2025 Edexcel IGCSE Chemistry 模擬試題連答案詳解

The process is called diffusion. When the crystal dissolves, the particles of potassium manganate(VII) move randomly and collide with water molecules, causing them to spread out from the region of high concentration at the bottom to the region of lower concentration throughout the beaker.

M1: Identify process as diffusion (1 mark) M2: Explain that dissolved particles move randomly or spread from high to low concentration (1 mark)

Relative atomic mass of X = \(\frac{(69 \times 60.0) + (71 \times 40.0)}{100} = \frac{4140 + 2840}{100} = \frac{6980}{100} = 69.8\)

M1: Correct mathematical expression showing calculation of relative abundance: \(\frac{(69 \times 60) + (71 \times 40)}{100}\) (1 mark) M2: Correct evaluation to three significant figures: 69.8 (1 mark)

Methane consists of simple molecules. The covalent bonds within the molecules are very strong, but the forces of attraction between the molecules (intermolecular forces) are weak. As a result, very little thermal energy is needed to overcome these weak forces and separate the molecules during boiling.

M1: State that there are weak intermolecular forces / weak forces between molecules (1 mark) M2: State that little energy is required to overcome or break these forces (1 mark) [Reject: reference to breaking covalent bonds]

Chlorine is more reactive than iodine, so it displaces iodine from potassium iodide. The solution changes from colorless to brown because iodine (\(\text{I}_2\)) is formed.

M1: State color change from colorless to brown / orange-brown (1 mark) M2: Name product as iodine / \(\text{I}_2\) (1 mark)

Calculate the amount in moles of each element: Moles of Ti = \(\frac{1.44}{48} = 0.03\text{ mol}\). Moles of O = \(\frac{0.96}{16} = 0.06\text{ mol}\). Simplest whole-number ratio is Ti : O = \(\frac{0.03}{0.03} : \frac{0.06}{0.03} = 1 : 2\). Therefore, the empirical formula is \(\text{TiO}_2\).

M1: Correct calculation of moles of titanium (0.03) and oxygen (0.06) (1 mark) M2: Correct empirical formula \(\text{TiO}_2\) derived from the ratio (1 mark)

The heated copper reacts with the oxygen in the syringe/system to form solid copper(II) oxide: \(2\text{Cu} + \text{O}_2 \rightarrow 2\text{CuO}\). Since oxygen gas is converted into a solid compound, it is removed from the gas phase, causing the overall gas volume to decrease.

M1: Identify that copper reacts with oxygen (1 mark) M2: State that oxygen gas is removed from the gas mixture / converted to solid (copper oxide) (1 mark)

Enthalpy change \(\Delta H = -\frac{\text{Heat energy released}}{\text{Number of moles of reactant}} = -\frac{10.5\text{ kJ}}{0.050\text{ mol}} = -210\text{ kJ/mol}\). The sign is negative because the temperature increases, meaning the reaction is exothermic.

M1: Correct calculation of numerical value as 210 (1 mark) M2: Negative sign preceding the value: -210 \(\text{kJ/mol}\) (1 mark)

The fractionating column is hotter at the bottom and cooler at the top. Different fractions have different boiling points. Fractions with higher boiling points condense near the bottom where temperatures are higher, while fractions with lower boiling points rise further up and condense near the top where temperatures are lower.

M1: State that different fractions have different boiling points (1 mark) M2: Explain that the column has a temperature gradient / fractions condense at their respective boiling points at different heights (1 mark)

At a higher temperature, the ammonia gas particles gain more kinetic energy. This causes them to move and spread out at a faster speed, resulting in a higher rate of diffusion.

M1: Particles have more kinetic energy (1) M2: Particles move faster / at a higher speed (1)

The atomic number represents the number of protons, which is 15. The mass number is the total number of protons and neutrons. Therefore, the number of neutrons is 31 - 15 = 16.

M1: 15 protons (1) M2: 16 neutrons (1)

Carbon dioxide has a simple molecular structure. There are weak intermolecular forces of attraction between individual carbon dioxide molecules. Very little energy is required to break these weak forces during boiling.

M1: Weak intermolecular forces (between molecules) (1) M2: Little energy required to overcome/break these forces (1) [Reject: covalent bonds break]

\(M_r = 2 \times [14 + (4 \times 1)] + 32 + (4 \times 16) = 2 \times 18 + 32 + 64 = 36 + 32 + 64 = 132\).

