Mark Scheme

Section A — Multiple Choice Answer Key

Detailed Marking Notes

Question 1: Answer B (3 marks)

• Particles in gases move randomly at high speeds due to high kinetic energy
• Reject A: solid particles are closer together than liquid particles
• Reject C: all particles possess kinetic energy
• Reject D: particles vibrate more when heated

Question 2: Answer B (3 marks)

• Number of neutrons = nucleon number - proton number = 35 - 17 = 18
• Accept calculation shown as 35 - 17
• Reject A: this is the proton number

Question 3: Answer C (3 marks)

• Electronic configuration 2,8,7 indicates 7 electrons in outer shell
• Elements with 7 outer electrons are in Group VII (halogens)
• Reject D: Group VIII (noble gases) have 8 outer electrons (except helium)

Question 4: Answer C (3 marks)

• Sodium chloride is formed by transfer of electrons from sodium (metal) to chlorine (non-metal)
• This electron transfer produces ions held together by ionic bonding
• Accept electrostatic attraction between oppositely charged ions

Question 5: Answer A (3 marks)

• One pair of electrons is shared between H and Cl
• This constitutes a single covalent bond
• Reject D: no electron transfer occurs

Question 6: Answer B (3 marks)

• Measuring volume of gas at regular intervals shows how rate changes over time
• Gas production slows as reaction proceeds (reactant concentration decreases)
• Accept measuring mass loss at regular intervals
• Reject A: temperature measurement does not directly show reaction rate

Question 7: Answer C (3 marks)

• Bauxite is aluminium ore (Al₂O₃·2H₂O)
• Major bauxite mining occurs in Jamaica and Guyana (Caribbean context)
• Reject A: iron ore is haematite/magnetite

Question 8: Answer D (3 marks)

• Catalysts increase reaction rate without being consumed
• They lower activation energy
• Reject all other options decrease rate

Question 9: Answer B (3 marks)

• Calcium carbonate breaks down when heated into two products
• This is thermal decomposition
• Reject A: neutralization involves acid + base → salt + water

Question 10: Answer B (3 marks)

• Hydrocarbons contain only carbon and hydrogen
• C₂H₆ (ethane) contains only C and H
• Reject A, C, D: all contain oxygen

Question 11: Answer C (3 marks)

• Relative formula mass = sum of relative atomic masses of all atoms
• Accept Ar values are used
• Reject A: proton number ≠ relative atomic mass for most elements

Question 12: Answer A (3 marks)

• Law of conservation of mass: mass of oxygen = mass of product - mass of magnesium
• 10.0 - 6.0 = 4.0 g
• Accept calculation shown

Question 13: Answer C (3 marks)

• pH 3 is acidic (pH < 7)
• Acidic solutions have higher concentration of H⁺ than OH⁻
• Reject B: alkaline pH > 7

Question 14: Answer C (3 marks)

• Distillation evaporates water then condenses pure water, leaving salts behind
• Reject B: evaporation leaves water behind and concentrates salts
• Reject A: filtration cannot separate dissolved substances

Question 15: Answer A (3 marks)

• CO₂ turns limewater (calcium hydroxide solution) milky/cloudy
• Forms white precipitate of calcium carbonate
• Reject B: this tests for oxygen
• Reject C: this tests for hydrogen

Question 16: Answer C (3 marks)

• Calcium hydroxide (lime/slaked lime) is a base used to neutralize acidic soil
• Common agricultural practice in Caribbean farming
• Reject A: acid would make soil more acidic
• Reject B: fertilizer, not used for pH adjustment

Question 17: Answer B (3 marks)

• Transition metals occupy the d-block between Groups II and III
• Accept reference to d-block position

Question 18: Answer A (3 marks)

• Neutralization: H⁺ + OH⁻ → H₂O
• Reaction between acid (H⁺) and base (OH⁻) to form water
• Reject B: precipitation reaction
• Reject C: reduction/redox reaction
• Reject D: decomposition/electrolysis

Question 19: Answer B (3 marks)

• Crude oil is separated by fractional distillation using different boiling points
• Fractions collected at different temperatures
• Reject A: filtration separates insoluble solid from liquid

Question 20: Answer C (3 marks)

• Isotopes have same proton number but different neutron numbers
• Therefore different nucleon numbers
• Reject D: isotopes have identical chemical properties
• Reject B: nucleon number differs

Sample Answers with Examiner Commentary

Question 8 — Sample Answers

Grade I (Distinction) answer

Answer: D

The student correctly identifies that using a catalyst increases the rate of a chemical reaction. A catalyst is a substance that speeds up a chemical reaction by providing an alternative reaction pathway with lower activation energy. The catalyst itself remains unchanged at the end of the reaction and can be reused. Options A, B, and C all describe conditions that would decrease the reaction rate: lowering temperature reduces kinetic energy of particles, larger particles have smaller surface area to volume ratio reducing collision frequency, and lower concentration means fewer particles per unit volume leading to fewer collisions per second.

