We analysed every real Chemistry Paper 1 Higher Tier sitting AQA has made public since the pandemic disruption, including the actual questions students saw and the mark schemes examiners used. Below is what each recurring question type has asked, what the real data and diagrams showed, and a complete worked answer for each sitting we have. This is the closest you can get to seeing exactly what a full mark answer looks like without a real exam paper in front of you.
Questions © AQA, quoted for analysis. Diagrams and data described in our own words, not reproduced. Mark scheme content translated into plain English, not copied. PrepWise is independent and not endorsed by AQA.
Every sitting we have opens on the history of atomic structure. Each of these is a small, separately marked recall question, worth between 1 and 3 marks on its own, never a single big combined question.
It wants a description built only from features of THIS specific model: a positively charged sphere with electrons embedded in it, not any detail from the nuclear model that came later. Real question: June 2023 Q01.1, 2 marks, AO1, point marked.
The plum pudding model describes the atom as a ball of positive charge, with negative electrons embedded throughout it.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise atomic structure history questionsIt tests recall of the historical order the three particles were confirmed experimentally: electrons first (1897), then protons, then neutrons last (1932). Real question: June 2023 Q01.2, 1 mark in total for the full correct order, AO1, point marked.
Earliest: electrons. Then: protons. Latest: neutrons.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise atomic structure history questionsIt wants three separate, correctly worded differences other than the nucleus itself (already given), drawn from: how much empty space each model has, where the mass is concentrated, and whether the electrons and the central charge are separate or mixed together. Real question: May 2020 Q05.1, 3 marks, AO1, any three from a list mark scheme.
In the nuclear model, the atom is mostly empty space, but in the plum pudding model there is no empty space. In the nuclear model, the mass is concentrated in the nucleus, but in the plum pudding model the mass is spread out through the whole sphere. In the nuclear model, the electrons and the positive charge are in separate places, with electrons in orbits around the nucleus, but in the plum pudding model the electrons are embedded within the positive charge itself.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise atomic structure history questionsIt wants the specific correction Bohr made to Rutherford's nuclear model: he proposed electrons orbit the nucleus at fixed, specific distances (energy levels/shells) rather than anywhere around it. Real question: May 2020 Q05.2, 2 marks, AO1, point marked.
Bohr adapted the nuclear model by proposing that electrons orbit the nucleus, and that they exist at specific distances from the nucleus, in fixed energy levels or shells, rather than anywhere at all.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise atomic structure history questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
What does the atomic number of an element tell you?
Every sitting opens with small questions on how our model of the atom developed. Know the plum pudding model, the nuclear model, the Bohr model, and the order particles were discovered, cold.
Practise atomic structure history questionsThis is a reasoning question, not a recall question. It gives you a timeline and a claim, and rewards using the specific dates on the timeline to disprove the claim, not just asserting that it is wrong.
It wants you to use the given timeline (showing protons were discovered decades after 1869) to show atomic number could not have existed as a usable concept in 1869, when Mendeleev published his table. Real question: May 2020 Q05.3, 2 marks, AO3, point marked.
The student's suggestion cannot be correct because atomic number is the number of protons in an atom, and Figure 2 shows that protons were not established as a scientific concept until around 1920, decades after Mendeleev published his periodic table in 1869. Since the idea of atomic number depends on knowing about protons, a concept that did not exist until the 1920s, Mendeleev could not have used atomic number to reorder elements in 1869.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise periodic table development questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
John Newlands proposed the Law of Octaves in 1866. What did he notice about the elements?
This question rewards using given evidence to argue, not just recalling facts. Practise reading a timeline and using its dates to build a case.
Practise periodic table development questionsThis calculation appears in every single sitting we have, always using the same method: multiply each mass number by its percentage abundance, add the results, then divide by 100.
A straightforward weighted mean calculation using mass numbers 6 and 7, with abundances 7.6% and 92.4%.
