▶️ Answer/Explanation
(a)(i) simple distillation
Explanation: Simple distillation is used to separate a pure solvent (like water) from a solution (like sea water) by boiling the solution and then condensing the vapor. It works because water has a much lower boiling point than the dissolved salts.
(a)(ii) chromatography
Explanation: Chromatography, specifically paper chromatography, is used to separate mixtures of dyes or pigments based on their different solubilities and attractions to the paper and solvent. Each dye travels a different distance, allowing them to be identified.
(a)(iii) fractional distillation
Explanation: Fractional distillation is used to separate two or more miscible liquids with different boiling points, such as ethanol (boiling point ~78°C) and water (boiling point 100°C). A fractionating column allows for more efficient separation than simple distillation.
(b)
M1: dissolve
M2: solute
M3: solvent
Explanation: A saturated solution is one in which no more solute can dissolve at a given temperature. The substance that dissolves (salt) is the solute, and the substance it dissolves in (water) is the solvent.
▶️ Answer/Explanation
(a)(i) Na or K (Accept sodium/potassium)
Explanation: Metals are found on the left side and centre of the Periodic Table. Sodium (Na) and Potassium (K) are both Group 1 alkali metals.
(a)(ii) He (Accept helium)
Explanation: Noble gases are in Group 0 (Group 18). Helium (He) is the only noble gas shown in the table snippet.
(a)(iii) Br (Accept bromine)
Explanation: At room temperature, bromine (Br) is the only element shown that exists as a liquid. Chlorine (Cl) is a gas, and the others are solids.
(a)(iv) Na and Cl (Accept sodium and chlorine)
Note: Reject “bromide” or “chloride” as these refer to ions/compounds, not elements.
Explanation: Period 3 refers to the third row of the Periodic Table. From the given table, sodium (Na) and chlorine (Cl) are in Period 3.
(b) 2.8.1 (Accept 2,8,1 or 2 8 1; also allow a diagram showing the electron configuration)
Explanation: Sodium (Na) has atomic number 11. Its electron configuration is: 2 electrons in the first shell, 8 in the second, and 1 in the third, represented as \(2.8.1\).
▶️ Answer/Explanation
(a) Completing the Table:
Explanation: Propene is the alkene with three carbons. The empirical formula is the simplest whole-number ratio (\(CH_2\)). The general formula for alkenes is \(C_nH_{2n}\). The displayed formula shows the carbon-carbon double bond, which is characteristic of alkenes.
(b)(i) Unsaturated:
• (A compound) containing a carbon-carbon double bond.
Explanation: Unsaturated hydrocarbons, like alkenes, have at least one C=C bond, meaning they can undergo addition reactions.
(b)(ii) Test for Unsaturation:
• Add bromine water (orange/brown) to the compound.
• If unsaturated, the bromine water decolourises (turns colourless).
Explanation: The double bond in alkenes undergoes an addition reaction with bromine, forming a colourless dibromoalkane. This is a standard qualitative test.
(c)(i) Equation:
• \( CH_4 + Cl_2 \rightarrow CH_3Cl + HCl \)
Explanation: This shows the substitution of one hydrogen atom in methane with a chlorine atom.
(c)(ii) Reaction Type:
• D (substitution)
Explanation: A hydrogen atom is replaced (substituted) by a chlorine atom. This is characteristic of the reaction between alkanes and halogens.
(c)(iii) Condition:
• Ultraviolet (UV) light / sunlight.
Explanation: The reaction requires UV light to provide the energy to break the Cl–Cl bond, initiating the free-radical chain reaction.
(d)(i) Isomers:
• Molecules with the same molecular formula
• but different structural/displayed formulae (or different arrangement of atoms)
Explanation: Isomerism is a key concept in organic chemistry. Structural isomers have the same atoms connected in different ways.
(d)(ii) Displayed Formulae of \( C_2H_4Cl_2 \) Isomers:
Explanation: Isomer 1 has both chlorine atoms on the same carbon. Isomer 2 has one chlorine on each carbon. These are structural isomers.
▶️ Answer/Explanation
(a) \( 2\ \text{H}_2\text{O}_2 \rightarrow 2\ \text{H}_2\text{O} + \text{O}_2 \)
Balance the equation. Two molecules of hydrogen peroxide decompose to produce two molecules of water and one molecule of oxygen. Accept multiples or fractions (e.g., \( \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \frac{1}{2}\text{O}_2 \)).
