WJEC GCSE Chemistry Revision

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WJEC · GCSE Chemistry3410QSExam years · 2027

Assessment For 2027, two 1 hour 45 minutes written papers each contribute 45%, with Foundation or Higher chosen separately for each unit, and an untiered 30-mark practical assessment contributes 10%.

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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 paper tier

Foundation tier
Higher tier

Summary of assessment, PDF p. 3: Unit 1 tiered assessment; entry rules, PDF p. 31: different unit tiers may exceptionally be combined.

Unit 2 paper tier

Foundation tier
Higher tier

Summary of assessment, PDF p. 3: Unit 2 tiered assessment; entry rules, PDF p. 31: different unit tiers may exceptionally be combined.

Elements, compounds, mixtures, formulae, separation and chemical reactionsCore
Scope: Core · Unit 1, Topic 1.1, points (a)–(o): particle and substance definitions; symbols, formulae and simple molecule diagrams; ionic formulae; relative masses, percentage composition and yield; mixture separation and Rf; conservation, reaction evidence, word and balanced symbol equations.
Open revision guidance
What to know
An element contains one type of atom and cannot be chemically broken into simpler substances. A compound contains two or more different elements chemically joined in a fixed ratio; its properties differ from those of its constituent elements. In a mixture, particles are together but not chemically joined, so physical methods can separate them.
Use element symbols and subscripts to show atom counts: H₂O has two H atoms for every O atom. A simple molecule diagram needs a key. Ionic formulae are electrically neutral overall: Mg²⁺ with two Cl⁻ ions gives MgCl₂.
Relative molecular mass, Mr, is the sum of the relative atomic masses in a molecular formula; for an ionic substance, relative formula mass is found in the same way. Percentage by mass of an element = (mass contributed by that element in the formula ÷ Mr) × 100. Percentage yield = (actual yield ÷ theoretical yield) × 100.
Filtration separates an insoluble solid from a liquid; evaporation/crystallisation recovers a dissolved solid; distillation separates a solvent or liquids with different boiling points; chromatography separates dissolved substances by their different movement between mobile and stationary phases. Rf = distance moved by substance ÷ distance moved by solvent front.
A chemical reaction rearranges atoms: the total number of each kind of atom is conserved. Colour or temperature change and effervescence can be evidence, but observations should be interpreted in context. A balanced equation has equal numbers of each atom on both sides and conserves total relative mass.
In a molecule model, count each kind of atom to obtain the formula: two hydrogen atoms joined to one oxygen atom represent H₂O, while one carbon and two oxygen atoms represent CO₂. Different unjoined types of particle in one container represent a mixture rather than a single compound.
What to do in an assessment
For MgCl₂, combine one Mg²⁺ with two Cl⁻ ions so the net charge is zero; do not change ion charges to balance a formula. For a percentage-composition calculation in CO₂, Mr = 12 + (2 × 16) = 44, so carbon by mass is (12 ÷ 44) × 100 = 27.3%.
For a chromatography result where a spot moves 3.0 cm and the solvent front moves 7.5 cm, Rf = 3.0 ÷ 7.5 = 0.40 (no units). Keep the origin above the solvent and compare Rf values only under the same conditions.
Balance by changing coefficients, never subscripts: Mg + O₂ → MgO becomes 2Mg + O₂ → 2MgO. Check each element count after balancing. If actual product is 8.0 g and the theoretical yield is 10.0 g, percentage yield = (8.0 ÷ 10.0) × 100 = 80%.
For a simple reaction, conserve every element and count atoms before and after. A neutralisation is exothermic: acid + base makes salt and water and transfers energy to the surroundings; not every reaction releases heat.
In plain language
A formula tells you which atoms are present and in what ratio; a balanced equation shows those atoms being rearranged without disappearing.
Self-check
A dye spot travels 2.4 cm while the solvent front travels 6.0 cm. Find its Rf. Then balance H₂ + O₂ → H₂O.
Show the answer
Rf = 2.4 ÷ 6.0 = 0.40. The balanced equation is 2H₂ + O₂ → 2H₂O; both sides contain four H atoms and two O atoms.
View diagrams
Molecules and a mixture
The mixture contains separate H₂ and O₂ molecules; they are not bonded to each other.
Read an atomic model to distinguish a compound from a mixture and work out the formula from the number of each kind of atom.
Higher tier: mole, formula from reacting masses and equation-mass calculationsCore
Scope: Core · Unit 1 paper tier: any of Higher tier · Unit 1, Topic 1.1, points (p)–(r): calculate a compound formula from reacting masses; calculate reactant or product masses from a balanced equation; Avogadro constant, mole and conversions between grams and moles; separate bold Working Scientifically fragment on appropriate significant figures; separate bold Mathematical Skills fragment on rearranging equations in mole calculations. The three point outcomes and the two mathematical fragments are covered without duplicating source-leaf counts.
Open revision guidance
What to know
One mole contains the Avogadro constant number of specified particles. Use n = m/Mr for amount in moles, m = nMr for mass in grams, and N = nNA for particle number. Use the Avogadro constant value supplied or specified in the question.
A balanced chemical equation gives particle and mole ratios, not gram ratios. Convert a known mass to moles, use the coefficient ratio, then convert the required amount back to mass.
To determine a formula from reacting masses, convert each element's mass to relative moles by dividing by its Ar, then divide every amount by the smallest to obtain the simplest whole-number ratio. Use experimental reacting-mass data as instructed; do not confuse an empirical ratio with a molecular formula.
For this Unit 1.1 Higher-tier mole/calculation context, rearrange equations to isolate the unknown and round calculation answers to an appropriate number of significant figures. The same named significant-figure and equation-rearrangement skills also have one separate qualification-wide Appendix C row, gated by either unit being Higher.
What to do in an assessment
Worked mole example: Mr(NaCl) = 58.5, so 5.85 g contains n = 5.85 ÷ 58.5 = 0.100 mol. Conversely, 0.250 mol of CO₂ has mass m = 0.250 × 44.0 = 11.0 g.
Worked equation-mass example: 2Mg + O₂ → 2MgO. If 4.8 g Mg reacts, n(Mg) = 4.8 ÷ 24.0 = 0.200 mol. The 2:2 coefficient ratio gives 0.200 mol MgO; with Mr(MgO) = 40.0, mass MgO = 0.200 × 40.0 = 8.0 g.
For a formula-from-masses example, 2.4 g Mg and 1.6 g O give 2.4 ÷ 24 = 0.100 mol Mg and 1.6 ÷ 16 = 0.100 mol O, a 1:1 ratio, hence MgO. Rearrange equations carefully and keep units attached.
Significant-figure example: 4.8 ÷ 24 = 0.20 mol Mg to 2 significant figures. With measurements 4.80 g ÷ 24.0 g mol⁻¹, report 0.200 mol to 3 significant figures; retain the trailing zero to show the stated precision. For c = n/V, multiply both sides by V to make n the subject (n = cV), then substitute with compatible units.
In plain language
Use moles to compare numbers of particles. A balanced equation tells you the mole ratio; relative formula mass then links moles back to grams.
Self-check
How many moles are in 5.85 g NaCl if Mr = 58.5? In 2Mg + O₂ → 2MgO, what mass of MgO forms from 2.4 g Mg if Ar(Mg) = 24 and Mr(MgO) = 40?
Show the answer
n(NaCl) = 5.85 ÷ 58.5 = 0.100 mol. For Mg, n = 2.4 ÷ 24 = 0.100 mol; the 2:2 ratio gives 0.100 mol MgO, so m = 0.100 × 40 = 4.0 g.

Atomic structure, periodic trends, reactions and ion testsCore
Scope: Core · Unit 1, Topic 1.2, common points (a)–(f), (h)–(m), (o)–(q), (s)–(u), and specified practical: identify unknown ionic compounds by flame tests and chemical tests for ions.
Open revision guidance
What to know
An atom has a tiny positive nucleus of protons and neutrons with negatively charged electrons around it. Relative charges are +1, 0 and −1; relative masses are about 1, 1 and 1/1836. Atomic number is proton number; mass number is proton plus neutron number; isotopes have the same proton number but different neutron numbers. A neutral atom has equal protons and electrons.
For a neutral atom, electrons equal atomic number; neutrons = mass number − atomic number. The first 20 elements use shell structures filled 2, then 8, then 8, then 2. The Periodic Table is ordered by increasing atomic number, in groups and periods; metals are mainly left/centre, non-metals right, with metalloids between.
Order-of-magnitude comparisons use powers of ten: if an atom is about 1 × 10⁻¹⁰ m across and its nucleus about 1 × 10⁻¹⁵ m, the atom is about 10⁵ times wider (100,000 times). Compare the powers by dividing the values, not by subtracting the exponents without checking direction.
For the main groups, the outer-shell electron arrangement links an element to its group; the number of occupied shells gives its period. Group 1 metals form +1 ions and Group 7 halogens form −1 ions; electron loss or gain makes charged ions. Members of each group share characteristic properties: Group 1 metals become more vigorous in reactions with water down the group; Group 7 elements are coloured diatomic non-metals, with darker colour and higher melting/boiling points down the group.
Group 0 gases are very unreactive because their outer shell is full. Helium is used in balloons, neon in signs and argon in lamps or welding atmospheres; relate each use to a relevant property such as low density or low reactivity.
Alkali metals react with oxygen to form oxides, with water to form hydroxide and hydrogen, and with halogens to form ionic halides. Halogens also react with metals such as iron; chlorine and iodine have characteristic properties and uses, including chlorine for water disinfection and iodine as an antiseptic.
Hydrogen gives a squeaky pop with a lit splint. Flame colours help identify Li⁺ (crimson), Na⁺ (yellow), K⁺ (lilac), Ca²⁺ (brick red) and Ba²⁺ (apple green). Acidified silver nitrate identifies halides: chloride gives white AgCl, bromide cream AgBr and iodide yellow AgI precipitate.
Atoms contain a small central nucleus with protons and neutrons, with electrons outside it; this nuclear model replaced the earlier plum-pudding model after evidence from scattering. The atom is mostly empty space, while nearly all its mass is in the nucleus.
What to do in an assessment
For ²³₁₁Na, proton number is 11, neutron number is 23 − 11 = 12 and a neutral atom has 11 electrons. For Na⁺, one electron has been lost, so there are 10 electrons; the nucleus has not changed.
Examples of common reactions are 4Li + O₂ → 2Li₂O, 2Na + 2H₂O → 2NaOH + H₂ and 2Na + Cl₂ → 2NaCl. Iron reacts with chlorine to form iron(III) chloride: 2Fe + 3Cl₂ → 2FeCl₃. Count atoms on each side when writing a balanced equation.
Practical identification: label separate clean samples, use a clean wire loop for each flame test, and record the observed colour before comparing with a reference. For a halide test, dissolve a small sample, acidify with dilute nitric acid, then add silver nitrate and record the precipitate colour. Use separate aliquots and clean equipment to avoid cross-contamination.
Use the observation table to identify ions, not colour alone where two tests are available. Wear eye protection; handle hot loops and silver nitrate carefully, and dispose of chemical waste as directed.
In plain language
The number of protons identifies the element; electrons explain charge and chemical behaviour, while careful tests turn observable colours into evidence for an unknown ion.
Self-check
An atom has atomic number 17 and mass number 35. State its protons, neutrons and electrons if it forms Cl⁻. What precipitate does chloride make with acidified silver nitrate?
Show the answer
It has 17 protons and 35 − 17 = 18 neutrons. Cl⁻ has gained one electron, so it has 18 electrons. Chloride gives a white silver chloride precipitate, AgCl.
Higher tier: isotope abundance and reactivity trendsCore
Scope: Core · Unit 1 paper tier: any of Higher tier · Unit 1, Topic 1.2, points (g), (n) and (r): calculate relative atomic mass from isotopes; trends in Group 1 and Group 7 reactivity in terms of readiness to lose or gain an electron; chlorine, bromine and iodine displacement reactions. These points are bold/Higher tier only.
Open revision guidance
What to know
Relative atomic mass is the abundance-weighted mean of isotope masses: multiply each isotope mass by its percentage abundance, add the products and divide by 100. The percentages must total 100.
Group 1 reactivity increases down the group because the outer electron is farther from the nucleus and more shielded, so it is lost more readily. Group 7 reactivity decreases down the group because an incoming electron is farther from the nucleus and more shielded, so it is gained less readily.
In a halogen displacement reaction, a more reactive halogen displaces a less reactive halide from solution. Chlorine is more reactive than bromine, which is more reactive than iodine. These readiness-to-lose/gain explanations and outcomes are Higher-only here.
What to do in an assessment
Worked isotope example: an element has 75% isotope mass 35 and 25% isotope mass 37. Relative atomic mass = ((75 × 35) + (25 × 37)) ÷ 100 = 35.5.
In chlorine water plus potassium bromide, chlorine displaces bromine: Cl₂ + 2KBr → 2KCl + Br₂. Chlorine atoms gain electrons to form chloride while bromide ions lose electrons to form bromine. Balance atoms and charges in ionic equations if requested.
Use comparable observations from a displacement series to infer a reactivity order; a reaction occurs only when the added halogen is more reactive than the halogen in the salt.
In plain language
An isotope average depends on how common each isotope is. In a group trend, explain the changing ease of electron loss or gain, not just the observed order.
Self-check
An element is 20% isotope 10 and 80% isotope 11. Calculate its relative atomic mass. Will bromine displace chloride from potassium chloride?
Show the answer
Ar = ((20 × 10) + (80 × 11)) ÷ 100 = 10.8. No: bromine is less reactive than chlorine, so it cannot displace chloride.