Ar = ((6 x 7.6) + (7 x 92.4)) / 100 = (45.6 + 646.8) / 100 = 692.4 / 100 = 6.924, which rounds to 6.9 to 1 decimal place.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise relative atomic mass calculationsThe same method extended to three isotopes instead of two: mass numbers 20, 21, 22 with abundances 90.48%, 0.27%, 9.25%.
Ar = ((90.48 x 20) + (0.27 x 21) + (9.25 x 22)) / 100 = (1809.6 + 5.67 + 203.5) / 100 = 2018.77 / 100 = 20.1877, which rounds to 20.2 to 3 significant figures.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise relative atomic mass calculationsMass numbers 69 and 71 with abundances 60% and 40%, a cleaner ratio than most sittings.
Ar = ((69 x 60) + (71 x 40)) / 100 = (4140 + 2840) / 100 = 6980 / 100 = 69.8.
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Practise relative atomic mass calculationsMass numbers 203 and 205 with abundances 30% and 70%, testing the same method with larger mass numbers (this metal is thallium).
Ar = ((203 x 30) + (205 x 70)) / 100 = (6090 + 14350) / 100 = 20440 / 100 = 204.4.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise relative atomic mass calculationsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
What does the atomic number of an element tell you?
This calculation is guaranteed on every paper. Master the weighted-average method (mass number x abundance, sum, divide by 100) so it becomes automatic.
Practise relative atomic mass calculationsEvery sitting asks you to compare bonding types using named example compounds. You need all three bonding types (ionic, covalent, metallic) and the giant vs simple molecular distinction rock solid.
It wants you to classify and compare all three bonding types using these specific named compounds: CO2 is simple molecular, MgO is giant ionic, SiO2 is giant covalent.
Carbon dioxide and silicon dioxide are both made of atoms bonded by shared electron pairs, covalent bonds, since both are formed from two non-metal elements. Magnesium oxide, in contrast, is made up of ions, since it is formed from a metal and a non-metal, so electrons are transferred rather than shared: two electrons move from magnesium to oxygen.
Silicon dioxide and magnesium oxide are both giant structures, extending in all directions with many strong bonds (covalent in silicon dioxide, ionic in magnesium oxide), which is why both have very high melting points. Carbon dioxide, however, exists as small, separate molecules held together only by weak intermolecular forces between molecules, even though the covalent bonds within each CO2 molecule are strong.
This structural difference explains carbon dioxide's low melting and boiling point: only the weak intermolecular forces need overcoming for it to become a gas, whereas silicon dioxide and magnesium oxide need their many strong bonds broken throughout the whole giant lattice, which needs far more energy and gives them very high melting points.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise structure and bonding comparison questionsIt wants you to explain the state difference using molecule size and intermolecular force strength: methane's small molecules have weak intermolecular forces (so it's a gas), poly(ethene)'s huge molecules have much stronger intermolecular forces (so it's a solid).
Poly(ethene) is a solid at room temperature because its molecules are much larger than methane's molecules. This means poly(ethene) has stronger intermolecular forces between its molecules than methane does. Stronger intermolecular forces need more energy to overcome, so poly(ethene) has a much higher melting and boiling point than methane, meaning it is a solid at room temperature while methane, with its small molecules and weak intermolecular forces needing very little energy to overcome, is a gas.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise structure and bonding comparison questionsIt wants two separate explanations linked to graphite's actual structure: delocalised electrons explain conductivity, and the layered structure with weak forces between layers explains softness.
Graphite conducts electricity because each carbon atom in graphite only forms three covalent bonds, using only three of its four outer electrons. This leaves one electron per carbon atom delocalised, free to move through the structure. These delocalised electrons carry electrical charge, which is why graphite conducts electricity, unlike diamond where every electron is used in a covalent bond.
Graphite is soft and slippery because it has a layered structure made of interlocking hexagonal rings of carbon atoms, with only weak intermolecular forces between the layers and no covalent bonds holding one layer to the next. This means the layers can slide over each other easily, which is what makes graphite soft and useful as a lubricant.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise structure and bonding comparison questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
Why do giant covalent structures have very high melting points?