M1: Correct substitution of values, e.g., \(2 \times (14 + 4) + 32 + 64\) (1) M2: 132 (1)

When sodium is added to water, it reacts vigorously. Key observations include: the sodium melts into a spherical ball (due to the heat released), it floats/moves rapidly on the surface of the water, and there is rapid fizzing/effervescence as hydrogen gas is produced.

Any two observations from: Melts into a ball/sphere (1) Fizzes / effervesces / bubbles of gas (1) Floats / moves on the surface (1) Disappears / gets smaller (1) (Max 2 marks)

Dry, unpolluted air consists of approximately 78% nitrogen and 21% oxygen by volume, with the remaining 1% consisting of argon, carbon dioxide, and other noble gases.

M1: Nitrogen: 78% (accept 78% to 79%) (1) M2: Oxygen: 21% (accept 20% to 21%) (1)

The overall energy change, \(\Delta H\) = energy taken in to break bonds - energy released when bonds form. \(\Delta H = 1500 - 1850 = -350\text{ kJ/mol\}. Because the value of \)\Delta H\) is negative (more energy is released than taken in), the reaction is exothermic.

M1: \(-350\) (kJ/mol) (1) M2: Exothermic (dependent on a negative value or explanation that more energy is released than absorbed) (1)

Fermentation of glucose is carried out using yeast as a catalyst, which contains enzymes that catalyze the reaction. The process requires an optimum temperature of about \(30^\circ\text{C}\) to \(40^\circ\text{C}\) (typically \(37^\circ\text{C}\)) to prevent the enzymes from becoming denatured or inactive.

M1: Yeast / enzymes in yeast (1) M2: Any temperature or range within \(25^\circ\text{C}\) to \(40^\circ\text{C}\) (1)

Fractional distillation is designed to separate liquid mixtures where the components have boiling points that are close together. It uses a fractionating column which allows multiple condensation and evaporation cycles. Simple distillation cannot efficiently separate liquids with boiling points that differ by less than around 25 to 50 degrees Celsius.

M1: The boiling points of ethanol and water are close together / not far apart (1). M2: Simple distillation is only effective if there is a large difference in boiling points / fractional distillation allows multiple cycles of condensation and evaporation to separate them (1).

Both isotopes have the same number of protons (6) and electrons (6) because they represent the same element, carbon. However, they have different numbers of neutrons: Carbon-12 has 6 neutrons, whereas Carbon-13 has 7 neutrons.

M1: Similarity: Both have the same number of protons or electrons or atomic number (1). M2: Difference: Carbon-13 has one more neutron than Carbon-12 or they have different numbers of neutrons or different mass numbers (1).

Potassium is an extremely reactive Group 1 alkali metal. When added to water, it reacts exothermically, melting into a ball, floating, and moving quickly across the water surface. It burns with a characteristic lilac flame due to the ignition of the hydrogen gas produced. As potassium hydroxide (an alkali) is formed, the phenolphthalein indicator turns from colorless to pink.

M1: Any one correct observation: potassium melts into a sphere / potassium moves rapidly on the surface of the water / burns with a lilac flame / bubbles of gas or effervescence / potassium disappears (1). M2: Any second correct observation OR the solution turns pink (1).

Short-wavelength radiation from the Sun passes through the Earth's atmosphere and warms the Earth's surface. The Earth then re-radiates this energy as longer-wavelength infrared radiation. Greenhouse gases like carbon dioxide absorb this infrared radiation and re-emit it in all directions, including back towards the Earth, trapping heat and raising global temperatures.

M1: Greenhouse gases absorb thermal or infrared radiation re-radiated from the Earth's surface (1). M2: They re-emit this radiation back towards Earth or trap heat in the atmosphere (1). Note: Do not award marks for reflecting heat or absorbing UV radiation.

To calculate the relative formula mass of \(\text{CuSO}_4\cdot5\text{H}_2\text{O}\): Relative mass of \(\text{CuSO}_4 = 63.5 + 32 + (4 \times 16) = 159.5\). Relative mass of \(5\text{H}_2\text{O} = 5 \times ((2 \times 1) + 16) = 5 \times 18 = 90\). Total \(M_r = 159.5 + 90 = 249.5\).

M1: Correct calculation of the mass of \(\text{CuSO}_4\) as \(159.5\) OR the mass of \(5\text{H}_2\text{O}\) as \(90\) (1). M2: Correct final answer of \(249.5\) (1).

Diamond has a giant covalent macromolecular structure where each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement. Because covalent bonds are strong, a very large amount of thermal energy is needed to break these numerous bonds throughout the entire giant structure to melt it.

M1: Diamond has a giant covalent structure / lattice containing many strong covalent bonds (1). M2: A large amount of energy is required to break these strong bonds (1). Note: Reject any reference to intermolecular forces or ions.