Water supplies, solubility, hardness and the soap practicalCore
Scope: Core · Unit 1 §1.3(a–j), (k) methods and their advantages/disadvantages, (l) health benefits and scale; specified soap-solution practical
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What to know
Natural water can contain dissolved gases and ions, microorganisms and pollutants. A public supply is treated by sedimentation, filtration and chlorination. Sustainable management also reduces consumption and the environmental effects of abstraction, distribution and treatment.
Fluoridation has arguments for and against: discuss tooth-decay prevention alongside consent, exposure, cost and environmental/social considerations, and make a decision from evidence. Desalination can provide drinking water by distilling seawater, but heating and large-scale infrastructure use energy and resources. Distillation separates miscible liquids by their different boiling points.
To make a solubility curve, use a known mass of solvent, add a measured solute quantity, warm until it dissolves, then cool and record the temperature at which crystals first appear; repeat across concentrations/temperatures and plot solubility against temperature. Express solubility as grams of solute per 100 g of water so results can be compared.
Hardness is caused mainly by dissolved calcium and magnesium ions. Hard water forms scum and needs more soap to make a lather; soft water lathers with less soap. Temporary hardness is associated with hydrogencarbonates and can be removed by boiling; permanent hardness is associated with other dissolved salts and is not removed by boiling.
Hard water contains dissolved calcium or magnesium ions. Calcium in drinking water supports healthy bones and teeth; hard water can supply useful minerals, although scale reduces the efficiency of boiler and heating elements and more soap is needed.
Water-softening methods include boiling temporary hard water, adding sodium carbonate, and passing water through an ion-exchange resin. Boiling cannot remove permanent hardness; sodium carbonate and ion exchange can remove both types.
What to do in an assessment
For a soap test, measure equal volumes of each water sample into labelled tubes. Add the same soap solution in small measured portions, stopper and shake in the same way after each addition, and record the volume or drops needed for a persistent lather. Use clean apparatus and repeat; keep sample volume, soap concentration, shaking and temperature the same.
For an unknown, compare its measured soap result with known standards or use a supplied calibration graph. Example using a question's hypothetical calibration: 6 drops corresponds to 100 mg dm⁻³ hardness and 10 drops to 200 mg dm⁻³; an unknown requiring 8 drops is halfway between, so estimate 150 mg dm⁻³. Interpolate only when the supplied graph/data justify it; otherwise report relative hardness (for example, harder than the reference).
Solubility calculation: if 18.0 g of solute dissolves in 50.0 g of water at a stated temperature, solubility = (18.0 ÷ 50.0) × 100 = 36.0 g per 100 g water. Use the mass of water, not the combined mass of solution, as the denominator.
For a solubility graph, label both axes with units, plot the measured pairs, draw a smooth curve through the trend, then read the required value at the specified temperature. Distinguish a point on the curve (saturated solution) from below it (unsaturated).
Compare boiling, sodium carbonate and ion exchange by the hardness they remove, energy use, cost, maintenance and waste. Basic methods and their advantages/disadvantages are common to both tiers; explanations of how the methods work and the removal equations are in the Higher supplement.
In plain language
Hard water contains dissolved ions that use up soap before lather appears. A fair comparison measures how much soap the same volume of each water needs under the same conditions.
Self-check
Two equal water samples need 12 drops and 5 drops of the same soap solution to form persistent lather. Which is harder? In a solubility curve, what does a point on the curve represent?
Show the answer
The sample needing 12 drops is harder. A point on the curve represents a saturated solution at that temperature: the stated amount is dissolved at its solubility limit.
Higher: explain water softeningCore
Scope: Core · Unit 1 paper tier: any of Higher tier · Unit 1 §1.3(k), bold explanation of how softening methods work, and Higher overview word/balanced equations
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What to know
Boiling decomposes soluble calcium hydrogencarbonate to insoluble calcium carbonate, carbon dioxide and water: Ca(HCO₃)₂ → CaCO₃ + CO₂ + H₂O. Removing the precipitate removes calcium ions; calcium sulfate does not decompose this way, so permanent hardness remains.
Added sodium carbonate supplies carbonate ions that precipitate dissolved calcium or magnesium ions as insoluble carbonates. Example: CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl. Remove the precipitate rather than claiming the hardness ions have vanished.
An ion-exchange resin replaces calcium/magnesium ions with sodium ions. Regeneration with concentrated sodium chloride restores the resin; compare salt use, regeneration and sodium content against boiling energy and carbonate precipitate disposal.
What to do in an assessment
Write word and balanced symbol equations for removing temporary hardness by boiling and for precipitation using sodium carbonate. Identify the insoluble product and explain why the remaining water is softer.
Explain ion exchange as replacement of dissolved ions, not filtering particles that were already present. Distinguish the mechanism from the common-tier comparison of advantages and disadvantages.
In plain language
Higher adds the particle explanation and equations behind the softening methods, not the basic knowledge that those methods exist.
Self-check
Why does boiling soften water containing calcium hydrogencarbonate but not calcium sulfate?
Show the answer
Heating hydrogencarbonate produces insoluble calcium carbonate, which can be removed. Calcium sulfate does not decompose this way, so its calcium ions remain dissolved.

Earth structure, atmosphere, geological change and standard formCore
Scope: Core · Unit 1, Topic 1.4, points (a)–(j): Earth structure and plate boundaries; continental drift/plate tectonics; atmosphere through geological time, cycles and human impacts; atmospheric gases and gas tests. Mathematical Skills: standard form for Earth's age and geological timescales, plus graphing trends.
Open revision guidance
What to know
Earth has a solid iron core, a molten iron outer core, mantle and crust. Wegener's continental-drift idea preceded plate tectonics; evidence and later understanding support moving plates. At constructive boundaries plates move apart, at destructive boundaries they move towards one another, and at conservative boundaries they slide past.
Volcanic gases, including carbon dioxide and water vapour, contributed to the early atmosphere. The present atmosphere is mostly nitrogen and oxygen, with argon and other gases in smaller amounts. Photosynthesis removes carbon dioxide and releases oxygen; respiration and combustion use oxygen and release carbon dioxide. These processes help maintain atmospheric levels, while their balance and geological change matter.
Burning fossil fuels releases carbon dioxide and sulfur dioxide. Carbon dioxide contributes to global warming; sulfur dioxide contributes to acid rain. Measures include reducing fossil-fuel use/emissions and treating or preventing pollutant releases. Evaluate impacts and limits in the context given rather than assuming one measure solves every effect.
For geological evidence, matching coastlines and rocks/fossils contributed to Wegener's continental-drift idea; later evidence such as seafloor spreading supported the plate-tectonic model.
Atmospheric history includes water vapour condensing as Earth cooled, carbon dioxide dissolving in oceans and becoming locked in carbonate rocks, and photosynthesis removing more carbon dioxide while adding oxygen. The present atmosphere is approximately 78% nitrogen, 21% oxygen and 0.9% argon, with other gases in trace amounts.
Reducing greenhouse-gas emissions through energy efficiency and low-carbon energy can limit global warming; low-sulfur fuels or flue-gas desulfurisation can reduce sulfur dioxide and acid rain. Compare costs, effectiveness and environmental trade-offs when evaluating a measure.
Air is a source of nitrogen, oxygen, neon and argon. Oxygen relights a glowing splint; carbon dioxide turns limewater milky.
What to do in an assessment
Use standard form for geological timescales: 4.6 × 10⁹ years means 4,600,000,000 years. A stated interval of 3.8 × 10⁹ years is 3,800,000,000 years; their difference is (4.6 − 3.8) × 10⁹ = 0.8 × 10⁹ = 8 × 10⁸ years. Keep the power of ten consistent when subtracting.
On an atmosphere-composition or geological-time graph, label axes and units, plot the given data, describe the direction/pattern and cite data points as evidence. A trend does not alone prove a single cause.
For a gas test, use a small sample with suitable apparatus: a glowing splint relights in oxygen; bubble carbon dioxide through limewater and observe it turn milky. Avoid inhaling gases and follow practical safety directions.
In plain language
Earth's rocks, moving plates and atmosphere change over immense timescales. Standard form keeps those very large geological numbers readable and easier to compare.
Self-check
Write 7.2 × 10⁸ years as an ordinary number. Which gas relights a glowing splint, and which turns limewater milky?
Show the answer
7.2 × 10⁸ = 720,000,000 years. Oxygen relights a glowing splint; carbon dioxide turns limewater milky.