This comparison question tests all three bonding types at once. Know giant ionic, giant covalent, and simple molecular structures with real named examples, and always link structure back to the observed property.
Practise structure and bonding comparison questionsIonic bonding between a Group 1 or Group 2 metal and a Group 6 or 7 non-metal appears in nearly every sitting, always testing electron transfer and the resulting ion charges.
It wants a full diagram showing TWO sodium atoms transferring one electron each to ONE oxygen atom, since sodium needs to lose 1 electron and oxygen needs to gain 2, so the ratio must be 2:1.
Two sodium atoms are needed for each oxygen atom, because sodium is in Group 1 and needs to lose 1 outer electron to reach a full outer shell, while oxygen is in Group 6 and needs to gain 2 electrons. So each sodium atom transfers its single outer electron to the oxygen atom, meaning two electrons in total are transferred from the two sodium atoms into oxygen's outer shell.
This produces two sodium ions, each with a single positive charge (Na+), and one oxide ion with a 2- charge (O2-), since oxygen has gained two electrons overall. In the diagram, each ion is shown in square brackets with its charge marked outside, and only the outer shell electrons are drawn, using crosses for sodium's electrons and dots for oxygen's own electrons to show clearly which atom they came from.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise ionic bonding electron transfer questionsIt wants a written explanation (not a diagram) of the 1:1 electron transfer between magnesium and oxygen, since both need to move 2 electrons.
Magnesium is in Group 2 and has 2 outer electrons, which it loses to reach a full outer shell. Oxygen is in Group 6 and needs to gain 2 electrons to reach a full outer shell. So the magnesium atom loses its 2 outer electrons, and these are transferred to the oxygen atom.
Because magnesium loses exactly 2 electrons and oxygen gains exactly 2 electrons, magnesium ions (Mg2+) and oxide ions (O2-) are formed, and both ions now have a complete outer shell.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise ionic bonding electron transfer questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
Which combination of elements forms an ionic compound?
Ionic bonding questions always test the electron count logic. Know your groups, work out the electron ratio needed, and always state the resulting ion charges.
Practise ionic bonding electron transfer questionsMetallic bonding and its link to conductivity and alloy hardness is tested in nearly every sitting, sometimes as recall, sometimes using a data table to evaluate which metal suits a use.
This exact question type has repeated across sittings. It wants the three-part explanation: delocalised electrons exist, they carry charge, they move through the structure.
Metals have delocalised electrons, meaning electrons that are free to move rather than being fixed to one particular atom. These electrons carry electrical charge, and because they can move through the whole metal structure, they allow electricity to be conducted.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise metallic bonding and alloy questionsIn a metal, the outer electrons of each atom become delocalised, meaning they are no longer attached to one particular atom and are free to move throughout the whole structure. These delocalised electrons carry electrical charge, so as they move through the metal, they allow an electric current to flow.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise metallic bonding and alloy questionsIt wants you to link the different-sized atoms in an alloy to a disruption in the regular structure that restricts electron movement, reducing conductivity.
In alloys, atoms of different sizes distort the regular layers of the metal structure. This restricts the movement of the delocalised electrons through the structure, which is why alloys do not conduct electricity as well as pure metals.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise metallic bonding and alloy questionsIt wants the mechanical explanation: different-sized atoms in an alloy distort the regular layers, making it harder for layers to slide over each other than in a pure metal.
In an alloy, atoms of different sizes are mixed in with the metal atoms. This distorts the regular layers of atoms found in a pure metal. Because the layers are distorted rather than regular, they cannot slide over each other as easily as they can in a pure metal, which is what makes alloys harder.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise metallic bonding and alloy questionsIt wants you to use the actual numbers given (conductivity, density, cost) to justify a specific metal choice for each of the three named uses, then reach an overall judgement.