The test for chloride ions involves adding dilute nitric acid (to remove any carbonate ions that could produce a false positive precipitate) followed by silver nitrate solution. If chloride ions are present, they react with the silver ions to form a white precipitate of silver chloride (\(\text{AgCl}\)).

M1: Add dilute nitric acid and silver nitrate solution (1). M2: Observe a white precipitate (1). Note: Reject hydrochloric acid as the acid used.

Ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)) reacts with ethanoic acid (\(\text{CH}_3\text{COOH}\)) to form the ester ethyl ethanoate and water. The structural formula of ethyl ethanoate is \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) (or can be written as \(\text{CH}_3\text{CO}_2\text{CH}_2\text{CH}_3\)).

M1: Correct name ethyl ethanoate OR a partially correct formula (1). M2: Correct structural formula: \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) OR \(\text{CH}_3\text{CO}_2\text{CH}_2\text{CH}_3\) (1).

The process is diffusion. Dissolved potassium manganate(VII) particles and water molecules are in constant, random motion. This random movement causes the particles to spread out from where they are highly concentrated at the bottom of the beaker to where they are less concentrated, until they are evenly distributed.

M1: Identify the process as diffusion (1 mark). M2: Describe that particles move randomly OR move from a region of high concentration to low concentration (1 mark).

Isotopes are atoms of the same element that have the same atomic number (number of protons) but a different mass number. This is because they contain the same number of protons in their nucleus but a different number of neutrons.

M1: Same number of protons / same atomic number (1 mark). M2: Different number of neutrons / different mass number (1 mark). Reject: different number of electrons.

The test for carbon dioxide involves bubbling the gas through an aqueous solution of calcium hydroxide, commonly known as limewater. If carbon dioxide is present, it reacts with the calcium hydroxide to form an insoluble precipitate of calcium carbonate, which causes the limewater to turn cloudy or milky.

M1: Bubble the gas through limewater (1 mark). M2: Limewater turns cloudy / milky / white precipitate forms (1 mark).

The process is endothermic. A decrease in temperature indicates that heat energy has been taken in (absorbed) from the surroundings (the water and the thermometer) into the reacting system, resulting in a temperature drop.

M1: Identify the process as endothermic (1 mark). M2: State that heat energy is taken in or absorbed from the surroundings (1 mark).

(a) Increasing the concentration increases the number of acid particles in a given volume. This results in more frequent collisions between reactant particles, which increases the rate of reaction.

(b) Volume of gas produced in the interval = \(28\text{ cm}^3 - 15\text{ cm}^3 = 13\text{ cm}^3\). Time interval = \(40\text{ s} - 20\text{ s} = 20\text{ s}\). Average rate = \(13\text{ cm}^3 / 20\text{ s} = 0.65\text{ cm}^3/\text{s}\).

(c) If the temperature is increased, the rate of reaction increases, so the initial gradient of the curve will be steeper. Because the amounts of reactants are the same, the curve will level off at the same final volume of gas.

(d) The rate can also be increased by increasing the surface area of the solid reactant (e.g., using calcium carbonate powder instead of chips) and by adding a suitable catalyst.

(a) M1: Increasing concentration increases the number of particles per unit volume (1)
M2: This leads to more frequent successful collisions (1)
M3: Consequently, the rate of reaction increases (1)

(b) M1: Calculates volume change (13) and time change (20) (1)
M2: Correct calculation of 0.65 with units of cm\u00b3/s (or cm\u00b3 s\u207b\u00b9) (1)

(c) M1: The curve has a steeper initial gradient (1)
M2: The curve levels off at the same volume of gas (1)

(d) M1: Use of powdered calcium carbonate / larger surface area of calcium carbonate (1)
M2: Addition of a catalyst (1)

(a) To ensure all water is removed, the student should heat the crucible and its contents, weigh it, heat again, and re-weigh. This process of heating and weighing is repeated until the mass remains constant.

(b) Mass of anhydrous copper(II) sulfate = \(21.71\text{ g} - 18.52\text{ g} = 3.19\text{ g}\). Mass of water lost = \(23.51\text{ g} - 21.71\text{ g} = 1.80\text{ g}\).

(c) Moles of \(\text{CuSO}_4 = 3.19 / 159.5 = 0.020\text{ mol}\).
Moles of \(\text{H}_2\text{O} = 1.80 / 18.0 = 0.100\text{ mol}\).
Ratio of \(\text{CuSO}_4 : \text{H}_2\text{O} = 0.020 : 0.100 = 1 : 5\).
Therefore, \(x = 5\).

