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Key to Symbols
Key Definition: Essential terminology you must memorize word-for-word for the exam.
Exam Tip: Hints, common question formats, and advice on how to secure full marks.
Worked Example: Step-by-step mathematical calculations or balancing procedures.
Common Mistake: Misconceptions to avoid that frequently cost students marks.
Unofficial revision material for AQA GCSE Chemistry. For revision only. Not affiliated with, endorsed by, or connected to AQA or any exam board.
An atom consists of a central positive nucleus (protons and neutrons) surrounded by negative electrons in shells.
Particle
Relative Charge
Relative Mass
Proton
+1
1
Neutron
0
1
Electron
-1
1/2000 (negligible)
1.6 Atomic Size & Mass Distribution
Ion
An atom (or group of atoms) that has gained or lost electrons, giving it a charge.
Atomic Number (Z)
The number of protons in the nucleus of an atom. Identifies the element.
Mass Number (A)
The total number of protons and neutrons in the nucleus of an atom.
Atoms are tiny. The atomic radius is approximately 1 × 10-10 metres (0.1 nm). The radius of the nucleus is about 1 × 10-14 metres (10,000 times smaller than the atom).
Nearly all of the atom's mass is concentrated in the nucleus, as protons and neutrons have a relative mass of 1, whereas electron mass is negligible.
Exam Tip
Be prepared to compare the size of an atom to its nucleus using standard form. For example, 1 × 10-10 m compared to 1 × 10-14 m shows the atom is 10,000 (104) times larger than its nucleus.
Common Mistake
Do not confuse mass number (A) with atomic number (Z). Mass number (top, larger value) is protons + neutrons. Atomic number (bottom, smaller value) is protons only.
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Electrons occupy energy levels (shells). Capacities: 1st shell = 2, 2nd shell = 8, 3rd shell = 8. Electrons fill lowest available shells first.
Configurations show the electron structure: Sodium (11 electrons) is 2.8.1; Chlorine (17 electrons) is 2.8.7.
Key Fact: Periodic Connections
Group number = number of outer shell (valence) electrons. Period number = number of occupied electron shells.
1.9 The Periodic Table
Elements are arranged in order of atomic number. Early periodic tables were arranged by atomic weight. Mendeleev left gaps for undiscovered elements, predicting their properties. Metals (left/bottom) lose electrons to form positive ions. Non-metals (right/top) gain/share electrons.
1.10 Groups in the Periodic Table
Group 0: The Noble Gases
Helium, Neon, Argon. Full outer shells (stable electronic configuration). Monatomic and chemically inert. Boiling points increase down the group as relative atomic mass increases (stronger intermolecular forces require more energy to break).
Group 1: The Alkali Metals
Li, Na, K. 1 outer electron. Reactivity increases down the group as the outer electron is further from the nucleus, more shielded by inner shells, and lost more easily. React with water to form metal hydroxide and hydrogen:
2Na + 2H2O → 2NaOH + H2
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F, Cl, Br, I. 7 outer electrons. Diatomic molecules (X2). Reactivity decreases down the group as it is harder to attract an incoming electron due to increased distance and shielding. Boiling points increase down the group.
Halogen
Colour
State at Room Temp
Fluorine (F2)
Pale yellow
Gas
Chlorine (Cl2)
Green
Gas
Bromine (Br2)
Red-brown
Liquid
Iodine (I2)
Dark grey
Solid (purple vapour)
A more reactive halogen displaces a less reactive halide from its salt solution:
Cl2 + 2KBr → 2KCl + Br2
Observation: the solution turns orange-brown as bromine is displaced.
Transition Metals Chemistry Only
Central block of the periodic table. Compared to Group 1 metals: harder, denser, stronger, higher melting points, and much less reactive.
Metal atoms lose outer electrons to become positive ions (cations). Non-metal atoms gain these electrons to become negative ions (anions). They are held by strong electrostatic attractions.
Figure 2.1: Electron transfer in NaCl ionic bonding (dot-and-cross diagram).
2.3 Giant Ionic Lattice
Ionic compounds form a 3D giant ionic lattice of alternating positive and negative ions held by strong, multi-directional electrostatic forces.
Properties:
High melting/boiling points: Large amount of energy needed to break strong electrostatic attractions between oppositely charged ions.
Electrical conductivity: Do not conduct when solid (ions fixed in place). Conduct when molten or dissolved in water because ions are free to move and carry charge.
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Covalent bonding occurs when two non-metal atoms share one or more pairs of electrons to achieve stable full outer shells. These bonds are very strong.
Covalent substances can consist of small molecules (like water) or giant covalent structures (like diamond).
Figure 2.2: Covalent bonding in a water molecule (H₂O) showing shared electron pairs.
2.5 Small Molecule Covalent Substances
Substances with small covalent molecules (e.g. H2O, CO2, CH4, NH3) have very low melting and boiling points.
Bonding structure: Atoms within the molecules are joined by very strong covalent bonds. However, the forces between separate molecules (intermolecular forces) are very weak.
Melting/boiling: When these substances melt or boil, it is only the weak intermolecular forces that are broken, not the strong covalent bonds. Very little energy is needed.
Conductivity: Do not conduct electricity because the molecules are neutral (no overall charge) and have no free delocalised electrons or ions to carry charge.
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Polymers consist of very large molecules made of long chains of repeating units (monomers) joined by strong covalent bonds.
Intermolecular forces: Because polymer molecules are very large, the sum of the intermolecular forces between the chains is relatively strong.
State: This makes polymers solid at room temperature. However, they melt at lower temperatures than giant ionic or giant covalent structures because these forces are still weaker than chemical bonds.
2.7 Giant Covalent Structures
Giant covalent structures (macromolecules) contain huge networks of atoms joined by strong covalent bonds. They have very high melting/boiling points because breaking these structures requires breaking many strong covalent bonds.
Diamond: Each carbon atom forms 4 strong covalent bonds in a rigid tetrahedral structure. This makes diamond extremely hard and unable to conduct electricity (no free electrons).
Graphite: Each carbon forms 3 bonds, creating hexagonal layers. The fourth electron is delocalised and free to move along the layers, so graphite conducts electricity and heat. Weak forces between layers let them slide, making graphite soft and slippery.
Silicon dioxide (Silica, SiO2): Similar structure to diamond. Each silicon atom is bonded to 4 oxygen atoms, and each oxygen to 2 silicon atoms. Extremely high melting point and hard.
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Graphene: A single 2D layer of graphite, one atom thick. Very strong, light, and conduct electricity. Used in electronics and composite materials.
Fullerenes: Carbon molecules with hollow shapes (e.g. Buckminsterfullerene, C60, a sphere). Used for drug delivery, lubricants, and catalysts.
Carbon nanotubes: Cylindrical fullerenes. High tensile strength and electrical conductivity. Used in nanotechnology and structural materials.
2.9 Metallic Bonding
Metals consist of a giant lattice of positive metal ions surrounded by a "sea" of delocalised electrons. This is held together by strong electrostatic attractions.
Properties: High melting/boiling points. Delocalised electrons carry thermal energy and electrical charge, so metals are excellent conductors. Pure metals are malleable because layers of ions can slide over each other.
2.10 Alloys
Alloys are mixtures of a metal with other elements. Pure metals are soft because their atoms are arranged in regular layers that slide easily. In alloys, different-sized atoms distort these layers, making it harder for them to slide, which makes alloys harder than pure metals.
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The three states of matter are solid, liquid, and gas. State changes depend on the strength of forces between particles, which is determined by the bonding and structure of the substance.
Figure 2.3: Arrangement of particles in solid, liquid, and gas states.
2.12 NanoparticlesChemistry Only
Nanoparticles have a diameter of 1 to 100 nm. They have an extremely high surface area to volume ratio, giving them properties that are different from the bulk material.
SA:V Ratio: As particles decrease in size, their surface area to volume ratio increases dramatically, making them highly reactive.
Uses: Sun creams (ZnO nanoparticles block UV, transparent on skin), catalysts (large surface area), medicine (targeted drug delivery), and electronics.
Risks: May enter the body and cells (toxic effects), or accumulate in environments. Long-term risks are not yet fully understood.
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No atoms are lost or made during a chemical reaction, so the total mass of the products is always equal to the total mass of the reactants. This is why symbol equations must be balanced.
Apparent mass changes: If the mass seems to increase, a gaseous reactant has joined (e.g. a metal reacting with oxygen to form an oxide). If it seems to decrease, a gas has escaped (e.g. thermal decomposition of a metal carbonate releasing CO2).
Exam Tip
Apparent mass changes only occur in non-enclosed systems. In a sealed container the total mass never changes, because no gas can enter or leave.
Worked example
In 2Mg + O2 → 2MgO, if 48 g of magnesium reacts with 32 g of oxygen, the mass of magnesium oxide formed = 48 + 32 = 80 g.
3.2 Relative Formula Mass (Mr)
The relative formula mass (Mr) of a compound is the sum of the relative atomic masses (Ar) of the atoms in the numbers shown in the formula.
Calculating Mr of CaCO3
Find Ar values: Ca = 40, C = 12, O = 16.
Add values: Mr = 40 + 12 + (16 × 3) = 100.
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Chemical amounts are measured in moles (mol). One mole of any substance contains 6.02 × 1023 particles (Avogadro's constant). The mass of one mole of a substance in grams is equal to its relative formula mass.
moles = mass (g) / Mr
Calculating Moles from Mass
Calculate the number of moles in 44 g of CO2 (Mr = 44):
Moles = 44 g / 44 = 1.0 mol.
Calculating Mass from Moles
Calculate the mass of 0.2 mol of H2O (Mr = 18):
Mass = moles × Mr = 0.2 mol × 18 = 3.6 g.
3.4 Amounts in EquationsChemistry Only
Balanced chemical equations show the ratio of moles that react together. We can use these ratios to calculate the mass of a product formed or reactant needed.
Calculating Product Mass
Calculate mass of MgO formed from burning 12 g of Mg: 2Mg + O2 → 2MgO
Calculate moles of Mg: moles = 12 g / 24 = 0.5 mol.
Use equation ratio: 2 mol Mg forms 2 mol MgO (1:1 ratio), so 0.5 mol Mg forms 0.5 mol MgO.
Calculate mass of MgO (Mr = 40): mass = 0.5 mol × 40 = 20 g.
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In a reaction, the reactant that is completely used up is the limiting reactant. Any other reactants are in excess. The amount of product formed is directly proportional to the amount of limiting reactant used.