Aluminium is the best choice for overhead power cables, since it has by far the lowest density of the three metals (2.7 g/cm3 compared to 9.0 for copper and 10.5 for silver) and is the cheapest at just 1.50 pounds per kg, both important since overhead cables need to be light enough to be supported by pylons over long distances without needing very high-cost material.
Copper is the best choice for wiring in homes, since it has a high electrical conductivity, close to silver's, at a much lower cost, 7.00 pounds per kg compared to silver's 640.00 pounds per kg. Since household wiring needs a large total length of wire, using silver would be far too expensive, so copper's balance of good conductivity and low cost makes it the practical choice.
Silver is only justified for printed circuit boards, since although it is the most expensive metal by far, circuit boards only need very small amounts of material, so the high cost is not as significant, and silver's higher conductivity than copper (63.0 compared to 59.6) gives a genuine performance benefit where a small amount of extra conductivity matters most.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise metallic bonding and alloy questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
In metallic bonding, what are the electrons called that are free to move throughout the metal structure?
Metallic bonding recall questions repeat almost word for word across sittings. Learn the three-part delocalised electron explanation and the alloy distortion explanation exactly.
Practise metallic bonding and alloy questionsA titration concentration calculation appears on every single sitting, always following the same three-step method: moles of the known solution, use the equation ratio, then moles over volume for the unknown concentration.
This uses the equation Na2CO3 + 2HNO3 -> 2NaNO3 + CO2 + H2O, so a 1:2 mole ratio applies between carbonate and acid.
Moles of Na2CO3 = (25.0 / 1000) x 0.124 = 0.00310 mol. From the equation, 1 mole of Na2CO3 reacts with 2 moles of HNO3, so moles of HNO3 = 2 x 0.00310 = 0.00620 mol. Concentration of HNO3 = (0.00620 / 23.6) x 1000 = 0.263 mol/dm3, to 3 significant figures.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise titration concentration calculationsThis uses the equation 2HCl + Ba(OH)2 -> BaCl2 + 2H2O, a 2:1 ratio (acid to hydroxide), the reverse ratio direction from the carbonate example.
Moles of Ba(OH)2 = (23.50 / 1000) x 0.100 = 0.00235 mol. From the equation, 2 moles of HCl react with 1 mole of Ba(OH)2, so moles of HCl = 2 x 0.00235 = 0.00470 mol. Concentration of HCl = (0.00470 / 25.00) x 1000 = 0.188 mol/dm3.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise titration concentration calculationsThis uses a 3:1 ratio (NaOH to citric acid), the largest ratio seen across the sittings, testing whether students can handle mole ratios beyond simple 1:1 or 1:2.
Moles of citric acid = (13.3 / 1000) x 0.0500 = 0.000665 mol. From the equation, 3 moles of NaOH react with 1 mole of citric acid, so moles of NaOH = 3 x 0.000665 = 0.001995 mol. Concentration of NaOH = (0.001995 / 25.0) x 1000 = 0.0798 mol/dm3.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise titration concentration calculationsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
What is the purpose of a titration?
This calculation is worth several guaranteed marks on every paper. Always read the actual mole ratio from the given equation rather than assuming 1:1, and follow the same three-step method every time.
Practise titration concentration calculationsA full practical planning question appears in nearly every sitting, either for making a salt (excess solid, filter, crystallise) or for planning an experiment with proper control of variables.
It wants the full standard method: add zinc carbonate to dilute hydrochloric acid until in excess, filter to remove the unreacted excess, then heat/evaporate the filtrate and leave to crystallise.
Add zinc carbonate to dilute hydrochloric acid in a beaker, stirring as you go, continuing to add the zinc carbonate until it is in excess, shown by solid remaining and no more effervescence (fizzing) occurring. This ensures all the acid has reacted, since we cannot easily test for leftover acid but we can see when the carbonate stops reacting.
Filter the reaction mixture to remove the excess, unreacted zinc carbonate, leaving a clear solution of zinc chloride in the filtrate.
