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The mass number of an atom is defined as the total number of protons and neutrons in its nucleus. For this carbon atom:

\text{Mass number} = \text{number of protons} + \text{number of neutrons} = 6 + 6 = 12.

Award 1 mark for the correct answer: B (12).

Carbon dioxide is a simple molecular substance. It consists of small molecules held together by weak intermolecular forces, which require relatively little energy to overcome, resulting in a low boiling point. Sodium chloride is giant ionic, diamond is giant covalent, and copper is metallic.

Award 1 mark for the correct answer: B (Carbon dioxide).

Filtration is the standard experimental technique used to separate an insoluble solid, such as sand, from a liquid, such as water, because the solid remains on the filter paper as residue while the liquid passes through as filtrate.

Award 1 mark for the correct answer: B (Filtration).

The pH scale ranges from 0 to 14. Solutions with a pH between 0 and 3 are strongly acidic. Therefore, a solution with a pH of 2 is strongly acidic.

Award 1 mark for the correct answer: C (Strongly acidic).

During the electrolysis of molten lead bromide (\text{PbBr}_2), the positive lead ions (\text{Pb}^{2+}) migrate to the negative electrode (cathode). Here they gain electrons (reduction) to form lead atoms: \text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}.

Award 1 mark for the correct answer: A (Lead atoms).

Carbon is more reactive than iron on the reactivity series. Therefore, carbon can reduce iron oxide by displacing iron from the compound, forming carbon dioxide and leaving iron metal.

Award 1 mark for the correct answer: B (Carbon is more reactive than iron).

Transition metals typically form colored compounds, whereas Group 1 alkali metals typically form white or colorless compounds. Group 1 metals are soft, have low melting points, and react vigorously with water, which are not characteristic properties of transition metals.

Award 1 mark for the correct answer: A (They form colored compounds).

Chlorine is added to water during the final stages of treatment to sterilise it by killing harmful bacteria and other microbes, making the water safe to drink (potable).

Award 1 mark for the correct answer: C (To kill harmful bacteria and microbes).

Simple distillation separates a solvent from a solution. When heated, the water evaporates, enters the condenser where it cools and condenses back to a liquid, and is collected as pure water. The sodium chloride remains in the distillation flask.

1 mark for selecting the correct option (b).

The atomic number (bottom number, 5) represents the number of protons, which is also equal to the number of electrons in a neutral atom. The mass number (top number, 11) is the sum of protons and neutrons. Therefore, the number of neutrons is \(11 - 5 = 6\).

1 mark for selecting the correct option (b).

The reaction between an acid and a metal carbonate produces a salt, carbon dioxide, and water. Hydrochloric acid and copper carbonate react to produce copper chloride, carbon dioxide, and water: \(\text{CuCO}_3 + 2\text{HCl} \rightarrow \text{CuCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\).

1 mark for selecting the correct option (b).

Iron is more reactive than copper in the reactivity series. Therefore, iron displaces copper from copper sulfate to form iron sulfate (which is green in solution) and solid copper metal (which deposits as a brown solid on the nail).

1 mark for selecting the correct option (b).

To find the relative formula mass, sum the relative atomic masses of all atoms in the formula: \(M_r = 40 + 2 \times (16 + 1) = 40 + 2 \times 17 = 40 + 34 = 74\).

1 mark for selecting the correct option (c).

Pure chemical substances have a sharp, fixed melting point. If a substance melts over a range of temperatures, it is impure (a mixture of different substances).

1 mark for selecting the correct option (b).

The atomic number is the number of protons in an atom, which is 5. The mass number is the total number of protons and neutrons in the nucleus. Subtracting the atomic number from the mass number gives the number of neutrons: \(11 - 5 = 6\).

1 mark: State that there are 5 protons. 1 mark: State that there are 6 neutrons (derived from 11 - 5).

To separate an insoluble solid like sand from a liquid mixture, filtration is used. The sand remains on the filter paper as residue. To ensure the sand is clean, it is washed with distilled water to remove any salt solution. It is then left to dry in a warm place.

1 mark: Filter the mixture to collect the sand as residue on the filter paper. 1 mark: Wash the sand residue with distilled water and dry it (e.g. in a warm oven or air dry).

