S4 Chemistry Practice Paper: Atomic Structure And The Periodic Table, Bonding, Moles And Stoichiometry, Solutions, Acids And Salts, Redox And Rates (100 Marks)

S4 Chemistry Practice Paper: Atomic Structure and the Periodic Table, Bonding, Moles and Stoichiometry, Solutions, Acids and Salts, Redox and Rates (100 marks)

2026-11-13

S4 Chemistry paper with 24 parts on atomic structure, bonding, mole calculations, concentration, acids and salts, redox and rates; all values verified, with complete answers.

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Paper details　S4 Chemistry practice paper with 6 sections and 24 parts on atomic structure, bonding, mole calculations, solution concentration, acids, salts, redox and reaction rates.

Section A: Atomic Structure and the Periodic Table (20 marks)

Answer all four questions, showing working and units.

#	Question	Answer and marking notes
1	An atom of element X has mass number 39 and atomic number 19. (a) State its numbers of protons, neutrons and electrons. (b) State the neutron number of another isotope with mass number 40.	Answer: (a) 19 protons, 20 neutrons and 19 electrons; (b) 21 neutrons. Protons equal the atomic number 19; neutrons equal mass number minus protons, 39 – 19 = 20; a neutral atom has as many electrons as protons. For the isotope 40 – 19 = 21. (Marking: 1 each for the three particles, 1 for the isotope, 1 for reasoning)
2	Atom A has 19 electrons and atom B has 20 electrons. (a) Write the electronic arrangement of each. (b) State the group of each in the periodic table.	Answer: (a) A: 2, 8, 8, 1; B: 2, 8, 8, 2. (b) A is in Group I; B is in Group II. The number of outermost electrons equals the group number: A has 1 and B has 2 outermost electrons. (Marking: 1 for each arrangement, 1 for each group, 1 for reasoning)
3	An element has two isotopes of mass numbers 35 and 37 with abundances 75% and 25%. Calculate the relative atomic mass of the element.	Answer: 35.5. Relative atomic mass is the sum of mass number times abundance: 35 times 75% plus 37 times 25% equals 26.25 plus 9.25, giving 35.5. (Marking: 2 for each product, 1 for the sum and conclusion)
4	Magnesium has atomic number 12 and chlorine has atomic number 17. (a) Write the electronic arrangement of each atom. (b) State how many electrons each gains or loses to form an ion, and write the formula of magnesium chloride.	Answer: (a) Mg: 2, 8, 2; Cl: 2, 8, 7. (b) Magnesium loses 2 electrons to form Mg²⁺ (2, 8) and chlorine gains 1 electron to form Cl⁻ (2, 8, 8). One Mg²⁺ needs 2 Cl⁻, so the formula is MgCl₂. (Marking: 2 for arrangements, 1 for each ion, 1 for the formula)

Section B: Chemical Bonding and Structure (16 marks)

In explanations, name the particles and forces involved.

#	Question	Answer and marking notes
5	Identify the main type of bonding (ionic or covalent) in each substance: (a) sodium chloride, (b) water, (c) magnesium oxide, (d) carbon dioxide.	Answer: (a) ionic; (b) covalent; (c) ionic; (d) covalent. Bonding between a metal and a non-metal is generally ionic (NaCl, MgO), while bonding between non-metals is covalent (H₂O, CO₂). (1 mark each)
6	For water and ammonia molecules, (a) how many O-H single covalent bonds and how many lone pairs on the oxygen atom does water have? (b) How many N-H single covalent bonds and lone pairs on nitrogen does ammonia have?	Answer: (a) water has 2 O-H bonds and 2 lone pairs on oxygen; (b) ammonia has 3 N-H bonds and 1 lone pair on nitrogen. Oxygen has 6 outer electrons, 4 remain after sharing two, giving 2 lone pairs; nitrogen has 5, 2 remain after sharing three, giving 1 lone pair. (1 mark each)
7	Sodium chloride melts at about 801 °C, does not conduct electricity when solid, but conducts when molten or in aqueous solution. Explain these three observations.	Answer: sodium chloride is ionic, held in a three-dimensional lattice by strong electrostatic attraction between Na⁺ and Cl⁻, hence the high melting point (about 801 °C). In the solid the ions are fixed in the lattice and cannot move, so it does not conduct; when molten or dissolved the ions are free to move, so it conducts. (1 mark per point, 1 for expression)
8	Ice melts at 0 °C, while diamond melts above 3500 °C. Both contain non-metal atoms. Explain the large difference in melting points.	Answer: ice is a molecular solid, so melting only overcomes weak forces between molecules (hydrogen bonds) and no covalent bond breaks, giving a low melting point. Diamond is a giant covalent structure in which each carbon is bonded to four others by strong covalent bonds, so melting requires breaking many covalent bonds and the melting point is very high. (2 for structure, 2 for what is broken)

Section C: Moles and Stoichiometry (16 marks)

Take the Avogadro constant as 6.02 × 10²³ per mol and the molar volume as 24 dm³/mol.