(d) Adding water to anhydrous copper(II) sulfate turns it from white to blue.

(a) M1: Heat the sample, weigh it, and repeat the heating (1)
M2: Repeat until a constant mass is obtained (1)

(b) M1: Correct calculation of anhydrous salt mass = 3.19 g (1)
M2: Correct calculation of water mass lost = 1.80 g (1)

(c) M1: Calculates moles of CuSO\u2084 = 3.19 / 159.5 = 0.020 mol (1)
M2: Calculates moles of H\u2082O = 1.80 / 18.0 = 0.100 mol (1)
M3: Finds simplest mole ratio of CuSO\u2084 : H\u2082O = 1 : 5 (1)
M4: States that x = 5 (1)

(d) M1: White to blue (1) (Both colors required)

(a) Polystyrene is an excellent thermal insulator compared to glass. Using a polystyrene cup minimizes heat exchange with the surrounding environment, resulting in a more accurate temperature change measurement.

(b) Moles of \(\text{CuSO}_4 = \text{volume in dm}^3 \times \text{concentration} = 0.0500\text{ dm}^3 \times 0.500\text{ mol/dm}^3 = 0.0250\text{ mol}\).
Moles of \(\text{Zn} = \text{mass} / A_r = 2.00 / 65.4 = 0.0306\text{ mol}\).
Since the reaction stoichiometry is 1:1, and the moles of zinc (0.0306 mol) exceed the moles of copper(II) sulfate (0.0250 mol), copper(II) sulfate is the limiting reactant.

(c) Temperature increase \(\Delta T = 34.2 - 19.5 = 14.7^\circ\text{C}\).
Mass of solution \(m = 50.0\text{ cm}^3 \times 1.00\text{ g/cm}^3 = 50.0\text{ g}\).
\(q = mc\Delta T = 50.0 \times 4.18 \times 14.7 = 3072.3\text{ J}\).

(d) First convert energy to kJ: \(3072.3\text{ J} = 3.0723\text{ kJ}\).
\(\Delta H = -q / \text{moles of limiting reactant} = -3.0723 / 0.0250 = -122.9\text{ kJ/mol}\) (or \(-123\text{ kJ/mol}\) to 3 significant figures). The reaction is exothermic, so the sign is negative.

(a) M1: Polystyrene is a heat insulator / poor conductor of heat (1)
M2: To reduce heat loss to the surroundings / ensure temperature rise is accurate (1)

(b) M1: Calculates moles of CuSO\u2084 = 0.0250 mol (1)
M2: Calculates moles of Zn = 0.0306 mol and concludes CuSO\u2084 is limiting because 0.0250 < 0.0306 (1)

(c) M1: Calculates \u0394T = 14.7\u00b0C (1)
M2: Calculates q = 50.0 x 4.18 x 14.7 = 3070 J or 3072.3 J (1)

(d) M1: Converts q to kJ (3.07 kJ) (1)
M2: Divides by moles of limiting reactant (0.0250) (1)
M3: Correct value of -123 kJ/mol (accept range -122.8 to -123.0) with negative sign (1)

(a) The balanced chemical equation with optional state symbols is: \(\text{CuO}(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{CuSO}_4(\text{aq}) + \text{H}_2\text{O}(\text{l})\).

(b) 1. Heat a measured volume of dilute sulfuric acid in a beaker.
2. Add copper(II) oxide to the warm acid, stirring continuously, until it is in excess (when some black solid remains unreacted at the bottom).
3. Filter the hot mixture using a funnel and filter paper to remove the unreacted copper(II) oxide.
4. Transfer the filtrate (copper(II) sulfate solution) to an evaporating basin and heat to evaporate some water until the crystallisation point is reached (or crystals begin to form on a cold glass rod).
5. Allow the solution to cool slowly so that large crystals form. Filter the crystals, rinse with a tiny amount of cold distilled water, and dry them by patting with filter paper.

(c) Copper(II) oxide is added in excess to ensure that all the sulfuric acid is completely neutralised and used up, ensuring a pure product without acid contamination. The excess insoluble copper(II) oxide is removed by filtration.