3.6 Concentration of Solutions
Concentration tells us how much solute is dissolved in a volume of solvent. It can be expressed in g/dm³ or mol/dm³ (where 1 dm³ = 1000 cm³).
concentration (g/dm³) = mass (g) / volume (dm³)
concentration (mol/dm³) = moles / volume (dm³) Chemistry Only
Concentration Calculation
Calculate concentration in g/dm³ when 5 g of salt is dissolved in 250 cm³ of water:
Convert volume to dm³: 250 / 1000 = 0.25 dm³.
Calculate concentration: 5 g / 0.25 dm³ = 20 g/dm³.
3.7 Percentage YieldChemistry Only
The yield is the amount of product obtained. Percentage yield compares actual yield to maximum theoretical yield:
percentage yield = (actual mass / theoretical mass) × 100
Why yield is less than 100%: Reaction may be reversible; some product lost in separation/transfer; side reactions may occur; reactants may not be pure.
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Metals react with oxygen to form metal oxides. This is an oxidation reaction because the metal gains oxygen.
Oxidation
Gain of oxygen (or loss of electrons).
Reduction
Loss of oxygen (or gain of electrons).
4.2 The Reactivity Series
Metals are arranged in order of reactivity based on their reactions with water and dilute acids. A more reactive metal displaces a less reactive metal from its compound.
Unreactive metals (like gold) are found as pure elements. Metals less reactive than carbon are extracted from their oxides by heating with carbon (carbon reduces the metal oxide by removing oxygen):
iron oxide + carbon → iron + carbon dioxide
2Fe2O3 + 3C → 4Fe + 3CO2
Metals more reactive than carbon must be extracted using electrolysis, which requires large amounts of energy.
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The pH scale (0 to 14) measures acidity or alkalinity. Acids produce H⁺ ions in solution; alkalis produce OH⁻ ions in solution. A pH of 7 is neutral.
Acid
A substance that releases H+ ions in aqueous solution (pH < 7).
Alkali
A soluble base that releases OH- ions in aqueous solution (pH > 7).
4.7 Strong & Weak AcidsChemistry Only
Strong acids: Completely ionise/dissociate in aqueous solution, releasing all H⁺ ions (e.g. HCl, HNO₃, H₂SO₄).
Weak acids: Only partially ionise in aqueous solution (e.g. ethanoic, citric, carbonic acids). The reaction is reversible:
CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)
pH scale relationship: As pH decreases by 1 unit, the hydrogen ion concentration increases by a factor of 10.
Strength vs Concentration: Acid strength refers to the degree of ionisation. Acid concentration refers to the mass of acid dissolved in a given volume of solution. A strong acid can be dilute, and a weak acid can be concentrated.
4.8 Electrolysis
Electrolysis is the decomposition of an electrolyte using electricity. Ions move to electrodes: positive ions (cations) move to the cathode (negative electrode) where they gain electrons (reduction). Negative ions (anions) move to the anode (positive electrode) where they lose electrons (oxidation).
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In aqueous solutions, water molecules ionise into H⁺ and OH⁻ ions alongside the compound's ions. Rules for products:
Cathode (-): Hydrogen is produced if the metal is more reactive than hydrogen. The metal is produced only if it is less reactive than hydrogen (e.g. copper).
Anode (+): Oxygen is produced (from OH⁻) unless halide ions (Cl⁻, Br⁻, I⁻) are present, in which case the halogen is formed.
Aqueous electrolysis cell showing inert electrodes, ion migration, and potential products.
4.10 Half EquationsChemistry Only
Write half equations to show reactions at electrodes (e.g. electrolysis of aqueous NaCl):
Anode (+): 2Cl-(aq) → Cl2(g) + 2e- (oxidation)
Cathode (-): 2H+(aq) + 2e- → H2(g) (reduction)
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Energy change = 678 - 862 = -184 kJ/mol (exothermic).
5.4 Chemical Cells & BatteriesChemistry Only
A simple chemical cell consists of two different metals in an electrolyte. The difference in reactivity creates a potential difference (voltage). A battery contains two or more cells connected in series.
Non-rechargeable cells: The chemical reactions stop when one of the reactants is used up (e.g. alkaline batteries).
Rechargeable cells: The chemical reactions can be reversed by connecting the cell to an external electrical current.
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Fuel cells are supplied by an external source of fuel (e.g. hydrogen) and oxygen. The fuel is oxidised electrochemically to produce a voltage continuously.
The overall reaction in a hydrogen fuel cell is the combustion of hydrogen to form water:
2H2(g) + O2(g) → 2H2O(l)
Simplified hydrogen fuel cell showing H₂ oxidation at the anode and O₂ reduction at the cathode.
Only waste product is water (clean at point of use)
No recharging needed, runs continuously with fuel
High energy efficiency
Disadvantages
Hydrogen is difficult and expensive to store
Hydrogen is often made from fossil fuels
Expensive platinum catalysts required
Limited refuelling infrastructure
Fuel Cells vs Rechargeable Batteries
Feature
Fuel Cell
Rechargeable Battery
Energy source
External fuel (H₂)
Stored chemicals inside
Running time
Continuous (with fuel supply)
Limited, needs recharging
Waste products
Water only
None during use
Portability
Needs fuel tank
Self-contained, portable
Exam Tip: In "evaluate" questions, discuss both sides (advantages and disadvantages) before giving a reasoned conclusion. Remember that hydrogen production itself may use fossil fuels.
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The rate of a chemical reaction tells us how fast reactants turn into products. It can be calculated as:
mean rate = amount of reactant used / time
mean rate = amount of product formed / time
Units are usually g/s, cm³/s, or mol/s. On a graph of amount against time, the rate at any moment equals the gradient of the tangent at that point. The reaction is fastest at the start and the line levels off when a reactant is used up.
Measuring rate (required practical)
Gas given off: collect it with a gas syringe and measure the volume over time.
Mass change: stand the flask on a balance and record the mass lost as gas escapes.
Cloudiness (turbidity): time how long a cross marked under the flask takes to disappear (e.g. sodium thiosulfate + hydrochloric acid).
Worked example
60 cm³ of gas is produced in 30 s. Mean rate = 60 / 30 = 2 cm³/s.
6.2 Collision Theory
Chemical reactions can only occur when reacting particles collide with each other with sufficient energy (the activation energy).
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Temperature: Particles move faster, colliding more frequently and with more energy. A higher proportion of collisions have energy ≥ activation energy.
Concentration / Pressure: More particles in a given volume, increasing collision frequency.
Surface Area: Breaking a solid reactant into smaller pieces increases exposed surface area, increasing collision frequency.
Rate curves showing how temperature, concentration, or surface area affect the rate. All curves reach the same final volume.
6.4 Catalysts
A catalyst changes the rate of a chemical reaction but is not used up. It provides an alternative pathway with a lower activation energy, increasing the proportion of successful collisions.
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In a reversible reaction, products can react to reform the original reactants. Shown by the double arrow (⇌).
A + B ⇌ C + D
If a reversible reaction is exothermic in one direction, it is endothermic in the opposite direction. The same amount of energy is transferred in both directions.
Key Examples
1. Ammonium chloride:
NH4Cl(s) ⇌ NH3(g) + HCl(g)
On heating, white ammonium chloride decomposes. On cooling, the gases recombine and white solid reforms higher up the tube.
2. Hydrated copper sulfate:
CuSO4·5H2O ⇌ CuSO4 + 5H2O
Forward (heating): blue crystals turn to white powder (endothermic). Reverse (add water): white powder turns blue (exothermic).
Key Fact: The energy transferred in the forward direction is exactly equal to the energy transferred in the reverse direction.
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In a closed system, dynamic equilibrium is reached when the forward and reverse reactions occur at exactly the same rate, and the concentrations of reactants and products remain constant.
Le Chatelier's Principle
If a system at equilibrium is subjected to a change in conditions, the system will adjust to counteract the change:
Temperature: If temperature is increased, the system moves in the endothermic direction to absorb heat. If decreased, it moves in the exothermic direction.
Pressure (gases): If pressure is increased, the system shifts to the side with fewer gas molecules to reduce pressure. If decreased, it shifts to the side with more molecules.
Concentration: If concentration of a reactant is increased, more products form to use it up.
Exam Tip: A catalyst does NOT change the position of equilibrium. It speeds up both forward and reverse reactions equally, so equilibrium is reached faster but the proportions of products and reactants stay the same.
Worked Example: Le Chatelier's Principle and the Haber Process
N2(g) + 3H2(g) ⇌ 2NH3(g) (forward reaction is exothermic)
1. Increase temperature: Equilibrium shifts LEFT (endothermic direction) to absorb the extra heat. Lower yield of NH3, but rate is faster.
2. Increase pressure: Left side has 4 moles of gas (1+3), right side has 2 moles. Equilibrium shifts RIGHT (fewer moles). Higher yield of NH3.
3. Iron catalyst: No effect on position of equilibrium or yield. Equilibrium is reached faster because both forward and reverse rates increase equally.
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Crude oil is a finite resource found in rocks, formed over millions of years from the remains of ancient biomass (mainly plankton). It is a mixture of hydrocarbons.
Hydrocarbons are compounds containing hydrogen and carbon atoms only. Most are alkanes, which are saturated hydrocarbons (each carbon has single covalent bonds only).
Alkanes general formula: CnH2n+2
The alkanes are a homologous series: a family of compounds with the same general formula, where each member differs from the next by CH2. Members have similar chemical properties and show a gradual trend in physical properties as the chain lengthens.
Alkane
Formula
State (room temp)
Methane
CH4
Gas
Ethane
C2H6
Gas
Propane
C3H8
Gas
Butane
C4H10
Gas
Crude oil as a feedstock
Crude oil is the main feedstock for the petrochemical industry, providing fuels and the raw materials for products such as polymers, solvents, lubricants and detergents.
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Crude oil is separated into fractions (groups of hydrocarbons with similar boiling points) by fractional distillation. The column is hot at the bottom and cold at the top.
Properties of hydrocarbons:
Short-chain: Low boiling point, low viscosity (runny), highly flammable. Condense at the cool top.
Long-chain: High boiling point, high viscosity (thick), low flammability. Condense at the hot bottom.
Figure 7.1: Tall fractionating column separating crude oil into fractions based on boiling points.
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Long-chain hydrocarbons are cracked into shorter, more useful alkanes and alkenes. Methods: catalytic cracking (high temperature, zeolite catalyst) and steam cracking (higher temperature, steam).
decane (C10H22) → octane (C8H18) + ethene (C2H4)
Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond (C=C).
Alkenes general formula: CnH2n
Test for Alkenes: Orange bromine water is decolourised (goes colourless) when mixed with an alkene (unsaturated), but stays orange with an alkane (saturated).
