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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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(a) Mendeleev vs Newlands [6 Marks]
M1: John Newlands arranged elements in order of increasing atomic weight and proposed the law of octaves, but this failed after calcium because elements in the same group had different properties [1].
M2: Mendeleev also arranged elements by atomic weight, but he swapped the order of some elements (e.g. tellurium and iodine) to ensure elements with similar chemical properties were placed in the same group [1].
M3: Mendeleev left gaps for elements that had not yet been discovered [1].
M4: Mendeleev used his table to predict the properties of these undiscovered elements [1].
M5: When these elements were discovered (e.g. gallium) and their properties matched Mendeleev's predictions, his periodic table was accepted by the scientific community [1].
M6: The later discovery of isotopes explained why ordering elements strictly by atomic weight was incorrect, validating Mendeleev's decision to swap elements [1].
Reject:
Any statement claiming Mendeleev ordered elements by atomic number.
(b) Chlorine relative atomic mass calculation [4 Marks]
M1: Calculates total mass contribution of isotopes: (35 * 75.77) + (37 * 24.23) [1].
M2: Sums isotope contributions: 2651.95 + 896.51 = 3548.46 [1].
M3: Divides by 100 to get relative atomic mass: 35.4846 [1].
M4: Correctly rounds to 2 decimal places: 35.48 [1].
Note:
An answer of 35.5 scores a maximum of 3 marks (for incorrect rounding).
(c) Electronic configuration of oxygen and sulfur [4 Marks]
M1: Oxygen electronic configuration is 2,6 [1].
M2: Sulfur electronic configuration is 2,8,6 [1].
M3: Both oxygen and sulfur have 6 electrons in their outer shell [1].
M4: The group number in the periodic table corresponds to the number of outer-shell electrons, so both are placed in Group 6 [1].
(d) Modern periodic table arrangement [2 Marks]
M1: Ordered by increasing atomic number / proton number [1].
M2: Early tables were ordered by atomic weight because protons, neutrons, and electrons had not yet been discovered [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.
Do not write outside this box
(a) Carbon dioxide dot-and-cross diagram [4 Marks]
M1: Carbon atom shares 4 electrons (needs double bonds on both sides) [1].
M2: Each Oxygen atom shares 2 electrons [1].
M3: Correct number of shared electrons in overlap regions (4 electrons in each overlap area, e.g. 2 dots and 2 crosses) [1].
M4: Non-bonding outer electrons shown correctly (4 non-bonding electrons on each oxygen atom; no non-bonding outer-shell electrons on the carbon atom) [1].
(b) Graphite vs Graphene comparison [6 Marks]
M1: Graphene is a single layer of carbon atoms arranged in hexagons, whereas graphite consists of many layers of graphene stacked together [1].
M2: In both structures, each carbon atom forms three strong covalent bonds with adjacent carbon atoms [1].
M3: Graphene is very strong because it has a continuous 2D network of strong covalent bonds which require a large amount of energy to break [1].
M4: Graphite is soft and slippery because its layers are held together by weak intermolecular forces, allowing layers to slide over each other easily [1].
M5: Because each carbon atom only uses three of its four outer electrons in bonding, each carbon has one delocalized electron [1].
M6: These delocalized electrons are free to move throughout the structure and carry electrical charge [1].
(c) Carbon dioxide properties [6 Marks]
M1: Carbon dioxide has a simple molecular structure [1].
M2: Covalent bonds within the molecules are strong [1].
M3: However, the intermolecular forces between individual molecules are very weak [1].
M4: Consequently, only a small amount of thermal energy is needed to overcome these weak intermolecular forces, giving a low boiling point [1].
M5: Carbon dioxide does not conduct electricity because it has no free-moving charged particles [1].
M6: All outer electrons are fixed in covalent bonds, so there are no free-moving ions or delocalized electrons [1].
Reject:
Any claim that covalent bonds break when carbon dioxide boils.
(d) Silicon dioxide vs Carbon dioxide [6 Marks]
M1: Carbon dioxide is simple molecular, whereas silicon dioxide is a giant covalent structure [1].
M2: In carbon dioxide, there are weak intermolecular forces. In silicon dioxide, there are no molecules, only a continuous 3D network of covalent bonds [1].
