Chemistry Revision: WJEC GCSE The Sciences (Double Award) · teaching from 2026 · first awarded 2028

This guide is for the replacement qualification, taught from 2026 and first awarded in 2028. The page shows Chemistry requirements plus shared skills and Unit 7; choose another science above to switch scope.

Qualification code: 3455QD

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Course details

WJEC · GCSE WJEC GCSE The Sciences (Double Award) · Chemistry3455QDExam years · 2028

Assessment Six 75-minute written exams: Units 1–3 have 56 marks and contribute 14.3% each; Units 4–6 have 64 marks and contribute 15.7% each. Unit 7 comprises two of three external enquiries, contributing 10%. Units 1–3 are first available in summer 2027; the full qualification is first awarded in summer 2028.

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Open a point to read what to know, how to use it in an assessment, a plain-language explanation and a self-check. Answers stay hidden until you choose to reveal them.

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Unit 1 tier

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Higher

§5.3: different tiers across units

Unit 2 tier

Foundation
Higher

§5.3: different tiers across units

Unit 3 tier

Foundation
Higher

§5.3: different tiers across units

Unit 4 tier

Foundation
Higher

§5.3: different tiers across units

Unit 5 tier

Foundation
Higher

§5.3: different tiers across units

Unit 6 tier

Foundation
Higher

§5.3: different tiers across units

Atomic Structure and Chemical CalculationsCore
Scope: Core · Topic 2.1 (core)
Open revision guidance
What to know
In the simple particle model, a solid's particles are close together and vibrate about fixed positions, commonly shown in a regular arrangement. Liquid particles remain close but move past each other; gas particles are widely separated and move randomly.
Melting (solid to liquid), boiling (liquid to gas), condensing (gas to liquid), and freezing (liquid to solid) are reversible physical changes.
At a stated pressure, below its melting point a substance is solid; between its melting and boiling points it is liquid; above its boiling point it is gas. At a transition temperature two states may coexist. For an illustrative substance with melting point 10°C and boiling point 80°C, its states at 0°C, 20°C and 100°C are solid, liquid and gas.
An element contains atoms with the same proton number; its isotopes can have different neutron numbers. Each element has a chemical symbol, such as O for oxygen.
The simplified atomic model has a small nucleus containing protons and, in most atoms, neutrons, with electrons represented in shells around it. Relative charges are proton +1, neutron 0 and electron −1. Relative masses are approximately 1 for a proton, 1 for a neutron and 1/1836 for an electron, often treated as negligible in school mass calculations.
Atomic number = proton count. Mass number = total protons + neutrons.
Atoms are neutral because proton count equals electron count.
Isotopes are atoms of the same element (same protons) with different neutron counts.
Relative atomic mass, Aᵣ, is the weighted mean mass of an element's atoms relative to one twelfth of the mass of a carbon-12 atom. The isotope abundances affect the mean, so it need not be a whole number. Aᵣ is a relative number, not a mass in grams.
Molecules (e.g., H2, O2) form via chemical bonds. Compounds have fixed formulas (e.g., H2O).
Relative formula mass (Mr) is the sum of Ar values for all atoms in a chemical formula.
Heating transfers energy to particles. During melting, particles can leave their fixed positions; during boiling, they separate much further. Cooling allows condensation and freezing. A change of state does not turn the particles into a different chemical substance or make individual particles grow.
Atomic models have changed with evidence: Dalton's solid-sphere model was followed by Thomson's model with electrons in positive material; Rutherford's scattering evidence supported a small positive nucleus and mostly empty space; Bohr used electron shells. Models represent evidence and can be revised, rather than being literal miniature pictures of atoms.
For a neutral atom with atomic number Z and mass number A, protons = Z, electrons = Z and neutrons = A − Z. Carbon-12 and carbon-14 each have six protons but have six and eight neutrons respectively.
A molecule is a discrete group of two or more chemically joined atoms: the smallest unit of a molecular substance. H₂ and O₂ are diatomic molecules of elements because each contains just one element. A compound contains at least two different elements chemically joined; H₂O is both a molecule and a compound, but ionic compounds such as NaCl are not collections of separate molecules.
What to do in an assessment
For a neutral sodium-23 atom with atomic number 11 and mass number 23, protons = 11, electrons = 11 and neutrons = 23 − 11 = 12. The electron count here is for the neutral atom, not a sodium ion.
Calculate Mr: For H2O (Ar: H=1, O=16). Mr = (2*1) + 16 = 18.
Calculate percentage by mass of O in H2O: (16/18) * 100 = 88.9%.
For each element, percentage by mass = (number of its atoms in the formula × its Aᵣ ÷ Mᵣ) × 100. Given Aᵣ values Ca = 40, C = 12 and O = 16, CaCO₃ has Mᵣ = 100: calcium is 40%, carbon is 12% and oxygen is (3 × 16 ÷ 100) × 100 = 48%. The percentages total 100%; count all three oxygen atoms.
Write formula for ionic compound: If given ions Na+ and Cl-, formula is NaCl.
Balance ionic charges in the formula, not by changing an ion's charge. Given Ca²⁺ and Cl⁻, two chloride ions are needed per calcium ion: CaCl₂. Given Mg²⁺ and NO₃⁻, write Mg(NO₃)₂; the brackets keep two whole nitrate ions.
Interpret a formula: H₂O has two hydrogen atoms for each oxygen atom in a molecule; CO₂ has one carbon and two oxygen atoms. In an ionic compound, a formula such as NaCl gives the simplest ion ratio, not a separate NaCl molecule.
Interpret a molecule model by its key: two identical joined circles may represent H₂, while differently labelled circles joined together may represent a compound. Colour and circle size are modelling choices, not the actual appearance of individual atoms.
In plain language
Particle models help explain matter and changes of state. Atoms of an element have the same proton number; compounds contain different elements chemically joined. Atomic and mass numbers let you work out an atom's particles, while chemical formulae describe composition and let you calculate relative formula masses and mass percentages.
Self-check
Calculate the Mr of Calcium Carbonate (CaCO3) given Ar: Ca=40, C=12, O=16.
Show the answer
Mr = 40 + 12 + (3*16) = 40 + 12 + 48 = 100.

Moles and Avogadro's NumberHigher
Scope: Higher · Unit 2 tier: Higher · Topic 2.1 (higher)
Open revision guidance
What to know
The mole, symbol mol, is the SI unit for amount of substance. The Avogadro number is approximately 6.02 × 10²³ entities in one mole.
Use n = m ÷ M and m = n × M, where n is amount in mol, m is mass in g and M is molar mass in g/mol. For school calculations, M has the same numerical value as the given Aᵣ or Mᵣ. Unlike relative mass, molar mass has units.
What to do in an assessment
Convert mass to amount: for 44 g of CO₂ with given Mᵣ = 44, the molar mass is 44 g/mol. Using n = m ÷ M gives 44 g ÷ (44 g/mol) = 1 mol.
Convert amount to mass: for 0.5 mol of carbon atoms with given Aᵣ = 12, the molar mass is 12 g/mol. Using m = n × M gives 0.5 mol × 12 g/mol = 6 g.
In plain language
A mole is an amount of substance containing approximately 6.02 × 10²³ specified entities, such as atoms, molecules or ions. State which entities are being counted. The molar mass connects an amount in moles to a measurable mass in grams.
Self-check
How many moles are in 80g of NaOH (Mr=40)?
Show the answer
The given Mᵣ of 40 corresponds to a molar mass of 40 g/mol. Using n = m ÷ M gives 80 g ÷ (40 g/mol) = 2 mol.

Structure and Properties of ElementsCore

Scope: Core · Topic 2.2 (core)

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What to know

Elements are arranged in the periodic table by increasing atomic number.
The table is organised into vertical columns called groups and horizontal rows called periods.
Electrons occupy shells. For Groups 1–7, the group number gives the number of outer-shell electrons. Group 0 has a full outer shell: helium has two; the other noble gases have eight.
Metals lie mainly on the left and centre; non-metals mainly on the right. Hydrogen is a non-metal despite its position on the left.
Group 0 (inert gases) have a full outer shell and are generally unreactive under ordinary conditions.
Many metals lose electrons and many non-metals gain electrons to form ions with full outer shells. Non-metals can also share electron pairs in covalent bonds; not every Group 1–7 atom forms an ion.
Group 1 atoms lose one electron to form +1 ions with a full outer shell. Group 7 atoms gain one electron to form −1 ions with a full outer shell. Proton numbers do not change: sodium goes from 11 protons and 11 electrons (2,8,1) to 11 protons and 10 electrons (2,8); chlorine goes from 17 protons and 17 electrons (2,8,7) to 17 protons and 18 electrons (2,8,8).
The specification's general word equations are metal + oxygen → metal oxide; metal + water → metal hydroxide + hydrogen; metal + halogen → metal halide. Use the product formula provided when balancing an oxygen reaction.
The Group 1 metal hydroxides formed with water dissolve to give alkaline solutions. On the usual school pH scale, an alkaline solution is above pH 7 and universal indicator is blue/purple; an acid is below pH 7 with red/orange/yellow indicator colours, while a neutral solution is pH 7 and green. Follow the supplied indicator chart rather than assuming every alkaline solution has the same pH.
For the first twenty elements, the period number gives the number of occupied shells. Using this course's main-group labels 1–7 and 0, outer electrons explain group position; helium is the two-electron full-shell exception.
Metals are generally good conductors of heat and electricity, shiny when freshly cut, malleable, ductile and sonorous (they ring when struck). Many are hard, dense and have high melting points. Non-metals are generally poor conductors with lower melting and boiling points, and solid non-metals are usually brittle. These are tendencies, not rules for every element: Group 1 metals are soft, mercury is liquid at room temperature, diamond has a very high melting temperature under suitable pressure and graphite conducts electricity.
Elements near the boundary between metals and non-metals may combine properties, such as a high melting point with poor electrical conductivity. For this course, recall that melting and boiling points decrease down Group 1 and increase down Group 7. The Group 7 trend helps explain why chlorine is a gas, bromine a liquid and iodine a solid at room temperature; it is separate from the reactivity trend.
Group 1 reactivity increases down the group; Group 7 reactivity decreases down the group. Knowing the directions is common content; explaining them in terms of electron attraction is Higher content.
Fresh Group 1 surfaces tarnish in air as they react with oxygen. In teacher-selected burning demonstrations or videos, lithium burns red (not brick red), sodium yellow/orange and potassium lilac; the course's oxide products are white solids. In teacher demonstrations with water, lithium floats and fizzes without melting; sodium reacts more vigorously, often melting into a moving ball; potassium is more vigorous still and may burn with a lilac flame. Describe visible fizzing, movement or flame colour as observations; identifying hydrogen is a conclusion supported by its gas test. These are hazardous demonstrations, not activities to attempt independently.