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Alkenes are highly reactive due to the C=C bond, reacting via addition reactions:
With halogens: E.g. ethene + bromine → dibromoethane.
With hydrogen (hydrogenation): E.g. ethene + H₂ → ethane (requires 150 °C, nickel catalyst).
With water (hydration): E.g. ethene + steam ⇌ ethanol (requires high temp, phosphoric acid catalyst).
7.6 Addition PolymerisationChemistry Only
Alkenes act as monomers. In addition polymerisation, many small alkene molecules join together to form polymers. The double bond opens up to form single bonds between monomers. Only one product forms.
7.7 AlcoholsChemistry Only
Alcohols contain the functional group -OH. The first four are methanol, ethanol, propanol, and butanol.
Reactions: Burn in air (combustion); react with sodium (releases hydrogen); oxidise to form carboxylic acids.
Carboxylic acids contain the functional group -COOH (e.g. ethanoic acid). They react with carbonates to produce a salt, water, and CO₂. They are weak acids (only partially ionise in solution).
React with alcohols (in presence of an acid catalyst) to form esters (functional group -COO-):
ethanol + ethanoic acid → ethyl ethanoate + water
7.9 Condensation PolymerisationChemistry Only
Involves monomers with two functional groups. When they react, they join together and lose small molecules such as water (unlike addition polymerisation which forms no by-products).
Example: Polyester from a dicarboxylic acid and a diol:
n HO-R-OH + n HOOC-R'-COOH → [O-R-O-CO-R'-CO]n + 2n H2O
7.10 Natural PolymersChemistry Only
Amino Acids: Have two different functional groups (basic amine group -NH₂ and acidic carboxylic acid group -COOH). Join by condensation polymerisation to form polypeptides and proteins.
DNA (Deoxyribonucleic acid): Large molecule essential for life. Consists of two polymer chains made from four different monomers (nucleotides) in a double helix.
Starch & Cellulose: Polymers made from glucose monomers.
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In chemistry, a pure substance is a single element or compound, not mixed with any other substance.
Melting/Boiling Points: Pure substances melt and boil at specific, sharp temperatures. Impure substances (mixtures) melt and boil over a range of temperatures, and at lower melting points / higher boiling points than the pure substance.
8.2 Formulations
A formulation is a mixture that has been designed as a useful product. Every component is carefully measured to ensure it has the required properties.
Examples: fuels, cleaning agents, paints, medicines, alloys, fertilisers, and foods.
8.3 Chromatography
Paper chromatography separates mixtures based on their relative solubility in a mobile phase (solvent) and attraction to a stationary phase (paper).
Rf = distance moved by substance / distance moved by solvent
A pure substance produces a single spot on a chromatogram. An impure substance produces multiple spots.
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Identify metal ions by adding sodium hydroxide (NaOH) solution:
Al3+, Ca2+, Mg2+: Form white precipitates. Only Al(OH)₃ dissolves in excess NaOH to form a clear solution.
Copper (Cu2+): Blue precipitate: Cu2+ + 2OH- → Cu(OH)2(s)
Iron(II) (Fe2+): Green precipitate: Fe2+ + 2OH- → Fe(OH)2(s)
Iron(III) (Fe3+): Brown precipitate: Fe3+ + 3OH- → Fe(OH)3(s)
8.7 Anion TestsChemistry Only
Carbonates (CO32-): Add dilute acid. Fizzing (effervescence) occurs because CO₂ is released. Test with limewater.
Halides (Cl-, Br-, I-): Add dilute nitric acid and silver nitrate. Chloride forms a white precipitate (AgCl); Bromide forms cream (AgBr); Iodide forms yellow (AgI).
Sulfates (SO42-): Add dilute hydrochloric acid and barium chloride. A white precipitate of barium sulfate forms.
8.8 Instrumental MethodsChemistry Only
Instrumental methods are rapid, sensitive, and accurate. Flame emission spectroscopy is an instrumental method used to analyze metal ions in solution. The sample is heated in a flame, and the emitted light is analyzed through a spectroscope to produce a line spectrum.
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For about 200 million years, the proportions of gases in the Earth's atmosphere have been roughly stable:
Nitrogen (N2): about 78%
Oxygen (O2): about 21%
Other gases: argon (~0.9%), carbon dioxide (~0.04%), and small, variable amounts of water vapour.
Test for oxygen
A glowing splint relights in a tube of oxygen.
Test for carbon dioxide
Bubbling CO2 through limewater turns it cloudy (milky).
The proportion of carbon dioxide is small but is now increasing because of human activity - mainly burning fossil fuels and deforestation (covered in 9.5 Greenhouse Gases).
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The Earth was formed 4.6 billion years ago. Theories suggest:
Volcanic Activity: Intense volcanic activity released gases that formed the early atmosphere - mainly carbon dioxide, with little or no oxygen, water vapour, nitrogen, and small amounts of methane and ammonia.
Ocean Formation: The Earth cooled and water vapour condensed to form the oceans. Large amounts of CO₂ dissolved in the oceans, reacting to form insoluble carbonate compounds.
9.3 How Oxygen Increased
Algae and plants produced oxygen by photosynthesis over billions of years:
carbon dioxide + water → glucose + oxygen
6CO2 + 6H2O → C6H12O6 + 6O2
As oxygen levels rose, animals evolved.
9.4 How Carbon Dioxide Decreased
CO₂ levels decreased because:
Photosynthesis absorbed CO₂.
CO₂ dissolved in oceans and formed carbonate precipitates.
Carbon was locked up in sedimentary rocks (e.g. limestone) and fossil fuels (coal, oil, gas) formed from dead plant and animal remains.
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Greenhouse gases (carbon dioxide, water vapour, methane) maintain temperatures on Earth high enough to support life.
The Greenhouse Effect: Short-wavelength radiation from the Sun passes through the atmosphere and heats the Earth. The Earth emits long-wavelength (infrared) radiation, which is absorbed by greenhouse gases in the atmosphere, trapping heat.
Human activities (burning fossil fuels, deforestation, agriculture) have increased CO₂ and methane levels, leading to global climate change.
9.6 Pollutants from Fuels
Combustion of fuels releases pollutants that harm health and the environment:
Humans use Earth's resources for warmth, shelter, food, and transport. Resources are finite (will run out, e.g. metal ores, fossil fuels) or renewable (reformed at a rate similar to use, e.g. timber).
Sustainable development
Development that meets the needs of current generations without compromising the ability of future generations to meet their own needs.
10.2 Potable Water
Potable water is water that is safe to drink. It is not pure water in a chemical sense (which contains H₂O molecules only) as it contains low levels of dissolved salts and microbes.
Treatment of fresh water (groundwater/rivers):
Filtration: Passed through filter beds to remove insoluble solids.
Sterilisation: Treated with chlorine, ozone, or UV light to kill microbes.
If fresh water is scarce, desalination of salty water (seawater) is used by distillation or reverse osmosis. Both require large amounts of energy.
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Sewage, agricultural, and industrial waste water must be treated before release to prevent pollution:
Screening: Removes large solid objects.
Sedimentation: Settles into solid sludge (sinks) and liquid effluent (floats).
Aerobic digestion: Effluent is treated with aerobic bacteria to break down organic matter.
Anaerobic digestion: Sludge is digested by anaerobic bacteria (produces biogas).
10.4 Alternative Metal ExtractionChemistry Only
Copper ores are finite. Phytomining and bioleaching extract copper from low-grade ores, avoiding traditional mining waste.
Phytomining: Plants are grown on soil containing copper compounds. The plants absorb copper and accumulate it in their tissues. They are harvested and burned. The ash containing copper compounds is dissolved in acid to make a leachate, and copper is displaced by scrap iron or extracted by electrolysis.
Bioleaching: Bacteria are used to produce acidic solutions (leachate) containing copper ions from the ore. Copper is then extracted using displacement or electrolysis.
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An LCA assesses the environmental impact of a product over its entire life cycle:
Extracting and processing raw materials.
Manufacturing and packaging.
Use and operation during its lifetime.
Disposal at the end of its life (incineration, landfill, recycling).
LCAs are not purely objective; while energy, water, and waste can be quantified, impacts like visual pollution require subjective value judgements.
10.6 Reduce, Reuse, Recycle
Reduces the use of limited resources, energy consumption, waste, and environmental impacts. Glass, metals, and plastics are commonly recycled by melting and reshaping.
10.7 CorrosionChemistry Only
Corrosion is the destruction of materials by chemical reactions with substances in the environment (e.g. rusting of iron). Rusting requires both oxygen and water.
Rusting equation: iron + oxygen + water → hydrated iron(III) oxide
Prevention: Barriers (paint, grease, electroplating); sacrificial protection (connecting to a more reactive metal like zinc, which corrodes instead of the iron); galvanising (coating iron in zinc, acting as both barrier and sacrificial protection).
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Bronze: Copper + tin. Hard and corrosion-resistant.
Brass: Copper + zinc. Malleable, used for musical instruments.
Gold alloys: Gold + silver/copper/zinc. Harder than pure gold (measured in carats, where 24 carat is 100% pure).
Steels: Iron + carbon/other metals. High-carbon steel is strong but brittle. Low-carbon steel is softer and easily shaped. Stainless steel (with chromium/nickel) is corrosion-resistant.
Polymers: Thermosoftening polymers melt when heated (contain individual chains with weak intermolecular forces). Thermosetting polymers do not melt when heated (contain cross-links between chains).
10.9 The Haber ProcessChemistry Only
Manufactures ammonia (NH₃) from nitrogen (from air) and hydrogen (from natural gas). Ammonia is used mainly to make fertilisers (e.g. ammonium nitrate, NH4NO3).
Temperature: ~450 °C. The forward reaction is exothermic, so a lower temperature would give a higher yield, but the rate would be too slow.
Pressure: ~200 atm. High pressure shifts equilibrium to the right (fewer gas moles), increasing yield. Very high pressures are expensive and hazardous.
Catalyst: iron. Speeds up both forward and reverse reactions equally. Does not change the yield, but equilibrium is reached faster.
The yield per pass is only about 15%. Unreacted nitrogen and hydrogen are recycled back through the reactor to improve overall conversion. The ammonia is cooled, liquefied, and removed.
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Unofficial revision material for AQA GCSE Chemistry. For revision only. Not affiliated with, endorsed by, or connected to AQA or any exam board.
Required Practicals HubChemistry Made Easy
AQA GCSE Chemistry Required Practicals
This hub provides quick-links to all key Required Practicals (RPs) featured in the AQA specification. Click any practical to jump directly to its revision notes.