(b) Relights a glowing splint/spill
Oxygen supports combustion. A glowing wooden splint will burst into flames when inserted into oxygen.
(c) Speeds up/increases rate of the reaction
A catalyst provides an alternative reaction pathway with a lower activation energy, increasing the frequency of successful collisions without being consumed. IGNORE references to lowering activation energy in this answer.
(d)(i)-(iii)
Graph plotted from the table, with a circle around the point at 8 minutes (55 cm³) as it is anomalous. A smooth curve of best fit is drawn through the other points. The curve should show an increasing gradient at first (fast reaction) that gradually levels off as the reaction finishes (oxygen volume becomes constant at 80 cm³).
(d)(iv) Took the reading too soon/before 8 minutes (OR misread the volume of oxygen)
The volume at 8 minutes (55 cm³) is lower than the trend suggests. This could be because the student recorded the volume before the 8-minute mark, or misread the measuring cylinder.
(d)(v) 29 or 30 cm³
A vertical line should be drawn from 3 minutes on the time axis up to the curve of best fit. A horizontal line from this intersection to the volume axis gives the volume. The expected value from a correctly drawn graph is 29 or 30 cm³. The student must show the construction lines on their graph.
(e)(i)
A curve should be drawn on the same grid that is less steep than the original curve, showing a slower initial rate. The curve should level off at 40 cm³ (half the final volume of the original experiment, as the concentration of hydrogen peroxide is halved, but the volume of solution is the same, so half the amount of reactant is present).
(e)(ii) An explanation that links the following three points:
1. The reaction is slower.
2. There are fewer hydrogen peroxide particles/molecules in the same volume (OR particles are further apart/less crowded).
3. This leads to fewer collisions per unit time between reactant particles (OR less frequent collisions).
Halving the concentration reduces the number of reactant particles in a given volume. According to collision theory, the rate of reaction depends on the frequency of successful collisions. Fewer particles mean fewer collisions, hence a slower rate. IGNORE “less chance of a collision”.
▶️ Answer/Explanation
(a)(i) 6 (six)
Explanation: The molecular formula is CH\(_3\)OH. This represents 1 Carbon, 4 Hydrogens (3 from CH\(_3\) and 1 from OH), and 1 Oxygen atom. Total = 1 + 4 + 1 = 6 atoms.
(a)(ii) One of the following:
• Methanol does not contain only carbon and hydrogen.
• Methanol contains (an atom of) oxygen.
Explanation: A hydrocarbon is defined as a compound containing only hydrogen and carbon atoms. Methanol contains oxygen, so it is not a hydrocarbon; it is an alcohol.
(b)(i) A covalent bond is a shared pair of electrons between two atoms. Alternatively: It is an electrostatic attraction between two nuclei and the shared pair(s) of electrons between them.
Explanation: This definition is from specification point 1.44: “know that a covalent bond is formed between atoms by the sharing of a pair of electrons.”
(b)(ii) A correct dot-and-cross diagram for methanol (CH\(_3\)OH). Key features:
- Central carbon atom with four outer electrons (e.g., crosses).
- Carbon shares one pair (single bond) with oxygen (oxygen can use dots for its outer electrons).
- Carbon shares one pair with each of the three hydrogen atoms.
- Oxygen shares its remaining pair with the fourth hydrogen atom.
- Oxygen also has two lone pairs (non-bonding electrons).
Note: The diagram should show only outer-shell electrons. A typical representation has the carbon atom in the center bonded to three H atoms and one O atom. The O atom is also bonded to an H atom and has two lone pairs.
(c)(i) Step-by-step calculation for empirical formula:
- Assume 100 g of compound, so masses are: C = 38.7 g, H = 9.7 g, O = 51.6 g.
