\mathrm{Ga}^{3+}\) | 31 | 38 | 28 |
(e)(i) Isotopes are atoms of the same element that have the same number of protons (same atomic number) but different numbers of neutrons.
(e)(ii) Relative atomic mass = \(\frac{(3 \times 203) + (7 \times 205)}{10} = \frac{609 + 1435}{10} = \frac{2044}{10} = 204.4\) (to one decimal place).
(e)(iii) The two isotopes have identical chemical properties because they have the same electronic configuration (same number and arrangement of electrons), and chemical behavior depends on electron structure, not neutron number.
Question 3
Copper(II) sulfate has the formula CuSO4. Aqueous copper(II) sulfate is a blue solution.
A sample of aqueous copper(II) sulfate is made by adding excess copper(II) oxide, CuO, to hot dilute sulfuric acid, H2SO4.
(a) Complete the symbol equation for this reaction. Include state symbols.
CuO(……) + H2SO4(……) → CuSO4(……) + …… (l)
(b) State one observation which shows that copper(II) oxide is added in excess.
(c) Describe how aqueous copper(II) sulfate can be separated from the reaction mixture.
(d) Crystals of hydrated copper(II) sulfate can be obtained from aqueous copper(II) sulfate by crystallisation.
(i) State what is meant by the term hydrated.
(ii) Write the formula of hydrated copper(II) sulfate.
(iii) Describe how this crystallisation is done.
(e) Aqueous copper(II) sulfate undergoes electrolysis using graphite electrodes.
(i) State why aqueous copper(II) sulfate conducts electricity.
(ii) Give two reasons why the electrodes are made of graphite.
(iii) Describe how the appearance of the electrolyte changes during the electrolysis of aqueous copper(II) sulfate.
(iv) Describe what is seen at the cathode during the electrolysis of aqueous copper(II) sulfate.
(v) Write the ionic half-equation for the reaction at the anode.
(vi) State two differences seen if the electrolysis is repeated using copper electrodes instead of graphite electrodes.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 7.3 — Preparation of salts (Parts (a)-(d))
• Topic 4.1 — Electrolysis (Parts (e)(i)-(vi))
▶️ Answer/Explanation
(a) CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)
Copper(II) oxide is an insoluble base which reacts with sulfuric acid in a neutralisation reaction. The products are a salt (copper sulfate) dissolved in water, and water itself. State symbols are solid, aqueous, aqueous, and liquid respectively.
(b) The black solid (copper(II) oxide) remains unreacted at the bottom of the flask / no more solid appears to dissolve.
When the base is in excess, all the acid has been used up. The leftover solid simply sits at the bottom because it cannot react further, clearly indicating that the reaction is complete.
(c) Filtration.
Filtration separates the insoluble excess copper(II) oxide (the residue) from the aqueous copper(II) sulfate solution (the filtrate). The solution is poured through filter paper, allowing the liquid to pass through while trapping the solid.
(d)(i) Hydrated means a crystalline substance that contains water molecules chemically bonded within its crystal structure (water of crystallisation).
These water molecules are an integral part of the crystal lattice, often affecting the colour of the compound (e.g., blue copper sulfate vs white anhydrous copper sulfate).
(d)(ii) CuSO4•5H2O
Hydrated copper(II) sulfate crystals contain five moles of water of crystallisation for every one mole of copper(II) sulfate. This is the common blue crystalline form seen in laboratories.
(d)(iii) Warm the solution gently to evaporate some water until a saturated solution forms (a crystal film appears on the surface), then leave the solution to cool slowly and undisturbed.
Heating concentrates the solution, and upon cooling, the solubility decreases, forcing the dissolved salt to form regular, pure crystals. Slow cooling allows for larger, more perfect crystals to grow.
(e)(i) It contains freely moving mobile ions (Cu2+ and SO42-).
When copper(II) sulfate dissolves in water, it dissociates into its constituent ions. These charged particles are free to move throughout the solution, allowing them to carry electric current when a potential difference is applied.
(e)(ii) Graphite is a good conductor of electricity and it is inert (does not react with the electrolyte or the products).
As a form of carbon, graphite has delocalised electrons allowing conductivity. Its inertness ensures it does not interfere with the electrolysis reactions, making it a reliable, non-consumable electrode.
(e)(iii) The blue colour of the electrolyte fades (becomes paler).
The blue colour is due to hydrated Cu2+ ions. As electrolysis proceeds, Cu2+ ions are removed from the solution and deposited as copper metal on the cathode, reducing the concentration of blue Cu2+ ions in solution.
(e)(iv) A pink/brown solid layer of copper metal is deposited on the cathode.
At the cathode (negative electrode), copper ions gain electrons in a reduction reaction: Cu2+(aq) + 2e– → Cu(s). This forms a visible coating of solid, metallic copper which has a characteristic pinkish-brown colour.