What to do in an assessment

Draw electronic structures for elements 1-20 (e.g., Sodium atomic number 11 is 2,8,1).
Use a main-group position in both directions: magnesium has electronic structure 2,8,2, so it is in Period 3 and Group 2. An element in Period 3 and Group 7 has three occupied shells and seven outer electrons: 2,8,7, identifying chlorine, atomic number 17.
Test for hydrogen gas using a burning splint (squeaky pop test).
Identify Lithium (red), Sodium (yellow), and Potassium (lilac) using flame tests.
Identify halide ions (chloride, bromide, iodide) using silver nitrate solution.
Balance equations such as: 2Na + Cl2 → 2NaCl.
For the first twenty neutral atoms, count electrons from the atomic number and use the school shell model: H 1; He 2; Li 2,1; Be 2,2; B 2,3; C 2,4; N 2,5; O 2,6; F 2,7; Ne 2,8; Na 2,8,1; Mg 2,8,2; Al 2,8,3; Si 2,8,4; P 2,8,5; S 2,8,6; Cl 2,8,7; Ar 2,8,8; K 2,8,8,1; Ca 2,8,8,2. Draw one dot per electron and check the total.
For these first twenty atoms, fill the first shell with up to two electrons, then use eight in the second and eight in the third in this school model. Potassium and calcium start the fourth shell. This is a model for these elements, not a claim that the third shell can never hold more than eight electrons.
Laboratory tests require teacher supervision and eye protection. For a teacher-provided small hydrogen sample, a lighted splint at the tube mouth gives a squeaky pop. Do not put a flame into a container holding a reacting alkali metal.
Flame tests use salts, not pieces of the reactive metals: a clean teacher-provided loop carrying a little salt solution is placed in the blue, non-luminous burner flame. Lithium gives crimson red, sodium yellow and potassium lilac. Clean the loop between samples using the teacher's procedure and check that it gives no sample colour before testing another salt; sodium contamination can obscure other colours.
Halide test: acidify a fresh sample with the teacher's dilute nitric acid, then add silver nitrate solution. Chloride gives a white precipitate, bromide cream and iodide yellow. Hydrochloric acid would add chloride and interfere with the test. Follow the teacher's chemical-waste procedure.
Word examples: lithium + oxygen → lithium oxide; sodium + water → sodium hydroxide + hydrogen; sodium + chlorine → sodium chloride.
In Unit 2, complete the given product formulae and coefficients, then balance atoms: Li + O₂ → Li₂O becomes 4Li + O₂ → 2Li₂O; Na + H₂O → NaOH + H₂ becomes 2Na + 2H₂O → 2NaOH + H₂. Never alter a correct formula's subscripts to balance it. Unit 5 requires writing these balanced equations without scaffolding.

In plain language

The periodic table arranges elements by atomic number. Main-group elements with the same outer-shell electron count have similar chemical properties. At this level, reactions are explained using atoms gaining, losing or sharing electrons towards full outer shells; the familiar Group 0 elements already have full outer shells and are generally unreactive.

Self-check

Which products form when sodium reacts with water?

Show the answer

Sodium hydroxide and hydrogen gas.

View diagrams

Electronic structures

Fill inner shells first. This table gives the school model for the first 20 elements.

Atomic number	Element	Symbol	Shell configuration
1	Hydrogen	H	1
2	Helium	He	2
3	Lithium	Li	2, 1
4	Beryllium	Be	2, 2
5	Boron	B	2, 3
6	Carbon	C	2, 4
7	Nitrogen	N	2, 5
8	Oxygen	O	2, 6
9	Fluorine	F	2, 7
10	Neon	Ne	2, 8
11	Sodium	Na	2, 8, 1
12	Magnesium	Mg	2, 8, 2
13	Aluminium	Al	2, 8, 3
14	Silicon	Si	2, 8, 4
15	Phosphorus	P	2, 8, 5
16	Sulfur	S	2, 8, 6
17	Chlorine	Cl	2, 8, 7
18	Argon	Ar	2, 8, 8
19	Potassium	K	2, 8, 8, 1
20	Calcium	Ca	2, 8, 8, 2

Neutral Na · Z = 11

Every electron is drawn; this is a schematic shell diagram.

Neutral Cl · Z = 17

Every electron is drawn; this is a schematic shell diagram.

For the first 20 elements, this is the school shell model. It does not mean the third shell has a universal maximum of eight, and the circles are not literal electron paths.

First-twenty neutral-atom configurations, with complete worked sodium and chlorine shell models. The circles represent a school model of energy levels, not literal electron paths.

Trends in Group ReactivityHigher
Scope: Higher · Unit 2 tier: Higher · Topic 2.2 (higher)
Open revision guidance
What to know
Group 1 reactivity increases down the group because distance and shielding reduce attraction to the outer electron, making it easier to lose.
Group 7 reactivity decreases down the group because distance and shielding reduce attraction to an incoming outer-shell electron, making it harder to gain.
What to do in an assessment
Compare lithium (2,1) with potassium (2,8,8,1). Both lose one outer electron, but potassium's outer electron is further from the nucleus and more shielded, so it is lost more readily: potassium is more reactive.
Compare fluorine (2,7) with chlorine (2,8,7). Both gain one electron, but chlorine has an extra occupied shell and more shielding; its nucleus attracts an incoming outer-shell electron less effectively. Chlorine gains an electron less readily and is less reactive.
In plain language
For Groups 1 and 7, explain these reactivity trends through the readiness to lose or gain an electron. Down either group there are more occupied shells: the outer shell is further from the nucleus and inner electrons provide more shielding. These effects outweigh the increase in positive nuclear charge for the electron involved in these reactions.
Self-check
Why does reactivity increase as you go down Group 1?
Show the answer
Down Group 1, the outer electron is further from the nucleus and more shielded by inner electrons. Despite the increased positive nuclear charge, attraction to that outer electron is weaker, so it is lost more readily and reactivity increases.

Chemical Reactions, Rates, and CatalystsCore
Scope: Core · Topic 2.3 (core)
Open revision guidance
What to know
Physical changes do not form new substances: melting ice and freezing the water reverse the state change while the substance remains water. Chemical reactions form new substances. Reversibility alone is not a reliable way to distinguish them because some chemical reactions are reversible, while physically breaking an object is not easily reversed.
Chemical reaction indicators include temperature change, gas production (effervescence), colour change, and precipitate formation.
Exothermic reactions transfer heat energy to the surroundings, whose temperature rises; endothermic reactions take heat energy from the surroundings, whose temperature falls. Energy is transferred, not created or destroyed.
The Law of Conservation of Mass states atoms are not created or destroyed in reactions.
For a reaction proceeding to completion, it stops when a limiting reactant is fully consumed, even if another reactant remains in excess.
Reaction progress is monitored by measuring changes in reactant/product quantity over time.
Rate is affected by temperature, concentration, and surface area, explained by collision theory.
Catalysts increase reaction rates without being consumed overall, allowing suitable industrial reactions to operate at lower temperatures with less heating.
As reactants are consumed, fewer reacting particles are available and successful collisions become less frequent, so the rate normally falls. When a limiting reactant is used up, that reaction stops.
Raising temperature makes particles move faster, increasing collision frequency and the fraction with enough energy to react. Higher concentration puts more particles in a given volume. Breaking a solid into smaller pieces exposes more surface for collisions; it does not change the total mass of an equal-mass sample.
Scientists develop catalysts that make useful reactions faster with lower energy demand or less waste. Sustainability also depends on catalyst manufacture, scarce materials, recovery and lifetime, not just the reaction rate.
What to do in an assessment
Use the Law of Conservation of Mass to balance equations, ensuring equal atom counts on both sides.
Apply collision theory: increased temperature raises particle kinetic energy; higher concentration increases collision frequency; larger surface area exposes more particles.
Interpret reaction graphs on the same scales: product quantity rises and reactant quantity remaining falls. A steeper rise or fall indicates a faster rate; compare steepness, not just the sign of the slope.
Compare rates on the same graph scales: a steeper rise in product quantity indicates a faster rate; a horizontal final section means no more product is being measured. Check that the sensor is still working before attributing a plateau to the reaction ending. Numerical rate from a straight-line graph is in the Higher lesson.
Balance the partially constructed example Mg + O₂ → MgO by counting atoms: 2Mg + O₂ → 2MgO has two magnesium and two oxygen atoms on each side.
Gas collection: in the teacher's suitable acid–carbonate reaction, start timing as reactants mix and record gas-syringe volume at fixed intervals. Keep reactant amounts and apparatus constant between trials except for the chosen factor. Check for leaks and never improvise a sealed pressure vessel.
Mass loss: use the teacher's safe gas-producing reaction on a balance and record mass at fixed times. A suitable loose cotton-wool plug prevents splashes while allowing gas to escape; do not seal the flask. The measured decrease is gas leaving the apparatus, not atoms disappearing.
Precipitation: in the teacher's dilute sodium-thiosulfate and acid reaction, sulfur particles make the mixture cloudy. Place the flask over a cross, start timing when the supplied reactants mix, and record when the cross is no longer visible. Keep solution depth, cross and viewing conditions constant and repeat because the endpoint is subjective. The reaction also releases irritating sulfur dioxide: follow the teacher's ventilation and eye-protection instructions; do not prepare it at home.
Where available, a gas-volume sensor, digital balance or data logger records measurements at regular intervals. Check calibration, units and the sampling interval rather than assuming a digital reading is automatically accurate.
Test temperature with the teacher's controlled baths and fixed concentrations and surface area; test concentration using measured dilutions while holding total volume and temperature constant; test surface area using equal masses of the same solid as larger chips and smaller pieces. Repeat and compare the complete curves.
Catalyst investigation: compare equal samples of the teacher's dilute hydrogen peroxide with and without a small teacher-provided manganese(IV) oxide sample. Record oxygen volume at fixed times with the approved gas-collection apparatus, holding peroxide concentration, volume and temperature constant. Wear eye protection and follow the supplied handling precautions; peroxide and catalyst powders are not harmless household ingredients.
A control without the catalyst helps attribute a changed rate to it. Recovering a suitable solid catalyst, or showing it can act again, supports that it is not used up overall; this does not mean it cannot become contaminated or less effective in real use.
In plain language
Chemical reactions form new substances; they are not all irreversible. We follow their progress by measuring reactant loss or product formation over time. Higher temperature, greater reactant concentration or greater exposed solid surface area can increase successful collisions. A suitable catalyst speeds up a reaction without being used up overall, allowing some industrial processes to operate with less heating.
Self-check
In a closed system, 10 g of reactant A reacts completely with 5 g of reactant B, leaving no excess reactant. What is the total mass of all products, including gases?
Show the answer
15 g of products, including any gases: 10 + 5 = 15. In this closed system all the reactants have reacted, and atoms are neither created nor destroyed.