RP 1: Soluble Salts
Prepare a pure, dry sample of a soluble salt (copper sulfate) from an insoluble oxide or carbonate.
Fill in the boxes at the top of this page with your details if required.
Answer all questions in the spaces provided.
Do all rough work in this book. Cross through any work you do not want to be marked.
In all calculations, show clearly how you work out your answer.
A Periodic Table is provided as a separate insert if required.
Calculators may be used.
Information for Candidates
This practice paper is designed to support student revision for the GCSE Chemistry examinations. It contains questions covering atomic structure, bonding, quantitative chemistry, chemical changes, and energy changes. The marks for individual questions and parts of questions are shown in round brackets.
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A student sets up a computer simulation to model Rutherford's alpha particle scattering experiment. They record the path of 10,000 positive alpha particles fired at a thin gold foil:
9,880 alpha particles pass straight through the foil with no deflection.
118 alpha particles are deflected by small angles.
2 alpha particles bounce back towards the source.
(3)
(a) Explain how these simulation results provide evidence for the nuclear model of the atom. Link each conclusion to a specific observation from the data.
(2)
(b) In 1932, James Chadwick discovered a new subatomic particle. State the name of this particle and explain why it was discovered much later than protons and electrons.
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 1: Atomic Structure & Bonding
Potassium is a Group 1 alkali metal and chlorine is a Group 7 halogen.
(c) Write the electronic configuration of:
(1)
(i) A potassium atom (atomic number = 19)
(1)
(ii) A chlorine atom (atomic number = 17)
(3)
(d) Explain, in terms of electronic configurations, why potassium is more reactive than sodium (atomic number = 11).
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 1: Atomic Structure & Bonding
Question 2 (Synoptic Target B)[6 Marks] Mark scheme →
Heated sodium metal reacts vigorously with chlorine gas to form the compound sodium chloride.
(4)
(a) Describe, in terms of electron transfer, how sodium atoms and chlorine atoms react to form sodium chloride. You must include the electronic configurations of the atoms and the resulting ions in your explanation.
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 2: Bonding & Properties
(2)
(b) Write the balanced chemical equation for the reaction of sodium metal with chlorine gas, including state symbols.
Question 3 (Synoptic Target A)[8 Marks] Mark scheme →
A student sets up an electrolysis experiment using carbon electrodes. They test the electrical conductivity of two samples: Sample X (solid copper(II) sulfate) and Sample Y (aqueous copper(II) sulfate).
(3)
(a) State which sample, X or Y, will conduct electricity. Explain this difference in conductivity by referring to the structure and bonding of copper(II) sulfate.
During the electrolysis of aqueous copper(II) sulfate, chemical changes occur at the electrodes as shown in the diagram.
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 2: Bonding & Properties
(b) During the electrolysis:
(3)
(i) Explain why copper metal forms at the negative electrode (cathode). Write a half-equation for this process.
(2)
(ii) Describe the chemical change at the positive electrode (anode) and write a half-equation for this process.
Carbon exists as different allotropes, including diamond and graphite. Both have giant covalent structures but exhibit very different physical properties.
(6)
(a) Compare the structure and bonding of diamond and graphite, and explain how these structures relate to their hardness and electrical conductivity.
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 2: Bonding & Properties
A student is given a table of physical properties for three unidentified substances, A, B, and C:
Substance
Melting Point / °C
Boiling Point / °C
Electrical Conductivity as Solid
Electrical Conductivity as Liquid
Solubility in Water
Substance A
1610
2230
Does not conduct
Does not conduct
Insoluble
Substance B
801
1413
Does not conduct
Conducts
Soluble
Substance C
-182
-161
Does not conduct
Does not conduct
Insoluble
The student claims that Substance A is a metal, Substance B is a simple molecular compound, and Substance C is a giant covalent structure.
(6)
(b) Evaluate the student's claims using the data in the table. For each substance, state whether the claim is correct or incorrect, and explain why by analyzing the physical properties.
Nanoparticles of titanium dioxide are used in some modern sunscreens.
(1)
(c) (i) Explain why nanoparticles have different properties compared to bulk materials like titanium dioxide powder.
(1)
(ii) State one potential risk of using nanoparticles in consumer cosmetics.
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(a) State what is meant by the term limiting reactant.
(2)
(b) A student prepared a sample of copper(II) carbonate precipitate. After filtering the reaction mixture, the student washed the precipitate with distilled water and dried it.
Explain why the actual yield of the dried copper(II) carbonate might be:
Lower than the calculated theoretical maximum yield.
Higher than the calculated theoretical maximum yield.
(5)
(c) Iron can be produced by reducing iron(III) oxide with carbon.
The balanced equation for the reaction is:
2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)
A student reacts 24.0 g of iron(III) oxide (Fe2O3) with 5.40 g of carbon (C).
Show by calculation which reactant is the limiting reactant, and calculate the maximum theoretical yield of iron (Fe) in grams.
Relative atomic masses (Ar): C = 12; O = 16; Fe = 56
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 3: Quantitative Chemistry
(3)
(d) Calculate the percentage atom economy for the reaction to produce iron. Give your answer to 3 significant figures.
Equation: 2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)
Relative atomic masses (Ar): C = 12; O = 16; Fe = 56
(1)
(e) State one reason why chemical companies aim to use reactions with a high atom economy.
(a) State two conditions that must be kept constant for 1 mole of any gas to occupy a volume of 24.0 dm3.
(b) When carrying out a titration, a student performs a rough titration first before carrying out further runs to obtain concordant titres.
(1)
(i) Explain the purpose of performing a rough titration first.
(1)
(ii) Explain what is meant by the term concordant titres in terms of experimental precision.
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 3: Quantitative Chemistry
(6)
(c) A student carries out a titration to find the concentration of a sodium hydroxide (NaOH) solution.
The student titrates 25.0 cm3 of the sodium hydroxide solution against a standard solution of sulfuric acid (H2SO4) of concentration 0.0500 mol/dm3.
The mean volume of sulfuric acid required to neutralise the sodium hydroxide is 20.0 cm3.
The balanced equation for the reaction is:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Calculate the concentration of the sodium hydroxide solution in g/dm3. Give your answer to 3 significant figures.
Relative atomic masses (Ar): H = 1; O = 16; Na = 23; S = 32
(3)
(d) A student reacts 0.243 g of magnesium ribbon with an excess of dilute hydrochloric acid to produce hydrogen gas.
The equation for the reaction is:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
Calculate the volume of hydrogen gas produced in cm3 at room temperature and pressure (RTP).
Assume 1 mole of gas occupies 24.0 dm3 at RTP.
Relative atomic mass (Ar): Mg = 24.3
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(a) Iron is extracted from iron(III) oxide in a blast furnace by heating with carbon. Aluminium is extracted from aluminium oxide using electrolysis.
Explain why carbon can be used to extract iron but cannot be used to extract aluminium.
A student adds zinc powder to copper(II) sulfate solution. A displacement reaction occurs.
(2)
(b) (i) Write a balanced ionic equation for this reaction. Include state symbols.
(1)
(ii) State which species is oxidised and explain this in terms of electron transfer.
Dilute hydrochloric acid and dilute ethanoic acid are both acids.
(2)
(c) (i) Explain the difference between a strong acid (such as hydrochloric acid) and a weak acid (such as ethanoic acid) in terms of their ionisation in aqueous solution.
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 4: Chemical Changes
(2)
(c) (ii) The concentration of hydrogen ions in an acid determines its pH. The mathematical relationship between pH and hydrogen ion concentration is given by:
pH = -log10[H+]
Using this equation, prove that decreasing the pH of an acid by exactly 1 unit (for example, from pH 3 to pH 2) corresponds to a tenfold (10 times) increase in the concentration of hydrogen ions.
A student carries out an experiment to prepare a pure, dry sample of copper(II) sulfate crystals. They add copper(II) oxide to dilute sulfuric acid.
(1)
(a) The student heats the dilute sulfuric acid gently in a beaker before adding the copper(II) oxide. State why the acid is warmed.
(b) The copper(II) oxide is added until it is in excess.
(1)
(i) Describe what the student would observe when the copper(II) oxide is in excess.
(1)
(ii) Explain why it is necessary to add an excess of copper(II) oxide.
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 4: Chemical Changes
(3)
(c) Describe the remaining steps the student must take to obtain a pure, dry sample of copper(II) sulfate crystals from the mixture.
(2)
(d) The theoretical yield of copper(II) sulfate crystals for this experiment was calculated to be 5.0 g. The student obtained a mass of 4.2 g of dry crystals.
Suggest two reasons why the actual yield was lower than the theoretical yield.
A student investigates the electrolysis of aqueous sodium chloride (brine) using inert carbon electrodes.
(2)
(a) Name the product formed at the negative electrode (cathode) and write a balanced half-equation to represent its formation.
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 4: Chemical Changes & Topic 5: Energy Changes
(b) A gas is produced at the positive electrode (anode).
(2)
(i) Identify this gas and explain why it is formed in preference to hydroxide ions.
(2)
(ii) Write a balanced half-equation for the discharge of chloride ions at the anode.
(b) When a solid is dissolved in water in an insulated cup, the temperature of the water rises.
State whether this process is exothermic or endothermic, and explain this in terms of energy transfer between the reacting chemicals and the water.
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 5: Energy Changes
(2)
(c) A student wants to measure the temperature change when different masses of ammonium chloride are dissolved in water.
Describe how the student could carry out this investigation to obtain valid results, mentioning a key piece of apparatus used to reduce heat loss to the surroundings.
(2)
(d) Explain how a catalyst increases the rate of a chemical reaction.
(3)
(e) Describe how the activation energy and the overall energy change of an exothermic reaction are represented on an energy profile diagram. In your answer, refer to the relative energy levels of the reactants and products.
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GCSE Chemistry
Practice Paper 1 - Higher Tier
Topic 5: Energy Changes
The reaction between hydrogen gas and chlorine gas is represented by the following equation:
H2(g) + Cl2(g) → 2HCl(g)
The table below lists the bond energies for the bonds involved in this reaction:
Bond
Bond Energy / kJ/mol
H-H
436
Cl-Cl
242
H-Cl
431
(f) Use the values in the table to answer the following questions.
(1)
(i) Calculate the energy required to break all the bonds in the reactants.
(1)
(ii) Calculate the energy released when the bonds in the products are formed.
(2)
(iii) Calculate the overall energy change for the reaction and state whether the reaction is exothermic or endothermic.