- Convert masses to moles:
Moles of C = \( \frac{38.7}{12.0} = 3.225 \)
Moles of H = \( \frac{9.7}{1.0} = 9.7 \)
Moles of O = \( \frac{51.6}{16.0} = 3.225 \) - Divide by the smallest number of moles (3.225):
C : \( \frac{3.225}{3.225} = 1 \)
H : \( \frac{9.7}{3.225} \approx 3 \)
O : \( \frac{3.225}{3.225} = 1 \) - Ratio C : H : O = 1 : 3 : 1
Empirical formula = CH\(_3\)O
(c)(ii) Step to find molecular formula:
- Calculate empirical formula mass of CH\(_3\)O:
C: 12.0, H\(_3\): 3.0, O: 16.0 → Total = 12 + 3 + 16 = 31.0 - Divide relative molecular mass by empirical formula mass:
\( \frac{62}{31} = 2 \) - Multiply empirical formula by this integer:
(CH\(_3\)O) × 2 = C\(_2\)H\(_6\)O\(_2\)
Molecular formula = C\(_2\)H\(_6\)O\(_2\)
Explanation: This compound could be ethylene glycol, a common organic compound.
▶️ Answer/Explanation
(a)(i) halogens
Detail: This is the collective name for Group 7 elements (e.g., fluorine, chlorine, bromine, iodine, astatine).
(a)(ii) pale green
Detail: Chlorine gas has a distinctive pale yellow-green colour.
(a)(iii)
• Test: Place damp (or moist) blue litmus paper into the gas.
• Result: The litmus paper is bleached (turns white).
Detail: Chlorine is a powerful bleaching agent. It reacts with water to form hypochlorous acid, which bleaches the dye in litmus paper. The paper may initially turn red (as chlorine forms an acidic solution) before being bleached.
(b)
• Test:
- Add dilute nitric acid to the test solution.
- Then add silver nitrate solution.
• Result: A pale yellow precipitate forms.
Detail: Nitric acid is added first to remove any carbonate or hydroxide ions that might interfere by forming precipitates with silver ions. Silver nitrate reacts with iodide ions \( (I^-) \) to form insoluble silver iodide \( (AgI) \), which is pale yellow. For chloride ions \( (Cl^-) \) the precipitate is white \( (AgCl) \), and for bromide ions \( (Br^-) \) it is cream \( (AgBr) \).
(c)
The order of reactivity is: chlorine > bromine > iodine.
• First reaction: Chlorine solution + potassium bromide solution.
– Observation: The solution turns orange (or yellow-brown).
– Explanation: Chlorine \( (Cl_2) \) is more reactive than bromine. It displaces bromine from potassium bromide \( (KBr) \).
– Chemical Equation: \( Cl_2(aq) + 2KBr(aq) \rightarrow 2KCl(aq) + Br_2(aq) \)
– The orange/yellow-brown colour is due to the formation of dissolved bromine \( (Br_2) \). This shows chlorine is more reactive than bromine.
• Second reaction: Bromine solution + potassium iodide solution.
– Observation: The solution turns brown.
– Explanation: Bromine \( (Br_2) \) is more reactive than iodine. It displaces iodine from potassium iodide \( (KI) \).
– Chemical Equation: \( Br_2(aq) + 2KI(aq) \rightarrow 2KBr(aq) + I_2(aq) \)
– The brown colour is due to the formation of dissolved iodine \( (I_2) \). This shows bromine is more reactive than iodine.
• Conclusion: Chlorine displaces bromine, and bromine displaces iodine. Therefore, the reactivity order is chlorine > bromine > iodine.
Detail: This is an example of a displacement (or redox) reaction. A more reactive halogen will displace a less reactive halogen from its halide salt in solution. Reactivity decreases down Group 7 due to increasing atomic size and decreasing ability to gain an electron (electronegativity).
▶️ Answer/Explanation
(a)(i) \(\text{Zn (s)} + \text{H}_2\text{SO}_4 \text{ (aq)} \rightarrow \text{ZnSO}_4 \text{ (aq)} + \text{H}_2 \text{ (g)}\)
Details: Zinc is a solid (s), sulfuric acid is aqueous (aq), zinc sulfate forms an aqueous solution (aq), and hydrogen is a gas (g).
(a)(ii) Effervescence / bubbles / fizzing.
Details: The reaction produces hydrogen gas, which is seen as bubbles forming on the surface of the zinc and rising through the solution.
(b)(i) An explanation that links the following two points:
• To make sure all of the acid reacts.
• So that a pure zinc sulfate solution is obtained / pure zinc sulfate crystals are obtained.
Details: Adding excess zinc ensures that the sulfuric acid is the limiting reactant. This prevents any residual acid from contaminating the product and ensures all sulfate ions are used to form zinc sulfate.