(a) M1: Correct reactants and products (CuO + H\u2082SO\u2084 -> CuSO\u2084 + H\u2082O) (1)
M2: Fully balanced equation with no errors (1)

(b) M1: Heat/warm sulfuric acid (1)
M2: Add copper(II) oxide until it is in excess / stops dissolving (1)
M3: Filter the mixture to remove excess copper(II) oxide (1)
M4: Heat the copper(II) sulfate solution to evaporate water to crystallisation point / until crystals start to form (1)
M5: Allow to cool, then filter off crystals and dry them with filter paper/warm oven (1)

(c) M1: Added in excess to ensure all sulfuric acid is completely reacted (1)
M2: Removed by filtration (1)

(a) Crude oil is vaporised by heating before entering the fractionating column. The column has a temperature gradient, being hot at the bottom and cooler at the top. As the vapours rise, they cool and condense at different heights depending on their boiling points. Hydrocarbons with high boiling points (long chains) condense near the bottom, while those with lower boiling points (short chains) condense near the top.

(b) (i) Catalytic cracking requires a silica, alumina, or zeolite catalyst, and a temperature of \(600 - 700^\circ\text{C}\) (accept range \(500 - 750^\circ\text{C}\)).
(ii) The balanced equation is: \(\text{C}_{12}\text{H}_{26} \rightarrow \text{C}_8\text{H}_{18} + 2\text{C}_2\text{H}_4\).

(c) Ethene is a hydrocarbon because it contains only carbon and hydrogen atoms. It is unsaturated because it contains at least one carbon-carbon double bond (\(\text{C}=\text{C}\)), meaning it does not contain the maximum possible number of hydrogen atoms per carbon atom.

(a) M1: Crude oil is heated/vaporised (1)
M2: The fractionating column is hotter at the bottom and cooler at the top / has a temperature gradient (1)
M3: Fractions/vapours condense at their respective boiling points (1)
M4: Shorter chains/lower boiling points condense higher up, or vice versa (1)

(b) (i) M1: Catalyst: Silica / alumina / aluminosilicate / zeolite (1)
M2: Temperature: 600\u2013700 \u00b0C (accept any value or range within 500\u2013750 \u00b0C) (1)

(ii) M1: Writes 2C\u2082H\u2084 on the product side (1)

(c) M1: Hydrocarbon: compound of hydrogen and carbon only (1)
M2: Unsaturated: contains a carbon-carbon double bond (C=C) (1)

(a) During heating, the red-brown copper turnings react with oxygen and are coated in black copper(II) oxide, so they turn black.

(b) The chemical equation for the reaction is: \(2\text{Cu} + \text{O}_2 \rightarrow 2\text{CuO}\).

(c) Gases expand when heated. If the volume is measured while the apparatus is hot, the gas volume will be artificially high, leading to an underestimation of the volume of oxygen that reacted.

(d) Volume of oxygen reacted = \(100.0\text{ cm}^3 - 79.0\text{ cm}^3 = 21.0\text{ cm}^3\).
Percentage of oxygen = \((21.0 / 100.0) \times 100 = 21.0\%\).

(e) The main gas remaining is nitrogen. Nitrogen is extremely unreactive because it contains a very strong triple covalent bond between the nitrogen atoms, which requires a large amount of energy to break.

(a) M1: Red-brown solid/metal turns black (1)

(b) M1: Correct reactant and product formulas (Cu + O\u2082 -> CuO) (1)
M2: Balanced equation (2Cu + O\u2082 -> 2CuO) (1)

(c) M1: Gases expand when heated / occupy a larger volume (1)
M2: Measuring when hot would give an inaccurately high reading / incorrect volume (1)

(d) M1: Calculates volume change of 21.0 cm\u00b3 (1)
M2: Calculates percentage = 21.0% (1)

(e) M1: Nitrogen (1)
M2: Unreactive because of a strong triple covalent bond (1)

At \(40\,^{\circ}\text{C}\), the bromine molecules have greater kinetic energy on average. Consequently, they move at a higher average speed, leading to a faster rate of diffusion.

M1: (At higher temperatures) the bromine molecules/particles have more kinetic energy (1 mark). M2: Therefore, the particles move faster / at a greater speed (1 mark).

Using the formula: \(\text{Volume} = \text{amount in moles} \times \text{molar volume}\). Substituting the given values: \(\text{Volume} = 0.125\text{ mol} \times 24\text{ dm}^3/\text{mol} = 3\text{ dm}^3\).

M1: Correct calculation setup: \(0.125 \times 24\) (1 mark). M2: Correct evaluation: \(3\) / \(3.0\) (\(\text{dm}^3\)) (1 mark).

Excess copper(II) oxide is added to make sure that all the sulfuric acid is completely reacted and neutralized. This prevents unreacted sulfuric acid from remaining in the solution and contaminating the copper(II) sulfate crystals.

M1: To ensure all of the sulfuric acid reacts / is neutralized (1 mark). M2: So that the resulting solution contains only copper(II) sulfate / prevents the salt from being contaminated with unreacted acid (1 mark).