Rates, collision theory, enzymes and rate investigationsCore
Scope: Core · Unit 1, Topic 1.5, common points (a)–(c), (d) catalyst definition/increase in rate/remaining unchanged, and (e)–(f); both specified practicals; common mathematics for means, tables, line graphs, slopes, cube areas and volumes.
Open revision guidance
What to know
Rate can be measured by gas volume produced, loss of mass, or formation of a precipitate (including data logging). A steeper product-versus-time graph means a faster rate. For a straight segment, rate = change in measured quantity ÷ change in time.
Higher temperature gives particles more kinetic energy and more frequent successful collisions; greater concentration (or gas pressure) places particles closer together, increasing collision frequency; greater surface area exposes more particles for collisions. These explain the rate changes in the common specification outcomes.
A catalyst increases reaction rate while remaining chemically unchanged. Enzymes are biological catalysts for particular reactions under particular conditions; uses span food manufacture, biological processes and industry—for example, amylase in starch processing, proteases in detergents and enzymes in food production.
The specified investigations are: vary a rate factor and collect gas; and time the reaction of dilute hydrochloric acid with sodium thiosulfate as a marked cross disappears beneath the cloudy mixture.
What to do in an assessment
Gas-collection method: use a sealed flask and gas syringe (or an appropriate displacement setup), start the timer promptly, record gas volume at regular intervals, and stop when the reaction ends. For a fair test change one factor at a time; keep reactant volumes/concentrations and temperature controlled unless they are the independent variable. Check for leaks and repeat.
Sodium thiosulfate method: put a measured volume/concentration of sodium thiosulfate in a flask over a black cross; add a measured volume of dilute hydrochloric acid, start timing and stop when the cross is no longer visible. Keep total volume, cross/viewing conditions and temperature consistent; vary the chosen concentration or temperature systematically. Use rate ∝ 1/time for comparable endpoints, repeat, and average concordant results. Wear eye protection, use small quantities in good ventilation, and do not inhale sulfur dioxide produced.
Record readings in a table with variable names, values and units in the headings; for example, time / s and gas volume / cm³. Plot the independent variable on the x-axis, dependent variable on the y-axis, use evenly spaced scales, label units, plot points accurately and draw a best-fit line or curve.
Worked mean: for gas volumes of 18, 20 and 22 cm³ after 20 s, mean volume = (18 + 20 + 22) ÷ 3 = 20 cm³ and mean rate = 20 ÷ 20 = 1.0 cm³ s⁻¹.
Worked straight-line gradient: if a linear section runs from (0 s, 0 cm³) to (30 s, 45 cm³), rate = (45 − 0) ÷ (30 − 0) = 1.5 cm³ s⁻¹. A section with a larger gradient represents a faster rate.
Cube comparison: a cube of side 1 cm has surface area 6 × 1² = 6 cm² and volume 1³ = 1 cm³, so SA:V = 6:1. A 3 cm cube has area 6 × 3² = 54 cm² and volume 3³ = 27 cm³, so SA:V = 2:1. Smaller pieces expose more surface area per unit volume, so generally react faster.
For the specified sodium thiosulfate/dilute hydrochloric acid practical, formation of sulfur makes the cross disappear. Keep solution depth, total volume and temperature controlled when varying concentration, use 1/time to compare rate, and work with ventilation because sulfur dioxide is produced.
In plain language
A reaction is faster when more useful collisions happen each second. Measure a changing amount against time, and make a fair comparison by changing only the factor you are testing.
Self-check
A reaction makes 30 cm³ gas in 15 s. Find its mean rate. Which cube has more surface area per unit volume: side 1 cm or side 3 cm?
Show the answer
Mean rate = 30 ÷ 15 = 2.0 cm³ s⁻¹. The 1 cm cube has the larger SA:V (6:1 compared with 2:1), so it has more exposed surface per unit volume.
Higher tier: catalyst energy and tangent ratesCore
Scope: Core · Unit 1 paper tier: any of Higher tier · Unit 1, Topic 1.5, point (d), bold lower-energy requirement for a successful collision and the topic's Higher-tier mathematical skill to draw/use a tangent to a curve as rate of change. The separate qualification-wide Appendix C tangent-slope leaf is covered by the shared either-unit Higher math card.
Open revision guidance
What to know
The common catalyst outcome is that a catalyst increases rate and remains chemically unchanged. The additional bold Higher-tier detail is that it lowers the energy required for a collision to be successful.
The gradient of a tangent to a curved quantity-versus-time graph estimates the rate at that instant. This tangent method is Higher-only; the common card's average and straight-line gradient checks are not.
What to do in an assessment
Draw a tangent touching the curve at the required time without cutting across the curve nearby. Choose two well-separated points on the tangent, read their coordinates, and calculate gradient = change in y ÷ change in x with the correct units. A gas-volume/time tangent is in cm³ s⁻¹.
Worked tangent example: if two points on the tangent are (20 s, 32 cm³) and (40 s, 56 cm³), gradient = (56 − 32) ÷ (40 − 20) = 24 ÷ 20 = 1.2 cm³ s⁻¹. This estimates the instantaneous rate at the contact time, not the average over the whole experiment.
In plain language
A catalyst makes a greater fraction of collisions energetic enough to succeed. A tangent zooms in on one moment of a changing reaction.
Self-check
Two points on a tangent are (10 s, 18 cm³) and (30 s, 42 cm³). Find the gradient and state what it represents.
Show the answer
(42 − 18) ÷ (30 − 10) = 24 ÷ 20 = 1.2 cm³ s⁻¹. It estimates the instantaneous rate at the time where the tangent touches the curve.

Limestone reactions, uses, quarrying and thermal-stability practicalCore
Scope: Core · Unit 1, Topic 1.6, points (a)–(e), and specified practical: investigate thermal stabilities of calcium carbonate, copper(II) carbonate and sodium carbonate. Topic overview requires word and balanced symbol equations for described reactions.
Open revision guidance
What to know
Limestone is calcium carbonate, CaCO₃; quicklime is calcium oxide, CaO; slaked lime is calcium hydroxide, Ca(OH)₂. Heating metal carbonates decomposes them to a metal oxide and carbon dioxide; the relative thermal stability trend is compared using the named carbonates.
The limestone cycle includes thermal decomposition, the exothermic reaction of quicklime with water, and limewater reacting with carbon dioxide. Word and balanced symbol equations should show the atoms conserved.
Limestone is used in iron and steel production, road building, neutralising acidic soil and making cement. Quarrying has social and economic benefits such as employment and raw materials, alongside environmental impacts such as landscape, habitat, dust, noise and transport.
In the specified comparison, copper(II) carbonate decomposes most readily, calcium carbonate requires stronger heating, and sodium carbonate is the most thermally stable of these three. Less stable carbonate gives carbon dioxide more readily on heating.
Limestone can be cut for buildings; heating and processing it contributes to cement, which binds sand and gravel in concrete. Glass manufacture also uses limestone. Distinguish limestone itself from cement and concrete rather than treating their names as interchangeable.
What to do in an assessment
Thermal-stability practical: place equal small masses of calcium carbonate, copper(II) carbonate and sodium carbonate in separate labelled hard-glass test tubes; heat each for the same measured time with comparable heating. Test any gas with limewater and record observations. Use eye protection, a holder and heatproof mat; allow apparatus to cool and avoid inhaling dust or gases.
To compare stability fairly, keep sample mass, particle size, apparatus and heating time/strength as similar as possible. A carbonate that decomposes more readily under the same conditions is less thermally stable; do not infer an exact temperature from a simple qualitative test.
Equations: calcium carbonate → calcium oxide + carbon dioxide; CaCO₃(s) → CaO(s) + CO₂(g). Quicklime + water → slaked lime; CaO(s) + H₂O(l) → Ca(OH)₂(aq). Limewater + carbon dioxide → calcium carbonate + water; Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l).
In plain language
Heating limestone makes quicklime and carbon dioxide. Adding water makes slaked lime; these reactions explain both the limestone cycle and several useful materials.
Self-check
Write a balanced symbol equation for the thermal decomposition of calcium carbonate and name quicklime.
Show the answer
CaCO₃(s) → CaO(s) + CO₂(g). Quicklime is calcium oxide.