(b)(ii) Diagram should show:
• A filter funnel containing filter paper.
• A suitable container to collect the filtrate (e.g., beaker, conical flask, evaporating basin).
Details: The mixture of zinc sulfate solution and excess solid zinc is poured through the filter paper. The solution (filtrate) passes through, while the unreacted zinc solid (residue) is left on the filter paper.
(c)(i)
Calculation:
\(M_r = 65 + 32 + (4 \times 16) + (7 \times 18)\)
\(= 65 + 32 + 64 + 126\)
\(= 287\)
Therefore, \(M_r = 287\).
Details: Atomic masses: Zn = 65, S = 32, O = 16, H = 1. For \(\text{ZnSO}_4\cdot7\text{H}_2\text{O}\): ZnSO₄ part = 65+32+64=161; 7H₂O part = 7×(2+16)=126; Total = 161+126=287.
(c)(ii)
From the balanced equation: \(\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\)
1 mole of \(\text{H}_2\text{SO}_4\) produces 1 mole of \(\text{ZnSO}_4\cdot7\text{H}_2\text{O}\).
Moles of acid used = 0.0200 mol.
Therefore, maximum moles of crystals = 0.0200 mol.
Mass = moles × \(M_r\) = \(0.0200 \times 287 = 5.74 \, \text{g}\).
This is approximately 6 g.
Details: The calculation shows the theoretical maximum yield based on the limiting reactant (sulfuric acid). Using the \(M_r\) from part (i), the mass is 5.74 g, which rounds to about 6 g as stated.
(c)(iii)
Percentage yield = \(\frac{\text{actual mass}}{\text{theoretical mass}} \times 100\%\)
Theoretical mass from (ii) = 5.74 g.
Actual mass = 4.28 g.
Percentage yield = \(\frac{4.28}{5.74} \times 100\%\)
\(= 0.7456 \times 100\% = 74.6\%\) (to three significant figures).
Details: The yield is less than 100% due to practical losses during steps like filtration, transfer, and crystallization. 74.6% is a reasonable yield for a school laboratory preparation.
▶️ Answer/Explanation
(a)(i)
1. (Burns with a) bright white flame / intense white light.
2. Formation of a white powder / white solid / white ash.
Detailed Answer: Magnesium burns vigorously in oxygen with a characteristic brilliant white flame, producing magnesium oxide as a fine white solid (ash).
(a)(ii) Magnesium gains oxygen.
Alternative answers: Magnesium loses electrons / The oxidation state (of Mg) increases from 0 to +2.
Detailed Answer: Oxidation is defined as the gain of oxygen or loss of electrons. In this reaction, each magnesium atom gains an oxygen atom to form MgO.
(b)(i) A reaction that gives out / releases heat (energy) to the surroundings.
Detailed Answer: In an exothermic reaction, the energy released from forming new bonds is greater than the energy needed to break the old bonds, resulting in a net release of thermal energy.
(b)(ii) \[ 2Mg + CO_2 \rightarrow 2MgO + C \]
Detailed Answer: The magnesium is so reactive that it can reduce carbon dioxide, removing the oxygen to form magnesium oxide and leaving behind carbon (soot).
(b)(iii) The magnesium would continue to burn in the carbon dioxide / The reaction is highly exothermic, releasing a large amount of heat.
Detailed Answer: Carbon dioxide extinguishers work by smothering fires, but magnesium can strip oxygen from CO₂ itself. Using a CO₂ extinguisher on a magnesium fire would therefore be ineffective and could even provide more reactant for the vigorous, heat-producing reaction.
(c) An explanation giving two linked changes. The student should:
First Change:
• M1: Lift and replace the lid repeatedly during heating.
• M2: (To allow) oxygen/air to enter the crucible (to ensure all magnesium reacts completely).
Second Change:
• M3: Reheat and reweigh / heat to constant mass.
• M4: (To make sure) that all the magnesium has reacted / that the reaction is complete.
Detailed Answer: The initial low yield is likely because the lid was kept on, limiting the oxygen supply and causing incomplete combustion (some unreacted Mg may remain). Lifting the lid allows oxygen in. Furthermore, a single heating may not be enough; reheating until the mass no longer increases ensures the conversion of all Mg to MgO.
▶️ Answer/Explanation
(a)(i) B (5)
- A (pH 1) is incorrect as it is the pH of a strong acid.
