Last reviewed: 7 November 2026
Quick answer: Practise mole calculations, ionic equations, bonding, gases, rates and electrolysis with eight original questions and explained answers. This is a 2026 revision edition, not a reproduction of the 2026 or 2025 examination. All eight worked answers below are free to read.
Original practice, not an examination paper. No verified 2025 paper or official answers are supplied here. The year in this page title identifies the revision edition only. Use the existing Chemistry past-paper practice page for the site’s separate paper-format resources; check each resource’s description before using it.
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How to use these worked questions
Attempt each task before reading its answer. Keep your working or first draft, compare it with the explanation, then record one correction to practise again. These are individual exercises, not a complete timed paper. No official marks, grade prediction or claim of exam coverage is attached.
Questions with worked answers
1. Moles and reacting masses
Question: Magnesium reacts with dilute hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. Calculate the amount of magnesium in 4.8 g, the amount of HCl required and the mass of MgCl₂ formed. Use Mg = 24, H = 1 and Cl = 35.5; assume complete reaction.
Worked answer: Amount of Mg = mass ÷ molar mass = 4.8 ÷ 24 = 0.20 mol. The equation requires two moles of HCl for every mole of Mg, so HCl required = 0.40 mol. One mole of Mg produces one mole of MgCl₂, so product amount = 0.20 mol. Molar mass of MgCl₂ = 24 + 2(35.5) = 95 g/mol. Product mass = 0.20 × 95 = 19 g. The product can weigh more than the original metal because chlorine from the acid is incorporated.
Check your work: Do not multiply the magnesium amount by two when finding magnesium chloride: the 2 belongs to HCl, not the product.
2. Concentration and neutralisation
Question: 25.0 cm³ of 0.10 mol/dm³ sodium hydroxide exactly neutralises 20.0 cm³ of hydrochloric acid. Find the acid concentration.
Worked answer: HCl + NaOH → NaCl + H₂O. Convert the base volume to dm³: 25.0 ÷ 1000 = 0.0250 dm³. Amount of NaOH = 0.10 × 0.0250 = 0.00250 mol. The ratio is 1:1, so the acid also contains 0.00250 mol. Acid volume = 0.0200 dm³; concentration = 0.00250 ÷ 0.0200 = 0.125 mol/dm³. This calculation assumes that the stated endpoint represents complete neutralisation.
Check your work: Keep both volumes in the same consistent units when using ratios, or convert to dm³ before using n = cV.
3. Ionic precipitation equation
Question: Aqueous silver nitrate is mixed with aqueous sodium chloride. Name the precipitate and write the net ionic equation with state symbols.
Worked answer: The white precipitate is silver chloride. The net ionic equation is Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Sodium and nitrate ions remain dissolved and are spectator ions, so they are not included in the net ionic equation. The total charge is zero on both sides. This is a precipitation reaction, not displacement of sodium metal.
Check your work: Writing AgCl(aq) contradicts the observation that an insoluble solid has formed.
4. Structure and electrical conduction
Question: Explain why graphite conducts electricity but diamond does not, although both are made of carbon.
Worked answer: Each carbon atom in graphite forms three covalent bonds within a layer. Electrons not used in these bonds are delocalised and can move along the layers to carry charge. In diamond, each carbon forms four covalent bonds in a three-dimensional network; there are no mobile electrons available for electrical conduction under ordinary conditions. Both contain strong covalent bonds, so bond strength alone does not explain their different conductivity.
Check your work: Do not say that graphite conducts because its layers slide. Sliding helps explain lubrication, not electrical conduction.
5. Gas volume
Question: A dry gas occupies 120 cm³ at 300 K. Calculate its volume at 350 K if the pressure and amount of gas remain constant.
Worked answer: At constant pressure, V₁/T₁ = V₂/T₂. Therefore V₂ = 120 × 350 ÷ 300 = 140 cm³. Kelvin temperatures are required because the relationship uses an absolute temperature scale. The answer is larger than 120 cm³, consistent with expansion on heating. Real gases only approximately follow this relationship, but the stated conditions give the usual school-level model.
Check your work: Never put Celsius temperatures directly into a proportional gas-law calculation.
6. Reaction rate
Question: Equal masses of calcium carbonate powder and large marble chips react separately with excess acid of the same concentration and temperature. Compare the initial rate and final amount of carbon dioxide.
Worked answer: The powder reacts faster initially because it exposes more surface area to the acid, allowing more collisions per second at reacting surfaces. If both samples have the same purity and react completely, they produce the same total amount of carbon dioxide: the same amount of calcium carbonate is present and the acid is in excess. A faster rate does not mean a greater final yield.
Check your work: Hold concentration and temperature constant when explaining a surface-area comparison.
7. Electrolysis
Question: During electrolysis of molten lead(II) bromide using inert electrodes, what forms at each electrode? Give the half-equations.
Worked answer: At the negative cathode, Pb²⁺ gains electrons: Pb²⁺ + 2e⁻ → Pb. Lead forms. At the positive anode, bromide ions lose electrons: 2Br⁻ → Br₂ + 2e⁻. Bromine forms. Reduction is gain of electrons and oxidation is loss of electrons. This is a theoretical question; bromine and lead compounds are hazardous, so it is not an instruction for a home experiment.
Check your work: For an electrolytic cell the cathode is negative; distinguish electrode signs from the definitions of oxidation and reduction.
8. Separation and purity
Question: A mixture contains sand and sodium chloride. Explain how to recover dry sodium chloride crystals without assuming that sand dissolves.
Worked answer: Add sufficient water to dissolve the sodium chloride, stir, then filter to remove the insoluble sand. The filtrate contains dissolved salt. Concentrate the filtrate and allow crystallisation, then separate and dry the crystals. Filtration alone cannot remove dissolved sodium chloride because its particles pass through filter paper. The description is a revision explanation, not a practical instruction to carry out unsupervised.
Check your work: Identify what is in the residue and filtrate at each stage rather than listing apparatus without a purpose.
What to revise next
If an answer was wrong, identify whether the gap was subject knowledge, interpretation of the instruction or the method used. Redo that question without looking, then try a different question on the same topic. A correct final answer does not replace a clear explanation where the task asks for one.
Read the Chemistry revision guide and return to the paper-format practice resource. The eight examples on this page remain free. If you want additional paid practice, a KSh 20 Day Pass is available. It does not purchase a verified 2025 examination paper or official marking scheme.
Frequently asked questions
Are these actual 2025 or 2026 examination questions?
No. They are original practice questions. The 2026 label is the revision edition, not the source year of an examination paper.
Are the worked answers free?
Yes. All eight questions and explanations on this page are free without a login.
Is this a complete examination paper?
No. It is a focused collection of exercises. It does not reproduce the structure or full coverage of an examination.
Does a Day Pass buy an official marking scheme?
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Reference for concepts
OpenStax textbook reference. This is a reference for general concepts, not a source of examination questions or an endorsement of these exercises.
Revision edition prepared 7 November 2026. Examples and model responses are original CBC Guide practice.