Bonding, structures, material properties, nanoscience and smart materialsCore
Scope: Core · Unit 2, Topic 2.1, common points (a)–(i) and (k): properties and models of metallic, ionic, simple molecular and giant covalent structures; dot-and-cross representations; carbon forms; nano-scale silver/titanium dioxide uses; smart materials. Mathematical Skills: unit prefixes, powers of ten and order-of-magnitude comparisons.
Open revision guidance
What to know
Metals contain a lattice of positive ions surrounded by delocalised electrons. The mobile electrons carry charge and thermal energy; layers of ions can slide while attraction to the electron sea remains, so metals conduct and can be shaped.
Ionic compounds contain oppositely charged ions in a giant lattice. Strong electrostatic attractions take much energy to overcome, so melting points are high. Solid ions are fixed and do not conduct, but mobile ions in a melt or aqueous solution carry current.
Simple molecular covalent substances have strong covalent bonds inside molecules but weaker intermolecular attractions between molecules. They tend to have low melting/boiling points and do not conduct because they lack mobile charged particles. Giant covalent structures have many strong covalent bonds throughout.
Diamond has four covalent bonds per carbon and is very hard; graphite has three bonds per carbon in layers with delocalised electrons, so layers slide and graphite conducts along them. Graphene is one strong, light carbon layer that conducts; carbon nanotubes are strong, conductive rolled sheets useful in composite materials; fullerenes are hollow cages with potential uses in drug delivery, catalysts and lubricants. These structures account for differing properties despite all being carbon.
A material's behaviour can change at nanoscale: silver nanoparticles are used for antimicrobial surfaces/coatings; titanium dioxide nanoparticles are used in sunscreens and photocatalytic/self-cleaning surfaces. Bulk silver or titanium dioxide does not automatically have the same behaviour as nanoscale particles.
Smart materials respond to a stimulus: thermochromic pigments change colour with temperature; photochromic pigments respond to light; polymer gels change volume; shape-memory alloys and polymers return toward a set shape when conditions change.
What to do in an assessment
Use a dot-and-cross diagram to track outer electrons from the original atoms. In ionic bonding, show electron transfer and brackets/charges around the resulting ions; in covalent bonding, show shared pairs. The diagrams model outer electrons rather than literal paths.
Convert scale units before comparing: 1 mm = 10⁻³ m, 1 μm = 10⁻⁶ m and 1 nm = 10⁻⁹ m. A 100 nm particle is 0.1 μm; it is 10⁻⁷ m because 100 × 10⁻⁹ = 10⁻⁷. An atom about 0.1 nm across is about 100 times narrower than a 10 nm particle.
When explaining a property, name the structure, the particle/bond that can or cannot move, and how that causes the observed behaviour. Do not say that covalent bonds between molecules are weak: it is the intermolecular attraction that is weaker in a simple molecular substance.
In plain language
The particles and bonds in a material determine whether it is hard, flexible, has a high melting point or can carry an electric current.
Self-check
Why does solid sodium chloride not conduct but molten sodium chloride does? Convert 250 nm to metres.
Show the answer
In the solid lattice ions are fixed; when molten the ions move and carry charge. 250 nm = 250 × 10⁻⁹ m = 2.5 × 10⁻⁷ m.
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.
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.
Compare electron transfer in a giant ionic compound with shared electron pairs in simple covalent molecules.
Higher tier: risks and future uses of nanoscienceCore
Scope: Core · Unit 2 paper tier: any of Higher tier · Unit 2, Topic 2.1, point (j): possible risks of silver and titanium dioxide nanoparticles and potential future developments in nanoscience. This outcome is bold/Higher tier only.
Open revision guidance
What to know
Nanoparticles may behave differently from the same material in bulk because a much larger fraction of their atoms is at the surface. This creates useful applications but also means effects on people and ecosystems must be investigated.
Possible concerns include inhalation or skin exposure during manufacture/use and release into water or soil, where very small particles may move differently from larger particles. The size and likelihood of a risk depend on the material, coating, dose and route of exposure; distinguish a possible hazard from a demonstrated harm.
Future nanoscience could provide targeted delivery, sensors, coatings or catalysts, but evaluate evidence, containment, long-term effects, benefits and uncertainty before recommending an application.
What to do in an assessment
For a risk evaluation, state the possible benefit, identify who or what could be exposed, explain the plausible route and effect, then weigh evidence, uncertainty and control measures. Do not claim that all nanoparticles are either safe or harmful.
For silver or titanium dioxide, link the proposed use to the material's nanoscale property, then consider exposure during production and disposal as well as during the intended use.
In plain language
A useful new material still needs careful safety testing: ask how people or the environment could encounter it and what the evidence actually shows.
Self-check
A nanoparticle coating is proposed for a product. Give one possible benefit and two questions needed to evaluate risk.
Show the answer
A possible benefit is an antimicrobial or self-cleaning surface, depending on the material. Ask whether particles can be released and inhaled or enter water/soil, and what reliable evidence shows about effects at realistic exposure levels.

Acids, alkalis, salt preparation and titrationCore
Scope: Core · Unit 2, Topic 2.2, common points (a)–(d), (e) neutralisation concept without the bold ionic-equation detail, and (f)–(j), (m)–(o); specified practicals: prepare soluble-salt crystals from an insoluble base/carbonate and titrate strong acid against strong base with indicator.
Open revision guidance
What to know
The pH scale describes acidity/alkalinity: acids are below 7, neutral solutions are about 7 and alkalis are above 7. Acids in water contain H⁺ ions; alkalis in water contain OH⁻ ions. Neutralisation makes water and a salt.
Do not confuse concentration with strength. Concentrated/dilute describes how much acid is present per volume; strong/weak describes the degree of ionisation in water. A strong acid ionises almost completely; a weak acid only partly. Hydrochloric acid and ethanoic acid both react with suitable metals to release hydrogen and form salts, but at the same concentration a strong acid initially supplies more H⁺ and reacts faster.
Acid + metal → salt + hydrogen; metals above hydrogen in the reactivity series react with dilute acids, while less-reactive metals such as copper do not. Acid + base/alkali → salt + water; acid + carbonate → salt + water + carbon dioxide. Hydrochloric acid makes chlorides, nitric acid makes nitrates and sulfuric acid makes sulfates.
A carbonate fizzes with dilute acid; the gas turns limewater milky. Sulfate ions give a white precipitate of barium sulfate when a sample is acidified and barium chloride solution is added. Insoluble salts can be made by mixing two suitable soluble-salt solutions so a precipitate forms.
Hydrochloric acid with sodium hydroxide, and dilute sulfuric acid with copper(II) oxide, both give exothermic neutralisation reactions. For a base that does not dissolve, use excess solid and filter; for two soluble reactants, use titration to avoid excess reactant.
What to do in an assessment
Specified soluble-salt preparation: warm dilute sulfuric acid, add copper(II) oxide in small portions until some remains unreacted, then filter off excess solid. Gently evaporate part of the filtrate, leave it to cool and crystallise, then collect, rinse with a little cold distilled water and dry the copper(II) sulfate crystals. Equation: CuO + H₂SO₄ → CuSO₄ + H₂O.
Specified titration: rinse/fill a burette with one solution and use a pipette to transfer a measured volume of the other strong solution into a conical flask. Add a few drops of a suitable indicator, record the initial burette reading, titrate to the colour endpoint while swirling, and record the final reading. Do a rough titre then repeat until concordant titres are obtained; use eye protection and handle acids/alkalis carefully.
To prepare a pure soluble salt by titration, first use indicator to find the exact neutralising volumes, then repeat with the same measured volumes but no indicator; the resulting solution contains the salt without indicator contamination. At common tier, titration compares relative concentrations; the Higher supplement shows the mole calculation.
For an insoluble salt by precipitation, mix measured solutions, filter the solid, wash it with distilled water and dry it. Example: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq); net ionic equation Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Avoid contamination by using clean apparatus.
In plain language
Acids provide hydrogen ions; alkalis provide hydroxide ions. Choose the acid and reactant to make the salt you want, then separate and purify it.
Self-check
What salt forms when nitric acid reacts with copper(II) oxide? Name the gas released when an acid reacts with a carbonate and give its test.
Show the answer
Copper(II) nitrate forms (with water). Carbon dioxide is released and turns limewater milky.
Higher tier: ionic neutralisation and concentration calculationsCore
Scope: Core · Unit 2 paper tier: any of Higher tier · Unit 2, Topic 2.2, bold ionic neutralisation detail in point (e), point (k) concentration in mol dm⁻³, and point (l) balanced-equation neutralisation calculations. These are Higher-tier-only source requirements.
Open revision guidance
What to know
The bold Higher-tier ionic detail is H⁺(aq) + OH⁻(aq) → H₂O(l). The complete ionic equation shows the particles that react; spectator ions are omitted.
Concentration c = n/V when amount n is in moles and volume V is in dm³, giving c in mol dm⁻³. Convert cm³ to dm³ by dividing by 1000. Use the balanced chemical equation for the mole ratio: equal solution volumes do not imply equal reacting amounts or concentrations.
For a titration calculation, determine moles of the known solution, use the balanced equation to find the moles of the unknown, then divide by its volume in dm³. This concentration calculation and source ionic-equation fragment are Higher-only.
What to do in an assessment
Worked 1:1 example: 25.0 cm³ of 0.100 mol dm⁻³ HCl contains n = cV = 0.100 × 0.0250 = 0.00250 mol HCl. It neutralises 0.00250 mol NaOH. If the NaOH volume is 20.0 cm³ = 0.0200 dm³, its concentration is 0.00250 ÷ 0.0200 = 0.125 mol dm⁻³.
Worked non-1:1 check: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. One mole of sulfuric acid reacts with two moles of sodium hydroxide. Write the coefficient ratio before calculating and keep units in every volume conversion.
In plain language
Use the balanced equation to count reacting moles, then use the volume to work out how concentrated the solution is.
Self-check
What volume in dm³ is 15.0 cm³? How many moles are in 15.0 cm³ of 0.200 mol dm⁻³ acid?
Show the answer
15.0 cm³ = 0.0150 dm³. n = cV = 0.200 × 0.0150 = 0.00300 mol.