Higher Supplement: Rate Calculation from GraphsHigher
Scope: Higher · Unit 2 tier: Higher · Topic 2.3 (higher)
Open revision guidance
What to know
Use two well-separated points on the straight section, not just the final graph height: a rise from 10 to 30 cm³ between 20 and 60 s gives (30 − 10) ÷ (60 − 20) = 0.5 cm³/s.
For a straight-line graph representing a constant rate, the slope (gradient) equals the rate of reaction.
What to do in an assessment
To calculate rate from a straight line: identify two points (x1, y1) and (x2, y2) on the line. Calculate the change in y (rise) divided by the change in x (run). Gradient = (y2 - y1) / (x2 - x1); use its magnitude for a rate of loss.
If a straight tangent is supplied on a curved quantity–time graph, find its gradient using two points on that line to obtain the rate at the touching point. You are not required to draw a tangent in this unit; do not substitute the endpoints of the whole curve for points on the supplied straight line.
In plain language
A straight section of a quantity–time graph has a constant gradient. For product formed or reactant consumed, divide the change in quantity by elapsed time and state the units. If the graph plots reactant quantity remaining, its gradient is negative: report the positive magnitude as the rate of loss.
Self-check
A straight-line section rises from 0 cm³ at 0 s to 20 cm³ at 40 s. What is the rate?
Show the answer
0.5 cm3/s. Calculated as 20 / 40 = 0.5.

Earth's Resources and Atmospheric ChangeCore
Scope: Core · Topic 2.4 (core)
Open revision guidance
What to know
Raw materials are unprocessed Earth resources; some are used directly, most are processed to create materials.
Finite resources (like metal ores) are limited and deplete, whereas renewable resources (like wood) can be replaced.
Water treatment involves filtration to remove solids and sterilization (e.g., using chlorine) to kill microbes.
Chemically pure water is a single substance, water; fresh water normally also contains dissolved minerals and salts. Safe drinking water does not have to be chemically pure.
Distillation separates seawater by boiling and condensing water vapor, leaving dissolved salts behind.
Atmosphere composition: ~78% Nitrogen, ~21% Oxygen, ~0.04% Carbon Dioxide.
A carbon footprint is the total greenhouse gases attributable to the human activity being assessed, not just carbon dioxide. Different gases, including methane, are expressed as a mass equivalent of carbon dioxide (kgCO₂eq) using the factors supplied.
Greenhouse gases absorb and re-emit some outgoing infrared radiation. The natural greenhouse effect helps keep Earth warm; higher greenhouse-gas concentrations strengthen the effect and increase mean global temperature.
A resource is renewable if natural processes replace it on a useful human timescale; that does not make unlimited harvesting sustainable. Compare extraction damage, transport, energy demand, water use, waste and the rate of replacement when judging a material's use.
Fresh water is needed for households, irrigation, farming and industry. The water cycle does not guarantee clean water where and when people need it. Sewage, fertilisers and industrial pollutants can damage supplies; climate change can worsen shortages through changing rainfall, drought and evaporation.
Filtering and disinfecting public water remove particles and harmful microbes but do not make it chemically pure. Distillation can produce pure water in the school model, but seawater desalination needs energy and management of the concentrated waste. Do not drink a school-produced sample.
Analytical methods measure contaminants rather than judging safety by appearance. Improved sensors, treatment, leak detection, water reuse and efficient irrigation can support sustainable water management; their costs and environmental effects must still be assessed.
The quoted atmospheric percentages are approximate dry-air values; most of the remaining approximately 1% is argon, with other trace gases. Water vapour varies. Photosynthesis removes carbon dioxide and releases oxygen; respiration and combustion use oxygen and release carbon dioxide. Human activity can change the balance, so these processes do not guarantee constant carbon dioxide levels.
The early atmosphere is explained through models using geological and chemical evidence: volcanic gases, cooling and oceans, and later photosynthetic organisms help explain major changes. Details are uncertain because direct measurements from early Earth do not exist. These models provide awareness and context; this unit does not require memorising a formation sequence. Recall the evolution of green plants and the role of photosynthesis in producing atmospheric oxygen.
Carbon dioxide and methane occur naturally. Fossil-fuel combustion and deforestation increase atmospheric carbon dioxide; livestock, landfill and fossil-fuel leaks add methane. Human activity has substantially increased greenhouse-gas concentrations over the past century.
A warmer atmosphere stores more energy and can contain more water vapour, increasing risks of heat extremes, heavy rainfall and other extreme weather. Effects differ by region and include sea-level rise, habitat changes, health risks and pressure on food and water supplies.
The international target aims to limit long-term warming to 1.5°C above the pre-industrial mean. This is not a guarantee of success or a claim that every individual year has the same temperature.
Reduce a footprint by saving energy, improving insulation, using lower-emission transport where practical, reducing avoidable consumption and waste, and choosing suitable lower-carbon energy supplies. Effects depend on the activity and its whole life cycle.
Net zero means balancing remaining greenhouse-gas emissions with removals, not stopping all activity or producing no emissions. Carbon offsetting attempts to compensate through reductions or removals elsewhere; check that they are additional, lasting and not counted twice. It does not automatically cancel the harm of any emission.
Techniques such as cleaner electricity generation, efficiency improvements and carbon capture continue to develop. Climate science uses many connected lines of evidence; judging the whole issue from one cold day or one isolated graph oversimplifies it.
What to do in an assessment
Perform filtration: Pass mixture through filter paper in a funnel; residue stays, filtrate passes.
For water containing non-volatile dissolved salts, use the teacher's distillation apparatus: heat the solution, cool its water vapour in the condenser and collect the condensed water as distillate. The salts remain behind. Distillation does not automatically remove every possible volatile contaminant.
Identify Oxygen: Relight a glowing splint.
Identify Carbon Dioxide: Bubble gas through limewater; it turns cloudy/milky.
Calculate Carbon Footprint: Use mass equivalent (kgCO2eq) formulas provided.
All practical separation and gas tests are teacher-supervised. To separate sand and salt: add water to dissolve the salt, filter to retain sand, then evaporate water from the filtrate to recover salt. To recover the water instead, use the teacher's distillation apparatus and collect condensed vapour. Do not heat an improvised sealed container.
Filtration separates an insoluble solid; it does not remove dissolved salt. Evaporation recovers a dissolved solid but loses the solvent. Distillation collects a solvent by evaporation and condensation, leaving non-volatile dissolved substances behind.
Paper chromatography: draw a pencil baseline, apply a small spot of the teacher's suitable sample, and stand the paper in the chosen solvent with the baseline above the liquid. Let the solvent rise, then mark its front before it dries. Pencil does not dissolve and travel as ink may.
Compare spots with known samples run under the same conditions. Multiple spots indicate more than one separated component; a matching position supports identification but is not universal proof for every possible chemical.
Use the provided expression: Rᶠ = distance from baseline to the centre of a substance's spot ÷ distance from baseline to solvent front. You need to use it, not recall the expression from memory for this unit. Constructed example: a spot travels 3.0 cm and solvent 5.0 cm, giving 3.0 ÷ 5.0 = 0.60. Both distances start at the same baseline; Rᶠ has no units and depends on the solvent and other conditions.
For a teacher-provided oxygen sample, insert a glowing splint: it relights. For carbon dioxide, pass the supplied sample through limewater: it becomes milky. Keep equipment clean and follow the teacher's directions for handling gases.
Constructed footprint calculation using supplied, illustrative factors: 30 kWh × 0.20 kgCO₂eq/kWh = 6 kgCO₂eq; 20 km × 0.10 kgCO₂eq/km = 2 kgCO₂eq. Total for these two activities = 8 kgCO₂eq. These are not claims about current UK emission factors or a person's entire footprint.
In plain language
We use Earth's resources to build our world, but these resources are often finite. We must treat water to make it safe by removing solids and germs. Our atmosphere, a mixture of gases, is changing because human activities release greenhouse gases like CO2 and methane, which trap heat and cause global warming. We need to measure our impact using carbon footprints and consider sustainability to protect the environment.
Self-check
Which gas is identified by bubbling it through limewater, and what is the positive observation?
Show the answer
Carbon dioxide; the limewater turns cloudy or milky.

Higher Concepts: Climate Feedback LoopsHigher
Scope: Higher · Unit 2 tier: Higher · Topic 2.4 (higher)
Open revision guidance
What to know
Positive feedback loops amplify the initial change. Their strength depends on the processes involved; positive does not mean beneficial or prove unlimited acceleration.
Example: Warming melts polar ice (lowering Earth's albedo), leading to more solar radiation absorption, which causes further warming.
What to do in an assessment
Explain how climate change risk involves complex interdependencies where one warming factor triggers another.
In plain language
A positive feedback reinforces an initial change. In climate examples, warming can trigger a change that causes further warming. Positive describes the direction of the feedback, not whether its effects are beneficial.
Self-check
Explain the concept of a positive feedback loop in the context of climate change.
Show the answer
An initial warming can reduce ice cover. Darker exposed surfaces absorb more incoming sunlight than reflective ice, causing further warming. This reinforces the initial change: a positive feedback.