M3: In silicon dioxide, each silicon atom is bonded to four oxygen atoms, and each oxygen atom is bonded to two silicon atoms [1].
M4: Melting silicon dioxide requires breaking many strong covalent bonds throughout the giant lattice [1].
M5: Breaking these covalent bonds requires a very large amount of energy [1].
M6: Thus, silicon dioxide has a very high melting point, whereas carbon dioxide has a very low melting point because only weak intermolecular forces need to be overcome [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.
Do not write outside this box
(a) Limiting reactant & ammonia mass [8 Marks]
M1: Calculates moles of N2: Moles = 56.0 / 28 = 2.0 mol [1].
M2: Calculates moles of H2: Moles = 15.0 / 2 = 7.5 mol [1].
M3: Uses stoichiometric ratio: 1 mole of N2 reacts with 3 moles of H2, so 2.0 mol N2 requires 6.0 mol H2 [1].
M4: States H2 is in excess: Since 7.5 mol H2 is present (which is greater than the required 6.0 mol), H2 is in excess [1].
M5: Identifies limiting reactant: Nitrogen (N2) is the limiting reactant [1].
M6: Relates limiting reactant moles to product moles: 1 mol N2 produces 2 mol NH3, so 2.0 mol N2 produces 4.0 mol NH3 [1].
M7: Calculates Mr of NH3: Mr = 14 + (3 * 1) = 17 [1].
M8: Calculates mass of NH3: Mass = 4.0 * 17 = 68.0 g (Accept 68 g) [1].
Note:
Award full marks for a final answer of 68.0 g with correct working. If limiting reactant logic is incorrect, apply ECF.
(b) Titration calculation [9 Marks]
M1: Converts volume of H2SO4 to dm3: 18.5 cm3 = 0.0185 dm3 [1].
M2: Calculates moles of H2SO4: Moles = 0.125 * 0.0185 = 0.0023125 mol [1].
M3: Relates stoichiometry: 1 mole H2SO4 reacts with 2 moles KOH, so moles of KOH = 0.0023125 * 2 = 0.004625 mol [1].
M4: Converts volume of KOH to dm3: 25.0 cm3 = 0.0250 dm3 [1].
M5: Calculates concentration of KOH in mol/dm3: Moles / Volume = 0.004625 / 0.0250 = 0.185 mol/dm3 [1].
M6: Calculates the Mr of KOH: Mr = 39 + 16 + 1 = 56 [1].
M7: Calculates concentration of KOH in g/dm3: 0.185 * 56 = 10.36 g/dm3 [1].
M8: Rounds concentration in mol/dm3 to 3 significant figures: 0.185 mol/dm3 [1].
M9: Rounds concentration in g/dm3 to 3 significant figures: 10.4 g/dm3 [1].
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Chemistry Made Easy is an independent resource. Not affiliated with or endorsed by AQA, Pearson Edexcel, or the IBO.
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(c) Percentage yield [3 Marks]
M1: States/uses formula: Percentage Yield = (Actual Yield / Theoretical Yield) * 100 [1].
M2: Substitutes values: (51.0 / 68.0) * 100 [1].
M3: Correct answer rounded to 3 significant figures: 75.0% [1].
(d) Gas volume calculation [2 Marks]
M1: States/uses formula: Volume = Moles * 24.0 [1].
M2: Correct calculation and answer: 2.00 * 24.0 = 48.0 dm3 (Accept 48 dm3) [1].
(e) Reasons for less than 100% yield [3 Marks]
M1: The reaction may be reversible and thus does not go to completion [1].
M2: Some product may be lost during separation / transfer (e.g. left on filter paper or in beaker) [1].
M3: Side reactions may occur or reactants may not be pure [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.
Do not write outside this box
(a) Electrolysis of dilute H2SO4 [8 Marks]
M1: Electrolyte contains H+ and OH- ions from water, and H+ and SO4(2-) ions from sulfuric acid [1].
M2: Cathode (Negative Electrode): Positive H+ ions migrate to the cathode [1].
M3: Hydrogen ions gain electrons (are reduced) to form hydrogen gas: 2H+ + 2e- → H2 [1].
M4: Anode (Positive Electrode): Negative OH- and SO4(2-) ions migrate to the anode [1].