Reactivity, extraction, electrolysis and metal applicationsCore
Scope: Core · Unit 2, Topic 2.3, common points (a)–(j), (l)–(n), point (p) electroplating, and (q); specified practicals: determine relative reactivities by displacement and investigate aqueous electrolysis/electroplating. Includes aqueous copper(II) chloride and water electrode equations, which are common.
Open revision guidance
What to know
Most metals are obtained from ores in Earth's crust; very unreactive metals such as gold can occur native. More reactive metals are generally harder and more energy-intensive to extract. A more reactive metal displaces a less reactive metal from its compound.
Oxidation can mean gain of oxygen and reduction loss of oxygen. In electron terms, oxidation is loss of electrons and reduction is gain. The thermit reaction is a competition example: Fe₂O₃ + 2Al → 2Fe + Al₂O₃.
In a blast furnace, coke burns (C + O₂ → CO₂), then carbon dioxide reacts with more carbon (CO₂ + C → 2CO). Carbon monoxide reduces iron(III) oxide (Fe₂O₃ + 3CO → 2Fe + 3CO₂). Limestone decomposes (CaCO₃ → CaO + CO₂); calcium oxide neutralises silica impurity to form slag (CaO + SiO₂ → CaSiO₃).
Electrolysis uses electricity to decompose molten or dissolved ionic substances. In molten lead(II) bromide, Pb²⁺ + 2e⁻ → Pb at the cathode and 2Br⁻ → Br₂ + 2e⁻ at the anode. Alumina is dissolved in molten cryolite to lower the operating temperature during aluminium extraction; carbon electrodes are consumed as oxygen reacts with them.
In aqueous copper(II) chloride electrolysis, copper forms at the negative cathode (Cu²⁺ + 2e⁻ → Cu) and chlorine forms at the positive anode (2Cl⁻ → Cl₂ + 2e⁻). In acidified-water electrolysis, cathode: 2H⁺ + 2e⁻ → H₂; anode: 2H₂O → O₂ + 4H⁺ + 4e⁻; overall 2H₂O(l) → 2H₂(g) + O₂(g).
During aluminium extraction, Al³⁺ + 3e⁻ → Al at the cathode; oxygen formed at the anode reacts with the carbon electrode, so the anodes gradually need replacing. Steel is used in buildings/bridges; low-density aluminium is useful in aircraft; copper's conductivity suits electrical wiring; strong, corrosion-resistant titanium is used in aircraft and medical implants. Transition metals are generally strong, dense, good conductors and often catalysts; they can form ions with different charges. An alloy is a mixture of molten metals whose properties can be changed by composition.
Electroplating coats an object with a thin layer of another metal. Extraction choices depend on ore location, energy/fuel costs, plant siting, greenhouse emissions, recycling and the value of the product.
Adding water to iron or copper is not the same as exposing a more reactive metal such as sodium, which reacts vigorously with cold water. Copper does not displace hydrogen from dilute hydrochloric acid; magnesium does. Use the observed reaction to place a metal in a reactivity series.
What to do in an assessment
Specified displacement practical: clean equal-sized metal strips, place each in labelled solutions of metal salts at the same concentration and temperature, and record colour change, coating or temperature evidence over a fixed time. For example, an iron nail in blue copper(II) chloride solution becomes coated with copper as the blue colour fades; iron is more reactive than copper. Repeat and compare; keep exposed surface area and solution conditions fair.
Specified electrolysis/electroplating practical: connect inert electrodes to a low-voltage DC supply, immerse them in the selected solution without letting them touch, and observe products at each electrode. For copper plating, the object to be plated is the cathode, the coating metal is the anode and the electrolyte contains ions of the coating metal. Record polarity and observations; use eye protection and avoid touching electrodes while powered.
For aqueous copper(II) chloride, use the common half-equations Cu²⁺ + 2e⁻ → Cu and 2Cl⁻ → Cl₂ + 2e⁻ to explain the copper coating and chlorine formation. Chlorine is hazardous: use small quantities with suitable ventilation and follow teacher/laboratory instructions.
In plain language
A more reactive metal can take the place of a less reactive one. Electrolysis uses a current to move ions to electrodes, where they gain or lose electrons.
Self-check
Write the cathode equation for aqueous copper(II) chloride electrolysis. In a copper-plating setup, which electrode is the object being plated?
Show the answer
Cu²⁺(aq) + 2e⁻ → Cu(s). The object being plated is the cathode (negative electrode).
Higher tier: hydroxide ion tests and industrial electrolysisCore
Scope: Core · Unit 2 paper tier: any of Higher tier · Unit 2, Topic 2.3, point (k) precipitation tests for Cu²⁺, Fe²⁺ and Fe³⁺ with aqueous hydroxide; point (o) competing ions in aqueous electrolysis including sodium chloride; and the bold detail of point (p), copper purification and sodium hydroxide/hydrogen/chlorine manufacture. These source fragments are Higher-tier only.
Open revision guidance
What to know
Adding aqueous sodium hydroxide identifies some metal ions by precipitate colour: Cu²⁺ gives blue Cu(OH)₂; Fe²⁺ gives green Fe(OH)₂ which can turn brown on standing; Fe³⁺ gives brown Fe(OH)₃. These identification tests are Higher-only.
In concentrated aqueous sodium chloride (brine), water competes with Na⁺ at the cathode and chloride competes with OH⁻ at the anode. Hydrogen forms at the cathode; chlorine forms at the anode; sodium hydroxide remains in solution. This competing-ion prediction is Higher-only; the separate common aqueous copper(II) chloride example stays on the common card.
Copper purification uses an impure copper anode, a pure copper cathode and copper(II) sulfate solution. Copper dissolves from the anode as Cu²⁺ and deposits as pure copper at the cathode; less-reactive impurities collect as anode sludge. Manufacture of sodium hydroxide by brine electrolysis also produces hydrogen and chlorine.
What to do in an assessment
Precipitation equations: Cu²⁺ + 2OH⁻ → Cu(OH)₂(s); Fe²⁺ + 2OH⁻ → Fe(OH)₂(s); Fe³⁺ + 3OH⁻ → Fe(OH)₃(s). Add the reagent dropwise to separate small samples and record colour; do not use one tube for several unknowns.
Brine half-equations: cathode, 2H₂O + 2e⁻ → H₂ + 2OH⁻; anode, 2Cl⁻ → Cl₂ + 2e⁻. Overall: 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + H₂(g) + Cl₂(g). The sodium ion remains in solution with hydroxide.
Copper purification: anode Cu(s) → Cu²⁺(aq) + 2e⁻; cathode Cu²⁺(aq) + 2e⁻ → Cu(s). Because one copper atom dissolves for each copper ion deposited, the solution's copper-ion concentration is approximately maintained.
Higher-tier reacting-mass link: Fe₂O₃ + 2Al → 2Fe + Al₂O₃. Mr(Fe₂O₃) = (2 × 56) + (3 × 16) = 160. From 16.0 g Fe₂O₃, n = 16.0 ÷ 160 = 0.100 mol; the 1:2 equation ratio gives 0.200 mol Fe, so mass Fe = 0.200 × 56 = 11.2 g (assuming sufficient aluminium and complete reaction).
In plain language
At each electrode, competing particles are discharged according to the solution and conditions. Track which ions are present and write the electron transfer at each electrode.
Self-check
What precipitate colour identifies Fe³⁺ with aqueous hydroxide? Name the two gases made during brine electrolysis.
Show the answer
Fe³⁺ gives a brown iron(III) hydroxide precipitate. Brine electrolysis produces hydrogen and chlorine.

Exothermic and endothermic reactions, profiles and bond energiesCore
Scope: Core · Unit 2, Topic 2.4, points (a)–(d): energy transfer and temperature changes; exothermic/endothermic profiles; activation energy; calculate overall reaction energy from bond-energy data and classify the reaction.
Open revision guidance
What to know
An exothermic reaction transfers energy to the surroundings, so the surroundings warm; products are at lower chemical energy than reactants. An endothermic reaction takes energy from the surroundings, which cool; products are at higher chemical energy.
An energy profile plots energy against reaction progress. The vertical difference between reactants and products represents the overall energy change; activation energy is the energy needed to reach the peak from the reactants.
Breaking bonds requires energy; forming bonds releases energy. Overall energy change = total energy needed to break reactant bonds − total energy released when product bonds form. A negative result is exothermic; a positive result is endothermic.
What to do in an assessment
For H₂ + Cl₂ → 2HCl, suppose the supplied bond-energy table gives H–H = 436, Cl–Cl = 243 and H–Cl = 431 kJ mol⁻¹. Bonds broken = 436 + 243 = 679; bonds formed = 2 × 431 = 862. ΔE = 679 − 862 = −183 kJ mol⁻¹, so the reaction is exothermic.
Count every bond from the balanced equation, including coefficients. Do not multiply the same bond twice because it appears on both sides; break reactant bonds and form product bonds as separate lists.
On a profile, label reactants, products and activation energy. For an exothermic profile draw products lower than reactants; for an endothermic profile draw them higher.
The specified displacement-temperature practical can use zinc and copper(II) sulfate: Zn + CuSO₄ → ZnSO₄ + Cu. Measure the solution temperature before adding zinc and the highest temperature after stirring; compare rises using controlled volumes/concentrations and account for heat lost to the surroundings.
In plain language
A reaction releases energy overall if making product bonds releases more than breaking the reactant bonds needs.
Self-check
Bond breaking takes 500 kJ mol⁻¹ and bond formation releases 650 kJ mol⁻¹. Find ΔE and classify the reaction.
Show the answer
ΔE = 500 − 650 = −150 kJ mol⁻¹. The negative value means exothermic.
View diagrams
Exothermic energy profile
Qualitative exothermic profile: the products lie at a lower energy level than the reactants, so energy is released. Activation energy Eₐ is measured from the reactant level to the peak; net energy change ΔH runs from reactants to products and is negative. The horizontal axis is reaction progress, not time. No numerical energies are implied.
Endothermic energy profile
Qualitative endothermic profile: the products lie at a higher energy level than the reactants, so energy is absorbed. Activation energy Eₐ is measured from the reactant level to the peak; net energy change ΔH runs from reactants to products and is positive. The horizontal axis is reaction progress, not time. No numerical energies are implied.
Compare reactant/product energy levels and the activation-energy rise for exothermic and endothermic reactions.