Bonding, Structure, and PropertiesCore
Scope: Core · Topic 5.1 (core)
Open revision guidance
What to know
Metals consist of a giant lattice of positive ions in a sea of delocalized electrons [5.1.2].
Alloys are mixtures of a metal with other elements, commonly prepared by mixing molten materials. A pure metal has characteristic properties under specified conditions; changing alloy composition adjusts hardness, corrosion resistance and other properties.
Ionic bonds are strong electrostatic attractions between oppositely charged ions [5.1.3].
Covalent bonds involve sharing electron pairs to achieve a full outer shell [5.1.4].
Covalent bonding is usual in the simple non-metal elements and compounds studied here: atoms share electron pairs rather than transferring them. Count shared electrons around both atoms in the full-shell model—two for hydrogen and usually eight for the other atoms in these examples.
Simple covalent molecules have weak intermolecular forces [5.1.4].
Metals usually conduct electricity because delocalised electrons can move, and these electrons help transfer thermal energy. Strong metallic attractions generally give high melting and boiling points. Layers can slide while bonding remains, explaining malleability (shaping into sheets) and ductility (drawing into wires). Their closely packed structure helps explain generally high density, although atomic mass also matters and some metals have low density.
Ionic solids usually have high melting and boiling points because many strong attractions must be overcome. They are brittle when displaced layers bring like charges together. Fixed ions cannot carry charge in a solid, but mobile ions conduct when molten or dissolved.
Simple molecular substances generally have low melting and boiling points because separating molecules overcomes relatively weak intermolecular forces, not the strong covalent bonds within each molecule. Without mobile charged particles they do not normally conduct electricity.
Some molecules have double covalent bonds, sharing two electron pairs. Knowing this is common content; producing their dot-and-cross diagrams is Higher.
Diamond and graphite have giant covalent structures, unlike simple molecules. Diamond's strong three-dimensional network makes it hard; graphite has strong bonding within layers, weaker attractions between layers and mobile electrons, so it is soft and conducts.
New alloys are developed for specialist properties. Nanomaterials and smart materials have useful properties linked to bonding and structure; examples include high-surface-area catalysts and materials that respond to temperature. Scientists investigate exposure, toxicity and environmental persistence rather than assuming every new material is harmless. These and the diamond/graphite examples provide context; their specific names, properties and uses are not a required memorised list.
What to do in an assessment
Draw dot and cross diagrams for ionic compounds showing electron transfer [5.1.3].
Draw dot and cross diagrams for covalent molecules showing electron sharing [5.1.4].
Explain physical properties like conductivity and melting points using structural models [5.1.2, 5.1.3, 5.1.4].
Ionic diagram example: sodium transfers its one outer electron to chlorine. Draw bracketed Na⁺ and Cl⁻ with charges and a full outer shell for chloride, distinguishing the transferred electron with a cross. The resulting attraction acts throughout a giant lattice, not a separate NaCl molecule.
Single covalent example: each hydrogen contributes one electron to the shared pair in H₂, giving each a filled first shell of two. For methane, carbon shares one pair with each of four hydrogens; keep dots/crosses consistent with their originating atoms.
In plain language
Use bonding and structural models to explain properties. Metallic bonding involves positive ions and delocalised electrons; ionic bonding attracts oppositely charged ions; covalent bonding shares electron pairs. The full-outer-shell model explains the familiar examples in this course, rather than being a rule that every possible molecule must obey.
Self-check
Why does an alloy have different properties than a pure metal?
Show the answer
In the simple pure-metal model, layers can slide while metallic bonding remains. In many solid alloys, differently sized atoms disrupt the regular arrangement and can hinder sliding. Composition changes strength and other properties; this does not mean every alloy is always harder than every pure metal [5.1.2].
View diagrams
Sodium chloride: ionic transfer
Before transfer, sodium is 2,8,1 and chlorine is 2,8,7. The sodium outer electron transfers to chlorine. Separate bracketed ions result: Na⁺ has configuration 2,8 and Cl⁻ has 2,8,8. Dots identify sodium-origin electrons; crosses identify chlorine-origin electrons. This is electron transfer, not a shared covalent pair.
Hydrogen · H₂
Two hydrogen atoms contribute one electron each to one shared pair. The dot comes from the left H atom and the cross from the right H atom; each hydrogen now has a full first shell.
Hydrogen chloride · HCl
One H electron and one Cl electron form one shared pair. Chlorine contributes six more electrons as three lone pairs; all seven of its outer-shell electrons are shown.
Water · H₂O
Oxygen shares one electron with each H, making two shared pairs. Its four remaining outer-shell electrons make two lone pairs.
Methane · CH₄
Carbon shares one electron with each of four H atoms, making four shared pairs. All four carbon outer-shell electrons and all four hydrogen electrons are shown; there are no lone pairs.
In dot-and-cross diagrams, a dot is an electron from the atom named in the explanation and a cross is from the other atom. The symbols show electron origin, not a charge. Shared pairs belong between bonded atoms; lone pairs stay on one atom.
The dot-and-cross and ionic shell sketches show outer-shell electrons only; inner-shell electrons are omitted from those sketches. Full sodium and chlorine configurations are printed beside the ions. Displayed structural formulae instead draw every atom, including each hydrogen, and every covalent bond.
Compare electron transfer with shared pairs. Dots and crosses distinguish electron origins; these are bonding models, not literal electron paths or scale drawings.

Advanced Covalent BondingHigher
Scope: Higher · Unit 5 tier: Higher · Topic 5.1 (higher)
Open revision guidance
What to know
Some molecules contain double bonds where two pairs of electrons are shared between two atoms to achieve a full outer shell [5.1.4].
What to do in an assessment
Produce dot and cross diagrams for molecules containing double bonds (e.g., O2 or CO2) [5.1.4].
For O₂, each oxygen starts with six outer electrons: share two pairs between the atoms and leave two lone pairs on each oxygen. Each oxygen then has eight electrons around it, counting shared electrons.
For CO₂, draw O=C=O with two shared pairs in each C–O bond. Carbon has four shared pairs and no lone pairs; each oxygen has two shared pairs and two lone pairs. Check the total of 16 outer electrons.
In plain language
While many molecules share a single pair of electrons, some require sharing two pairs to reach stability. This is represented in dot and cross diagrams as two overlapping circles containing four electrons in total between the atoms [5.1.4].
Self-check
How many electrons are shared in a double bond?
Show the answer
Four electrons (two pairs) are shared between the two atoms [5.1.4].
View diagrams
Oxygen · O₂
The two oxygen atoms share two pairs, forming a double bond. Each oxygen keeps two lone pairs: 12 outer-shell electrons are shown altogether.
Carbon dioxide · CO₂
Carbon forms a double bond to each oxygen: four shared pairs in total. Each oxygen has two lone pairs; carbon has none. All 16 outer-shell electrons are shown.
In dot-and-cross diagrams, a dot is an electron from the atom named in the explanation and a cross is from the other atom. The symbols show electron origin, not a charge. Shared pairs belong between bonded atoms; lone pairs stay on one atom.
The dot-and-cross and ionic shell sketches show outer-shell electrons only; inner-shell electrons are omitted from those sketches. Full sodium and chlorine configurations are printed beside the ions. Displayed structural formulae instead draw every atom, including each hydrogen, and every covalent bond.
Higher drawing skill: each double bond has two shared electron pairs. Include the unshared outer electrons; the labels do not replace counting the electrons.

Acids, Bases, and Salt IdentificationCore
Scope: Core · Topic 5.2 (core)
Open revision guidance
What to know
Acids produce hydrogen ions (H+), while alkalis produce hydroxide ions (OH-).
The acid–oxide/hydroxide neutralisations studied here form salt and water and release energy. An acid reacting with a carbonate also produces carbon dioxide.
Common metal oxides such as copper(II) oxide act as bases. An alkali is a base that dissolves in water and provides hydroxide ions.
General equations: Acid+Base→Salt+Water; Acid+Alkali→Salt+Water; Acid+Carbonate→Salt+Water+CO2; Acid+Metal→Salt+Hydrogen.
Salt naming: Hydrochloric acid forms chlorides, nitric acid forms nitrates, and sulfuric acid forms sulfates.
Carbon dioxide is identified by bubbling it through limewater, which turns milky/cloudy.
Carbonate ions (CO3 2-) are identified by adding dilute acid; bubbles of CO2 indicate a carbonate.
Sulfate ions (SO4 2-) are identified using barium chloride solution, forming a white precipitate of barium sulfate.
Calcium and barium ions are identified using flame tests: calcium gives brick red; barium gives apple green.
A reactivity series orders metals by relative reactivity and includes hydrogen and carbon as useful comparison entries. Use the supplied series; recalling the whole list is not required.
Use the supplied reactivity series to predict dilute-acid reactions. Appropriate metals above hydrogen, such as magnesium and zinc, release hydrogen with dilute hydrochloric acid; copper does not. This rule is not a claim that every metal behaves identically with every acid.
For the hydrogen test, use only the teacher's small gas sample and approved lighted-splint method; a squeaky pop indicates hydrogen. Never apply a flame to a gas-generating reaction vessel.
What to do in an assessment
On the usual school pH scale, below 7 is acidic, 7 neutral at room temperature and above 7 alkaline. Universal indicator commonly changes from red/orange in acids through green near neutral to blue/purple in alkalis; compare with its supplied colour chart.
Typical universal-indicator ranges are red around pH 1–2, orange around 3–4, yellow around 5–6, green near 7 and blue/purple above 7, with the exact shade checked against the chart. pH describes the sample: concentration also matters, so pH alone does not establish whether an acid is chemically strong or weak.
To measure pH, use separate clean samples. Calibrate a probe with appropriate buffers, rinse between samples and wait for a stable reading; or add a consistent amount of universal indicator and compare its chart. pH and colour are not the same quantity, and a probe gives more precise numerical readings.
Write balanced symbol equations for acid-base and acid-metal reactions, such as: 2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g).
For that metal reaction, the word equation is magnesium + hydrochloric acid → magnesium chloride + hydrogen. Use the general reaction and the acid's salt name to construct equations, not simply memorise one example.
Soluble-salt method: gently warm the teacher's dilute acid, add insoluble base or carbonate in small portions until some remains unreacted, then filter off the excess solid. Gently concentrate the filtrate, allow it to cool and crystallise, then filter and dry the crystals. Do not boil it dry.
Copper sulfate example: sulfuric acid + copper(II) oxide → copper(II) sulfate + water; H₂SO₄ + CuO → CuSO₄ + H₂O. Excess black oxide ensures the acid has reacted; filtration removes that excess, not the dissolved salt.
Titration method: use a pipette and filler for a measured volume of strong base and add indicator. Record the initial burette reading for the strong acid, add while swirling and finish dropwise at the endpoint. Read the meniscus at eye level; delivered volume = final − initial reading. Repeat to obtain consistent volumes.
For pure salt crystals, repeat with the measured reacting volumes but without indicator, then concentrate and crystallise. Indicator would contaminate the crystals. A constructed reading change from 2.00 to 27.20 cm³ delivers 25.20 cm³, not 27.20 cm³.
Use fresh portions for different ion tests, known controls, goggles and the teacher's disposal protocol. The Group 1 flame and Group 7 halide methods are taught in Unit 2.2; do not mix the acid or reagents used for different tests.
Sulfate test: acidify a fresh portion with dilute hydrochloric acid, then add barium chloride solution. A white barium sulfate precipitate supports sulfate identification; acid removes carbonate interference.
Group 2 flame tests use a clean loop and teacher-provided salts: calcium gives brick red and barium apple green. Record these specific colour names, not just red or green. Clean between samples and compare known controls; contamination can mask colours.
Plan identification by listing possible cations and anions, selecting relevant tests and using fresh portions and controls. A brick-red flame supports calcium; acidified silver nitrate giving a white precipitate in another portion supports chloride. Together they support calcium chloride, rather than identifying the entire salt from one test.
Precipitation example: barium chloride + sodium sulfate → barium sulfate + sodium chloride; BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl. For chloride: silver nitrate + sodium chloride → silver chloride + sodium nitrate; AgNO₃ + NaCl → AgCl + NaNO₃. The insoluble products are the observed precipitates.
Acid plus carbonate releases gas; bubble it through limewater, which turns cloudy if carbon dioxide forms. Use safe gas-transfer equipment, never improvise a sealed reaction vessel.
When acid is added to alkali, pH falls towards the endpoint; adding alkali to acid makes pH rise. Temperature can rise as neutralisation releases energy, with a maximum near the reacting proportions under controlled conditions. Heat loss, dilution and addition size affect the measured maximum. Antacids neutralise excess stomach acid; making ammonium salts for fertilisers is an industrial neutralisation use.
Slow evaporation or cooling can produce larger salt crystals than rapid evaporation. This is a comparison under controlled conditions, not a reason to boil a solution dry.
Use only the teacher's approved dilute-acid procedure and metal samples. Do not put metals more reactive than magnesium into acid for this school investigation.
In plain language
Acids and bases neutralise each other to form salts and water. Acids are substances that release hydrogen ions in water, and they react with metals to produce hydrogen gas. Indicators show pH changes, helping us identify solutions as acidic, neutral, or alkaline. We can identify specific chemicals by testing for their ions, such as using barium chloride for sulfates or flame tests for specific group 2 metals.
Self-check
What is the result of reacting hydrochloric acid with calcium carbonate?
Show the answer
The reaction produces calcium chloride, water, and carbon dioxide gas (2HCl + CaCO3 → CaCl2 + H2O + CO2).