The reaction between ethanol and ethanoic acid is an esterification reaction, catalyzed by concentrated sulfuric acid. Ethyl ethanoate is suitable for perfumes because it is volatile and has a pleasant, sweet, fruity smell.

M1: Concentrated sulfuric acid (accept \(\text{H}_2\text{SO}_4\), reject dilute sulfuric acid) (1 mark). M2: Sweet smell / fruity smell / volatile / easily vaporized (1 mark).

Bond breaking is an endothermic process that absorbs energy, while bond making is an exothermic process that releases energy. The combustion of methane is exothermic because the total energy released during the formation of new bonds in the products is greater than the total energy absorbed to break the existing bonds in the reactants.

M1: More energy is released when bonds are made / formed (in the products) (1 mark). M2: Than is taken in / absorbed when bonds are broken (in the reactants) (1 mark).

Potassium iodide is a colourless solution. Chlorine is more reactive than iodine, so it displaces iodine from potassium iodide, forming potassium chloride and aqueous iodine. Aqueous iodine is brown, so the final colour of the mixture is brown.

M1: Initial colour: colourless (1 mark). M2: Final colour: brown / orange-brown / yellow (1 mark).

Industrial catalytic cracking of hydrocarbons requires a catalyst of alumina (aluminium oxide) or silica (silicon dioxide), or zeolites, and a temperature in the range of \(600\,^{\circ}\text{C}\) to \(700\,^{\circ}\text{C}\).

M1: Silica / alumina / aluminosilicate / zeolite (accept chemical formulae \(\text{SiO}_2\) / \(\text{Al}_2\text{O}_3\)) (1 mark). M2: \(600\,^{\circ}\text{C}\) to \(700\,^{\circ}\text{C}\) (accept any value or range within \(500\,^{\circ}\text{C}\) to \(800\,^{\circ}\text{C}\)) (1 mark).

The relative atomic mass is the weighted average of the masses of the isotopes. \(A_r = \frac{(69 \times 60.1) + (71 \times 39.9)}{100} = \frac{4146.9 + 2832.9}{100} = \frac{6979.8}{100} = 69.798\). Rounding to one decimal place gives \(69.8\).

M1: Correct calculation setup: \(\frac{(69 \times 60.1) + (71 \times 39.9)}{100}\) (or showing equivalent values: \(41.469 + 28.329\)) (1 mark). M2: Correct final answer to one decimal place: \(69.8\) (1 mark). (Award 2 marks for correct final answer with no working).

As temperature increases, the average kinetic energy of the particles increases. Because the ammonia particles have more kinetic energy, they move at a higher average speed, which increases the rate at which they diffuse through the air.

M1: Particles gain kinetic energy / have more kinetic energy (1)
M2: Particles move faster / move with greater speed (1)

Using the formula \(Q = mc\Delta T\):
\(Q = 50.0\text{ g} \times 4.18\text{ J/g/}^\circ\text{C} \times 5.5\text{ }^\circ\text{C} = 1149.5\text{ J}\).
Rounded to 3 significant figures, this is \(1150\text{ J}\).

M1: Correct substitution of values into the equation: \(50.0 \times 4.18 \times 5.5\) (1)
M2: Evaluation to get \(1149.5\text{ J}\) or \(1150\text{ J}\) (1)

Ethyl propanoate is an ester formed from the condensation reaction between ethanol (which provides the ethyl group) and propanoic acid (which provides the propanoate group).

M1: Ethanol (1)
M2: Propanoic acid (1)

During the chemical reaction, oxygen gas from the air reacts with phosphorus. Since oxygen gas is converted into solid phosphorus oxide, the number of moles of gaseous particles in the sealed tube decreases, resulting in a reduction in gas volume.

M1: Oxygen (gas) is used up / reacts with phosphorus (1)
M2: The product (phosphorus oxide) is a solid / gas is turned into a solid (1)

Ammonium ions react with warm sodium hydroxide solution to produce ammonia gas. Ammonia is alkaline and will turn damp red litmus paper blue, confirming the presence of ammonium ions.

M1: Add sodium hydroxide (solution) and heat / warm (1)
M2: Gas produced turns damp red litmus paper blue (1) (Reject: blue litmus paper)

An increase in concentration means there are more reactant particles in the same volume of solution. This results in the particles being closer together, leading to more frequent collisions (more collisions per second) and therefore a higher rate of reaction.