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GCSE Chemistry
Periodic Table of the Elements
Insert
Group 1
Group 2
Transition Metals
Group 3
Group 4
Group 5
Group 6
Group 7
Group 0
KEY
relative atomic mass
H
atomic symbol
name
atomic (proton) number
* Lanthanides
** Actinides
1HHydrogen1
4HeHelium2
7LiLithium3
9BeBeryllium4
11BBoron5
12CCarbon6
14NNitrogen7
16OOxygen8
19FFluorine9
20NeNeon10
23NaSodium11
24MgMagnesium12
27AlAluminium13
28SiSilicon14
31PPhosphorus15
32SSulfur16
35.5ClChlorine17
40ArArgon18
39KPotassium19
40CaCalcium20
45ScScandium21
48TiTitanium22
51VVanadium23
52CrChromium24
55MnManganese25
56FeIron26
59CoCobalt27
59NiNickel28
63.5CuCopper29
65ZnZinc30
70GaGallium31
73GeGermanium32
75AsArsenic33
79SeSelenium34
80BrBromine35
84KrKrypton36
85.5RbRubidium37
88SrStrontium38
89YYttrium39
91ZrZirconium40
93NbNiobium41
96MoMolybdenum42
98TcTechnetium43
101RuRuthenium44
103RhRhodium45
106PdPalladium46
108AgSilver47
112CdCadmium48
115InIndium49
119SnTin50
122SbAntimony51
128TeTellurium52
127IIodine53
131XeXenon54
133CsCesium55
137BaBarium56
139La*Lanthanum57
178.5HfHafnium72
181TaTantalum73
184WTungsten74
186ReRhenium75
190OsOsmium76
192IrIridium77
195PtPlatinum78
197AuGold79
201HgMercury80
204TlThallium81
207PbLead82
209BiBismuth83
209PoPolonium84
210AtAstatine85
222RnRadon86
223FrFrancium87
226RaRadium88
227Ac**Actinium89
267RfRutherfordium104
268DbDubnium105
269SgSeaborgium106
270BhBohrium107
269HsHassium108
278MtMeitnerium109
281DsDarmstadtium110
282RgRoentgenium111
285CnCopernicium112
286NhNihonium113
289FlFlerovium114
289McMoscovium115
293LvLivermorium116
294TsTennessine117
294OgOganesson118
140CeCerium58
141PrPraseodymium59
144NdNeodymium60
145PmPromethium61
150SmSamarium62
152EuEuropium63
157GdGadolinium64
159TbTerbium65
162.5DyDysprosium66
165HoHolmium67
167ErErbium68
169TmThulium69
173YbYtterbium70
175LuLutetium71
232ThThorium90
231PaProtactinium91
238UUranium92
237NpNeptunium93
244PuPlutonium94
243AmAmericium95
247CmCurium96
247BkBerkelium97
251CfCalifornium98
252EsEinsteinium99
257FmFermium100
258MdMendelevium101
259NoNobelium102
266LrLawrencium103
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Fill in the boxes at the top of this page with your details if required.
Answer all questions in the spaces provided.
Do all rough work in this book. Cross through any work you do not want to be marked.
In all calculations, show clearly how you work out your answer.
A Periodic Table is provided as a separate insert if required.
Calculators may be used.
Information for Candidates
This practice paper is designed to support student revision for the GCSE Chemistry examinations. It contains questions covering the rate and extent of chemical change, organic chemistry, chemical analysis, chemistry of the atmosphere, and using resources. The marks for individual questions and parts of questions are shown in round brackets.
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A student investigates how the concentration of hydrochloric acid affects the rate of reaction with marble chips (calcium carbonate). The equation for the reaction is:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
The student measures the volume of carbon dioxide gas produced over 100 seconds using a gas syringe. The results are shown in the graph below.
(1)
(a) State the independent variable in this investigation.
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GCSE Chemistry
Practice Paper 2 - Higher Tier
Topic 6: Rates of Reaction
(1)
(b) State one control variable that must be kept constant to ensure the results are valid.
(2)
(c) Use Curve A to calculate the mean rate of reaction between 0 and 20 seconds. Show your working and state the unit.
(3)
(d) Draw a tangent to Curve B at 30 seconds and use it to calculate the rate of reaction at this time. Show your working and state the unit.
(3)
(e) Explain, in terms of collision theory, why Curve A has a faster rate of reaction than Curve B.
(2)
(f) The student repeats the experiment using the same mass of powdered calcium carbonate instead of marble chips. Sketch a curve on the graph on Page 2 to show the expected results for this reaction.
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Reversible reactions can reach a state of dynamic equilibrium.
(2)
(a) State two conditions required for a system to reach dynamic equilibrium.
Nitrogen dioxide gas (NO₂) exists in dynamic equilibrium with dinitrogen tetroxide gas (N₂O₄) according to the equation:
2NO₂(g) ⇌ N₂O₄(g) ΔH is negative
Nitrogen dioxide is a brown gas and dinitrogen tetroxide is a colourless gas.
(2)
(b) Predict and explain the effect of increasing the concentration of nitrogen dioxide gas (NO₂) on the position of equilibrium.
(2)
(c) Predict and explain the effect of increasing the temperature on the colour of the equilibrium mixture.
(2)
(d) A catalyst is added to the mixture. Explain the effect of the catalyst on the position of equilibrium and on the time taken to reach equilibrium.
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This resource is an independent educational tool created to support student revision. It is completely independent and is not endorsed by, affiliated with, or sponsored by any official examination board. All trademarked terms are used under Nominative Fair Use purely for descriptive compatibility indexing. Licensed for individual personal use only. Chemistry Made Easy is an independent resource. Not affiliated with or endorsed by AQA, Pearson Edexcel, or the IBO.
Cracking is an industrial process used to break down larger hydrocarbons into smaller, more useful molecules.
(2)
(a) Explain why cracking long-chain hydrocarbons is economically useful for the petrochemical industry.
(2)
(b) Describe the difference in conditions between catalytic cracking and steam cracking.
(1)
(c) Complete the chemical equation for the cracking of dodecane (C₁₂H₂₆):
C₁₂H₂₆ → C₈H₁₈ + 2 .....
(2)
(d) Alkenes are unsaturated hydrocarbons. Describe the chemical test to distinguish between alkanes and alkenes, stating the positive result.
(1)
(e) Draw the displayed structural formula of the product formed when ethene (C₂H₄) reacts with bromine (Br₂).
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GCSE Chemistry
Practice Paper 2 - Higher Tier
Topic 7: Organic Chemistry
Ethanol is an alcohol that can be produced using different methods.
(2)
(f) Fermentation of sugar solutions using yeast produces ethanol. State two reaction conditions required for this process to happen efficiently.
(4)
(g) Ethanol can also be produced by reacting ethene with steam in the presence of a catalyst. Compare this industrial hydration method with fermentation in terms of raw materials, process type, and product purity.
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A student carries out paper chromatography to analyse food dyes used in an unknown sweet sample. The chromatogram is shown below.
(1)
(a) Explain why the student should draw the baseline in pencil rather than ink.
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GCSE Chemistry
Practice Paper 2 - Higher Tier
Topic 8: Chemical Analysis
(2)
(b) Calculate the Rf value of spot A. Show your working.
(2)
(c) Deduce which of the food dyes (A, B, or C) are present in the unknown Y. Explain your reasoning.
(2)
(d) A bottle of fruit juice is labelled as "100% pure fruit juice". Explain the difference between this everyday use of "pure" and the chemical definition of a pure substance.
(3)
(e) Describe how the melting point of an impure sample of a chemical compound differs from that of a pure sample.
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Qualitative tests are used to identify ions in unknown compounds.
(2)
(a) State the flame colour observed when testing:
(i) Sodium ions: .....................................................................
(ii) Calcium ions: ....................................................................
(3)
(b) Describe the test to confirm the presence of sulfate ions in a solution. Include the reagents used and the observation.
(2)
(c) A student tests an unknown compound Z. They find that:
Adding sodium hydroxide solution produces a green precipitate.
Adding dilute hydrochloric acid followed by barium chloride solution produces a white precipitate.
Identify compound Z.
(1)
(d) State one advantage of using instrumental methods (such as flame emission spectroscopy) rather than chemical tests to identify ions.
(2)
(e) Describe the test for carbonate ions in a solid sample, stating the gas produced and the observation that confirms its presence.
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The composition of the Earth's atmosphere has evolved significantly over billions of years.
(1)
(a) State approximately how long the gas composition of the Earth's atmosphere has been similar to how it is today.
(2)
(b) State the approximate percentage of:
(i) Nitrogen in today's atmosphere: ................................
(ii) Oxygen in today's atmosphere: ...................................
(3)
(c) The early atmosphere contained high levels of carbon dioxide. Describe how the formation of oceans and the evolution of photosynthesising organisms reduced carbon dioxide levels.
(2)
(d) The early atmosphere is estimated to have contained 95.0% carbon dioxide. The modern atmosphere contains 0.04% carbon dioxide. Calculate the percentage decrease in carbon dioxide concentration relative to the early atmosphere.
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Greenhouse gases absorb radiation emitted from the Earth's surface and trap heat in the atmosphere.
(3)
(a) Explain the greenhouse effect in terms of short-wavelength and long-wavelength radiation.
(1)
(b) Name two greenhouse gases other than carbon dioxide.
(2)
(c) Define the term "carbon footprint" and state one action individuals can take to reduce theirs.
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GCSE Chemistry
Practice Paper 2 - Higher Tier
Topic 9: Chemistry of the Atmosphere
(6)
(d) Evaluate the actions that could be taken by individuals, governments, and industries to reduce carbon dioxide emissions. Include constraints and challenges faced by these groups in your evaluation.
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The provision of potable water is essential for human life.
(2)
(a) Define the term "potable water" and explain why it is not chemically pure water.
(2)
(b) In the UK, fresh water is treated to make it potable. Describe the two main steps in this treatment process and state the purpose of each step.
(2)
(c) Desalination of seawater is used in some countries. State one method used for desalination and explain why it requires high amounts of energy.
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GCSE Chemistry
Practice Paper 2 - Higher Tier
Topic 10: Using Resources
(6)
(d) A life cycle assessment (LCA) is carried out to evaluate the environmental impacts of two types of shopping bags: paper bags and polyethene plastic bags.
Evaluate the sustainability of these two types of bags, from raw material extraction to disposal. Conclude which type of bag is more sustainable, citing limitations of using LCAs.
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The Haber process is used to manufacture ammonia for fertilisers.
(2)
(a) The equation for the Haber process is:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH is negative
Explain why the compromise temperature of 450 °C is chosen. Refer to reaction rate and yield in your answer.
(1)
(b) Describe how ammonia is separated from unreacted nitrogen and hydrogen gases after leaving the reactor.