Hydrochloric acid reacts with sodium hydroxide to produce sodium chloride and water. The word equation is: hydrochloric acid + sodium hydroxide \(\rightarrow\) sodium chloride + water.

1 mark: Correct reactants: 'hydrochloric acid + sodium hydroxide' (accept in either order). 1 mark: Correct products: 'sodium chloride + water' (accept in either order).

In this reaction, copper oxide (\(\text{CuO}\)) loses oxygen to become copper (\(\text{Cu}\)). Since reduction is defined as the loss of oxygen, copper oxide is the substance that is reduced.

1 mark: Identify copper oxide (or \(\text{CuO}\)) as the substance reduced. 1 mark: Explain that it loses oxygen (to form copper).

Magnesium is in Group 2 and forms \(2+\) ions (\(\text{Mg}^{2+}\)). Oxygen is in Group 6 and forms \(2-\) ions (\(\text{O}^{2-}\)). To balance the charges, they combine in a 1:1 ratio, giving the formula \(\text{MgO}\).

1 mark: Correct formula of magnesium oxide (\(\text{MgO}\)). 1 mark: Correct charges on both ions (magnesium ion is 2+ or \(\text{Mg}^{2+}\) and oxide ion is 2- or \(\text{O}^{2-}\)).

Hydrogen ions (\(\text{H}^+\)) are positive and move to the negative electrode, which is the cathode. The test for hydrogen gas involves holding a lighted splint near the gas; it burns with a distinctive 'squeaky pop' sound.

1 mark: Identify cathode / negative electrode. 1 mark: Describe the test: apply a lighted splint and it makes a squeaky pop sound (both the lighted splint and the squeaky pop sound are required).

Phytoextraction is an eco-friendly method. It allows metals to be extracted from low-grade ores, conserving high-grade ores. It also causes much less environmental damage (less dust, noise, and destruction of habitats) compared to large-scale traditional mining.

1 mark for each of two valid advantages (max 2 marks): Conserves high-grade ores (or uses low-grade ores) / Decontaminates polluted soil / Less environmental damage or less visual/noise/dust pollution / Lower energy requirement or lower carbon footprint.

A pure substance contains only one chemical species, so it does not separate and produces a single spot. A mixture contains multiple substances which separate, producing two or more spots at different positions on the chromatogram.

1 mark: A pure substance produces only one spot on the chromatogram. 1 mark: A mixture produces more than one spot / multiple spots.

To separate pure water from a salt solution, simple distillation is used. In this process, the mixture is heated, and water evaporates to form steam (gas). This steam then enters the condenser where it is cooled, causing it to change state back into a liquid (condensation).

- Mark 1: State that the process is simple distillation (1)
- Mark 2: State that the change of state in the condenser is gas to liquid / vapour to liquid / condensation (1)

The atomic number represents the number of protons in an atom, which is 5. The mass number is the sum of protons and neutrons. Therefore, the number of neutrons is calculated as: mass number minus atomic number: \(11 - 5 = 6\).

- Mark 1: Identify that there are 5 protons (1)
- Mark 2: Calculate and state that there are 6 neutrons (1)

Phenolphthalein is a chemical indicator that is colourless in acidic conditions (such as pH 2) and turns a distinct pink (or magenta/red) colour in alkaline conditions (such as pH 12).

- Mark 1: State that the indicator is colourless in acidic solution / pH 2 (1)
- Mark 2: State that the indicator is pink / red / magenta in alkaline solution / pH 12 (1) (do not accept purple)

In this extraction, iron oxide reacts with carbon to produce iron and carbon dioxide. Reduction is chemically defined in terms of oxygen as the loss of oxygen. Since oxygen is removed from the iron oxide to leave pure iron, the iron oxide has been reduced.

- Mark 1: State that oxygen is removed or lost from the iron oxide (1)
- Mark 2: State that reduction is the loss of oxygen (1)

The formula \(\text{Ca(OH)}_2\) contains one calcium atom, two oxygen atoms, and two hydrogen atoms.

We calculate the relative formula mass as follows:
\(M_r = 40 + [2 \times (16 + 1)]\)
\(M_r = 40 + [2 \times 17]\)
\(M_r = 40 + 34 = 74\).