(e)(v) 4OH–(aq) → 2H2O(l) + O2(g) + 4e–
At the anode (positive electrode), hydroxide ions from the water are preferentially discharged over sulfate ions. They lose electrons to form oxygen gas and water. You would observe bubbles of colourless oxygen gas forming at the anode.
(e)(vi) 1. The blue colour of the solution does not fade. 2. The anode (positive electrode) dissolves/get smaller, and no oxygen gas is produced at the anode.
With copper electrodes, the anode is active. Copper atoms from the anode lose electrons (Cu → Cu2+ + 2e–) and go into solution, maintaining the Cu2+ concentration, so the colour stays blue and the anode corrodes.
Question 4
When magnesium nitrate is heated strongly, magnesium oxide is formed.
(a) The equation for this reaction is shown.
\[ 2\mathrm{Mg(NO_3)_2} \rightarrow 2\mathrm{MgO} + 4\mathrm{NO_2} + \mathrm{O_2} \]
(i) State the change in oxidation number of nitrogen, N, in this reaction.
(ii) Identify the element which is oxidised in this reaction.
(iii) Calculate the volume of NO2 gas, at r.t.p., formed when 7.40 g of Mg(NO3)2 is heated.
Use the following steps.
- Calculate the Mr of Mg(NO3)2.
- Calculate the number of moles of Mg(NO3)2 used.
- Determine the number of moles of NO2 formed.
- Calculate the volume of NO2 gas, in cm3, at r.t.p.
(b) Magnesium oxide, MgO, is an ionic compound.
Complete the dot-and-cross diagram in Fig. 4.1 of the ions in magnesium oxide.
Give the charges on each of the ions.
(c) Oxygen is a covalent molecule.
Complete the dot-and-cross diagram in Fig. 4.2 of a molecule of oxygen. The inner shells have been drawn.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 6.4 — Redox (Parts (a)(i), (a)(ii))
• Topic 3.3 — The mole and the Avogadro constant (Part (a)(iii))
• Topic 2.4 — Ions and ionic bonds (Part (b))
• Topic 2.5 — Simple molecules and covalent bonds (Part (c))
▶️ Answer/Explanation
(a)(i) from +5 to +4
Topic 6.4 — Redox
In nitrate (NO₃⁻), nitrogen has oxidation number +5 (oxygen is –2, total –6, so N = +5). In NO₂, oxygen is –2 each, so N = +4.
(a)(ii) oxygen
Topic 6.4 — Redox
Oxygen changes from –2 in nitrate to 0 in O₂ — an increase in oxidation number, so oxygen is oxidised.
(a)(iii)
Topic 3.3 — The mole and the Avogadro constant
Mr of Mg(NO₃)₂ = 24 + (14 + 48)×2 = 148
Moles of Mg(NO₃)₂ = 7.40 g / 148 g/mol = 0.0500 mol
Mole ratio: 2 mol Mg(NO₃)₂ → 4 mol NO₂ → 0.0500 mol → 0.100 mol NO₂
Volume at r.t.p. = 0.100 mol × 24,000 cm³/mol = 2400 cm³
(b) Topic 2.4 — Ions and ionic bonds
Mg loses two electrons → Mg²⁺ (2,8 with 2+ charge). O gains two electrons → O²⁻ (2,8 with 2– charge). Dot-and-cross must show full outer shells and charges.
(c) Topic 2.5 — Simple molecules and covalent bonds
Each O atom has 6 outer electrons. They form a double bond (4 shared electrons) and each retains two lone pairs (4 non-bonding). Inner shells already drawn.
Question 5
Hydrogen is the first element of the Periodic Table.
(a) Hydrogen is used in fuel cells to produce electricity in vehicles.
(i) Name the substance which combines with hydrogen in a fuel cell.
(ii) Give one advantage and one disadvantage of using fuel cells instead of gasoline in vehicle engines.
(b) Hydrogen gas can be made from petroleum by a two-step procedure.
step 1 Petroleum is separated into different components.
step 2 Large molecules obtained in step 1 are converted into smaller molecules including hydrogen gas.
(i) Name the process used in step 1.
(ii) Name the process used in step 2.
(c) Organic compounds contain hydrogen atoms.
Calculate the number of hydrogen atoms in 44.0 g of the ester methyl propanoate, CH3CH2COOCH3.
One mole of CH3CH2COOCH3 contains \(6.02 \times 10^{23}\) molecules.
Give your answer in standard form.
(d) For each of the homologous series shown, name a member that contains six hydrogen atoms.
– alkanes
– alkenes
– alcohols
– carboxylic acids
(e) Unsaturated alkenes are converted into saturated alkanes by reaction with hydrogen gas.
(i) State why alkenes and alkanes are hydrocarbons.
(ii) State why alkenes are unsaturated.
(iii) Name the catalyst needed to convert alkenes into alkanes.