Mark: 3/3 marks

Examiner commentary: This is an exemplary multiple-choice response that demonstrates thorough understanding beyond simply selecting the correct option. The student shows command of particle collision theory and explains why the catalyst works (lower activation energy, alternative pathway). The systematic elimination of incorrect options with scientific reasoning demonstrates distinction-level comprehension. The student correctly uses key terminology: activation energy, collision frequency, surface area to volume ratio, concentration.

Grade III (Pass) answer

Answer: D

I chose D because a catalyst makes reactions go faster. Catalysts are not used up in the reaction so they can be used again. The other options would slow the reaction down because decreasing temperature makes particles move slower and using bigger pieces means less surface area.

Mark: 3/3 marks

Examiner commentary: This answer earns full credit as the correct option is selected. The justification provided is adequate and shows acceptable understanding of catalysts and factors affecting rate. However, the explanation lacks the depth and precision expected at distinction level—no mention of activation energy, collision theory is implicit rather than explicit, and the explanation of surface area is abbreviated. For multiple-choice questions, marks are awarded for the correct answer selection only, but the explanation reveals mid-range conceptual understanding.

Grade V (Near miss) answer

Answer: A

I think the answer is A because when you increase the temperature the reaction goes faster, so decreasing the temperature would also increase the rate but more slowly. Or maybe D because catalysts speed things up, but I think catalysts are only used in factories not in normal reactions.

Mark: 0/3 marks

Examiner commentary: This response demonstrates a critical misconception about the relationship between temperature and reaction rate—the student incorrectly believes decreasing temperature increases rate "but more slowly," which is contradictory and scientifically incorrect. The student shows uncertainty and changes between options A and D without committing to the correct answer D. The misunderstanding about catalysts (believing they are only industrial) reveals limited practical knowledge. This illustrates why incorrect option selection in multiple-choice results in zero marks regardless of partial knowledge shown. To improve, the student needs to understand that decreasing temperature always decreases rate, and that catalysts are used in many contexts including biological systems (enzymes).

Question 12 — Sample Answers

Grade I (Distinction) answer

Answer: A (4.0 g)

Using the law of conservation of mass: Mass of reactants = Mass of products

Mass of magnesium + Mass of oxygen = Mass of magnesium oxide 6.0 g + mass of oxygen = 10.0 g

Therefore: mass of oxygen = 10.0 g - 6.0 g = 4.0 g

This makes chemical sense because oxygen has been added to magnesium during combustion to form the oxide, so the product mass must be greater than the original magnesium mass. The difference (4.0 g) represents the oxygen that combined with the magnesium.

Mark: 3/3 marks

Examiner commentary: This answer demonstrates excellence in multiple ways. The student correctly applies the law of conservation of mass and shows clear mathematical working with proper units throughout. The algebraic rearrangement is explicit and logical. Importantly, the student provides a chemical context for the answer, explaining why the oxide is heavier than the original metal and what the difference represents. This synthesis of mathematical calculation with chemical understanding exemplifies distinction-level thinking. The answer is complete, well-structured, and would earn full credit with additional commendation.

Grade III (Pass) answer

Answer: A (4.0 g)

The mass of oxygen = 10.0 - 6.0 = 4.0 g

I subtracted the mass of magnesium from the mass of magnesium oxide to find the mass of oxygen that was added.

Mark: 3/3 marks

Examiner commentary: This answer correctly identifies option A and shows the essential calculation with appropriate units. The student demonstrates understanding that subtraction is required and recognizes that oxygen was "added" to form the oxide. However, the response lacks the sophistication of the distinction answer—no explicit reference to conservation of mass, no equation setup, and minimal chemical reasoning. While this earns full marks in multiple-choice format, the limited explanation indicates secure mid-grade understanding rather than exceptional grasp. The student should develop their ability to cite scientific principles explicitly.

Grade V (Near miss) answer

Answer: D (16.0 g)

The formula for magnesium oxide is MgO. Magnesium has mass 24 and oxygen has mass 16. So 16.0 g of oxygen reacted. The question asks for mass of oxygen so I used the atomic mass from the periodic table.

Mark: 0/3 marks

Examiner commentary: This answer reveals a fundamental misunderstanding of the question requirements. The student has recalled the relative atomic mass of oxygen (16) from the periodic table but has not recognized this is a practical mass calculation based on experimental data, not a theoretical calculation from formulae. The student has ignored the given masses (6.0 g and 10.0 g) entirely and instead attempted to use Ar values directly as actual masses. This represents a common CSEC-level error: confusing relative atomic mass with actual mass in grams. To improve, the student must learn to distinguish between: (1) calculations using Ar/Mr for molar relationships, and (2) calculations using conservation of mass with given experimental masses. The student needs practice interpreting what the question is actually asking for.