Higher Level Identification of Metal IonsHigher
Scope: Higher · Unit 5 tier: Higher · Topic 5.2 (higher)
Open revision guidance
What to know
Metal ions can be identified by the colour of the precipitate formed when sodium hydroxide solution is added.
Copper(II) ions form a blue precipitate (copper(II) hydroxide).
Iron(II) ions form a green precipitate (iron(II) hydroxide).
Iron(III) ions form a brown precipitate (iron(III) hydroxide).
What to do in an assessment
Add sodium hydroxide dropwise to separate fresh portions of teacher-provided solutions and compare with known controls. Copper(II) gives blue, iron(II) green and iron(III) brown hydroxide precipitates. Iron(II)'s precipitate can turn brown on standing in air, so note the fresh observation.
Copper(II) sulfate + sodium hydroxide → copper(II) hydroxide + sodium sulfate; CuSO₄ + 2NaOH → Cu(OH)₂ + Na₂SO₄. Iron(II) sulfate + sodium hydroxide → iron(II) hydroxide + sodium sulfate; FeSO₄ + 2NaOH → Fe(OH)₂ + Na₂SO₄.
Iron(III) chloride + sodium hydroxide → iron(III) hydroxide + sodium chloride; FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl. Construct balanced word/symbol representations; ionic precipitation equations are not required.
Combine cation and anion evidence to identify a salt: a fresh blue hydroxide precipitate and positive sulfate test support copper(II) sulfate. One test alone usually identifies an ion, not every part of a compound.
In plain language
Higher chemistry uses sodium hydroxide to test for copper(II), iron(II) and iron(III) ions. Their fresh hydroxide precipitates have different colours. Compare known controls and combine evidence; an observation supports ion identification, not the identity of every possible unknown compound.
Self-check
Which colour precipitate forms when sodium hydroxide is added to a solution containing iron(III) ions?
Show the answer
A brown precipitate, often described as orange-brown.

Reactivity of Metals and Extraction PrinciplesCore
Scope: Core · Topic 5.3 (core)
Open revision guidance
What to know
Relative reactivity of metals is demonstrated by competition reactions with metal oxides and displacement reactions.
Displacement reactions occur when a more reactive metal displaces a less reactive metal from a solution of its salt.
Oxidation is the gain of oxygen; reduction is the loss of oxygen.
Some unreactive metals, such as gold, can occur native as elements; most metal sources are compounds in ores. An ore contains enough useful material for extraction to be economically worthwhile, not simply any metal-containing compound.
Reactive metals are extracted using electrolysis; moderately reactive metals (below carbon in reactivity series) are extracted by reduction with carbon.
During electrolysis, ions must be in a molten or dissolved state to be mobile.
Ions are attracted to oppositely charged electrodes: positive ions to the negative cathode, and negative ions to the positive anode.
At the cathode, positive metal ions gain electrons and form metal atoms. This process is common content; electron definitions of redox and writing electrode half-equations are Higher.
Metal-ore supplies are finite and unevenly distributed. As rich ores are depleted, processing poorer ores moves more rock and often uses more energy and water and produces more waste per unit of metal.
Lithium and cobalt have high demand for batteries and emerging technologies despite limited accessible resources. Extraction decisions involve cost, land and water impacts, transport and working conditions, not only whether a reaction works.
Recycling metals reduces demand for new ore and can save energy, but collection, sorting and processing also have costs. Consider sustainability from extraction through product design, use, repair, reuse and final recycling or disposal.
What to do in an assessment
Write balanced symbol equations for metal displacement, such as: Fe(s) + CuSO4(aq) -> FeSO4(aq) + Cu(s).
The corresponding word equation is iron + copper(II) sulfate → iron(II) sulfate + copper. Use the identities of reactants and products, not just the observation of a colour change.
A simplified direct-carbon equation is 2Fe₂O₃ + 3C → 4Fe + 3CO₂. A key furnace reaction is Fe₂O₃ + 3CO → 2Fe + 3CO₂: iron oxide loses oxygen and is reduced, while carbon monoxide gains oxygen and is oxidised.
Competition example: magnesium + copper(II) oxide → magnesium oxide + copper; Mg + CuO → MgO + Cu. More reactive magnesium takes oxygen from copper oxide. Such hot reactions are teacher-controlled, not home investigations.
Predict displacement using the series: iron displaces copper from copper sulfate, but copper does not displace iron from iron sulfate. Keep atoms balanced and recognise a no-reaction case rather than inventing products.
In plain language
Relative reactivity helps predict displacement and choose extraction methods. Carbon can reduce suitable metal oxides below it in the series; more reactive metals generally require electrolysis. In a molten ionic compound the mobile ions move to electrodes, where reactions form products.
Self-check
Why is carbon used to extract iron from iron(III) oxide?
Show the answer
Iron is below carbon in the reactivity series, so its oxide can be reduced using carbon. In a blast furnace, carbon also produces carbon monoxide, an important reducing agent; removing oxygen leaves the metal.

Advanced Redox and CalculationsHigher
Scope: Higher · Unit 5 tier: Higher · Topic 5.3 (higher)
Open revision guidance
What to know
Oxidation is the loss of electrons (OIL); reduction is the gain of electrons (RIG).
In electrolysis, positive metal ions gain electrons at the cathode (reduction). Negative non-metal ions lose electrons at the anode (oxidation).
Half-equations represent the change at specific electrodes, e.g., for molten lead bromide: Pb2+ + 2e- -> Pb (cathode) and 2Br- -> Br2 + 2e- (anode).
For simplest formula, divide each reacting mass in g by its molar mass in g/mol, then divide all mole amounts by the smallest. Multiply to remove simple fractions if necessary; a mole ratio is not automatically the original mass ratio.
For reacting masses, convert known mass to moles, use coefficients in the balanced equation for the mole ratio, then multiply target moles by its molar mass. State sufficient reactant and complete-reaction assumptions; actual yield may be lower.
What to do in an assessment
Worked extraction maximum: with Ar(Al)=27 and Ar(O)=16, Mr(Al₂O₃)=102. Its aluminium mass fraction is 54/102, so 100 g oxide could supply 100 × 54/102 ≈ 52.9 g aluminium if extraction is complete with no losses.
Worked simplest formula: 2.4 g magnesium and 1.6 g oxygen give 2.4/24 = 0.10 mol Mg and 1.6/16 = 0.10 mol O. Divide by the smaller amount to get 1:1, so MgO. Use atom amounts for oxygen here, not O₂ molecules.
Worked reacting mass: 2Mg + O₂ → 2MgO. Given 12 g Mg and sufficient oxygen, 12/24 = 0.50 mol Mg forms 0.50 mol MgO. Mr(MgO)=40, so maximum product mass is 20 g.
Write half-equations for ionic compounds in electrolysis.
In plain language
Higher chemistry defines oxidation and reduction by electron loss and gain and represents electrode reactions with half-equations. Calculations use balanced equations to predict theoretical reacting masses under stated assumptions; practical yields can be lower.
Self-check
Write the half-equation for the oxidation of chloride ions (Cl-) at the anode.
Show the answer
2Cl- -> Cl2 + 2e-