(3)
(c) As metal ores become scarce, alternative methods of metal extraction are developed. Describe the process of phytomining to extract copper from low-grade ores.
(2)
(d) State one advantage and one disadvantage of phytomining compared to traditional open-cast mining.
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GCSE Chemistry
Periodic Table of the Elements
Insert
Group 1
Group 2
Transition Metals
Group 3
Group 4
Group 5
Group 6
Group 7
Group 0
KEY
relative atomic mass
H
atomic symbol
name
atomic (proton) number
* Lanthanides
** Actinides
1HHydrogen1
4HeHelium2
7LiLithium3
9BeBeryllium4
11BBoron5
12CCarbon6
14NNitrogen7
16OOxygen8
19FFluorine9
20NeNeon10
23NaSodium11
24MgMagnesium12
27AlAluminium13
28SiSilicon14
31PPhosphorus15
32SSulfur16
35.5ClChlorine17
40ArArgon18
39KPotassium19
40CaCalcium20
45ScScandium21
48TiTitanium22
51VVanadium23
52CrChromium24
55MnManganese25
56FeIron26
59CoCobalt27
59NiNickel28
63.5CuCopper29
65ZnZinc30
70GaGallium31
73GeGermanium32
75AsArsenic33
79SeSelenium34
80BrBromine35
84KrKrypton36
85.5RbRubidium37
88SrStrontium38
89YYttrium39
91ZrZirconium40
93NbNiobium41
96MoMolybdenum42
98TcTechnetium43
101RuRuthenium44
103RhRhodium45
106PdPalladium46
108AgSilver47
112CdCadmium48
115InIndium49
119SnTin50
122SbAntimony51
128TeTellurium52
127IIodine53
131XeXenon54
133CsCesium55
137BaBarium56
139La*Lanthanum57
178.5HfHafnium72
181TaTantalum73
184WTungsten74
186ReRhenium75
190OsOsmium76
192IrIridium77
195PtPlatinum78
197AuGold79
201HgMercury80
204TlThallium81
207PbLead82
209BiBismuth83
209PoPolonium84
210AtAstatine85
222RnRadon86
223FrFrancium87
226RaRadium88
227Ac**Actinium89
267RfRutherfordium104
268DbDubnium105
269SgSeaborgium106
270BhBohrium107
269HsHassium108
278MtMeitnerium109
281DsDarmstadtium110
282RgRoentgenium111
285CnCopernicium112
286NhNihonium113
289FlFlerovium114
289McMoscovium115
293LvLivermorium116
294TsTennessine117
294OgOganesson118
140CeCerium58
141PrPraseodymium59
144NdNeodymium60
145PmPromethium61
150SmSamarium62
152EuEuropium63
157GdGadolinium64
159TbTerbium65
162.5DyDysprosium66
165HoHolmium67
167ErErbium68
169TmThulium69
173YbYtterbium70
175LuLutetium71
232ThThorium90
231PaProtactinium91
238UUranium92
237NpNeptunium93
244PuPlutonium94
243AmAmericium95
247CmCurium96
247BkBerkelium97
251CfCalifornium98
252EsEinsteinium99
257FmFermium100
258MdMendelevium101
259NoNobelium102
266LrLawrencium103
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Practice Paper 1 - Master Mark Scheme (Unofficial)
Total Marks: 100
Version 1.0
General Marking Guidance
Acceptable Answers: Mark schemes are prepared by subject specialists. They indicate the points required to gain marks. Alternative wording or symbols that express the same chemical meaning should be accepted.
Bolding: Bold chemical terms are key elements that must be present in the student's answer to score the mark.
Error Carried Forward (ECF): ECF applies to mathematical calculations. If a student makes an early arithmetic error, they lose that specific mark but can score full marks for subsequent steps that apply correct chemical calculations to their incorrect value.
Reject Boxes: These specify incorrect chemical concepts or terminology that negate the mark if included.
Ignore: Refers to details that are irrelevant and neither score nor penalise.
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M1: Most alpha particles (9,880) passing straight through shows that the atom is mostly empty space [1].
M2: A tiny fraction of alpha particles (2) bouncing back shows that the mass and positive charge is concentrated in a tiny, dense center / nucleus [1].
M3: The deflection of positive alpha particles (118) confirms that this nucleus has a positive charge, causing electrostatic repulsion [1].
Reject:
Any reference to electrons being inside the nucleus.
(b) Chadwick's discovery [2 Marks]
M1:Neutron [1].
M2: It has no electrical charge / is neutral, meaning it did not interact with or get deflected by electric or magnetic fields, making it difficult to detect [1].
Reject:
Vague statements that neutrons are inside the nucleus (as this does not explain the difficulty of detection).
(c) Electronic configurations [2 Marks]
A1: (i) Potassium atom (atomic number = 19): 2,8,8,1 [1].
A2: (ii) Chlorine atom (atomic number = 17): 2,8,7 [1].
(d) Potassium vs Sodium reactivity [3 Marks]
M1: Potassium has more electron shells / the outer electron in potassium is further from the nucleus [1].
M2: There is weaker electrostatic attraction between the nucleus and the outer electron in potassium / there is more shielding [1].
M3: Therefore, the outer electron is lost more easily in potassium than in sodium [1].
Reject:
Mention of gaining or sharing electrons. Mention of "stable octet" or "full outer shell" as the driving force.
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GCSE Chemistry Mark Scheme
Topic 1: Atomic Structure & Bonding
Topic 1: Atomic Structure & Bonding
Question 2 (Topic 1) - Group 1 Reactivity & Electron Configurations [6 Marks] ↑ Question
(a) Sodium chloride electron transfer [4 Marks]
M1: Sodium atom (2,8,1) loses its one outer electron to form a sodium ion, Na+ (2,8) [1].
M2: Chlorine atom (2,8,7) gains this one electron to form a chloride ion, Cl- (2,8,8) [1].
M3: The reaction forms oppositely charged ions, which attract each other [1].
M4: These ions are held together in a giant ionic lattice by strong electrostatic forces of attraction [1].
Reject:
Any reference to sharing of electrons / covalent bonding.
(b) Sodium and chlorine equation [2 Marks]
M1: Correct formulas and balancing: 2Na + Cl2 → 2NaCl (Accept Na + 1/2 Cl2 → NaCl) [1].
M2: Correct state symbols: 2Na(s) + Cl2(g) → 2NaCl(s) [1].
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GCSE Chemistry Mark Scheme
Topic 2: Bonding & Properties
Topic 2 Total: 22 Marks
Question 3 (Topic 1) - Transition Metals vs Group 1 Metals [8 Marks] ↑ Question
(a) Conductivity of Solid vs Aqueous Copper(II) Sulfate [3 Marks]
A1: Sample Y (aqueous copper(II) sulfate) conducts electricity, whereas Sample X (solid copper(II) sulfate) does not [1].
M1: In solid copper(II) sulfate (Sample X), the ions are locked in fixed positions in a giant ionic lattice and cannot move [1].
M2: In aqueous copper(II) sulfate (Sample Y), the lattice is broken and the ions are free to move and carry charge throughout the solution [1].
Reject:
Any reference to delocalized electrons carrying charge in copper(II) sulfate.
(b) (i) Cathode process [3 Marks]
M1: Positive copper ions (Cu2+) are attracted to the negative electrode / cathode [1].
M2: The copper ions gain electrons / are reduced to form copper atoms [1].
A1: Equation: Cu2+ + 2e- → Cu [1].
(b) (ii) Anode process [2 Marks]
M3: Hydroxide ions (OH-) from the water are attracted to the positive electrode / anode, where they lose electrons / are oxidised to form oxygen gas and water [1].
Sulfate ions are discharged (hydroxide ions are discharged preferentially).
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Bonding Structure: In diamond, each carbon atom is covalently bonded to four other carbon atoms in a rigid tetrahedral lattice. In graphite, each carbon atom is covalently bonded to three other carbon atoms in flat hexagonal layers.
Delocalized Electrons: In diamond, all four outer electrons are localized in covalent bonds; there are no delocalized electrons. In graphite, each carbon atom has one delocalized electron free to move between layers.
Hardness: Diamond is extremely hard because it has a continuous 3D network of strong covalent bonds requiring high energy to break. Graphite is soft/slippery because layers are held by weak intermolecular forces allowing layers to slide.
Electrical Conductivity: Diamond is an insulator because there are no free ions or delocalized electrons to carry charge. Graphite is a conductor because its delocalized electrons can flow throughout the structure.
Level 3 (5-6 Marks): A detailed, coherent comparison covering all four aspects: bonding structure, presence of delocalized electrons, hardness, and electrical conductivity for both diamond and graphite. Clearly links structural features to the macro-properties.
Level 2 (3-4 Marks): Relevant structural features identified for both allotropes, with some explanation linking them to hardness and/or conductivity. May lack detail in orbital bonding descriptions.
Level 1 (1-2 Marks): Simple statements about the properties of diamond and graphite, lacking clear comparison or detailed structural explanation.
Reject:
Breaking covalent bonds when melting or sliding layers of graphite.
Reference to molecules in diamond or graphite.
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GCSE Chemistry Mark Scheme
Topic 2 & Topic 3
Topic 2 & Topic 3
(b) Evaluation of Student's Claims for A, B, C [6 Marks]
M1: Substance A claim is incorrect [1].
M2: Metals have high melting points but they conduct electricity as both solids and liquids. Substance A does not conduct in either state, indicating it is a giant covalent structure [1].
M3: Substance B claim is incorrect [1].
M4: Simple molecular compounds have low melting points and do not conduct electricity in any state. Substance B has a high melting point and conducts electricity when liquid, indicating it is a giant ionic lattice [1].
M5: Substance C claim is incorrect [1].
M6: Giant covalent structures have extremely high melting and boiling points. Substance C has very low melting and boiling points and does not conduct, indicating it is a simple molecular substance [1].
(c) Nanoparticles [2 Marks]
M1: (i) Nanoparticles have a much higher surface area to volume ratio compared to the bulk material [1].
M2: (ii) Nanoparticles are small enough to pass through skin cells / cell membranes into the body, which could cause cellular damage / unknown long-term toxicity [1].
Reject:
Vague statements like "they are dangerous" or "they react too fast".
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The reactant that is completely used up / completely consumed during the reaction (which limits the amount of product that can be formed) [1].
Reject:
"the reactant that runs out first" or "the reactant with the smaller mass" or "the reactant that reacts the fastest".
(b) Actual yield discrepancies [2 Marks]
M1 (Lower yield):Loss of product during physical transfer stages / filtration / washing [1].
M2 (Higher yield): The precipitate is still wet / retains water / was not dried completely (so the water adds to the measured mass) [1].