Heat the filtrate using a water bath or electric heater until it reaches crystallisation point, then leave the solution to cool and finish crystallising. Finally, pat the crystals dry using filter paper.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise practical planning questionsIt wants the practical steps of measuring a fixed volume of acid, recording starting temperature, adding a known mass of carbonate, recording the highest temperature reached, then repeating with different masses while controlling other variables.
Measure a fixed volume of hydrochloric acid using a measuring cylinder and pour it into a suitable insulated container, such as a polystyrene cup, then measure and record the initial temperature of the acid using a thermometer.
Add a known mass of sodium carbonate, measured using a balance, to the acid and stir. Measure and record the highest temperature reached by the mixture. Repeat this whole investigation using different masses of sodium carbonate each time.
To make the investigation a fair test, use the same starting temperature, the same volume of hydrochloric acid, and the same concentration of hydrochloric acid every time, changing only the mass of sodium carbonate.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise practical planning questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
Which of the following is the correct method for making copper sulfate crystals from copper oxide and sulfuric acid?
Practical planning questions reward a logically sequenced method with named equipment and controlled variables. Learn the standard salt-making method and the standard fair-test structure cold.
Practise practical planning questionsThis shorter planning question asks you to design a fair comparison test that tells three metals apart using their reactivity, rather than planning a full salt preparation.
It wants a comparative method: add each metal to the same volume/concentration of acid, and compare the rate of bubbling or temperature change, since copper does not react at all while magnesium reacts fastest.
I will add each metal, X, Y and Z, in turn to separate but identical samples of dilute hydrochloric acid. To keep this a fair test I will control four variables: the same volume and concentration of acid each time, the same mass of each metal, the same particle size or form of each metal (for example all three cut as strips of the same surface area, or all three used as powder), and the acid starting at the same room temperature for each test.
To compare the reactions quantitatively rather than just predicting the outcome, I will measure the temperature of each acid sample every thirty seconds using a thermometer for two minutes, and separately time how long it takes for effervescence (bubbling) to stop in each sample, using a stopwatch. Recording numbers this way, rather than just describing what I expect to see, means the results can actually be compared and are repeatable.
Copper is below hydrogen in the reactivity series, so I expect no bubbles and no temperature rise in that sample, identifying it as copper. Of the remaining two metals, the one that shows a faster rate of bubbling and a bigger temperature increase is the more reactive metal, magnesium, since it is above iron in the reactivity series. The slower, cooler reaction identifies iron.
As a further check, once the reactions have finished I will look at the colour of the resulting solution. Iron reacts with hydrochloric acid to form iron chloride solution, which is pale green, while magnesium forms magnesium chloride solution, which is colourless. This gives me a second, independent way to distinguish iron from magnesium in case the bubbling or temperature results are too close to call.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise practical planning questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
Which of the following is the correct method for making copper sulfate crystals from copper oxide and sulfuric acid?
This shorter planning question rewards a fair comparison test with named conditions and a clear predicted outcome for each metal. Know the reactivity series and how to describe a measurable comparison.
Practise practical planning questionsAqueous electrolysis is tested in every sitting, always requiring you to work out which ion is actually discharged (not just which ions are present) using the standard preferential discharge rules.
It wants the full mechanism: water molecules break down to produce H+ and OH- ions, the OH- ions move to the positive electrode where they are discharged (lose electrons) to form oxygen.
Water molecules break down to produce hydrogen ions and hydroxide ions. Because the hydroxide ions are negatively charged, they are attracted to and move towards the positive electrode.
At the positive electrode, the hydroxide ions are discharged, losing electrons (oxidised), to produce oxygen molecules.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise electrolysis product prediction questionsIt wants you to explain that chlorine is discharged at the positive electrode and hydrogen at the negative electrode, leaving sodium ions and hydroxide ions behind in solution, which together form sodium hydroxide.
At the negative electrode, hydrogen ions are discharged to produce hydrogen gas, and at the positive electrode, chloride ions are discharged to produce chlorine gas.