M1: More particles in a given volume / unit volume (1)
M2: More frequent collisions / more collisions per unit time (1) (Do not accept 'more collisions' without a time reference)

The balanced chemical equation is: \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\).
This shows a 1:1 mole ratio between \(\text{CaCO}_3\) and \(\text{CO}_2\).
Therefore, \(0.20\text{ mol}\) of \(\text{CO}_2\) gas is produced.
\(\text{Volume of } \text{CO}_2 = 0.20\text{ mol} \times 24\text{ dm}^3/\text{mol} = 4.8\text{ dm}^3\).

M1: State or use 1:1 mole ratio to show \(0.20\text{ mol}\) of \(\text{CO}_2\) is produced (1)
M2: Multiply by 24 to get \(4.8\text{ } (\text{dm}^3)\) (1)

When ethanol is oxidised, the orange dichromate(VI) ions (\(\text{Cr}_2\text{O}_7^{2-}\)) are reduced to green chromium(III) ions (\(\text{Cr}^{3+}\)), resulting in a color change from orange to green.

M1: Orange (starting color) (1)
M2: to green (final color) (1)

Use the formula \(Q = m c \Delta T\), where \(m = 150\text{ g}\), \(c = 4.18\text{ J/g/}^{\circ}\text{C}\), and \(\Delta T = 12.0\text{ }^{\circ}\text{C}\). This gives \(Q = 150 \times 4.18 \times 12.0 = 7524\text{ J}\). To convert Joules to kilojoules (\(kJ\)), divide by \(1000\): \(Q = \frac{7524}{1000} = 7.524\text{ kJ}\). Rounding to three significant figures gives \(7.52\text{ kJ}\).

M1: Substitution of values into \(Q = mc\Delta T\) (e.g., \(150 \times 4.18 \times 12.0\)) to get \(7524\text{ J}\) (1 mark). M2: Conversion of \(J\) to \(kJ\) to get \(7.52\text{ kJ}\) or \(7.524\text{ kJ}\) (1 mark). Accept: \(7.524\), \(7.52\), \(7.5\).

First, calculate the moles of \(\text{NaOH}\) used: \(\text{moles of NaOH} = \frac{25.0}{1000} \times 0.120 = 0.0030\text{ mol}\). Second, use the \(2:1\) stoichiometry of the reaction to find the moles of \(\text{H}_2\text{SO}_4\): \(\text{moles of H}_2\text{SO}_4 = \frac{0.0030}{2} = 0.0015\text{ mol}\).

M1: Calculation of moles of \(\text{NaOH}\): \(\frac{25.0}{1000} \times 0.120 = 0.0030\text{ mol}\) (1 mark). M2: Division of moles of \(\text{NaOH}\) by 2 to get \(0.0015\text{ mol}\) (1 mark). Accept: \(1.5 \times 10^{-3}\). Allow ECF for M2 if M1 is divided by 2.

(a)
1. Measure \(50.0\text{ cm}^3\) of copper(II) sulfate solution using a measuring cylinder and pour it into a polystyrene cup.
2. Place the polystyrene cup inside a beaker for stability and insulation.
3. Use a thermometer to measure the initial temperature of the solution every minute for 3 minutes.
4. At the 4th minute, add the excess zinc powder and stir the mixture continuously.
5. Record the temperature at regular intervals and find the maximum temperature reached to calculate the maximum temperature rise.

(b)
Mass of solution, \(m = 50.0\text{ cm}^3 \times 1.00\text{ g/cm}^3 = 50.0\text{ g}\)
Temperature change, \(\Delta T = 34.5\text{ }^\circ\text{C} - 18.5\text{ }^\circ\text{C} = 16.0\text{ }^\circ\text{C}\)
Using \(Q = mc\Delta T\):
\(Q = 50.0\text{ g} \times 4.18\text{ J/g/}^\circ\text{C} \times 16.0\text{ }^\circ\text{C} = 3344\text{ J}\)

(c)
Moles of \(\text{CuSO}_4 = \text{concentration} \times \text{volume} = 0.500\text{ mol/dm}^3 \times 0.0500\text{ dm}^3 = 0.0250\text{ mol}\)
Since zinc is in excess, \(\text{CuSO}_4\) is the limiting reactant.
\(\Delta H = -\frac{Q}{\text{moles}} = -\frac{3344\text{ J}}{0.0250\text{ mol}} = -133760\text{ J/mol}\)
Convert to \(\text{kJ/mol}\):
\(\Delta H = -134\text{ kJ/mol}\) (or \(-133.76\text{ kJ/mol}\))