Reject:
"human error" or "spillage" without explaining that product is lost.
"impure reactants" as an explanation for higher mass unless specifically linked to precipitation.
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Moles of Fe2O3 = 24.0 / 160 = 0.150 mol
Moles of C = 5.40 / 12 = 0.450 mol [1]
*(Both moles must be correct to award this mark)*
M2: Determine stoichiometric ratio requirements:
According to the equation, 2 mol of Fe2O3 reacts with 3 mol of C (a 1:1.5 ratio).
Therefore, 0.150 mol of Fe2O3 requires (0.150 x 1.5) = 0.225 mol of C [1].
*(Alternatively, show that 0.450 mol of C requires 0.300 mol of Fe2O3)*
M3: Draw limiting reactant conclusion:
Since 0.450 mol of C is available and only 0.225 mol is required, carbon is in excess. This proves that iron(III) oxide (Fe2O3) is the limiting reactant [1].
M4: Determine moles of Fe product:
From the equation, 2 mol of Fe2O3 produces 4 mol of Fe (a 1:2 ratio).
Therefore, moles of Fe produced = 0.150 mol x 2 = 0.300 mol [1].
A1: Calculate theoretical yield mass:
Theoretical yield of Fe = 0.300 mol x 56 g/mol = 16.8 g [1].
Error Carried Forward (ECF) Policy:
If the student calculates incorrect molar masses or incorrect moles in M1, award M2, M3, M4, and A1 as ECF based on their values, provided the chemical logic is mathematically correct.
If a mathematical error in M2 leads to the wrong limiting reactant conclusion in M3, award M4 and A1 as ECF based on their identified limiting reactant.
Reject:
Selecting carbon as the limiting reactant simply because its starting mass (5.40 g) is smaller than the iron(III) oxide mass (24.0 g) without calculating moles.
Intermediate rounding to 1 significant figure that results in an incorrect final yield.
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GCSE Chemistry Mark Scheme
Topic 3: Quantitative Chemistry
Topic 3: Quantitative Chemistry
(d) Percentage Atom Economy Calculation [3 Marks]
M1: Calculate relative masses using balancing coefficients:
Total Mr of desired product (4Fe) = 4 x 56 = 224
Total Mr of all reactants (2Fe2O3 + 3C) = (2 x 160) + (3 x 12) = 320 + 36 = 356 [1]
*(Both values must be correct to award this mark)*
M2: Setup atom economy equation:
Percentage atom economy = (224 / 356) x 100 [1]
A1: Final answer rounded to 3 significant figures:
= 62.9% [1]
*(Accept 62.92% or 63% only if it is clearly derived from correct work)*
Error Carried Forward (ECF):
If incorrect total Mr values are calculated in M1, award M2 and A1 as ECF based on those values.
Reject:
Calculation of atom economy using single formula masses (e.g. Mr of Fe / (Mr of Fe2O3 + Mr of C)) which ignores the balancing coefficients from the equation.
(e) Aim of high atom economy [1 Mark]
To minimise the production of waste / make the process more sustainable / save money on raw materials / reduce the cost of waste disposal [1].
Reject:
"increases the percentage yield" or "makes the reaction faster" or "increases the rate of reaction".
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To find the approximate volume of acid required to neutralise the alkali so that in subsequent runs the acid can be added dropwise near the end point [1].
Reject:
"to get the correct answer" or "to prevent errors".
(ii) Concordant titres definition [1 Mark]
Titres that are within 0.10 cm3 of each other, indicating that the experimental measurements are highly precise / reproducible [1].
Reject:
"results that are close" or "similar results" without specifying the concordancy range of 0.10 cm3.
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GCSE Chemistry Mark Scheme
Topic 3: Quantitative Chemistry
Topic 3 Total: 25 Marks
(c) Unstructured Titration Calculation [6 Marks]
*(Full marks awarded for the correct final answer of 3.20 g/dm3 even with minimal working)*
M1: Calculate moles of sulfuric acid:
Volume of H2SO4 = 20.0 / 1000 = 0.0200 dm3
Moles of H2SO4 = 0.0200 dm3 x 0.0500 mol/dm3 = 0.00100 mol [1]
M2: Determine moles of sodium hydroxide:
Moles of NaOH = 0.00100 mol x 2 = 0.00200 mol [1]
*(Based on the 1:2 stoichiometric ratio in the equation)*
Concentration of NaOH = 0.0800 mol/dm3 x 40 g/mol = 3.20 g/dm3 [1]
*(Accept 3.2 g/dm3 or 3.20 g/dm3. Final answer must be rounded to 3 significant figures)*
Error Carried Forward (ECF):
Full ECF applies at each step.
If the student fails to use the 1:2 stoichiometric ratio (uses a 1:1 ratio instead, scoring 0 for M2), the moles of NaOH = 0.00100 mol. Concentration in mol/dm3 = 0.0400 mol/dm3. Final concentration in g/dm3 = 1.60 g/dm3. This incorrect pathway will score 5 marks out of 6 (losing only M2).
Reject:
Using the Mr of sulfuric acid (98) instead of sodium hydroxide (40) to convert the concentration.
Dividing the moles of NaOH by the volume of H2SO4 (20.0 cm3) in step M4.
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GCSE Chemistry Mark Scheme
Topic 3 & Topic 4
Topic 3 & Topic 4
(d) Magnesium Reacting with Acid Gas Volume [3 Marks]
M1: Calculate moles of Mg:
Moles of Mg = 0.243 g / 24.3 g/mol = 0.0100 mol [1]
M2: Determine moles of H2 gas and calculate volume in dm3:
Moles of H2 = 0.0100 mol (due to 1:1 stoichiometric ratio)
Volume of H2 = 0.0100 mol x 24.0 dm3/mol = 0.240 dm3 [1]
A1: Convert volume to cm3:
Volume of H2 = 0.240 dm3 x 1000 = 240 cm3 [1]
Error Carried Forward (ECF):
If the student calculates incorrect moles of Mg in M1, allow ECF for M2 and A1.
If the student uses the wrong stoichiometric ratio (e.g. 1:2, leading to 0.0200 mol of H2 and 480 cm3 of H2), award M1 and allow ECF for M2 and A1, scoring 2 marks out of 3.
Reject:
Giving the final answer in dm3 (0.240 dm3) instead of cm3.
Using the molar mass of magnesium chloride (95.3 g/mol) in any step.
Zinc / Zn is oxidised because it loses electrons (to form Zn2+) [1].
Reject:
"zinc gains positive charge" without reference to electron loss.
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GCSE Chemistry Mark Scheme
Topic 4: Chemical Changes
Topic 4 Total: 23 Marks
(c) Acid ionisation and pH logarithmic relationship [4 Marks]
(i) Strong vs Weak acid ionisation [2 Marks]
M1: Hydrochloric acid / strong acid is completely ionised / dissociated in aqueous solution [1].
M2: Ethanoic acid / weak acid is only partially ionised / dissociated in aqueous solution [1].
Let initial pH be pH1 and new pH be pH2 = pH1 - 1.
[H+]2 / [H+]1 = 10-pH2 / 10-pH1
[H+]2 / [H+]1 = 10-(pH1 - 1) / 10-pH1 = 10-pH1 + 1 / 10-pH1 = 101 = 10 [1].
Therefore, decreasing the pH by 1 unit increases the hydrogen ion concentration by a factor of 10.
*(Accept alternative logically complete mathematical proofs using logarithm rules)*
To increase the rate of reaction / make the reaction faster [1].
Reject:
"to start the reaction" or "to dissolve the copper oxide" on its own.
(b) Excess reactant details [2 Marks]
(i) Observation when copper oxide is in excess [1 Mark]
A black solid remains at the bottom of the beaker (that does not dissolve upon stirring) [1].
(ii) Purpose of excess reactant [1 Mark]
To ensure that all of the sulfuric acid reacts / is completely neutralised [1].
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GCSE Chemistry Mark Scheme
Topic 4: Chemical Changes
Topic 4 Total: 23 Marks
(c) Crystallisation procedure [3 Marks]
M1:Filter the mixture (to remove the excess unreacted copper(II) oxide) [1].
M2:Heat the filtrate / solution in an evaporating basin to evaporate some of the water until the crystallisation point is reached [1].
Reject:
"heat to dryness" or "evaporate all the water".
M3:Leave the solution to cool and crystallise, then filter / decant the crystals and pat them dry (with filter paper / in a warm oven) [1].
(d) Discrepancies in actual vs theoretical yield [2 Marks]
Any two reasons from:
Some product was lost during transfer steps (e.g. left on filter paper / beaker walls) [1].
Some crystals remained dissolved in the cold filtrate and were not collected [1].
Incomplete crystallisation occurred [1].
The reaction did not go to 100% completion (e.g. if heating was insufficient) [1].
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The minimum energy that colliding particles must have in order to react / start a reaction [1].
(b) Exothermic dissolution explanation [2 Marks]
M1:Exothermic [1].
M2: Chemical energy is transferred as thermal energy to the surroundings / water (or energy is released to the surroundings) [1].
(c) Temperature change investigation design [2 Marks]
M1: Measure a fixed volume of water into a polystyrene cup (which acts as an insulator / reduces heat loss) [1].
M2: Record the initial temperature of the water, add a weighed mass of ammonium chloride, stir, and record the maximum or minimum temperature reached (to calculate the change) [1].
(d) Catalyst mechanism [2 Marks]
M1: A catalyst provides an alternative reaction pathway [1].
M2: This pathway has a lower activation energy (so more particles have enough energy to react when they collide) [1].
(e) Exothermic energy profile representation [3 Marks]
M1:Reactants are drawn at a higher energy level than the products [1].
M2: Activation energy is shown as a vertical arrow pointing upwards from the reactants level to the highest point / peak of the curve [1].
M3: Overall energy change is shown as a vertical arrow pointing downwards from the reactants level to the products level [1].
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GCSE Chemistry Mark Scheme
Topic 5: Energy Changes
Topic 5 Total: 14 Marks
(f) Bond Energy Calculations [4 Marks]
(i) Reactants bond breaking energy [1 Mark]
Calculation: 436 + 242 = 678 (kJ/mol) [1].
(ii) Products bond making energy [1 Mark]
Calculation: 2 x 431 = 862 (kJ/mol) [1].
(iii) Overall energy change and reaction classification [2 Marks]
M2:Exothermic, because the overall energy change is negative / energy is released / more energy is released in bond-making than is absorbed in bond-breaking [1].