- Mark 1: Show correct working, e.g. \(40 + (16 + 1) \times 2\) or \(40 + 32 + 2\) (1)
- Mark 2: Provide the correct final value of 74 (1)

Transition metals like copper, iron, and nickel typically have high melting points, high densities, and are very hard and strong. Conversely, Group 1 alkali metals have very low melting points, low densities, and are very soft.

- Mark 1: State high melting point / high boiling point (1)
- Mark 2: State high density / high strength / hard (1)

(Accept any two correct physical properties that contrast transition metals with alkali metals; do not accept chemical properties such as forming coloured compounds or catalysis).

Molten lead bromide contains positive lead ions (\(\text{Pb}^{2+}\)) and negative bromide ions (\(\text{Br}^{-}\)). The positive lead ions are attracted to the negative electrode, which is the cathode, where they gain electrons to form lead metal. Bromine is a liquid at room temperature.

- Mark 1: Identify that lead is produced at the cathode / negative electrode (1)
- Mark 2: State that bromine is a liquid at room temperature (1)

To balance the chemical equation, we compare the number of atoms of each element on both sides:
- Left side: \(1\text{ Na}\), \(2\text{ H}\), \(1\text{ O}\)
- Right side: \(1\text{ Na}\), \(3\text{ H}\), \(1\text{ O}\)

By putting a coefficient of 2 in front of \(\text{NaOH}\), we get an even number of hydrogens on the right:
\(\dots\text{Na} + \dots\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\)
Now we have \(4\text{ H}\) and \(2\text{ O}\) on the right. We balance this by putting a 2 in front of \(\text{H}_2\text{O}\) on the left:
\(\dots\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\)
Finally, we have \(2\text{ Na}\) on the right, so we place a 2 in front of \(\text{Na}\) on the left:
\(2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\).

- Mark 1: Show partially correct balancing (e.g. correct 1:1 ratio of Na to NaOH, such as 2 Na and 2 NaOH, or oxygen balanced) (1)
- Mark 2: Fully correct balancing coefficients: 2 Na, 2 \(\text{H}_2\text{O}\), and 2 NaOH (1)

To obtain dry sand from the mixture, the student must use filtration:

1. Pour the mixture into a filter funnel lined with filter paper. The sand will be left on the filter paper as residue, while the sodium chloride solution passes through as the filtrate.
2. Wash the residue (sand) with a small amount of distilled water to remove any remaining salt solution, then leave the sand in a warm oven or on a windowsill to dry.

1 mark: Filter the mixture to separate the sand (as residue).
1 mark: Wash the sand with distilled water and dry it (e.g., in an oven or warm place).

The atomic number represents the number of protons in the nucleus of an atom. Therefore, the lithium atom has 3 protons.
The mass number represents the total number of protons and neutrons. To find the number of neutrons, subtract the atomic number from the mass number: \(7 - 3 = 4\) neutrons.

1 mark: 3 protons.
1 mark: 4 neutrons.

When an acid reacts with a metal carbonate, the products are a salt, water, and carbon dioxide gas.

1. The gas produced is carbon dioxide.
2. The chemical test for carbon dioxide is to bubble the gas through limewater. If carbon dioxide is present, the limewater will turn cloudy (milky).

1 mark: Identify the gas as carbon dioxide.
1 mark: Test: bubble the gas through limewater and observe it turning cloudy/milky.

A metal can be extracted from its oxide using carbon reduction only if the metal is less reactive than carbon.

1. Carbon is more reactive than iron, meaning carbon can displace iron from iron oxide, reducing the iron oxide to iron.
2. Aluminium is more reactive than carbon, so carbon is not reactive enough to displace aluminium from its oxide.

1 mark: Explain that carbon is more reactive than iron (allowing reduction of iron oxide).
1 mark: Explain that carbon is less reactive than aluminium (meaning it cannot reduce aluminium oxide).

Pure copper has a regular lattice structure where all atoms are of the same size, allowing the layers of atoms to slide over each other easily when a force is applied.

In brass, zinc atoms of a different size are introduced into the structure. This distorts the regular arrangement of the copper atoms. As a result, the layers can no longer slide over each other easily, making the alloy much harder than the pure metal.

1 mark: Mention that zinc atoms have a different size to copper atoms, distorting the regular layers/structure.
1 mark: Explain that this prevents the layers of atoms from sliding over each other easily.