#	Question	Answer and marking notes
9	Find the number of moles in (a) 8 g of oxygen O₂ (molar mass 32 g/mol) and (b) 11 g of carbon dioxide CO₂ (molar mass 44 g/mol).	Answer: (a) 0.25 mol; (b) 0.25 mol. n = m divided by M: oxygen 8 divided by 32 is 0.25 mol; carbon dioxide 11 divided by 44 is 0.25 mol. (2 marks each)
10	How many oxygen molecules and how many oxygen atoms are there in 0.25 mol of oxygen gas? (Take the Avogadro constant as 6.02e+23 per mol.)	Answer: 1.51 times 10^23 molecules and 3.01 times 10^23 atoms. Number of particles equals n times NA, that is 0.25 times 6.02e+23, giving 1.51 times 10^23; each O₂ has 2 oxygen atoms, so multiply by 2. (2 marks molecules, 2 marks atoms)
11	At room temperature and pressure the molar volume of a gas is 24 dm³/mol. Find (a) the volume occupied by 0.5 mol of carbon dioxide and (b) the number of moles in 6 dm³ of hydrogen at the same conditions.	Answer: (a) 12 dm³; (b) 0.25 mol. V equals n times 24, so 0.5 times 24 gives 12 dm³; reversing, n equals 6 divided by 24, that is 0.25 mol. (2 marks each)
12	Hydrogen burns as 2H₂ + O₂ -> 2H₂O. If 4 mol of hydrogen burns completely, find (a) the moles of oxygen needed, (b) the moles of water formed and (c) the mass of water formed in grams. (Molar mass of water is 18 g/mol)	Answer: (a) 2 mol; (b) 4 mol; (c) 72 g. The equation gives H₂ : O₂ : H₂O = 2 : 1 : 2, so oxygen is 4 divided by 2, that is 2 mol, and water is 4 mol; mass is 4 times 18, that is 72 g. (1 mark each, 1 for the ratio)

Section D: Solutions and Concentration (16 marks)

Give units with every answer; express concentration in mol/dm³ or as a mass percentage.

#	Question	Answer and marking notes
13	0.5 mol of sodium chloride is dissolved in water to make 250 cm³ of solution. Find the molarity in mol/dm³.	Answer: 2 mol/dm³. Volume is 250 cm³, that is 0.25 dm³; concentration equals n divided by V, 0.5 divided by 0.25, giving 2 mol/dm³. (2 for conversion, 2 for calculation)
14	25 cm³ of hydrochloric acid of concentration 2 mol/dm³ is diluted with water to 100 cm³. Find the concentration of the diluted solution.	Answer: 0.5 mol/dm³. The number of moles of solute is unchanged on dilution, so c₁V₁ equals c₂V₂ and c₂ is 2 times 25 divided by 100, giving 0.5 mol/dm³. (2 for conservation, 2 for calculation)
15	20 g of potassium nitrate is dissolved in 80 g of water. Find (a) the total mass of the solution and (b) the concentration as a mass percentage.	Answer: (a) 100 g; (b) 20%. Total mass is 20 plus 80, that is 100 g; percentage is solute divided by solution times 100%, 20 divided by 100, giving 20%. (2 marks each)
16	At 20 °C 36 g of a substance dissolves in 100 g of water. (a) What mass dissolves in 200 g of water? (b) If 100 g is added to 200 g of water, what mass stays undissolved?	Answer: (a) 72 g; (b) 28 g. The mass dissolved is proportional to the mass of water: 36 divided by 100 times 200 gives 72 g; undissolved is 100 minus 72, that is 28 g. (2 marks each)

Section E: Acids, Bases and Salts (16 marks)

Balance every equation and include state symbols where appropriate.