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Practice Paper 2 - Master Mark Scheme (Unofficial)
Total Marks: 100
Version 1.0
General Marking Guidance
Acceptable Answers: Mark schemes are prepared by subject specialists. They indicate the points required to gain marks. Alternative wording or symbols that express the same chemical meaning should be accepted.
Bolding: Bold chemical terms are key elements that must be present in the student's answer to score the mark.
Error Carried Forward (ECF): ECF applies to mathematical calculations. If a student makes an early arithmetic error, they lose that specific mark but can score full marks for subsequent steps that apply correct chemical calculations to their incorrect value.
Reject Boxes: These specify incorrect chemical concepts or terminology that negate the mark if included.
Ignore: Refers to details that are irrelevant and neither score nor penalise.
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M1: Any one from: temperature of acid / mass of marble chips / surface area / size of marble chips / volume of acid [1].
(c) Mean rate calculation [2 Marks]
M1: Volume at 20s is 48 cm³ (allow 47-49 cm³). Rate = 48 / 20 [1].
M2: = 2.4 cm³/s (or 2.4 cm³ s⁻¹) [1].
Error Carried Forward (ECF):
Apply ECF to M2 for incorrect volume value read from Curve A (e.g. 50 / 20 = 2.5 cm³/s).
(d) Tangent rate calculation [3 Marks]
M1: Tangent drawn correctly to Curve B at 30 seconds [1].
M2: Calculation of gradient from drawn tangent using dy/dx (e.g., change in volume / change in time) [1].
M3: Value between 0.65 to 0.75 cm³/s with correct unit cm³/s [1].
(e) Collision theory explanation [3 Marks]
M1: Curve A has a higher concentration, so there are more acid particles / H⁺ ions per unit volume [1].
M2: This increases the frequency of collisions (or particles collide more frequently) [1].
M3: Resulting in a higher frequency of successful collisions (or more successful collisions per second) [1].
Reject:
Particles have more energy / particles move faster (unless specifically related to temperature).
(f) Powered CaCO3 curve [2 Marks]
M1: Curve starts steeper than Curve A, indicating a faster initial rate [1].
M2: Curve levels off at the same final volume of 80 cm³ [1].
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M2: The rate of the forward reaction equals the rate of the reverse reaction [1].
(b) Concentration effect [2 Marks]
M1: The equilibrium position shifts to the right / forward direction [1].
M2: Because the system acts to oppose the increase in nitrogen dioxide / reactant concentration (by converting it into product) [1].
(c) Temperature effect [2 Marks]
M1: The mixture turns brown / darker brown [1].
M2: Because the forward reaction is exothermic, so increasing the temperature shifts equilibrium in the endothermic direction (to the left / producing more brown NO₂ gas) [1].
(d) Catalyst effect [2 Marks]
M1: There is no effect on the position of equilibrium [1].
M2: Because the catalyst increases the rate of both forward and reverse reactions equally / by the same amount [1].
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This resource is an independent educational tool created to support student revision. It is completely independent and is not endorsed by, affiliated with, or sponsored by any official examination board. All trademarked terms are used under Nominative Fair Use purely for descriptive compatibility indexing. Licensed for individual personal use only. Chemistry Made Easy is an independent resource. Not affiliated with or endorsed by AQA, Pearson Edexcel, or the IBO.
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GCSE Chemistry Mark Scheme
Practice Paper 2 - Higher Tier
Topic 7: Organic Chemistry
Question 4 (Topic 7) - Cracking, Alkenes & Alcohols [14 Marks] (continued on next page) ↑ Question
(a) Economic usefulness of cracking [2 Marks]
M1: Short-chain hydrocarbons are in higher demand as fuels [1].
M2: Alkenes are produced, which are used to make polymers / plastics / chemicals [1].
(b) Cracking conditions [2 Marks]
M1: Catalytic cracking uses high temperatures and a catalyst / zeolite [1].
M2: Steam cracking uses high temperatures and steam [1].
(c) Balancing equation [1 Mark]
M1: 2 C₂H₄ (accept ethene) [1].
(d) Test for alkenes [2 Marks]
M1: Add bromine water [1].
M2: Solution turns from orange / brown to colourless / decolourises [1].
Reject:
Turns clear (instead of colourless).
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M4: Reasonable evaluation linking these differences to sustainability (e.g. fermentation is more sustainable because it uses renewable resources, but hydration is more efficient and does not require land for crops) [1].
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M1: Pencil is insoluble in the solvent / will not run or bleed up the paper [1].
(b) Rf calculation [2 Marks]
M1: Distance moved by spot A = 3.5 cm. Distance moved by solvent front = 7.0 cm. Rf = 3.5 / 7.0 [1].
M2: = 0.50 (or 0.5, accept no unit) [1].
(c) Deducing dye presence [2 Marks]
M1: Dyes A and C are present in Y [1].
M2: Because Y contains spots that travel the exact same distance / have the same Rf values as spots in A and C [1].
(d) Purity definitions [2 Marks]
M1: Everyday: nothing has been added (it is natural) [1].
M2: Chemical: consists of only a single element or compound [1].
(e) Impurities and melting point [3 Marks]
M1: Impurities lower the melting point [1].
M2: Impurities cause the substance to melt over a wider temperature range / broad range (instead of a single temperature) [1].
M3: Pure substances melt at a sharp, specific temperature [1].
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Adding sulfuric acid (this adds sulfate ions and invalidates the test).
(c) Identifying compound Z [2 Marks]
M1: Cation: green precipitate with NaOH is iron(II) / Fe²⁺ [1].
M2: Anion: white precipitate with barium chloride is sulfate / SO₄²⁻. Compound is iron(II) sulfate [1].
(d) Instrumental method advantage [1 Mark]
M1: Any one from: more accurate / more sensitive (can detect tiny amounts) / faster / can identify mixtures of ions [1].
(e) Carbonate ion test [2 Marks]
M1: Add dilute hydrochloric acid (and bubble gas produced through limewater) [1].
M2: Carbon dioxide is produced / limewater turns cloudy / milky [1].
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Question 8 (Topic 9) - Greenhouse Effect & Carbon Footprints [12 Marks] (continued on next page) ↑ Question
(a) Greenhouse mechanism [3 Marks]
M1:Short-wavelength radiation from the sun passes through the atmosphere [1].
M2: The Earth's surface absorbs this and re-emits it as long-wavelength / infrared radiation [1].
M3: Greenhouse gases absorb the long-wavelength radiation, trapping the energy in the atmosphere [1].
(b) Greenhouse gases [1 Mark]
M1:Methane and water vapour [1].
(c) Carbon footprint definition [2 Marks]
M1: The total amount of carbon dioxide and other greenhouse gases emitted over the full life cycle of a product, service, or event [1].
M2: Walk/cycle instead of drive / insulate homes / eat less meat / reduce electricity usage [1].
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Individual actions: Walk/cycle instead of driving, reduce home heating/insulate, install solar panels, eat less meat.
Individual constraints: High initial cost of solar/insulation, lack of public transport infrastructure, lifestyle choices.
Government actions: Invest in renewable energy (wind/solar), tax carbon emissions, subsidise green technology.
Government constraints: Political unpopularity of green taxes, high national debt limits funding, economic competition.
Industrial actions: Improve efficiency, transition to carbon-neutral raw materials, carbon capture and storage (CCS).
Industrial constraints: Cost of CCS, technological barriers, need to remain profitable for shareholders.
Level 3 (5-6 marks): A detailed, balanced evaluation covering at least two groups. Explains several actions and constraints. Concludes with a justified decision on which actions are most critical.
Level 2 (3-4 marks): A description of actions and some constraints. Shows logical reasoning but lacks depth or a clear final conclusion.
Level 1 (1-2 marks): Mentions basic actions (e.g., "recycle more" or "use electric cars") without connecting them to constraints or an overall evaluation.
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GCSE Chemistry Mark Scheme
Practice Paper 2 - Higher Tier
Topic 10: Using Resources
Question 9 (Topic 10) - Water Resources & Treatment [12 Marks] (continued on next page) ↑ Question
(a) Potable water definition [2 Marks]
M1: Water that is safe to drink [1].
M2: It contains dissolved salts / minerals, whereas pure water only contains water molecules [1].
(b) UK water treatment [2 Marks]
M1:Filtration: passes water through sand/gravel beds to remove insoluble solids [1].
M2: Requires large amounts of energy to boil water (distillation) or pressurise water through membranes (reverse osmosis) [1].
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Raw materials: Plastic uses non-renewable crude oil (finite, high transport footprint). Paper uses trees (renewable, requires land and deforests if unmanaged).
Manufacturing: Plastic requires high energy cracking/polymerisation. Paper requires boiling wood pulp and large volumes of clean water.
Lifespan/usage: Plastic is strong, waterproof, and reusable. Paper is weak when wet and rarely reused.
Disposal: Plastic is non-biodegradable, causes litter, fills landfills, but can be recycled. Paper is biodegradable, recyclable, and releases CO₂/methane if it rots in landfills.
LCA limitations: Hard to put exact numerical values on visual pollution, loss of habitats, or forest damage (subjective value judgments).
Level 3 (5-6 marks): Balanced analysis across all four life cycle stages. Evaluates both paper and plastic. Mentions the subjective nature of LCAs and concludes with a clear, justified decision on which bag is more sustainable.
Level 2 (3-4 marks): Describes some stages of the LCA for both bags. Includes some comparison but has minor omissions and lacks a clear, well-argued conclusion.
Level 1 (1-2 marks): Disjointed points (e.g. "plastic doesn't rot" and "paper comes from trees") without an evaluation or structured comparison.
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GCSE Chemistry Mark Scheme
Practice Paper 2 - Higher Tier
Topic 10: Using Resources
Question 10 (Topic 10) - Haber Process & Alternative Metal Extraction [8 Marks] ↑ Question
(a) Compromise temperature [2 Marks]
M1: Low temperature gives a high yield because the forward reaction is exothermic, but the rate is too slow [1].
M2: 450 °C is a compromise to obtain a reasonable yield at a fast rate [1].
(b) Ammonia separation [1 Mark]
M1: The mixture is cooled; ammonia has a higher boiling point so it condenses / liquefies and is removed [1].
(c) Phytomining process [3 Marks]
M1: Plants are grown on soil containing low-grade copper ores and absorb copper ions [1].
M2: The plants are harvested and burned to produce ash [1].
M3: The ash is reacted with sulfuric acid to make a copper sulfate solution (leachate), from which copper is displaced using scrap iron [1].
(d) Phytomining evaluation [2 Marks]
M1: Advantage: less destructive to land / conserves high-grade ores / carbon-neutral when growing [1].
M2: Disadvantage: very slow process / requires land that could grow crops [1].
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