Sodium is in Group 1 of the periodic table and has one electron in its outer shell. During the reaction with chlorine, the sodium atom transfers its outer electron to the chlorine atom.

1. The sodium atom loses one electron.
2. By losing this negatively charged electron, it forms a positively charged sodium ion (\(\text{Na}^+\)) with a stable full outer shell.

1 mark: State that the sodium atom loses one electron.
1 mark: State that it forms a positive ion (or \(\text{Na}^+\)).

1. Multiply the number of atoms of each element by its relative atomic mass: \(Mg = 1 \times 24 = 24\); \(N = 2 \times 14 = 28\); \(O = 6 \times 16 = 96\). 2. Add these values together: \(24 + 28 + 96 = 148\).

Award 1 mark for calculating correct total mass of nitrogen (28) or oxygen (96) in the formula. Award 1 mark for showing addition of all component relative atomic masses: \(24 + 28 + 96\). Award 1 mark for the correct final answer of 148.

1. State the formula for \(R_f\): \(R_f = \frac{\text{distance moved by spot}}{\text{distance moved by solvent front}}\). 2. Substitute the values: \(R_f = \frac{3.2}{8.0}\). 3. Calculate the final value: \(R_f = 0.40\).

Award 1 mark for recall of correct formula or correct substitution of values (3.2 and 8.0). Award 1 mark for correct calculation of fraction (0.4). Award 1 mark for expressing the final answer to an appropriate precision (0.4 or 0.40).

1. Convert the volume from \(\text{cm}^3\) to \(\text{dm}^3\): \(250 / 1000 = 0.25\text{ dm}^3\). 2. Use the concentration formula: \(\text{mass} = \text{concentration} \times \text{volume}\). 3. Substitute values: \(\text{mass} = 40 \times 0.25 = 10\text{ g}\).

Award 1 mark for converting volume to \(\text{dm}^3\) (0.25). Award 1 mark for multiplying their volume by 40. Award 1 mark for correct final answer of 10 (g).

1. State the percentage yield formula: \(\text{Percentage yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100\). 2. Substitute the values: \(\frac{9.5}{12.5} \times 100\). 3. Calculate final value: \(0.76 \times 100 = 76\%\).

Award 1 mark for dividing 9.5 by 12.5. Award 1 mark for multiplying by 100. Award 1 mark for the correct final answer of 76 (%).

1. Calculate moles of each element: \(\text{moles of Cu} = \frac{4.0}{64} = 0.0625\text{ mol}\), \(\text{moles of O} = \frac{0.5}{16} = 0.03125\text{ mol}\). 2. Divide by the smallest number to find the ratio: \(\text{ratio of Cu} = \frac{0.0625}{0.03125} = 2\), \(\text{ratio of O} = \frac{0.03125}{0.03125} = 1\). 3. The empirical formula is \(Cu_2O\).

Award 1 mark for calculating correct moles of either copper (0.0625) or oxygen (0.03125). Award 1 mark for finding the simplest whole number ratio of 2:1. Award 1 mark for the correct empirical formula \(Cu_2O\).

1. Calculate the mass of zinc in the sample: \(5.00\text{ g} - 3.80\text{ g} = 1.20\text{ g}\). 2. Calculate the proportion of zinc: \(\frac{1.20}{5.00} = 0.24\). 3. Convert to percentage: \(0.24 \times 100 = 24\%\).

Award 1 mark for calculating correct mass of zinc (1.20 g). Award 1 mark for dividing the mass of zinc by the total mass (5.00 g). Award 1 mark for the correct final answer of 24 (%).

1. Use the weighted average formula: \(A_r = \frac{(\text{mass of isotope 1} \times \% \text{ abundance}) + (\text{mass of isotope 2} \times \% \text{ abundance})}{100}\). 2. Substitute values: \(A_r = \frac{(35 \times 75.0) + (37 \times 25.0)}{100} = \frac{2625 + 925}{100} = \frac{3550}{100}\). 3. Calculate final value: \(35.5\).

Award 1 mark for showing correct products of mass and abundance: \((35 \times 75)\) and \((37 \times 25)\). Award 1 mark for dividing the sum by 100. Award 1 mark for the correct final answer of 35.5.