- C (pH 7) is incorrect as it is the pH of a neutral solution.
- D (pH 9) is incorrect as it is the pH of a weak alkali.
- Weak acids (like carbonic acid formed from CO₂) typically have a pH between 4 and 6.
(a)(ii) D (acids are proton donors)
- A is incorrect because alkalis (bases) contain OH⁻ ions, not acids.
- B is incorrect because acids are not typically described as electron donors in the Brønsted-Lowry definition used at IGCSE.
- C is incorrect because bases are proton acceptors, not acids.
- This is a key definition from the Brønsted-Lowry theory of acids and bases (Specification point 2.36).
(b)(i) (Thermal) decomposition
This is a reaction where a single compound breaks down into two or more simpler substances when heated (Specification point 2.12).
(b)(ii) \[ \text{PbCO}_3 \rightarrow \text{PbO} + \text{CO}_2 \]
This is a balanced thermal decomposition equation. Metal carbonates decompose to form a metal oxide and carbon dioxide gas.
(c)(i) Liquid
Since the boiling point of SiCl₄ (58 °C) is above room temperature (approx. 20-25 °C) and its melting point (-69 °C) is far below, it exists as a liquid at room temperature.
(c)(ii)
An explanation linking structure and bonding:
- Silicon dioxide (SiO₂) has a giant covalent (macromolecular) structure (Specification point 1.49).
- In this structure, many strong covalent bonds hold all the atoms together in a rigid, continuous network.
- To melt SiO₂, a large amount of energy is required to break these numerous strong covalent bonds.
- Silicon(IV) chloride (SiCl₄) has a simple molecular structure (Specification point 1.47).
- The atoms within each SiCl₄ molecule are held by strong covalent bonds, but the forces between the molecules (intermolecular forces) are weak.
- To melt SiCl₄, only a small amount of energy is needed to overcome these weak intermolecular forces, not to break the covalent bonds within the molecules.
- Therefore, SiO₂ has a much higher melting point than SiCl₄ because breaking the strong covalent bonds in a giant structure requires significantly more energy than overcoming the weak forces between simple molecules.
This directly assesses understanding of Specification points 1.47, 1.49, and 1.50, comparing simple vs. giant covalent structures.
▶️ Answer/Explanation
(a)(i)
potassium hydroxide + hydrochloric acid → potassium chloride + water
This is a neutralisation reaction, forming a salt (potassium chloride) and water.
(a)(ii)
• To mix the two solutions thoroughly so that all the reactant particles come into contact.
• To ensure the mixture has a uniform temperature throughout.
• This allows the heat energy from the exothermic reaction to be released quickly and uniformly, ensuring an accurate measurement of the maximum temperature change.
(b)
Mean temperature = \( \frac{17.8 + 18.4}{2} = \frac{36.2}{2} = 18.1 \)
mean temperature = 18.1 °C
The average of the two initial temperatures is used as the starting temperature for calculating the temperature rise.
(c)(i)
From the given temperature rise of 5.2 °C and the provided answer key:
mean temperature at start in °C = 17.2
temperature at end in °C = 22.4
These values are consistent: \( 22.4 – 17.2 = 5.2 \).
(c)(ii)
Volume of mixture = \( 25 + 25 = 50 \) cm³
Since mass of 1 cm³ = 1 g, total mass, \( m = 50 \) g
Specific heat capacity, \( c = 4.2 \, \text{J g}^{-1} °C^{-1} \)
Temperature rise, \( \Delta T = 5.2 \) °C
\( Q = mc\Delta T \)
\( Q = 50 \times 4.2 \times 5.2 \)
\( Q = 1092 \, \text{J} \approx 1100 \, \text{J} \) (to about 1100 J)
The calculation shows the heat energy change is approximately 1100 J.
(c)(iii)
Heat change for 0.020 mol KOH = 1092 J (from (c)(ii))
Heat change for 1 mol KOH = \( \frac{1092}{0.020} = 54600 \, \text{J} \)
Convert to kJ: \( 54600 \, \text{J} = 54.6 \, \text{kJ} \)
The reaction is exothermic (temperature increased), so \( \Delta H \) is negative.
\( \Delta H = \mathbf{-54.6 \, \text{kJ mol}^{-1}} \)
Enthalpy change is negative because heat is released to the surroundings.