This leaves sodium ions and hydroxide ions remaining in the solution, since neither of these was discharged. Together these form sodium hydroxide, which is the alkaline solution produced.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise electrolysis product prediction questionsIt wants the completed half equation for aluminium formation, and the reason sodium is not produced: sodium is more reactive than aluminium, so aluminium ions are preferentially discharged.
Al3+ + 3e- -> Al. Sodium is not a product of the electrolysis because sodium is more reactive than aluminium, so aluminium ions are discharged in preference to sodium ions at the negative electrode.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise electrolysis product prediction questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
When sodium chloride (NaCl) is dissolved in water, which four types of ion are present in the solution?
Electrolysis questions always come down to the preferential discharge rules. Learn them as a checklist: is the metal more reactive than hydrogen? Is a halide present? Answer those two questions and you know both products.
Practise electrolysis product prediction questionsDisplacement and redox questions appear on every sitting, often using unnamed metals A, B, C, D and their reactions to test whether you can reason about reactivity from evidence rather than recall alone.
It wants Fe2O3 identified as the reduced substance, with the reason being that it loses oxygen during the reaction.
Iron oxide (Fe2O3) is reduced, because it loses oxygen during the reaction, becoming iron metal.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise redox and displacement reasoning questionsIt wants you to identify electron loss/gain for the redox question (A loses electrons, B+ gains electrons), then use the ionic charges shown in the given table's formulae to work out which metal has a 3+ ion, matching aluminium.
This is a redox reaction because A loses electrons (is oxidised) while B+ gains electrons (is reduced), happening simultaneously in the same reaction, which is what makes a reaction 'redox' (reduction and oxidation together).
Metal C could be aluminium. This is because Table 6 shows metal C forms the compound C(NO3)3, and since the nitrate ion has a 1- charge, three nitrate ions are needed to balance a metal ion with a 3+ charge. Aluminium forms Al3+ ions, matching this exactly.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise redox and displacement reasoning questionsIt wants you to read the given displacement results for metals A, B, C and D and work out which one is the most reactive, since the metal with the greatest tendency to form positive ions is simply the most reactive metal.
Metal D has the greatest tendency to form positive ions. Table 6 shows that metal C did not react with a solution containing D ions, meaning C cannot displace D, so D must be more reactive than C. Since C is already known to be more reactive than A, and A is more reactive than B, D sits above all the others in the reactivity order.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise redox and displacement reasoning questionsIt wants you to link each observation to what actually happened chemically: grey crystals are silver (displaced out of solution), blue solution shows copper ions now in solution, meaning copper displaced silver, so copper must be more reactive.
The pale grey crystals forming are solid silver, which has been displaced out of the solution by the copper. The solution turning blue shows that copper ions are now present in the solution, since copper compounds are typically blue.
Because copper has displaced silver from the solution, meaning copper has reacted to form ions while silver has been produced as the metal, this shows copper is more reactive than silver, and therefore silver is less reactive than copper.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise redox and displacement reasoning questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
Which statement correctly describes a displacement reaction?
Redox and displacement questions test reasoning from evidence as much as recall. Learn OIL RIG for electron transfer, and the oxygen-based version (loses oxygen = reduced) for extraction reactions specifically.
Practise redox and displacement reasoning questionsAn atom economy calculation appears in nearly every sitting, always using the same formula: (Mr of desired product / total Mr of all reactants) x 100.
It wants tin metal (the useful product, Mr 119) divided by the total Mr of the reactants (SnO2 plus carbon, using the given atomic masses C=12, O=16, Sn=119).
Mr of SnO2 = 119 + (2 x 16) = 151. Total Mr of reactants = 151 + 12 (carbon) = 163. Percentage atom economy = (119 / 163) x 100 = 73.0% to 2 significant figures.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise atom economy calculationsThis reverses the usual calculation: given the atom economy percentage, you must work backwards to find the unknown metal's Ar using algebra.