Crude oil, fuels, common organic formulae, plastics and ethanolCore
Scope: Core · Unit 2, Topic 2.5, common points (a)–(l), (o)–(v), (x), and the simple-alcohol formula fragment of point (w), plus specified practical: determine energy released per gram by a fuel. Simple alkane/alkene formulae and simple alcohol formulae are common; Higher-only complex naming/isomer/IR fragments are separate.
Open revision guidance
What to know
Crude oil is a finite complex mixture of hydrocarbons formed over millions of years from remains of simple marine organisms. Fractional distillation separates it into fractions: each is a mixture of hydrocarbons (mainly alkanes) with a similar boiling range. Shorter chains have lower boiling points, lower viscosity and greater volatility/flammability; longer chains are more viscous, less volatile and harder to ignite. Petrol is useful for road transport, kerosene for aviation and diesel for heavy transport; long-chain residues are less suitable as fuels and include bitumen for road surfacing.
The oil industry affects employment, transport, energy security, international trade and prices; extraction/spills damage habitats and combustion releases greenhouse gases. Complete combustion in plentiful oxygen makes carbon dioxide and water; incomplete combustion can make carbon monoxide and/or carbon. Carbon monoxide is poisonous. Hydrogen burns to water (2H₂ + O₂ → 2H₂O); it produces no carbon dioxide at point of use, but production, storage, transport and energy source affect its overall benefits.
The fire triangle is fuel, oxygen and heat: remove one to extinguish/prevent a fire. Cracking breaks some long-chain hydrocarbons into shorter alkanes and alkenes, which are useful as fuels and as plastic monomers. General formulae: alkanes CₙH₂ₙ₊₂; alkenes CₙH₂ₙ.
For simple molecules, methane CH₄, ethane C₂H₆, propane C₃H₈ and butane C₄H₁₀ are alkanes; ethene C₂H₄ has structure CH₂=CH₂ and propene C₃H₆ has structure CH₂=CHCH₃. Alkenes add hydrogen to form alkanes and add bromine; bromine water changes from orange/brown to colourless in the presence of an alkene.
Addition polymerisation joins many alkene monomers into a long chain: ethene makes poly(ethene), propene makes poly(propene), chloroethene makes PVC and tetrafluoroethene makes PTFE. Poly(ethene) is used for packaging, poly(propene) for containers/ropes, PVC for pipes/cable insulation and PTFE for non-stick/chemical-resistant coatings. Plastics are useful and durable; non-biodegradability and landfill pressure make reuse/recycling and careful management of finite oil resources important.
Yeast ferments sugar without oxygen in warm conditions to ethanol and carbon dioxide: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Ethanol is used in alcoholic drinks, as a solvent and as a fuel. Drinks support economic activity but can cause health and social harms; bioethanol development also involves land, food, cost and environmental trade-offs. Ethanol can be microbially oxidised to ethanoic acid: CH₃CH₂OH + O₂ → CH₃COOH + H₂O.
Fractional distillation separates crude oil by boiling range: refinery gases are bottled for heating, petrol fuels cars, kerosene fuels aircraft, diesel fuels engines, fuel oil fuels ships/heating, and bitumen is used for roads. Long molecules have higher boiling points and are more viscous; cracking makes shorter alkanes and useful alkenes.
What to do in an assessment
For a crude-oil distillation diagram, label the hottest region near the bottom, cooler region toward the top, and show vapours rising and fractions condensing over their boiling ranges. Match short-chain, low-boiling fractions near the top and long-chain, high-boiling fractions lower down.
Specified fuel-energy practical: weigh a spirit burner with fuel, heat a known mass of water in a supported metal can, record the water temperature change, then reweigh the burner. Estimate energy transferred to water using q = m c ΔT, with c(water) = 4.18 J g⁻¹ °C⁻¹; energy per gram fuel = q ÷ mass of fuel burnt. Example: 100 g water rises 20.0 °C while 0.50 g fuel burns: q = 100 × 4.18 × 20.0 = 8360 J, so measured energy transfer is 8360 ÷ 0.50 = 16 720 J g⁻¹ (16.7 kJ g⁻¹).
Reduce heat loss with a draught shield, lid and insulation; keep flame-to-can distance and water mass consistent, stir before reading temperature, and extinguish the burner safely. The measured value is usually lower than the fuel's full chemical energy because some heat warms the apparatus or escapes.
Write balanced combustion and addition equations. Examples: C₂H₄ + H₂ → C₂H₆ and C₂H₄ + Br₂ → C₂H₄Br₂. Complete ethanol combustion is C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. The polymerisation patterns are nCH₂=CH₂ → [–CH₂–CH₂–]ₙ; nCH₂=CHCH₃ → [–CH₂–CH(CH₃)–]ₙ; PVC [–CH₂–CHCl–]ₙ and PTFE [–CF₂–CF₂–]ₙ.
Acidified potassium dichromate(VI) can test for ethanol: orange solution turns green as ethanol is oxidised. It is hazardous; use the prescribed small-scale procedure and eye protection.
Make ethanol by fermenting glucose with yeast in warm, anaerobic conditions: glucose → ethanol + carbon dioxide. Distillation separates ethanol from the mixture because of different boiling points. Aerobic oxidation can turn ethanol into ethanoic acid; compare this acid with hydrochloric acid at the same concentration.
Show ethene/propene addition reactions by opening the C=C bond. Bromine water is decolourised by an alkene. Compare displayed formulas with models so a carbon has four bonds and hydrogen one.
In plain language
Crude oil is separated into useful mixtures; cracking changes large molecules into smaller ones, while addition reactions let alkenes make other molecules and plastics.
Self-check
A fuel burns 0.40 g and heats 80 g water by 15 °C. Estimate energy transferred per gram using c = 4.18 J g⁻¹ °C⁻¹. What happens to bromine water with an alkene?
Show the answer
q = 80 × 4.18 × 15 = 5016 J; per gram = 5016 ÷ 0.40 = 12 540 J g⁻¹ (12.5 kJ g⁻¹). Bromine water decolourises from orange/brown to colourless.
View diagrams
Alkanes: displayed formulae
Methane · CH₄
Every carbon and hydrogen atom and every C-C and C-H bond is shown; each carbon has four covalent bonds.
Ethane · C₂H₆
Every carbon and hydrogen atom and every C-C and C-H bond is shown; each carbon has four covalent bonds.
Propane · C₃H₈
Every carbon and hydrogen atom and every C-C and C-H bond is shown; each carbon has four covalent bonds.
Butane · C₄H₁₀
Every carbon and hydrogen atom and every C-C and C-H bond is shown; each carbon has four covalent bonds.
Every carbon and hydrogen atom is labelled, and every C-H and C-C covalent bond is drawn.
Simple alkenes: displayed formulae
Ethene · C₂H₄
Every carbon and hydrogen atom is shown, with every covalent bond drawn.
Propene · C₃H₆
Every carbon and hydrogen atom is shown, with every covalent bond drawn.
A single line is one covalent bond; two parallel lines are a double bond. Every carbon and hydrogen atom is shown, with every covalent bond drawn.
Alcohols: displayed formulae
Methanol · CH₄O
Every atom is shown: the -OH group has explicit C-O and O-H bonds, and all hydrogen atoms are included.
Ethanol · C₂H₆O
Every atom is shown: the -OH group has explicit C-O and O-H bonds, and all hydrogen atoms are included.
Propan-1-ol · C₃H₈O
Every atom is shown: the -OH group has explicit C-O and O-H bonds, and all hydrogen atoms are included.
Butan-1-ol · C₄H₁₀O
Every atom is shown: the -OH group has explicit C-O and O-H bonds, and all hydrogen atoms are included.
The C-O and O-H bonds in each hydroxyl group are explicit; all carbon-bound and oxygen-bound hydrogen atoms are drawn.
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.
Displayed formulae show every atom and bond in common alkanes, alkenes and simple alcohols. Ethene and propene are common examples available to Foundation learners; the gallery's but-1-ene/but-2-ene positional comparison is Higher-only and is not required for Foundation.
Higher tier: complex names, isomers, alcohol structures and IR spectraCore
Scope: Core · Unit 2 paper tier: any of Higher tier · Unit 2, Topic 2.5, bold points (m) and (n), Higher-only positional-isomer fragment of point (w), and point (y): isomerism/naming of more complex alkanes and alkenes, alcohol positional isomers, and infrared evidence for organic functional groups.
Open revision guidance
What to know
Structural isomers have the same molecular formula but different atom connectivity. Butane and 2-methylpropane are C₄H₁₀ isomers; but-1-ene and but-2-ene are C₄H₈ positional isomers because the double bond is in a different position.
Name the longest carbon chain, number it to give the double bond or substituent the lowest possible position, then identify substituents and the -ane/-ene ending. Examples: CH₃CH(CH₃)CH₃ is 2-methylpropane; CH₃CH=CHCH₃ is but-2-ene.
Alcohol positional isomers have the same molecular formula but the –OH group is attached at a different carbon: propan-1-ol CH₃CH₂CH₂OH and propan-2-ol CH₃CH(OH)CH₃ are both C₃H₈O. The simple alcohol formulae themselves are common; the positional-isomer comparison is Higher-only.
Infrared spectroscopy can show bond absorptions that indicate a likely functional group: alkane C–H near 2850–3000 cm⁻¹; C=C around 1620–1680 cm⁻¹; C=O around 1680–1750 cm⁻¹; an alcohol O–H is broad around 3200–3600 cm⁻¹; a carboxylic-acid O–H is very broad across roughly 2500–3300 cm⁻¹. Use the supplied spectrum/key and treat peaks as evidence for likely bonds, not a complete structural proof.
What to do in an assessment
For C₄H₈, compare but-1-ene CH₂=CHCH₂CH₃ with but-2-ene CH₃CH=CHCH₃. Both have formula C₄H₈ but the C=C position changes; the gallery displays both complete structures.
In an IR question, match a prominent absorption to the provided wavenumber range, state the bond/group indicated and combine it with other evidence. A broad O–H plus C=O evidence is consistent with a carboxylic acid; an O–H without C=O can indicate an alcohol.
When given a more complex alkane or alkene, count the longest chain first, number consistently, locate the multiple bond/substituent, and check that the name reproduces the stated molecular structure.
In plain language
Isomers have the same atom counts but different arrangements. IR peaks help identify bonds, so they narrow down which kind of organic molecule a sample may be.
Self-check
Name CH₃CH=CHCH₃ and state how it differs from but-1-ene. Which broad IR feature suggests an O–H bond?
Show the answer
CH₃CH=CHCH₃ is but-2-ene; its double bond is between carbons 2 and 3 rather than at carbon 1. A broad O–H absorption is in the approximately 3200–3600 cm⁻¹ region (carboxylic-acid O–H is especially broad at lower wavenumbers).
View diagrams
Alkenes: displayed formulae
Ethene · C₂H₄
Every atom and bond is shown; the double bond is between C1 and C2.
Propene · C₃H₆
Every atom and bond is shown; the double bond is between C1 and C2.
But-1-ene · C₄H₈
Every atom and bond is shown; the double bond is between C1 and C2.
But-2-ene · C₄H₈
Every atom and bond is shown; the double bond is between C2 and C3.
Every atom and bond is drawn. The double-bond position is shown explicitly, so but-1-ene and but-2-ene are different structures.
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 displayed-formula gallery explicitly distinguishes but-1-ene and but-2-ene by the position of the C=C bond.

Reversible reactions, sulfuric acid, ammonia and fertilisersCore
Scope: Core · Unit 2, Topic 2.6, common points (a)–(b), (d)–(j): reversible reactions; ammonia/Haber equation; ammonia test; contact process; sulfuric-acid uses and dehydration; nitrogenous-fertiliser equations/test; fertiliser benefits and environmental impacts.
Open revision guidance
What to know
A reversible reaction can proceed in both directions, shown by ⇌. In a closed system at dynamic equilibrium the forward and reverse reactions continue at equal rates, so measured amounts remain constant.
The Haber reaction makes ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Ammonia is alkaline; it turns damp red litmus paper blue.
The Contact process makes sulfuric acid in stages: burn sulfur, S + O₂ → SO₂; oxidise 2SO₂ + O₂ ⇌ 2SO₃ using a vanadium(V) oxide catalyst; absorb sulfur trioxide in sulfuric acid to form oleum, then add water to make sulfuric acid. Overall product stages: SO₃ + H₂SO₄ → H₂S₂O₇; H₂S₂O₇ + H₂O → 2H₂SO₄.
Sulfuric acid is used to make fertilisers, paints, dyes, fibres, plastics and detergents. Concentrated sulfuric acid is a dehydrating agent: it removes water from sugar, leaving black carbon, and from hydrated copper(II) sulfate, leaving white anhydrous CuSO₄.
Ammonium sulfate and ammonium nitrate are nitrogenous fertilisers made by neutralising ammonia solution. Equations: 2NH₃ + H₂SO₄ → (NH₄)₂SO₄; NH₃ + HNO₃ → NH₄NO₃. Fertilisers support crop growth, but runoff into waterways can cause excessive algal growth and oxygen depletion.
What to do in an assessment
A practical ammonium-ion test is to add aqueous sodium hydroxide and warm gently: ammonia is released and turns damp red litmus blue. Ionic equation: NH₄⁺ + OH⁻ → NH₃ + H₂O. Use a small sample and do not smell gases directly.
Concentrated sulfuric acid dehydrates sucrose: C₁₂H₂₂O₁₁ → 12C + 11H₂O. Hydrated copper(II) sulfate loses its water of crystallisation: CuSO₄·5H₂O → CuSO₄ + 5H₂O. These are dehydration changes, not neutralisation.
Evaluate fertiliser use by weighing increased crop yield against runoff, eutrophication, water treatment and the need to apply only the required amount.
In plain language
Reversible processes make important industrial chemicals. The same chemistry helps produce fertilisers, but their use must be managed to avoid polluting waterways.
Self-check
What observation identifies ammonia gas? Write the formula equation for making ammonium nitrate from ammonia and nitric acid.
Show the answer
Ammonia turns damp red litmus blue. NH₃ + HNO₃ → NH₄NO₃.
Higher tier: Haber conditions, yield and economic decisionsCore
Scope: Core · Unit 2 paper tier: any of Higher tier · Unit 2, Topic 2.6, point (c) in full: factors in choosing Haber-process conditions for economical ammonia production. The whole point is bold/Higher tier only; formal Le Chatelier's principle is not required.
Open revision guidance
What to know
Industrial Haber conditions are a compromise: about 450 °C, about 200 atmospheres and an iron catalyst. Higher pressure favours a greater ammonia yield and faster production but needs strong, expensive equipment; too high a temperature gives a faster reaction but a lower equilibrium yield for the exothermic forward reaction.
A lower temperature could improve equilibrium yield but make production too slow; a catalyst speeds the approach to equilibrium without changing the equilibrium yield. Unreacted nitrogen and hydrogen are recycled to improve overall use of feedstocks.
The decision is economic as well as chemical: balance yield per pass, rate/throughput, energy and compression costs, safety, equipment, catalyst and recycling. Le Chatelier's principle is not required for this point.
What to do in an assessment
When interpreting a supplied yield/conditions table, separate yield from rate and cost. Identify which condition gives a useful compromise, quote evidence from the table, then explain the trade-off (for example, a higher temperature may increase rate but reduce equilibrium yield).
For a production decision, include at least one benefit and one cost of changing pressure or temperature and explain why the chosen industrial conditions need not maximise a single factor.
In plain language
Industry chooses conditions that make ammonia quickly and economically, not simply the condition that gives the greatest yield in one pass.
Self-check
Why is the Haber process not run at the lowest possible temperature even though a lower temperature gives a higher equilibrium yield?
Show the answer
The reaction would proceed too slowly, reducing production rate. Industry balances yield with reaction rate, energy/equipment costs and safety; the catalyst helps reach equilibrium faster.