Part (a) [4 marks]:
- M1: Use of a polystyrene cup (or lid / draft shield / cotton wool insulation) to minimize heat loss [1]
- M2: Measure the initial temperature of the copper(II) sulfate solution before adding zinc [1]
- M3: Stir the mixture during the reaction [1]
- M4: Record the maximum temperature reached / measure temperature at regular intervals [1]

Part (b) [3 marks]:
- M1: Correct calculation of \(\Delta T = 16.0\text{ }^\circ\text{C}\) [1]
- M2: Correct substitution: \(Q = 50.0 \times 4.18 \times 16.0\) [1]
- M3: \(3344\text{ J}\) (Accept \(3.34\text{ kJ}\)) [1]

Part (c) [4.33 marks]:
- M1: Calculate moles of \(\text{CuSO}_4 = 0.0250\text{ mol}\) [1]
- M2: Divide \(Q\) (in kJ) by moles: \(\frac{3.344}{0.0250}\) [1]
- M3: Value of \(134\) (or \(133.76\)) [1]
- M4: Negative sign AND unit of \(\text{kJ/mol}\) (\(-134\text{ kJ/mol}\)) [1.33]

(a)
Fermentation:
- Advantage: Uses renewable resources (glucose/sugars) / requires low energy (low temperature/pressure).
- Disadvantage: Slow batch process / produces impure ethanol that requires fractional distillation.
Hydration of ethene:
- Advantage: Fast continuous process / produces highly pure ethanol.
- Disadvantage: Uses non-renewable crude oil / requires high energy (high temperature/pressure).

(b) \(\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}\)

(c)
(i) Potassium dichromate(VI) and dilute sulfuric acid.
(ii) The color changes from orange to green.
(iii) The displayed formula showing all atoms and all bonds:
H O
| //
H-C-C
| \
H O-H

Part (a) [4 marks]:
- M1: One valid advantage of fermentation (e.g. renewable raw materials, low temperature/pressure) [1]
- M2: One valid disadvantage of fermentation (e.g. batch process, slow rate, impure product requiring distillation) [1]
- M3: One valid advantage of hydration (e.g. continuous process, fast rate, pure product) [1]
- M4: One valid disadvantage of hydration (e.g. non-renewable raw material/crude oil, high temperature/pressure/energy cost) [1]

Part (b) [2 marks]:
- M1: Correct formulas for all reactants and products [1]
- M2: Correct balancing: \(\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}\) [1]

Part (c) [5.33 marks]:
- (i) Potassium dichromate((VI)) [1] AND dilute sulfuric acid [1] (Accept acidified potassium dichromate for [1] if only one line provided, but [2] for both distinct reagents).
- (ii) Orange [1] to green [1]
- (iii) Displayed formula of ethanoic acid showing all covalent bonds (must show the C-H, C-C, C=O, C-O, and O-H bonds explicitly) [1.33]

(a)
(i) The colorless solution turns orange (or yellow/orange/brown).
(ii) \(\text{Cl}_2\text{(aq)} + 2\text{Br}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{Br}_2\text{(aq)}\) (Allow \(\text{g}\) or \(\text{aq}\) for \(\text{Cl}_2\)).
(iii) Redox involves both oxidation (loss of electrons) and reduction (gain of electrons). Bromide ions (\(\text{Br}^-\)) lose electrons to form bromine molecules (\(\text{Br}_2\)), so they are oxidized. Chlorine molecules (\(\text{Cl}_2\)) gain electrons to form chloride ions (\(\text{Cl}^-\)), so they are reduced.

(b)
To test for bromide ions:
1. Add a few drops of dilute nitric acid (\(\text{HNO}_3\)) to the solution to remove carbonate impurities.
2. Add a few drops of silver nitrate solution (\(\text{AgNO}_3\)).
3. A cream precipitate (of silver bromide, \(\text{AgBr}\)) will form if bromide ions are present.

Part (a) [8 marks]:
- (i) Colorless [1] to orange/yellow/brown [1]
- (ii) M1: Correct ionic species: \(\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2\) [1]; M2: Correct state symbols: \(\text{(g)}\) or \(\text{(aq)}\) for chlorine, \(\text{(aq)}\) for all other species [1]
- (iii) M1: State that oxidation is loss of electrons and reduction is gain of electrons (or OIL RIG) [1]; M2: Identify that bromide ions lose electrons and are oxidized [1]; M3: Identify that chlorine gains electrons and is reduced [1]

Part (b) [3.33 marks]:
- M1: Add dilute nitric acid [1]
- M2: Add silver nitrate solution [1]
- M3: Formation of a cream precipitate [1]
- M4: Reject use of hydrochloric acid / sulfuric acid instead of nitric acid [0.33]