Mr of 3H2O (the unwanted byproduct) = 3 x ((2 x 1) + 16) = 3 x 18 = 54. Setting up the atom economy equation: (ArX / (ArX + 54)) x 100 = 77.3. Rearranging: 100 x ArX = 77.3 x (ArX + 54), so 100ArX = 77.3ArX + 4174.2, so 22.7ArX = 4174.2, so ArX = 184.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise atom economy calculationsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
Which method is used to extract a metal that is MORE reactive than carbon?
Atom economy always uses the same core formula. Learn to identify the actual useful product correctly, and be ready for questions that ask you to rearrange the formula rather than just plug numbers in.
Practise atom economy calculationsReaction profiles and bond energy calculations appear in every sitting. You need to draw exothermic profiles correctly (products below reactants) and use the bonds-broken-minus-bonds-formed method fluently.
It wants the missing product energy level drawn below the reactants (since the reaction is exothermic), with activation energy labelled from reactants to peak, and overall energy change labelled from reactants to products.
Since the reaction between hydrogen and chlorine is exothermic, the product level (2HCl) must be drawn below the level of the reactants (H2 + Cl2), showing that energy has been released overall. The activation energy is labelled as the vertical distance from the reactants' energy level up to the very top of the curve (the peak), and the overall energy change is labelled as the vertical distance from the reactants' level down to the products' level.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise reaction profile and bond energy calculationsIt wants the correct bonds-broken-minus-bonds-formed expression (accounting for 2 H-Cl bonds formed), and the explanation that more energy is released forming bonds than is needed to break them.
The correct expression is 436 + 346 - (2 x 432) kJ/mol, since one H-H bond and one Cl-Cl bond are broken, but TWO H-Cl bonds are formed (since the equation produces 2HCl, not 1).
Energy is needed to break the bonds in the reactants, and energy is released when the new bonds form in the products. This reaction releases energy to the surroundings because the energy released forming the new bonds is greater than the energy needed to break the original bonds.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise reaction profile and bond energy calculationsIt wants the full bonds-broken and bonds-formed totals set up as an equation using the given 1034 kJ/mol difference, then solved for X.
Bonds broken = 4(H-S) + 3(O=O) = 4(364) + 3(498) = 1456 + 1494 = 2950 kJ/mol. Since energy released forming bonds is 1034 kJ/mol greater than energy needed to break bonds, bonds formed = 2950 + 1034 = 3984 kJ/mol.
Bonds formed = 4(H-O) + 4X = 4(464) + 4X = 1856 + 4X. Setting this equal to 3984: 1856 + 4X = 3984, so 4X = 2128, so X = 532 kJ/mol.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise reaction profile and bond energy calculationsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
What does activation energy represent on a reaction profile?
Reaction profile and bond energy questions reward precision: check the actual bond counts in the equation before calculating, and always know whether you're drawing an exothermic or endothermic profile.
Practise reaction profile and bond energy calculationsHydrogen fuel cells appear as a distinct question, usually combining recall of advantages/disadvantages with a calculation linking energy released per mole to the volume of gas needed.
It wants two genuine, specific comparative advantages: faster refuelling than recharging, and greater range/no loss of efficiency in cold weather, or no toxic waste to dispose of.
Hydrogen fuel cells take less time to refuel than rechargeable cells take to recharge, meaning a car can be back on the road faster. Fuel cell vehicles can also travel further before needing to refuel than a rechargeable cell vehicle can travel before needing to recharge.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise hydrogen fuel cell questionsIt wants a correctly balanced half equation for either electrode: hydrogen oxidation at the negative electrode, or oxygen reduction at the positive electrode.
At the negative electrode: H2 -> 2H+ + 2e-.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise hydrogen fuel cell questionsIt wants unit conversion (MJ to kJ), then energy divided by energy-per-mole to get moles, then moles multiplied by molar gas volume to get the final volume.
Converting units: 58 MJ = 58,000 kJ. Moles of hydrogen needed = 58,000 / 290 = 200 mol.
Volume of hydrogen gas = 200 x 24 = 4800 dm3.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise hydrogen fuel cell questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
A simple electrochemical cell is made by placing two different metal electrodes into an electrolyte. Which statement correctly describes what affects the voltage produced?