1. Calculate the difference in solubility per \(100\text{ g}\) of water: \(40\text{ g} - 15\text{ g} = 25\text{ g}\). 2. Scale up for \(200\text{ g}\) of water: \(25\text{ g} \times \frac{200}{100} = 50\text{ g}\).

Award 1 mark for calculating the solubility difference per 100 g of water (25 g) OR calculating the masses dissolved in 200 g of water at either temperature (80 g or 30 g). Award 1 mark for showing a scaling factor of 2 (or multiplying by 200/100). Award 1 mark for the correct final answer of 50 (g).

To separate the mixture:
1. Add distilled water to the mixture in a beaker and stir using a glass rod. The copper(II) sulfate dissolves because it is soluble, while the sand remains as a solid because it is insoluble.
2. Filter the mixture. Place filter paper in a filter funnel over a conical flask. Pour the mixture through. The insoluble sand remains on the filter paper as the residue. The copper(II) sulfate solution passes through into the flask as the filtrate.
3. Wash the sand residue on the filter paper with a small volume of distilled water to wash away any remaining copper(II) sulfate solution. Leave the sand in a warm oven or a sunny windowsill to dry.
4. Pour the copper(II) sulfate filtrate into an evaporating basin. Heat the basin gently using a Bunsen burner to evaporate some of the water until crystals begin to form. Leave the remaining solution to cool slowly so large crystals grow.
5. Filter the crystals from the remaining liquid, rinse with cold distilled water, and dry them by patting with filter paper.

Level 1 (1-2 marks): Simple description of at least one separation technique (filtration or evaporation) but lacks detail or logical order.
Level 2 (3-4 marks): Clear description of both filtration and crystallisation, with some apparatus named or reasons given, but missing steps for drying or showing minor errors in sequence.
Level 3 (5-6 marks): Detailed and logical sequence to obtain both dry sand and dry crystals. Correct apparatus named and clear explanation/reasons for steps (dissolving, filtering, washing, evaporating/crystallising, drying).

Indicative content:
- Dissolving: Add water to beaker and stir (solubilities differ).
- Filtration: Filter funnel, paper, conical flask. Sand is residue, solution is filtrate.
- Washing residue: Wash sand with distilled water to remove salt solution, then dry.
- Crystallisation: Heat filtrate in evaporating basin, evaporate some water, cool slowly for crystals.
- Drying crystals: Filter, wash with cold water, dry with filter paper.

To prepare the zinc sulfate crystals:
1. Measure a set volume of dilute sulfuric acid into a beaker.
2. Warm the acid gently using a Bunsen burner to increase the rate of reaction.
3. Add zinc oxide powder to the acid a little at a time, stirring with a glass rod, until it is in excess (unreacted powder settled at the bottom). This ensures all the acid has reacted and been neutralised.
4. Filter the mixture using filter paper and a filter funnel into a conical flask. This removes the excess, unreacted zinc oxide as the residue, leaving a pure zinc sulfate solution as the filtrate.
5. Pour the zinc sulfate solution into an evaporating basin. Heat the solution gently to evaporate about half of the water until the crystallisation point is reached.
6. Leave the concentrated solution to cool slowly at room temperature, which allows large, well-formed zinc sulfate crystals to grow.
7. Decant or filter the crystals from any remaining liquid, wash them with a small amount of cold distilled water to remove impurities, and dry them by patting gently with filter paper.

Level 1 (1-2 marks): Identifies some basic steps (e.g., mixing reactants, filtering, or evaporating) but the method is incomplete, out of order, or lacks explanations.
Level 2 (3-4 marks): Describes most of the steps in a logical sequence (heating acid, adding excess zinc oxide, filtering, heating filtrate, crystallising). Explains the need for at least one step (e.g., why excess zinc oxide is added or why it is filtered).
Level 3 (5-6 marks): Gives a fully detailed, logical, and safe experimental method. Explains the reasons for key steps (excess zinc oxide to ensure all acid reacts; filtration to remove unreacted zinc oxide; partial evaporation/cooling to get larger/better crystals; drying).