Introduction to Alkanes and AlkenesCore
Scope: Core · Topic 5.4 (core)
Open revision guidance
What to know
For the simple open-chain alkanes studied here, the general formula is CₙH₂ₙ₊₂ with only single carbon–carbon bonds. For open-chain alkenes with one C=C, it is CₙH₂ₙ; the formula is not a rule for every molecule with multiple double bonds or rings.
Bromine water changes from orange-brown to colourless with an alkene under the standard test conditions; a simple alkane does not. This comparison is not proof that every other kind of substance that decolourises bromine is an alkene.
The C=C bond makes simple alkenes react readily by addition with halogens; simple alkanes are less reactive in this comparison and have no C=C for that addition.
Other organic families include alcohols, carboxylic acids and amino acids. They contain atoms such as oxygen or nitrogen as well as carbon and hydrogen, so are not hydrocarbons.
Crude oil is a finite, non-renewable resource
Crude oil is a mixture of hydrocarbons
Hydrocarbons contain only carbon and hydrogen atoms
Crude oil is separated by fractional distillation based on different boiling points
Physical properties change with chain length: longer chains have higher boiling points, higher viscosity, and are harder to ignite
Know the main fuel uses: petrol for many road vehicles, diesel for diesel-engine vehicles such as lorries and buses, and kerosene for jet aircraft. Exact fraction boiling points and chain-length ranges are not required recall.
Cracking is the thermal breakdown of large, less useful hydrocarbon molecules into smaller, more useful ones
Cracking is necessary to meet the high demand for smaller, more volatile fuels
Balanced symbol equations are used to represent cracking, ensuring the number of carbon and hydrogen atoms remains constant on both sides
Continued oil extraction supports fuels, materials, jobs and trade but uses a finite resource and carries spill, habitat, climate and health impacts. Debate involves local communities, national economies and international demand; compare alternatives and consequences rather than only purchase price.
Polymerisation joins small reactive monomers into a large polymer chain
Common plastics include polyethene, polypropene, PVC, and PTFE
Polyethene is light, tough and waterproof for packaging; polypropene is tough and useful for containers; PVC is durable and can be rigid for pipes or flexible with additives for insulation; PTFE has low friction and good chemical resistance for coatings and bearings. Properties depend on formulation.
Many common plastics have low density, resist water and corrosion and are poor thermal/electrical conductors. They may be rigid or flexible depending on structure and additives. Compare these properties with metals, glass or wood for a proposed use, rather than assuming every plastic behaves identically.
Microplastics are small plastic particles, generally below 5 mm, formed by fragmentation or manufactured at that size. They occur in water, soil and air and can be ingested by organisms; their effects are still under research.
Disposal of plastics causes significant environmental issues
Reusing and recycling help conserve finite natural resources
Complete hydrocarbon combustion produces carbon dioxide and water and transfers energy. Limited oxygen can also produce toxic carbon monoxide and soot particles.
The fire triangle consists of fuel, heat, and oxygen
Carbon dioxide contributes to enhanced greenhouse warming; soot damages respiratory health; carbon monoxide reduces blood's oxygen transport. Sulfur impurities can form sulfur dioxide, and high-temperature engines can form nitrogen oxides from air.
Sulfur dioxide and nitrogen oxides react with oxygen and water in the atmosphere to form acids that harm ecosystems, water bodies and buildings. Lower-sulfur fuels, flue-gas treatment and catalytic converters have greatly reduced relevant emissions in many regions, not eliminated the problem everywhere.
Compare fuel energy using energy absorbed by water divided by mass burned; measurements require the same conditions and an evaluation of heat loss.
Hydrogen can offer low-emission use and can be produced using renewable electricity, but production methods, energy demand, compressed storage and distribution matter. It can form explosive mixtures with air, so handling requires appropriate controls. Renewable hydrogen and fossil-fuel-derived hydrogen do not have the same footprint.
Bioethanol uses replenishable feedstocks but needs land, water and processing energy, can compete with food production and releases carbon dioxide when burned. Its life-cycle impact depends on how crops and fuel are produced.
Researchers continue to develop alternatives; assess benefits, drawbacks and full production-to-use impacts rather than calling any fuel automatically clean.
Exothermic reactions release energy to surroundings because more energy is released when bonds form than is required to break them.
Endothermic reactions absorb energy from surroundings because more energy is required to break bonds than is released when new bonds form.
Exothermic reactions are used in self-heating food cans and hand warmers.
Endothermic dissolving processes are used in some instant cold packs for injury treatment.
What to do in an assessment
Identify saturated hydrocarbons (alkanes) versus unsaturated hydrocarbons (alkenes).
For the teacher-supervised bromine-water test, compare small approved samples using the same reagent quantity and known controls. Shake as instructed: an alkene decolourises orange-brown bromine water; a simple alkane does not under these test conditions. Follow chemical-exposure and waste precautions, not a home experiment.
Represent addition with CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br. One bromine attaches to each of the two carbons and their bond becomes single. The product's name is given here; naming unfamiliar addition products is Higher.
Check carbon has four bonds and hydrogen one in a displayed formula. Condensed structural formulae group bonded atoms but must still show the C=C position.
Required straight-chain alkanes: methane CH₄; ethane C₂H₆, CH₃–CH₃; propane C₃H₈, CH₃–CH₂–CH₃; butane C₄H₁₀, CH₃–CH₂–CH₂–CH₃; pentane C₅H₁₂, CH₃–CH₂–CH₂–CH₂–CH₃.
Required common alkene examples: ethene C₂H₄, CH₂=CH₂, and propene C₃H₆, CH₂=CH–CH₃. There is no one-carbon alkene because C=C needs two carbons. Higher extends the named alkene structures to five carbons.
If asked for a displayed formula, draw every bond, including C–H bonds; these condensed examples do not display every bond. Carbon forms four bonds and hydrogen one.
Define crude oil and identify it as a mixture of hydrocarbons
Describe the relationship between chain length and the physical properties of hydrocarbons
Fractional distillation: heat crude oil so much of it vaporises, then feed it into a column hot at the bottom and cooler above. Vapours condense where the temperature is below their boiling range. Long-chain fractions condense lower down; shorter-chain fractions condense higher; some gases leave the top.
A fraction is still a mixture with a similar boiling range, not one pure compound. Longer chains generally have higher boiling points and viscosity, are less volatile and harder to ignite; under limited-oxygen conditions they tend to burn more smokily.
Explain why cracking is essential for the industry
Write a balanced equation for cracking, for example: C10H22 -> C8H18 + C2H4
Check the example: 10 carbon atoms become 8 + 2 and 22 hydrogen atoms become 18 + 4. Cracking is a chemical change requiring suitable high-temperature conditions, not separation by boiling.
Draw structural formulae for monomers and repeating units
Write symbol equations for addition polymerisation using 'n'
For polyethene, change CH₂=CH₂ into the repeating unit [–CH₂–CH₂–]ₙ. Draw brackets, an n outside and continuation bonds through both sides; n means many identical monomer molecules join without losing small molecules.
For polypropene, CH₂=CH–CH₃ gives [–CH₂–CH(CH₃)–]ₙ; chloroethene CH₂=CHCl gives PVC [–CH₂–CHCl–]ₙ; tetrafluoroethene CF₂=CF₂ gives PTFE [–CF₂–CF₂–]ₙ. Side groups stay attached while the former double bond becomes a backbone single bond.
Evaluate the challenges associated with the recycling process
Many plastics persist in landfill or fragment into microplastics in habitats; burning waste can release pollutants and carbon dioxide unless appropriately managed. Reuse extends product life and recycling can reduce new material demand, but neither makes waste impact-free.
Sorting mixed polymers, removing food/additives, degradation during reprocessing and the energy/cost of collection make recycling difficult. Microplastic effects remain under research; uncertainty is not proof of either no harm or universal harm.
Write word equations for hydrocarbon and hydrogen combustion
Write balanced symbol equations for hydrocarbon and hydrogen combustion
Experimentally determine energy released by burning fuels
Balanced examples: CH₄ + 2O₂ → CO₂ + 2H₂O; 2H₂ + O₂ → 2H₂O. Hydrogen contains no carbon, so its combustion itself produces no carbon dioxide; production and transport can still cause emissions.
Supervised calorimetry: measure the fuel-burner mass before and after heating a measured mass of water, and record the water's temperature rise. Keep water quantity, burner distance and apparatus consistent, repeat and compare energy per gram burned. Follow teacher-approved fire and hot-equipment precautions.
Constructed result using the supplied equation Q = mcΔT and supplied c = 4.2 J/(g °C): 100 g water warmed by 12°C absorbs 100 × 4.2 × 12 = 5040 J. If 0.50 g fuel burned, the calculator value for measured energy per gram is 5040/0.50 = 10 080 J/g = 10.08 kJ/g. To two significant figures this is 10 kJ/g. Heat loss and incomplete combustion make this an underestimate. The equation and water's specific heat capacity are used when supplied, not required recall.
Calculate the energy released by different fuels
Evaluate the pros and cons of hydrogen and bioethanol
Calculate overall energy change using bond energies: (Sum of energy to break bonds) - (Sum of energy to form bonds).
Identify reaction type: Negative result = exothermic, Positive result = endothermic.
Constructed paper example using supplied average bond energies in kJ/mol: H–H = 436, Cl–Cl = 243, H–Cl = 431. For H₂ + Cl₂ → 2HCl, break one H–H and one Cl–Cl and form two H–Cl bonds. Change = (436 + 243) − 2 × 431 = −183 kJ per mole of reaction as written, so exothermic. This is a data calculation, not an instruction to perform the reaction.
Missing-bond calculation using a supplied equation form: for that reaction with change −183 kJ/mol, D(H–Cl) = [436 + 243 − (−183)] ÷ 2 = 431 kJ/mol. Two product bonds must be counted; average bond energies estimate gaseous reaction changes.
Explain everyday heating/cooling uses through energy transfer, and distinguish an energy change from proof that new chemical substances have formed.
In plain language
Hydrocarbons contain only carbon and hydrogen. The simple alkanes here have only single bonds; simple alkenes have a carbon–carbon double bond and undergo addition reactions. In addition, the double bond becomes a single bond while atoms are added across it: the whole carbon chain does not split apart. Crude oil is a finite mixture made of long and short chains of carbon and hydrogen atoms called hydrocarbons. We use fractional distillation to separate this mixture because different sized molecules boil at different temperatures. Cracking converts surplus longer-chain hydrocarbons into smaller molecules in greater demand, including fuel molecules and alkenes for making polymers. Longer chains are not universally useless: some are valuable lubricants and other products. Plastics contain large polymer molecules made by joining many small monomers. Bonding, structure and additives influence properties, making different materials useful for packaging, piping or coatings. Many common plastics persist in the environment and can fragment into microplastics. Many common plastics do not biodegrade readily. Reuse and recycling can reduce waste and new raw-material demand, but sorting, contamination, energy use and material quality affect what can be recovered. Complete combustion of a hydrocarbon in sufficient oxygen produces carbon dioxide and water, transferring energy to the surroundings. Incomplete combustion can produce carbon monoxide and soot. The fire triangle represents fuel, oxygen and enough heat to sustain combustion. Fuel choices involve pollution, measured energy output and life-cycle trade-offs. Acid rain, fuel calorimetry and alternative-fuel benefits and drawbacks are common content in this specification, not Higher-only additions. Breaking reactant bonds takes energy; forming product bonds releases energy. If more is released than taken in, the reaction transfers energy to its surroundings and is exothermic. If more is taken in, it is endothermic. Temperature measurements provide evidence, but heat losses and other energy transfers also affect them. Energy-releasing processes are used in heat packs and self-heating containers. Energy-absorbing processes are used in some instant cold packs; many of these cool by dissolving a substance, rather than by making new covalent molecules.
Self-check
What happens to the colour of bromine water when it is added to a sample of propane compared to a sample of propene? How does increasing the carbon chain length of a hydrocarbon affect its viscosity? Write a balanced equation for the cracking of decane (C10H22) into octane (C8H18) and ethene. Represent the addition polymerisation of ethene to polyethene using symbol equations with 'n'. Why is recycling plastics considered a strategy to conserve finite natural resources? What are the three components of the fire triangle and the products of hydrocarbon combustion? Name two alternative fuels and one key consideration when using them. If breaking bonds takes 400kJ and forming new bonds releases 500kJ, what is the energy change and is it exothermic or endothermic? Give one everyday use of an endothermic process.
Show the answer
Propane (an alkane) will show no colour change (stays orange-brown). Propene (an alkene) will cause the bromine water to turn colourless. Increasing the chain length increases the viscosity, making the substance thicker and less likely to flow easily. C10H22 -> C8H18 + C2H4 n CH2=CH2 -> -[-CH2-CH2-]-n Recycling allows existing plastic materials to be reprocessed into new products, reducing the need to extract and refine more crude oil, which is a finite resource. Fuel, heat and oxygen. Complete hydrocarbon combustion forms carbon dioxide and water; heat describes energy transfer, not another chemical product. Hydrogen combustion avoids carbon dioxide at use but production may emit it and storage is difficult. Bioethanol can be made from renewable crops but growing them uses land, and its combustion still releases carbon dioxide. Energy change = 400kJ - 500kJ = -100kJ. Since the result is negative, it is an exothermic reaction. An instant cold pack used to treat sports injuries.
View diagrams
Hydrocarbons: displayed structures
The first five straight-chain alkanes and the two simple alkenes are fully displayed.
Methane · CH₄
Ethane · C₂H₆
Propane · C₃H₈
Butane · C₄H₁₀
Pentane · C₅H₁₂
Ethene · C₂H₄
Propene · C₃H₆
Every H and each C-H, C-C or C=C bond is explicit. These examples focus on displayed structures.
Addition to alkenes: ethene and bromine
The C=C bond becomes a C-C bond; one Br atom bonds to each carbon.
Reactants
→
Product · Higher name
1,2-dibromoethane (Higher-tier name)
Both sides contain 2 C, 4 H and 2 Br. The product is fully displayed.
Addition polymers
Each alkene monomer C=C becomes a single bond in the chain. n monomers form one idealised chain.
Ethene → Polyethene
n monomers form one idealised chain
Monomer
Repeat unit · n monomers
Propene → Polypropene
n monomers form one idealised chain
Monomer
Repeat unit · n monomers
Chloroethene → Polychloroethene · PVC
n monomers form one idealised chain
Monomer
Repeat unit · n monomers
Tetrafluoroethene → PTFE
n monomers form one idealised chain
Monomer
Repeat unit · n monomers
The chain bonds cross the repeat-unit brackets; each side group is retained unchanged.
Fully displayed common structures and reactions. Trace every atom and bond; repeat-unit continuation bonds lead to the rest of the polymer chain. Product names are explanatory labels, not additional common-tier recall requirements.