Fuel cell questions combine recall of genuine advantages over rechargeable cells with an energy-to-volume calculation. Always convert units carefully before dividing.
Practise hydrogen fuel cell questionsGroup trend explanations appear in nearly every sitting, always requiring the same shielding-and-distance logic, just applied in opposite directions for Group 1 (losing electrons more easily) and Group 7 (gaining electrons less easily).
It wants the full shielding-and-distance explanation applied to Group 1: rubidium's outer electron is further from the nucleus (more shells) than potassium's, so it is lost more easily.
Rubidium's outer shell electron is further from the nucleus than potassium's, since rubidium has more electron shells than potassium.
This means there is less (electrostatic) attraction between the nucleus and rubidium's outer electron than there is in potassium.
So the outer electron in rubidium is lost more easily than in potassium, which is why rubidium is more reactive than potassium.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise Group 1 and Group 7 reactivity trend questionsA more general version of the same question, asking for the trend across the whole group rather than a specific pair, testing the same shielding-and-distance logic.
Reactivity increases going down Group 1. This is because atoms get larger going down the group, with the outer electron further from the nucleus in each successive element.
Because the outer electron is further away, there is less attraction between the nucleus and the outer electron/shell, so the atom loses its outer electron more easily as you go down the group, which is why reactivity increases.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise Group 1 and Group 7 reactivity trend questionsIt wants the mirror-image explanation to Group 1: halogen atoms get larger down the group, so an incoming electron is attracted less strongly and gained less easily, decreasing reactivity.
Going down Group 7, the outer shell/electrons become further from the nucleus, since the atoms have more electron shells lower down the group.
So the nucleus has less attraction for the outer electrons, or equivalently, there is more shielding between the nucleus and an incoming electron, meaning an electron is gained less easily as you go down the group, which is why halogens become less reactive lower down.
Could you have written this? Every fact in this answer is drilled in our quizzes — the writing is the easy part once the evidence is automatic.
Practise Group 1 and Group 7 reactivity trend questionsThe topic changes by sitting — the mark scheme never does. Learn this once, then open your question above for that sitting’s sources and a full worked answer.
How many electrons do halogens have in their outer shell?
Group trend explanations use the same shielding-and-distance logic every time, just in opposite directions. Practise both Group 1 (losing electrons, reactivity increases) and Group 7 (gaining electrons, reactivity decreases) until you never mix them up.
Practise Group 1 and Group 7 reactivity trend questionsAcross the 4 sittings we have full papers for, these are the topics with the most exam appearances and marks at stake on Paper 1.
Nanoparticles as a standalone main topic (has appeared only as brief context) · Transition metals as a standalone main topic beyond general properties · Electrolysis of aluminium extraction from molten aluminium oxide as the SOLE focus of a full question (it has appeared as part of a wider electrolysis question)
These topics have not been the main focus of a Paper 1 question in the papers we analysed, but Paper 1 covers the first half of the specification, so make sure you have them ready too since they can appear in any future sitting.
The diagrams and data are described in our own words, not reproduced, and the worked answers are written entirely by us, aimed at the top mark scheme point for each question. They are not copied from AQA's own exemplar materials, since that would breach copyright, but they are built to hit exactly what the real mark scheme rewarded that year. PrepWise is independent of AQA and not endorsed by them.
Sometimes stems repeat closely, and calculation types like relative atomic mass, titration concentration, and atom economy return in some form almost every sitting. But you cannot rely on repeats alone, since the actual numbers and compounds change every time even when a question type is similar. Use this page to see which TYPES of question keep returning and make sure you can do the method cold, whatever the exact numbers turn out to be.
Chemistry Paper 1 does not use the long 12 or 16 mark essay-style questions found in some other subjects. Its longest single questions are 6-mark level of response questions (planning a method, or comparing structure and bonding) and multi-step calculations worth up to 6 marks total across several linked steps. We have still verified every claimed mark against the real mark scheme.
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