(iv) Explain why the conversion of alkenes into alkanes is an addition reaction.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 4.2 — Hydrogen-oxygen fuel cells (Parts (a)(i), (a)(ii))
• Topic 11.3 — Fuels / Petroleum (Parts (b)(i), (b)(ii))
• Topic 3.3 — The mole and the Avogadro constant (Part (c))
• Topic 11.2 — Naming organic compounds / Homologous series (Part (d))
• Topic 11.5 — Alkenes / Addition reactions (Parts (e)(i) to (e)(iv))
▶️ Answer/Explanation
(a)(i) Oxygen (O2). In a hydrogen fuel cell, hydrogen gas is combined with oxygen gas to produce electricity, with water as the only chemical product.
(a)(ii) Advantage: Water is the only product (no pollution / environmentally friendly). Disadvantage: Hydrogen is difficult to store (requires high pressure or low temperatures) / Fewer refuelling stations.
(b)(i) Fractional distillation. This process separates petroleum into different fractions based on their different boiling points.
(b)(ii) Cracking. This process breaks down large alkane molecules into smaller, more useful molecules (including alkenes and hydrogen) using high temperature and a catalyst.
(c) \(2.408 \times 10^{24}\) atoms.
1. Molar mass = (4×12)+(8×1)+(2×16) = 88 g/mol. Moles = 44/88 = 0.5 mol.
2. Each molecule has 8 H atoms. Moles of H atoms = 0.5 × 8 = 4.0 mol.
3. Number of atoms = 4.0 × (6.02 × 1023) = 2.408 × 1024.
(d) Alkanes: Ethane (C2H6). Alkenes: Propene (C3H6). Alcohols: Ethanol (C2H5OH). Carboxylic acids: Propanoic acid (C2H5COOH).
(e)(i) They contain only carbon and hydrogen atoms (they are compounds made of hydrogen and carbon only).
(e)(ii) They contain a carbon-carbon double bond (C=C) / they do not have the maximum number of hydrogen atoms.
(e)(iii) Nickel (Ni).
(e)(iv) Only one product is formed (the alkane) from the two reactants (alkene and hydrogen), with no other products / atoms are added across the double bond.
Question 6
Natural polyamides are polymers made from amino acid monomers.
(a) State the type of polymerisation reaction that occurs when natural polyamides form.
(b) State the term given to natural polyamides.
(c) An amino acid is represented as shown in Fig. 6.1.
Complete Fig. 6.2 to show the general structure of an amino acid.
Show all of the atoms and all of the bonds in the functional groups.
(d) Three different amino acids are represented as shown in Fig. 6.3.
Complete the diagram in Fig. 6.4 to show the part of the structure of the natural polyamide that forms when the three amino acids, A, B and C, combine. Show all of the atoms and all of the bonds in the linkages.
(e) A mixture of the three amino acids, A, B and C, can be separated and the amino acids identified using paper chromatography.
Complete the equation for \( R_f \).
\[R_f =\]
(f) A sample of the mixture of the three amino acids, A, B and C, is placed onto the baseline and a chromatogram is allowed to develop as shown in Fig. 6.5.
The finished chromatogram is shown in Fig. 6.6.
The amino acids, A, B and C, are colourless. Water is used as the solvent.
(i) Explain why the baseline is drawn in pencil.
(ii) State the type of substance used to make the colourless amino acids visible on the chromatogram in Fig. 6.6.
(iii) Explain why in Fig. 6.6 only two spots are seen from the mixture of three amino acids.
(iv) Suggest how the experiment can be changed to separate all three amino acids.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic 11.8 — Polymers (Parts a, b, c, d)
• Topic 12.3 — Chromatography (Parts e, f)
▶️ Answer/Explanation
(a) Condensation
Natural polyamides form by condensation polymerisation, where amino acids join together and eliminate a small molecule (water).
(b) Proteins
Natural polyamides are commonly known as proteins, which are essential biopolymers made from amino acid monomers.
(c) The general structure of an amino acid is:
This shows the central carbon atom bonded to an amino group (NH2), a carboxyl group (COOH), a hydrogen atom, and a variable R group.
(d) The completed polymer structure is:
This shows the peptide (amide) linkages (-CO-NH-) connecting the amino acids in a chain.
(e) \( R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}} \)
The Rf value is a ratio used to identify substances by comparing how far they move relative to the solvent front.
(f)(i) Pencil is insoluble in the solvent.
Pencil marks do not dissolve or move up the paper, preventing interference with the separated spots.
(f)(ii) A locating agent (e.g., ninhydrin).
A locating agent reacts with colourless substances to produce visible coloured spots.
(f)(iii) Two of the amino acids have the same (or very similar) Rf value.
When two different substances have identical Rf values in a given solvent, they co-migrate and appear as a single spot.
(f)(iv) Use a different solvent (or a mixture of solvents).
Changing the solvent system alters the partition coefficients, allowing all three amino acids to be resolved.