Appendix A: scientific thinking, evidence and decisionsCore
Scope: Core · Subject Content, Scientific Enquiry introductory definition/scope (printed pp. 6–7; PDF pp. 9–10); Appendix A opening requirement to develop enquiry through subject-content activities across all units; §1 Development of scientific thinking, all seven outcomes (printed pp. 34–35; PDF pp. 37–38).
Open revision guidance
What to know
Scientific enquiry is the broad process of asking questions, exploring ideas and seeking understanding about natural phenomena. It includes practical/investigation skills and curiosity-driven questioning, observing, gathering information and critical thinking, from forming a research question through carrying out experiments. It is not confined to one assessment: the specification expects its development through subject content across all units.
Models help explain observations but have assumptions and limits; scientific models and theories can change when new evidence is collected, scrutinised and communicated.
Choose methods that fit the question to solve a problem, make a prediction or build an explanation. A prediction should follow from a model or known chemistry and be testable. Explain both familiar results and unfamiliar observations by linking claims to evidence and chemical ideas.
Science can produce powerful explanations and useful technologies, but evidence can be incomplete, a model may not capture every factor, and applications can raise ethical issues. Everyday examples include water treatment and batteries; technological examples include catalysts, smart materials and medicines. Explain the chemistry behind an application rather than merely naming it.
Evaluate implications for people, communities, the economy and the environment. Separate evidence-supported effects from assumptions, recognise who benefits and who may bear costs, consider alternatives and uncertainty, and make a reasoned decision that responds to the evidence and arguments.
Evaluate risk in practical work and wider society. Distinguish a hazard from the chance and consequence of exposure; compare likelihood, severity and control measures. Risk perception can differ from the risk indicated by data, so identify both evidence and uncertainty.
Peer review lets other scientists examine methods, data and reasoning before results are trusted. Communicate methods and results clearly to the audience, using suitable language, diagrams, units and evidence; adapt technical depth without changing the findings.
What to do in an assessment
Worked evidence decision: a community is considering an extraction site. List evidence for the useful metal supply, employment and local income; compare it with habitat, water, dust, noise, transport and waste impacts. State which claims need data, weigh the alternatives and give a qualified recommendation rather than treating one benefit or cost as decisive.
When a model predicts a result, record the prediction before testing it, collect relevant observations, compare evidence with the prediction, explain agreement or disagreement and identify what the model cannot explain. Suggest how independent replication or peer review could test the conclusion.
For any practical or societal risk, name the hazard, describe the exposure route and potential consequence, then choose a proportionate control. Review whether the control reduces risk without making the investigation impossible.
In plain language
Use chemistry to explain and predict, test ideas with evidence, acknowledge what the evidence cannot show, and communicate a fair decision.
Self-check
A technology has a useful chemical application but possible environmental effects. What four elements should an evidence-based recommendation include?
Show the answer
Describe the benefit and relevant evidence; identify plausible environmental effects and who/what is exposed; assess uncertainty, risk and alternatives; then give a reasoned decision that weighs benefits and costs rather than asserting certainty.
Appendix A: experimental skills and strategiesCore
Scope: Core · Original Appendix A §2 experimental skills and strategies and §3 analysis and evaluation
Open revision guidance
What to know
An investigation is part of scientific enquiry: it is a planned, structured collection of data to answer a specific question, using experiments, observations or measurements to test a hypothesis or explore a problem. It includes evaluation, next steps and further work. Use a scientific explanation to form a testable hypothesis, then define the independent variable (the one deliberately changed), dependent variable (the outcome measured) and controlled variables (kept consistent). Use accurate chemistry vocabulary and define any symbol or unfamiliar term.
Plan a method that answers the question: choose substances, a suitable range, apparatus and techniques; state how to make/identify substances or collect observations; control other factors; plan repeats or representative sampling where suitable; and identify safety measures. Correctly set up apparatus and use it at a resolution suitable for the expected change.
Observe and measure with appropriate apparatus; record raw readings promptly with headings, units and suitable precision. Understand how a scientific quantity is determined, including any calculation linking a measured quantity to the reported result.
Present observations and data in a suitable table, diagram or graph. Translate between words, tables, graphs and numerical values without changing units or meaning; carry out the necessary mathematical analysis.
Represent spread or distributions, identify uncertainty and anomalies, and interpret primary data gathered directly or secondary data supplied by another source. Describe patterns/trends, distinguish observation from inference, draw a conclusion supported by the data and relevant chemistry, and relate the explanation back to the hypothesis.
Evaluate whether the method was objective and suitable; distinguish accuracy from precision, repeatability from reproducibility, and random from systematic error. Suggest a specific improvement or further investigation and explain how it would address a limitation. Communicate the scientific rationale for the design and method.
What to do in an assessment
Worked plan: test whether increasing hydrochloric-acid concentration changes the rate of magnesium reacting. Hypothesis: greater concentration gives a faster initial rate because more acid particles are present per unit volume and successful collisions occur more often. Change concentration; measure hydrogen volume at fixed times with a gas syringe; keep magnesium mass/surface area, acid volume, temperature, apparatus and timing consistent. Use a safe small quantity, eye protection and keep hydrogen away from ignition.
Choose a concentration range with several evenly spaced values, repeat each condition, record raw gas volumes against time, flag an anomalous reading rather than silently deleting it, and calculate a mean when repeats are comparable. Plot volume against time with units; compare initial gradients, state the pattern and use collision theory to explain it.
Evaluate a concrete limitation: if a bung is fitted too slowly, early hydrogen escapes and the measured initial rate is inaccurate. Practise a consistent start procedure or use apparatus that minimises the delay; repeat the affected runs. A better instrument resolution alone will not correct that systematic start delay.
A representative sample is one that fairly reflects the population being studied. Select sampling locations or items using a stated method and adequate coverage; do not choose only convenient or unusually clean samples.
Present a reasoned investigation report: state the question and scientific rationale, describe the apparatus and method, show tables/graphs/calculations with units, relate conclusions to the hypothesis, and explain uncertainty and limitations. Use verbal, diagrammatic, numerical and symbolic forms where they make the evidence clearer.
In plain language
A good investigation asks a clear question, changes one thing fairly, measures carefully, shows the evidence, explains it and improves the method.
Self-check
In the magnesium/acid investigation, identify the independent variable, dependent variable and two controlled variables. Give one reason to repeat each measurement.
Show the answer
Independent: acid concentration. Dependent: hydrogen volume or rate derived from it. Controls include magnesium mass/surface area, acid volume and temperature. Repeats help reveal anomalous readings and estimate the mean/variability, reducing the influence of random error.
Scientific language and unitsCore
Scope: Core · Original Appendix A §4, scientific vocabulary, quantities, SI/IUPAC units and nomenclature, prefixes and conversions; common fragments
Open revision guidance
What to know
Independent variable: the variable chosen to change. Dependent variable: the variable measured each time the independent variable changes. Controlled variable: a factor that could affect results and is therefore kept constant. Control experiment: a comparison run with the independent variable absent or held constant while other variables and methods stay the same.
Hypothesis: an idea that can be tested by experiment and may be supported or contradicted. Range: the lowest and highest values of the independent variable used. Valid conclusion: a finding supported by results from a well-designed investigation and sound scientific knowledge. Validity of experimental design: how suitable the method is for answering the stated question, including appropriate control of variables.
Resolution: the smallest change an instrument can show (for example, the smallest scale division). Anomaly: a result much larger or smaller than the rest of a set. True value: the value that would be obtained in ideal conditions. Uncertainty: the interval within which the true value is expected; express it with units, for example 20 °C ± 2 °C. Use the uncertainty formula supplied in an examination if required.
Measurement error: the difference between a measured value and the true value. Systematic error shifts readings consistently, often because of apparatus or method, and is not reduced by taking more repeats. Random error varies unpredictably between readings and can be reduced in influence by taking more measurements and calculating a mean.
Accuracy: closeness of a measurement to the true value. Inaccuracy: a result far from the true value. Precision: repeated measurements show little spread about their mean. Repeatability: precision when one learner/group repeats readings; repeatable means the same learner/group using the same method and apparatus obtains the same or similar results again.
Reproducibility: precision when readings are repeated by different learners/groups. Reproducible means different learners/groups obtain the same or similar results, potentially with different apparatus or methods. A hazard is a chemical or apparatus with potential to cause harm; risk is a step/exposure in which a hazard may cause danger; control measures reduce or prevent that risk while the work remains possible.
Use SI units and IUPAC chemical names where appropriate. Quantities are determined by a stated method: mass with a balance, volume with suitable glassware, temperature with a thermometer and concentration from amount per volume.
What to do in an assessment
Worked classification: when testing acid concentration, concentration is independent, hydrogen volume is dependent, and magnesium mass, acid volume and temperature are controlled. A thermometer marked every 1 °C has a resolution of 1 °C. If repeated results are 19.8, 20.0 and 20.1 °C, report their spread/mean rather than claiming perfect accuracy without knowing the true value.
For a hot apparatus, the heat source is a hazard; moving the tube while heating could expose a person to burns (the risk); using a holder, eye protection and a heatproof mat are control measures. State the hazardous agent and route/consequence, not just “be careful”.
When results are close to one another but all are offset because a balance was zeroed incorrectly, they are precise but inaccurate and affected by systematic error. Repeating the same flawed procedure will not remove the offset; zero/calibrate the balance.
Use prefixes and powers of ten: tera 10¹², giga 10⁹, mega 10⁶, kilo 10³, centi 10⁻², milli 10⁻³, micro 10⁻⁶ and nano 10⁻⁹. Convert units explicitly, for example 25 cm³ = 0.025 dm³ and 1000 mg = 1 g. Higher adds the separately bold significant-figures calculation requirement.
In plain language
These words let you say precisely what changed, what was measured, how trustworthy results are and how danger is controlled.
Self-check
Several readings are tightly grouped but a balance has a constant zero error. Are they necessarily accurate? What kind of error is present, and will more repeats remove it?
Show the answer
No. They may be precise but inaccurate. The constant offset is systematic error; more repeats do not remove it, so the balance/method must be corrected.