Higher Supplement: Isomerism and Addition ProductsHigher
Scope: Higher · Unit 5 tier: Higher · Topic 5.4 (higher)
Open revision guidance
What to know
Isomers are compounds that share the same molecular formula but possess different structural formulae.
Addition reactions of alkenes with halogens (like bromine) involve breaking the double bond to add the halogen atoms across the carbons.
Example of naming an addition product: ethene + bromine forms 1,2-dibromoethane.
What to do in an assessment
Draw structural isomers for a given molecular formula (e.g., butane C4H10 can be drawn as a straight chain or branched).
Name products of addition reactions by identifying the position of halogen atoms on the carbon chain.
Construct balanced chemical equations for the addition of halogens to alkenes.
For C₄H₈, but-1-ene is CH₂=CH–CH₂–CH₃, but-2-ene is CH₃–CH=CH–CH₃, and 2-methylpropene is CH₂=C(CH₃)₂. Each has four carbons and eight hydrogens but different connectivity; rotating the same drawing does not make a structural isomer.
The three C₅H₁₂ chain isomers are pentane CH₃–CH₂–CH₂–CH₂–CH₃, 2-methylbutane CH₃–CH(CH₃)–CH₂–CH₃, and 2,2-dimethylpropane C(CH₃)₄. Check five carbons and twelve hydrogens in each.
The five open-chain C₅H₁₀ structural alkene isomers are pent-1-ene CH₂=CH–CH₂–CH₂–CH₃; pent-2-ene CH₃–CH=CH–CH₂–CH₃; 2-methylbut-1-ene CH₂=C(CH₃)–CH₂–CH₃; 3-methylbut-1-ene CH₂=CH–CH(CH₃)–CH₃; and 2-methylbut-2-ene CH₃–C(CH₃)=CH–CH₃. Each has one C=C, five carbons and ten hydrogens. E/Z stereoisomers and skeletal formulae are not required.
For a name, draw the numbered parent chain, place the double bond and branches, then fill remaining carbon valencies with hydrogen. To name a structure, use the longest chain containing C=C and the lowest appropriate double-bond position.
Adding bromine to propene gives CH₂Br–CHBr–CH₃, named 1,2-dibromopropane; propene + chlorine gives 1,2-dichloropropane. The two halogens attach across the former double bond, not by replacing existing hydrogens.
In plain language
Structural isomers have the same molecular formula but different atom connectivity, not merely a different drawing orientation. Higher includes chain and double-bond-position isomers up to five carbons. Adding a halogen across an alkene changes C=C to C–C and attaches one halogen to each carbon; it does not replace the existing hydrogen atoms.
Self-check
Draw two structural isomers for C4H10 and explain why they are considered isomers.
Show the answer
Butane (CH3CH2CH2CH3) and 2-methylpropane (CH3CH(CH3)CH3). They are isomers because they have the same molecular formula (C4H10) but different structural arrangements.
View diagrams
Hydrocarbon isomers
Connectivity differs between structures; every atom and bond is displayed.
Pentane · C₅H₁₂
2-methylbutane · C₅H₁₂
2,2-dimethylpropane · C₅H₁₂
Pent-1-ene · C₅H₁₀
Pent-2-ene · C₅H₁₀
2-methylbut-1-ene · C₅H₁₀
3-methylbut-1-ene · C₅H₁₀
2-methylbut-2-ene · C₅H₁₀
C₅H₁₂: all three open-chain isomers. C₅H₁₀: all five open-chain alkene constitutional isomers. Rings, E/Z and skeletal notation are not included.
Higher: all three C₅H₁₂ alkane and all five open-chain C₅H₁₀ alkene constitutional isomers. Compare carbon connectivity and double-bond position; no E/Z or cyclic isomers are required here.

Scientific Enquiry FrameworkCore
Scope: Core · shared-enquiry; core
Open revision guidance
What to know
The independent variable is changed; the dependent variable is measured; controlled variables are kept constant. A control experiment provides a comparison with the independent variable absent or fixed.
A hypothesis is a testable explanation which may be supported or contradicted. Experimental validity means that the method answers the question; a valid conclusion is supported by the results.
Range identifies the lowest and highest values tested. Resolution is the smallest change an instrument can distinguish. An anomaly is markedly different from the other results.
The true value is the result in ideal conditions; measurement error is the difference between a reading and that value. An inaccurate reading is far from the true value.
Accuracy, precision, repeatability and reproducibility describe different aspects of data quality; precise results are not necessarily accurate.
Random errors cause unpredictable variation; systematic errors consistently shift readings. Repeats reduce random error but not systematic error.
Hazards, risks and control measures must be distinguished. Consider perceived risk alongside evidence about likelihood and consequences.
Scientific models have limitations, develop with evidence and are checked through peer review and clear communication.
Primary data are observations or measurements you collect; secondary data were collected by someone else. Evaluate a secondary source's methods and evidence, not just whether its conclusion agrees with yours. Check who produced it, when, its purpose and any relevant conflicts of interest.
Constructed comparison, not a published experiment: one group measures oxygen production rates of 0.8 and 1.2 cm³ per minute at two light settings. A second group's recorded rates at those settings are 0.9 and 1.3 cm³ per minute. Both show an increase of 0.4 cm³ per minute, but the second group's values are each 0.1 higher. Check plant size, temperature, carbon dioxide, apparatus calibration and measurement uncertainty before deciding whether the results disagree. Do not combine differently controlled experiments into a mean without justification.
Science can test predictions and estimate consequences, but cannot by itself decide which ethical priorities society should choose. For example, evidence about a fuel's emissions and cost informs a decision, while fairness and acceptable environmental harm also involve values.
What to do in an assessment
Turn a scientific explanation into a testable hypothesis and identify the independent, dependent and controlled variables.
Explain why the apparatus, sampling method, range, repeats and control experiment suit the question.
Record observations with units; translate between tables, calculations, diagrams and graphs.
Use evidence to draw a valid conclusion, discuss uncertainty and explain any anomalies.
Evaluate personal, social, economic and environmental implications, practical risks and ethical limitations; communicate a reasoned decision.
Explain a specific method improvement and a follow-up investigation rather than simply saying 'make it more accurate'.
For secondary data, record the original source and date, identify how the measurements were obtained, and check units, sample size, conditions and uncertainty before comparing datasets.
Choose a presentation suited to the question: a table preserves numerical readings, a graph shows relationships, and a labelled diagram explains an apparatus or structure. Describe the spread of repeated readings, not just their mean.
In plain language
A scientific enquiry tests an explanation against evidence. Start with a question, use scientific knowledge to predict what should happen, and choose a method that can answer that question. For example, predict that increasing light intensity increases a pondweed's photosynthesis rate. Change light intensity, measure oxygen produced in a fixed time and keep plant size, temperature and carbon dioxide supply constant. A control experiment, such as a comparison in darkness with other conditions unchanged, helps show whether light caused the difference. Choose apparatus with suitable resolution: a ruler marked every millimetre cannot distinguish changes smaller than one millimetre. Check that equipment is set up correctly and calibrated. Use representative samples rather than choosing only convenient organisms or results. Record units, the range of values tested and repeats in a table before plotting or calculating. An unusual reading is an anomaly: investigate and explain it rather than silently deleting it. Precision means little spread between readings; accuracy means closeness to the true value. Readings of 21.0, 21.1 and 21.0 degrees Celsius are precise, but inaccurate if a calibrated reference reads 19.0. A systematic offset remains when measurements are repeated; correct or replace the faulty apparatus. Random variation can be reduced by repeats and a mean. Repeatability concerns the same learner or group repeating the method; reproducibility concerns other learners or groups obtaining similar results. Uncertainty expresses the interval in which the true value is expected, such as 20 ± 2 degrees Celsius. Use the uncertainty formula supplied in an assessment rather than assuming a formula is always required from memory. A valid conclusion is supported by the collected evidence and a method suited to the question. Explain patterns, compare primary measurements with trustworthy secondary data and distinguish an association from proof of a cause. Suggest a specific improvement and say why it would help, for example collecting oxygen volume rather than counting bubbles of unequal size. Communicate the question, method, results and limitations so others can check the work. Peer review means other scientists critically examine methods and conclusions; new evidence can lead to changes in models and theories. Evaluate practical and wider risks using both likelihood and consequences. A hazard is something capable of harm; risk concerns the step where harm could occur; a control measure reduces that risk. For example, an irritant solution is a hazard, splashing during transfer is a risk, and eye protection and a supervised safe transfer method are controls. Follow your teacher's risk assessment; this guide is not permission to perform experiments at home. Scientific evidence can inform choices about health, costs and environmental effects, but ethical questions and people's values also matter.
Self-check
How do random and systematic errors differ, and why will repeating measurements not correct a systematic error?
Show the answer
Random errors vary unpredictably; repeats and a mean reduce their effect. A systematic error consistently shifts readings, for example a balance with a zero error. Repeating that faulty measurement does not remove the offset.
Essential Mathematical Skills in ScienceCore
Scope: Core · shared-maths; core
Open revision guidance
What to know
Decimal numbers, SI prefixes, unit conversions, ratios, fractions, percentages, significant figures, arithmetic means, probability and estimates are tools for interpreting scientific evidence.
A table or graph must communicate quantities and units, not only a pattern.
Substitution requires consistent units; a physical answer includes its unit.
For a square of side a: area = a². For a rectangle: area = length × width. For a triangle: area = base × perpendicular height/2.
For a circle: area = πr² and circumference = 2πr. For a cube: volume = a³ and surface area = 6a².
Recognise quantities and their units: length m, mass kg, time s, current A, speed m/s, force N, energy J, power W, voltage V and resistance Ω. A prefix changes the size of the unit: 1 kW = 1000 W and 1 mA = 0.001 A.
Because 1 cm = 0.01 m, 1 cm² = 0.0001 m² and 1 cm³ = 0.000001 m³. Do not use the length conversion factor unchanged for an area or volume.
The specification says the Unit 2 and Unit 5 papers include a periodic table on their final page, and the Unit 3 and Unit 6 papers include equations on page 2. Practise selecting and using the relevant information. This does not mean every equation used in every science paper is supplied, or promise an additional 2028 formula aid.
What to do in an assessment
Convert SI units and prefixes, including the different factors for lengths, areas and volumes.
Calculate ratios, fractions, percentages, probabilities and arithmetic means; check decimal answers against an order-of-magnitude estimate.
Construct and interpret tables, diagrams and experimental graphs, identifying axes, units, scales and patterns.
Substitute numerical values into equations with appropriate units and round final answers sensibly.
Calculate areas, circumference, cube surface area and cube volume; use angular measures in degrees.
In plain language
Use numbers with their units. Kilo means 1000 times the base unit, centi means one hundredth, milli means one thousandth and micro means one millionth. Thus 2500 milligrams = 2.5 grams and 3 centimetres = 0.03 metres. Area and volume conversions require squaring or cubing the length conversion: 1 cubic centimetre = 0.000001 cubic metres. Keep extra digits during a calculation and round the final answer to a justified number of significant figures; 0.004567 to two significant figures is 0.0046. Ratios compare quantities in the same units: 6 red beads to 3 blue beads gives 2:1. Red beads are 6/9 = 2/3 of the sample, or about 66.7%. If one bead is selected at random, its probability of being blue is 3/9 = 1/3. The mean of readings 8, 10 and 12 is (8 + 10 + 12)/3 = 10. An order-of-magnitude estimate uses rounded values: 198 × 51 is approximately 200 × 50 = 10000, so an answer near 100 is implausible. Record experimental data in a table with a heading and unit for each quantity. Plot the independent variable on the horizontal axis and the dependent variable on the vertical axis, using a sensible scale. For the pairs (time in seconds, distance in metres) (0,0), (2,4), (4,8), points lie on a rising straight line: at 3 seconds the distance is 6 metres. Read scales carefully and use a suitable best-fit line or curve rather than joining points arbitrarily. Translate a graph back into numerical values before substituting them into an equation. Substitute with consistent units: speed = distance/time = 12 metres/3 seconds = 4 metres per second. State the physical meaning of the answer, not just a number. For geometry, a rectangle's area is length × width; a triangle's area is base × perpendicular height/2; a cube of side a has volume a³ and surface area 6a². A 2 centimetre cube therefore has volume 8 cubic centimetres and surface area 24 square centimetres. A circle of radius 3 centimetres has area πr² = 9π square centimetres and circumference 2πr = 6π centimetres. Use angles in degrees and measure them from the stated reference line, such as the normal in a reflection diagram.
Self-check
A rectangle is 5 centimetres long and 4 centimetres wide. What is its area, including its unit?
Show the answer
Area = length × width = 5 × 4 = 20 square centimetres. The unit is cm² because two lengths are multiplied.
Higher Tier Mathematical SkillsHigher
Scope: Higher · Higher in either Unit 1 tier or Unit 2 tier or Unit 3 tier or Unit 4 tier or Unit 5 tier or Unit 6 tier · shared-maths; higher
Open revision guidance
What to know
Standard form is A × 10ⁿ with 1 ≤ A < 10.
Changing the subject preserves equality by doing the same operation on both sides.
The equation y = mx + c describes a straight line: m is its gradient and c its vertical-axis intercept.
A gradient and intercept have meanings and units determined by the plotted quantities.
What to do in an assessment
Express large and small scientific quantities in standard form.
Rearrange an equation step by step and check the answer by substitution.
Higher circuit example, substituting before changing the subject: given I = V/R, I = 0.20 A and V = 6.0 V, write 0.20 = 6.0/R. Multiply both sides by R to give 0.20R = 6.0, then divide by 0.20: R = 30 Ω. Check that 6.0/30 = 0.20 A.
Calculate a linear graph's gradient using change in vertical quantity divided by change in horizontal quantity.
Interpret slope and intercept in the physical context, with units; distinguish an instantaneous tangent gradient from a mean rate.
In plain language
Standard form writes a number as A × 10ⁿ, with 1 ≤ A < 10. For example, 450000 = 4.5 × 10⁵ and 0.00032 = 3.2 × 10⁻⁴. The exponent tells you how many places the decimal point moves; a negative exponent denotes a small number. Change the subject of an equation by applying the same operation to both sides. From v = d/t, multiply both sides by t to get vt = d, then divide by v to obtain t = d/v. If d = 24 metres and v = 6 metres per second, t = 24/6 = 4 seconds. Check by substituting back: 6 = 24/4. For a straight line y = mx + c, m is the gradient and c the vertical-axis intercept. Choose two widely separated points on the best-fit line, not necessarily two measured points. For (2,7) and (6,19), gradient = (19 − 7)/(6 − 2) = 3, and c = 7 − 3 × 2 = 1, so y = 3x + 1. If y is distance in metres and x is time in seconds, the slope means 3 metres per second and the intercept means a starting distance of 1 metre. For reaction data, a tangent's gradient describes the rate at one instant; choose two points on that tangent. Do not confuse this with a mean rate over the whole experiment.
Self-check
In y = 3x + 2, what is the vertical-axis intercept, and how can you check it?
Show the answer
The intercept is 2. Substitute x = 0: y = 3 × 0 + 2 = 2.