Indicative content:
- Heat acid (increases reaction rate).
- Add excess zinc oxide (ensures all acid reacts so salt is not contaminated with acid).
- Filter (removes excess insoluble zinc oxide).
- Evaporate water partially (concentrates solution to reach crystallisation point).
- Cool slowly (allows large crystals to form).
- Wash and dry (removes soluble impurities and gets dry crystals).

An endothermic reaction takes in thermal energy from the surroundings, which causes the temperature of the reaction mixture and the surroundings to decrease.

1 mark for the correct choice B. All other choices represent processes that are typically exothermic or unrelated to temperature decrease.

Group 1 alkali metals are soft, low-density metals that react vigorously with water to produce hydrogen gas and a metal hydroxide solution. Their reactivity increases down the group, and they form positive ions with a 1+ charge.

1 mark for identifying that they react with water to produce hydrogen gas.

Volcanic activity on the early Earth released large amounts of gases, with carbon dioxide being the dominant component of the early atmosphere, similar to the atmospheres of Mars and Venus today.

1 mark for selecting carbon dioxide.

Sodium ions, \( \text{Na}^+ \), produce a distinct yellow flame during a flame test. Lithium produces red, potassium produces lilac, and calcium produces orange-red.

1 mark for identifying the sodium ion.

The relative formula mass is calculated by adding the relative atomic masses of all the atoms in the formula: \(M_r = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100\).

1 mark for the correct calculation and value of 100.

Using larger lumps of calcium carbonate decreases the surface area exposed to the acid. This reduces the frequency of successful collisions, thereby decreasing the rate of the reaction.

1 mark for identifying that using larger lumps decreases the rate.

As you move down the fractionating column, temperatures increase and larger hydrocarbons are collected. Larger molecules have stronger intermolecular forces, leading to higher boiling points and higher viscosity (making them thicker and stickier).

1 mark for identifying that both boiling point and viscosity increase down the column.

At room temperature, chlorine is a pale green gas, bromine is a red-brown liquid, and iodine is a dark grey solid.

1 mark for identifying liquid as the physical state of bromine.

Bromine is a red-brown liquid at room temperature and pressure. Chlorine is a gas, and iodine is a solid.

Award 1 mark for selecting B (liquid). Reject all other options.

Reactions that transfer thermal energy to the surroundings, resulting in a temperature increase of the surroundings, are classified as exothermic reactions.

Award 1 mark for selecting B (exothermic).

The fractionating column is coolest at the top and hottest at the bottom. Hydrocarbons with low boiling points rise to the very top before they condense and are collected as gases or highly volatile liquids.

Award 1 mark for selecting C (at the top, where it is coolest).

Sodium ions produce a characteristic yellow flame when tested. Lithium gives red, potassium gives lilac, and calcium gives orange-red.

Award 1 mark for selecting B (sodium ion).

The relative formula mass is calculated by adding the relative atomic masses of all atoms in the formula:  \( M_r = (2 \times 1) + 16 = 18 \).

Award 1 mark for selecting B (18).

Noble gases have a full outer shell of electrons, which is a stable configuration, meaning they have no tendency to lose, gain, or share electrons to react.

Award 1 mark for selecting B (full outer shell of electrons).

Using powdered calcium carbonate instead of large chips increases the surface area to volume ratio, leading to more frequent successful collisions between reactant particles and increasing the rate of reaction.

Award 1 mark for selecting D (using powdered calcium carbonate).

As the Earth cooled, water vapour in the atmosphere condensed to form oceans. A vast amount of carbon dioxide dissolved in these newly formed oceans, significantly reducing its atmospheric level.

Award 1 mark for selecting C (carbon dioxide dissolved in oceans).

In a fractionating column, there is a temperature gradient where it is hottest at the bottom and coolest at the top. Hydrocarbons with the lowest boiling points do not condense in the column and exit from the very top. Bitumen has the highest boiling point and is collected at the bottom.

Award 1 mark for the correct answer (D). Reject A, B, C.

A catalyst works by providing an alternative reaction pathway that has a lower activation energy. This means more reactant particles have enough energy to react when they collide, which increases the rate of reaction. A catalyst remains chemically unchanged and is not used up in the reaction, and it does not increase the temperature or kinetic energy of the particles.

Award 1 mark for the correct answer (B). Reject A, C, D.