Appendix C: common mathematical skillsCore
Scope: Core · Original Appendix B, Mathematical Skills: common arithmetic, data, algebra, graph and geometry statements
Open revision guidance
What to know
Decimal form writes a number with a decimal point. Standard form is a × 10ⁿ with 1 ≤ a < 10; positive n represents a large value and negative n a small one. Ratios compare quantities; fractions describe parts of a whole; percentages are parts per hundred.
An arithmetic mean is the sum of measurements divided by their number. Tables need clear headings and units; diagrams/graphs need labels, suitable scales, units and a clear key where needed. An order-of-magnitude estimate compares powers of ten to judge approximate scale, not exact equality.
Substitute measured values into an equation, convert to compatible units, retain units through working and report the requested unit. Translate data between numerical, table and graph forms. Plot both variables from data, choose suitable axes/scale, and use a straight-line gradient as change in y divided by change in x with units.
For geometry, rectangle area = length × width; triangle area = ½ × base × perpendicular height; cube surface area = 6s²; cube volume = s³. Keep squared and cubed units (cm², cm³) distinct.
What to do in an assessment
Standard form: 325 000 = 3.25 × 10⁵; 0.00042 = 4.2 × 10⁻⁴. Order-of-magnitude check: 4.2 × 10⁻⁴ is nearest in scale to 10⁻⁴, not 10⁴.
Percent example: a sample contains 12 g carbon in a 44 g formula mass, so percentage carbon = 12 ÷ 44 × 100 = 27.3%. Mean example: repeated readings 18, 20 and 22 cm³ give (18+20+22)÷3 = 20 cm³.
Ratio/fraction example: if a reaction uses 0.20 mol of A for every 0.10 mol of B, the amount ratio A:B is 2:1. A chromatographic spot travelling 3.0 cm while the solvent front travels 7.5 cm gives the fraction 3.0/7.5 = 0.40 (two-fifths) as its Rf.
Substitution example: q = mcΔT, with 100 g water, c = 4.18 J g⁻¹ °C⁻¹ and ΔT = 20.0 °C, gives q = 100 × 4.18 × 20.0 = 8360 J. Check that the mass and specific heat units are compatible.
Graph example: if volume rises from 0 cm³ at 0 s to 45 cm³ at 30 s on a straight section, gradient = 45÷30 = 1.5 cm³ s⁻¹. Plot time on the x-axis and volume on the y-axis; use a suitable scale and distinguish the measured points from the best-fit line.
Geometry example: a rectangle 8 cm by 3 cm has area 24 cm²; a triangle of base 8 cm and perpendicular height 3 cm has area ½×8×3 = 12 cm². A cube of side 2 cm has surface area 6×2² = 24 cm² and volume 2³ = 8 cm³.
In plain language
Use the right number form, show how data are organised, extract meaning from a graph, and keep units and dimensions consistent.
Self-check
Write 0.00056 in standard form. Find the mean of 14, 16 and 18. A cube has side 3 cm: find surface area and volume.
Show the answer
5.6 × 10⁻⁴. Mean = (14+16+18)÷3 = 16. Surface area = 6×3² = 54 cm²; volume = 3³ = 27 cm³.
Appendix C: Higher-tier significant figures, rearrangement and tangent gradientCore
Scope: Core · Higher tier in either Unit 1 paper tier or Unit 2 paper tier · Original Appendix A §4 significant figures and Appendix B bold mathematical skills; Higher in either written unit
Open revision guidance
What to know
Significant figures start at the first non-zero digit: leading zeros locate the decimal point and do not count; zeros between significant digits and meaningful trailing zeros do count. Give the precision requested and retain a sensible number of digits through intermediate working.
Changing the subject means rearranging an equation to isolate a specified variable while preserving equality. Apply inverse operations to both sides, use brackets where necessary, and check the new expression has compatible units.
The gradient of a tangent to a curved graph estimates the instantaneous rate of change at the point of contact. This skill is Higher-only in Appendix C; the common straight-line gradient remains available to both tiers.
What to do in an assessment
Round 0.004786 to two significant figures: the first two significant digits are 4 and 7; the next is 8, so the 7 rounds up. Answer: 0.0048. A value such as 2.0 may be needed to communicate two significant figures.
Rearrange c = n/V: multiply both sides by V to obtain n = cV; divide by c to obtain V = n/c. For q = mcΔT, divide by mc to make ΔT the subject: ΔT = q/(mc).
For a tangent to a curved volume–time graph, draw a straight tangent at the specified time, select two widely spaced points on that tangent and calculate gradient = Δvolume/Δtime. If points (20 s, 32 cm³) and (40 s, 56 cm³) lie on it, rate = (56−32)/(40−20) = 1.2 cm³ s⁻¹. This is the rate at the tangent point, not the mean for the full reaction.
Qualification-wide selection: these three Appendix C skills are required when either unit is entered at Higher tier. Foundation in one unit does not cancel Higher in the other.
In plain language
When either written unit is Higher, round answers precisely, rearrange equations safely and use a tangent to measure the rate at one instant.
Self-check
Round 0.03986 to 3 significant figures. Rearrange v = u + at for t. A tangent rises 18 cm³ over 15 s; find its gradient.
Show the answer
0.0399. t = (v−u)/a. Gradient = 18÷15 = 1.2 cm³ s⁻¹.

Apply chemistry across four required contextsCore
Scope: Core · Introduction, qualification aims: learners apply specified chemistry knowledge, understanding and skills in theoretical, practical, industrial and environmental contexts (printed p. 5; PDF p. 8).
Open revision guidance
What to know
The specification expects transfer of chemistry knowledge, understanding and skills across four kinds of setting, not four unrelated bodies of facts. Identify which chemistry is relevant, interpret the information provided and adapt the reasoning to the question's setting.
Theoretical context: use a particle, bonding or reaction model to explain an unfamiliar property or predict what happens when conditions change. State the model-based reason and its limits.
Practical context: plan and interpret an experiment, select apparatus, control variables, record evidence and evaluate reliability/safety. A titration, electrolysis or fuel-energy measurement connects classroom chemistry with direct observations.
Industrial context: explain how reaction chemistry and process choices scale to manufacturing. Blast-furnace extraction, aluminium electrolysis, the Contact process and Haber ammonia production involve feedstocks, yield, rate, energy, safety and economic decisions.
Environmental context: assess effects and possible responses using evidence. Examples include greenhouse-gas emissions, acid rain, quarrying, metal recycling, plastic disposal and fertiliser runoff; weigh benefits, consequences, uncertainty and alternatives.
What to do in an assessment
Worked transfer example: for a proposed fertiliser application, use the equation/chemistry to identify the nutrient benefit, read any crop or runoff data, explain how dissolved ions can reach waterways, weigh yield against eutrophication and recommend an application supported by the evidence. The context changes the decision criteria, not the underlying chemistry.
Classify a question by what it asks you to do, not merely by a noun in its story: explain a particle model (theoretical); measure a reaction (practical); compare plant conditions and costs (industrial); evaluate emissions or water quality (environmental). Some questions legitimately combine contexts.
In plain language
A chemistry question may be about a model, an experiment, a factory or the environment; use the relevant science and evidence in each setting.
Self-check
A question asks you to interpret a factory's ammonia-yield data and compare energy cost with production rate. Which two contexts are combined?
Show the answer
Industrial context (manufacturing choices, yield, rate and cost) and theoretical context (explaining how reaction conditions affect yield/rate).

The original 2027 Unit 3 Practical AssessmentCore
Scope: Core · Original English WJEC GCSE Chemistry Version 2, March 2019, §2.3 Unit 3 Practical Assessment (printed p. 28; PDF p. 30): 10%, untiered, 30 marks; two WJEC-provided annual task options, one selected by centres; Section A (Obtaining results, 6 marks) and Section B (Analysing and evaluating results, 24 marks), their controls and 60-minute sessions. Summary of assessment (printed p. 2; PDF p. 3) gives 10%, 30 marks and external marking. The 2026 specification's revised Unit 3 is excluded from the 2027 cohort.
Open revision guidance
What to know
For 2027, Unit 3 is the original Practical Assessment, not the revised Unit 3 first awarded in 2028. The English original specification sets it at 10% of the qualification, untiered and 30 marks. Each year WJEC provides two tasks based on GCSE Chemistry content; learners submit one, selected by the centre. Each task consists of two sections.
Section A — Obtaining results (6 marks): learners may work in a group of no more than three to obtain results using a given experimental method. It is limited control: candidates may work together to obtain results but must provide their own responses. Section A is one 60-minute session; teacher help is normally not expected, except if equipment fails.
Section B — Analysing and evaluating results (24 marks): learners analyse and evaluate the data from Section A and need access to their Section A assessment. It is high control: candidates work individually, without teacher feedback or assistance, under formal supervision. Section B is one 60-minute session.
The assessment takes place in centres in the first half of the spring term (January–February), is recommended for the final year of study, and is externally marked by WJEC. The tasks change annually. Its practical performance is not a written Unit 1/2 paper and cannot be completed or certified by reading this guide.
What to do in an assessment
For Section A, follow the centre's given method, work safely with the permitted group and record results yourself; for Section B, analyse/evaluate those results independently under formal supervision without feedback. Allow one 60-minute session for each section.
The centre selects one of WJEC's two annual task options. The guide can teach relevant chemistry knowledge and enquiry concepts, but learners still need their centre's instructions, practical experience and the actual assessment task. This overview does not replace the assessment.
In plain language
The original assessment has a 6-mark results section and a 24-mark analysis/evaluation section. It is untiered, centre-run and externally marked; this overview is not a substitute for doing it.
Self-check
Is the original 2027 Unit 3 tiered? What are the Section A and Section B names and mark totals, and how is Section B controlled?
Show the answer
No; it is untiered and worth 30 marks overall. Section A, Obtaining results, is 6 marks. Section B, Analysing and evaluating results, is 24 marks; it is completed individually, without teacher feedback or assistance, under formal supervision. This guide cannot complete the practical assessment.

Earlier topic checklist

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Chemistry topic list

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