Unit 7 Scientific Enquiry AssessmentPractical assessment guidanceCore
Scope: Core · unit-7; core
Reading this guidance does not complete practical assessment. It is not counted as written-exam progress.
Open revision guidance
What to know
Two enquiries are chosen from three; each has a one-hour practical task and one-hour written task.
Task 1: collaborative practical work, 6 marks per enquiry, groups of at most three learners.
Task 2: independent written work under high control, 22 marks per enquiry.
This external assessment is marked by WJEC, totals 56 marks and contributes 10%; it is not counted in written-exam checklist progress.
Unit 7 is not tiered. The Foundation/Higher revision toggle does not change its assessment arrangements or register you for an exam.
Your centre chooses the enquiries and manages the official assessment arrangements. Use only the materials and assistance permitted by your teacher; never submit another learner's analysis as your own.
What to do in an assessment
Use suitable apparatus, the approved safe method and clear data recording during Task 1.
Use Task 1 evidence to produce your own analysis and evaluation in Task 2.
Apply the scientific enquiry framework and terminology to explain the quality and limitations of the evidence.
In plain language
Unit 7 is an external Scientific Enquiry assessment marked by WJEC, not another topic to tick as completed written-exam revision. Complete two enquiries from the three available. Each has a one-hour practical task worth 6 marks and a one-hour independent written task worth 22 marks: 28 marks per enquiry and 56 altogether. The unit contributes 10% of the qualification and is first available in spring 2028. For the practical task, work in a group of no more than three learners and record your own observations and measurements. The subsequent written task is completed independently under high control: use the evidence to analyse, explain and evaluate, without help from your group. Your teacher organises the assessment and its controls. Practise the enquiry skills above, but do not treat a tick in this guide as proof that the assessed tasks have been completed.
Self-check
How many marks are available across both enquiries, and may the group produce a shared answer for Task 2?
Show the answer
Each enquiry has 6 practical + 22 written = 28 marks. Two enquiries give 2 × 28 = 56 marks. Task 2 is independent, not a group answer.

Previous revision ticks

These are your ticks from the earlier guide. They do not count towards the new lessons, and the old content is not current teaching.

Earlier revision tick: Cells and transport
Cells and transport
Earlier revision tick: Active Transport Supplement
Active Transport Supplement
Earlier revision tick: Enzymes, Respiration, and Photosynthesis
Enzymes, Respiration, and Photosynthesis
Earlier revision tick: Advanced Energy Systems and Limiting Factors
Advanced Energy Systems and Limiting Factors
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Plant Transport Systems
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Mineral uptake
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Energy Transfer and Ecosystem Interactions
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Atomic Structure and Chemical Calculations
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Moles and Avogadro's Number
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Structure and Properties of Elements
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Trends in Group Reactivity
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Chemical Reactions, Rates, and Catalysts
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Higher Supplement: Rate Calculation from Graphs
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Earth's Resources and Atmospheric Change
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Higher Concepts: Climate Feedback Loops
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Motion, Graphs, and Stopping Distances
Earlier revision tick: Advanced Motion Graphs (Higher)
Advanced Motion Graphs (Higher)
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Energy Principles, Efficiency, and Power
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Forces, Newton's Laws, and Terminal Speed
Earlier revision tick: Advanced Applications of Newton's Laws
Advanced Applications of Newton's Laws
Earlier revision tick: Wave Properties and the Electromagnetic Spectrum
Wave Properties and the Electromagnetic Spectrum
Earlier revision tick: The Solar System, Observations, and Star Life Cycles
The Solar System, Observations, and Star Life Cycles
Earlier revision tick: Higher Physics: Stellar Equilibrium Dynamics
Higher Physics: Stellar Equilibrium Dynamics
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Human Impact on Ecosystems
Earlier revision tick: DNA Structure and Function
DNA Structure and Function
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Protein Synthesis and Control
Earlier revision tick: Variation, Natural Selection, and Gene Technology
Variation, Natural Selection, and Gene Technology
Earlier revision tick: Higher Concepts in Variation and Ethics
Higher Concepts in Variation and Ethics
Earlier revision tick: The Nervous System and Reflex Actions
The Nervous System and Reflex Actions
Earlier revision tick: Higher Reflex Arc Mechanisms
Higher Reflex Arc Mechanisms
Earlier revision tick: Disease and Human Defence Systems
Disease and Human Defence Systems
Earlier revision tick: Higher Immunity: Memory Cells
Higher Immunity: Memory Cells
Earlier revision tick: Bonding, Structure, and Properties
Bonding, Structure, and Properties
Earlier revision tick: Advanced Covalent Bonding
Advanced Covalent Bonding
Earlier revision tick: Acids, Bases, and Salt Identification
Acids, Bases, and Salt Identification
Earlier revision tick: Higher Level Identification of Metal Ions
Higher Level Identification of Metal Ions
Earlier revision tick: Reactivity of Metals and Extraction Principles
Reactivity of Metals and Extraction Principles
Earlier revision tick: Advanced Redox and Calculations
Advanced Redox and Calculations
Earlier revision tick: Introduction to Alkanes and Alkenes
Introduction to Alkanes and Alkenes
Earlier revision tick: Higher Supplement: Isomerism and Addition Products
Higher Supplement: Isomerism and Addition Products
Earlier revision tick: Fundamentals of Radioactive Decay and Half-life
Fundamentals of Radioactive Decay and Half-life
Earlier revision tick: Advanced Decay Equations and Kinetics
Advanced Decay Equations and Kinetics
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Wave Behaviour and Medical Imaging
Earlier revision tick: Electrical Circuits and Ohm's Law
Electrical Circuits and Ohm's Law
Earlier revision tick: Work Done and Energy Transfer
Work Done and Energy Transfer
Earlier revision tick: Gravitational and Kinetic Energy Calculations
Gravitational and Kinetic Energy Calculations
Earlier revision tick: Magnetic Fields, Electromagnetism, and Induction
Magnetic Fields, Electromagnetism, and Induction
Earlier revision tick: Higher Principles: Fleming's Rules and Transformer Operation
Higher Principles: Fleming's Rules and Transformer Operation
Earlier revision tick: Evidence for the Big Bang Model
Evidence for the Big Bang Model
Earlier revision tick: Cosmological Redshift and Universal Expansion
Cosmological Redshift and Universal Expansion
Earlier revision tick: Scientific Enquiry Framework
Scientific Enquiry Framework
Earlier revision tick: Essential Mathematical Skills in Science
Essential Mathematical Skills in Science
Earlier revision tick: Higher Tier Mathematical Skills
Higher Tier Mathematical Skills

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