M5: Hydroxide (OH-) ions are preferentially discharged because they are easier to oxidize than sulfate (SO4(2-)) ions [1].
M6: Hydroxide ions lose electrons (are oxidized) to form oxygen gas and water: 4OH- → O2 + 2H2O + 4e- (Accept 4OH- - 4e- → O2 + 2H2O) [1].
M7: The volume of hydrogen gas produced is double the volume of oxygen gas produced [1].
M8: This is because discharging 4OH- ions transfers 4 electrons, which reduces 4H+ ions to produce 2 molecules of H2 for every 1 molecule of O2 [1].
(b) Required Practical 3: Electrolysis of brine [6 Marks]
M1: Anode product: Chlorine gas (Cl2) [1].
M2: Cathode product: Hydrogen gas (H2) [1].
M3: Chlorine gas forms at the anode because chloride ions (Cl-) are halide ions, which are preferentially discharged over hydroxide ions [1].
M4: Hydrogen gas forms at the cathode because hydrogen ions (H+) are preferentially discharged over sodium ions (Na+) since hydrogen is less reactive than sodium [1].
M5: Sodium ions (Na+) and hydroxide ions (OH-) remain in the solution [1].
M6: The useful substance left in the solution is sodium hydroxide (NaOH) solution [1].
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Chemistry Made Easy is an independent resource. Not affiliated with or endorsed by AQA, Pearson Edexcel, or the IBO.
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(c) Strong vs weak acid pH change log proof [6 Marks]
M1: The pH scale is logarithmic: each change of 1 pH unit represents a 10-fold change in hydrogen ion concentration [H+] [1].
M2: The pH difference is: 5.00 - 2.00 = 3 units [1].
M3: Difference in hydrogen ion concentration is: 10^3 = 1000 times [1].
M4: The hydrochloric acid solution has a hydrogen ion concentration 1000 times greater than that of the weak acid solution [1].
M5: A strong acid is completely ionised / dissociated in aqueous solution, releasing all its hydrogen ions [1].
M6: A weak acid is only partially ionised / dissociated in aqueous solution, so only a small percentage of its molecules release hydrogen ions [1].
(d) Required Practical 1: Salt preparation [3 Marks]
M1: Balanced equation: CuO(s) + 2HCl(aq) → CuCl2(aq) + H2O(l) [1].
M2: Filtration: Filter the reaction mixture to remove excess copper(II) oxide (black solid residue) to obtain a blue copper(II) chloride solution as the filtrate [1].
M3: Crystallisation: Heat the filtrate gently in an evaporating basin to evaporate water until the crystallisation point is reached, then leave to cool and crystallise; filter the crystals and pat dry [1].
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Chemistry Made Easy is an independent resource. Not affiliated with or endorsed by AQA, Pearson Edexcel, or the IBO.
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(a) Energy profile diagram [6 Marks]
M1: Draw reactants level higher than products level [1].
M2: Labeled reactants line (Mg + 2HCl) and products line (MgCl2 + H2) [1].
M3: Upward arrow from reactants level to peak of curve represents activation energy (Ea) [1].
M4: Downward arrow from reactants level to products level represents overall energy change (ΔH) [1].
M5: Exothermic because the products are at a lower energy level than the reactants [1].
M6: Overall energy is released to the surroundings / overall energy change is negative [1].
(b) Bond energy calculation [8 Marks]
M1: Identifies bonds broken: 1 H-H bond and 1 Cl-Cl bond [1].
M2: Calculates total energy to break bonds: 436 + 242 = 678 kJ/mol [1].
M3: Identifies bonds made: 2 H-Cl bonds [1].
M4: Calculates total energy released in bond making: 2 * 431 = 862 kJ/mol [1].
M5: Calculates overall energy change: energy to break - energy to make = 678 - 862 = -184 kJ/mol [1].
M6: Obtains correct final value: -184 kJ/mol [1].
M7: States reaction is exothermic because overall energy change is negative [1].
M8: Explains that bond breaking is an endothermic process (requires energy) and bond making is an exothermic process (releases energy), and more energy is released than is required [1].
Reject:
Any unit other than kJ/mol or kJ.
Do not write outside this box
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.