#	Question	Answer and marking notes
17	A hydrochloric acid solution has pH 3. (a) Find its hydrogen ion concentration. (b) Compared with hydrochloric acid of the same molarity, does acetic acid (a weak acid) have a higher or lower pH? Explain.	Answer: (a) 0.001 mol/dm³; (b) acetic acid has the higher pH. pH 3 means [H⁺] equals 10⁻³ mol/dm³; a weak acid ionises only partially, so at the same molarity [H⁺] is smaller and pH is higher. (2 for (a), 1 for the choice, 1 for the explanation)
18	20 cm³ of 0.1 mol/dm³ hydrochloric acid exactly neutralises 25 cm³ of sodium hydroxide solution. Find the molarity of the sodium hydroxide solution.	Answer: 0.08 mol/dm³. Moles of HCl equal 0.1 times 20 divided by 1000, that is 0.002 mol; the ratio is 1 : 1, so NaOH is also 0.002 mol; concentration is 0.002 divided by 0.025 dm³, giving 0.08 mol/dm³. (2 for moles, 2 for concentration)
19	(a) Write the equation for making copper(II) sulphate solution from copper(II) oxide and dilute sulphuric acid. (b) Which method would you use to prepare insoluble barium sulphate? Just give the name of the method.	Answer: (a) CuO + H₂SO₄ -> CuSO₄ + H₂O; (b) precipitation, mixing solutions of a soluble barium salt and a soluble sulphate, for example BaCl₂ + Na₂SO₄ -> BaSO₄(s) + 2NaCl. Soluble salts are made from an acid with a metal oxide or carbonate; insoluble salts are made by precipitation. (2 for equation, 1 for method, 1 for reasoning)
20	(a) Which type of indicator can be used for a strong acid and strong base titration? (b) For a weak acid and strong base titration, why is phenolphthalein usually chosen rather than methyl orange?	Answer: (a) any suitable acid-base indicator such as methyl orange or phenolphthalein, because the equivalence point is near pH 7. (b) The equivalence point of a weak acid with a strong base is alkaline, around pH 8 to 9, where phenolphthalein changes colour; methyl orange changes in the acidic range and would change too early, causing error. (2 marks each)

Section F: Redox and Reaction Rates (16 marks)

For redox questions, state the changes in oxidation number.

#	Question	Answer and marking notes
21	In the reaction 2Fe + 3Cl₂ -> 2FeCl₃, (a) state the change in oxidation number of iron and of chlorine; (b) which substance is the oxidising agent? Explain.	Answer: (a) iron rises from 0 to +3, so it is oxidised; chlorine falls from 0 to -1, so it is reduced. (b) Chlorine is the oxidising agent because it gains electrons and is itself reduced, causing iron to be oxidised. (1 mark each in (a), 1 for the choice and 1 for the reason)
22	When magnesium reacts with dilute hydrochloric acid, each magnesium atom loses 2 electrons to form Mg²⁺. (a) Write the oxidation half equation for magnesium. (b) How many electrons are transferred when 0.5 mol of magnesium reacts completely?	Answer: (a) Mg -> Mg²⁺ + 2e⁻; (b) 6.02 times 10^23 electrons. Moles of electrons equal 0.5 times 2, that is 1 mol; the number is 1 times 6.02e+23, giving 6.02 times 10^23. (2 for the half equation, 2 for the number)
23	Zinc granules added to dilute sulphuric acid give hydrogen. Explain how each change affects the rate and give a reason: (a) grinding the granules into powder; (b) raising the temperature from 25 °C to 35 °C; (c) adding a little copper(II) sulphate solution.	Answer: (a) faster, because the larger surface area lets more particles contact the acid; (b) faster, because the particles have greater average kinetic energy, so both the collision frequency and the fraction of effective collisions rise; (c) faster, because copper in contact with zinc provides a path for electron transfer and lowers the energy barrier. (1 mark per reason, 1 for expression)
24	(a) How does a catalyst increase the rate of a reaction? (b) Does a catalyst change the enthalpy change or the position of equilibrium? Answer each in one sentence.	Answer: (a) a catalyst provides an alternative pathway of lower energy barrier, so the activation energy falls, the fraction of effective collisions rises and the rate increases; (b) it does not change the enthalpy change, because the energies of reactants and products are unchanged, and it does not shift the equilibrium position, only shortening the time needed to reach it. (2 marks each)

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