When sodium reacts with water, it melts into a sphere due to the heat produced, floats and moves quickly on the surface of the water, and fizzes because hydrogen gas is produced. It also gradually gets smaller until it completely disappears.

Award 1 mark for each correct observation, up to a maximum of 2 marks:
- floats
- melts / forms a ball
- fizzes / bubbles / effervescence
- moves on the surface
- gets smaller / disappears / dissolves

Use the equation: \(\text{mean rate of reaction} = \frac{\text{volume of gas produced}}{\text{time taken}}\). Substituting the given values: \(\text{mean rate of reaction} = \frac{80\text{ cm}^3}{40\text{ s}} = 2.0\text{ cm}^3\text{ s}^{-1}\).

1 mark for correct working: \(\frac{80}{40}\)
1 mark for correct evaluation: 2 (or 2.0)

Electromagnetic radiation from the Sun passes through the atmosphere and warms the Earth's surface. The Earth's surface then radiates longer-wavelength infrared radiation. Greenhouse gases absorb this infrared radiation and re-emit it in all directions, trapping heat in the atmosphere.

1 mark for: absorbing infrared / thermal / heat radiation emitted from the Earth's surface.
1 mark for: re-emitting / reflecting / trapping this heat back towards the Earth / in the atmosphere.

To calculate the relative formula mass, multiply the relative atomic mass of each element by the number of atoms present in the formula: \(2 \times \text{N} = 2 \times 14 = 28\), \(8 \times \text{H} = 8 \times 1 = 8\), \(1 \times \text{S} = 1 \times 32 = 32\), \(4 \times \text{O} = 4 \times 16 = 64\). Add these together: \(28 + 8 + 32 + 64 = 132\).

1 mark for correct working showing the addition of the relative atomic masses: \((2 \times 14) + (8 \times 1) + 32 + (4 \times 16)\) (or equivalent).
1 mark for correct final value: 132.

(i) Calcium ions produce an orange-red flame in a flame test.
(ii) To clean the wire loop, it is dipped into concentrated hydrochloric acid and then heated in a roaring blue Bunsen burner flame to remove any contaminating ions.

1 mark for (i): orange-red / brick-red (reject: red on its own).
1 mark for (ii): dip the loop in (hydrochloric) acid and heat in a Bunsen burner flame (accept: clean with acid and heat).

Chlorine is more reactive than iodine, so it displaces iodine from potassium iodide: \(\text{Cl}_2 + 2\text{KI} \rightarrow 2\text{KCl} + \text{I}_2\). The iodine released dissolves in the solution, causing the colourless solution to turn brown.

1 mark for stating the starting colour: colourless.
1 mark for stating the final colour: brown / orange-brown / yellow-brown.

(i) A reaction that increases the temperature of the surroundings is exothermic.
(ii) Bond breaking is endothermic and requires energy, whereas bond making is exothermic and releases energy. In an exothermic reaction, the energy released when new bonds form is greater than the energy required to break the original bonds.

1 mark for (i): exothermic.
1 mark for (ii): (more) energy is released in bond making than is taken in / used / needed for bond breaking.

(i) Kerosene (sometimes called paraffin) is the fraction used as aviation fuel.
(ii) The fractionating column is colder at the top and hotter at the bottom. Small molecules with low boiling points condense near the top, while larger molecules with high boiling points condense near the bottom. Therefore, boiling point increases as you go down the column.

1 mark for (i): kerosene / paraffin.
1 mark for (ii): (boiling point) increases / gets higher.

When sodium is added to water, it floats because its density is less than that of water. The heat released from the reaction is enough to melt the sodium, so it forms a spherical ball. It moves rapidly across the water surface due to the release of hydrogen gas, which causes visible fizzing. Eventually, the sodium is completely consumed and disappears.

Award 1 mark for each correct observation up to a maximum of 2 marks:
- Floats (1)
- Melts / forms a ball (1)
- Moves around the surface (1)
- Fizzes / bubbles / effervescence (1)
- Disappears / gets smaller / dissolves (1)

Reject: 'burns with a lilac flame' (this is for potassium).

At a higher temperature, particles gain kinetic energy and move faster. This increases the frequency of collisions (more collisions per second). Because more particles also have energy greater than or equal to the activation energy, a higher proportion of these collisions are successful.

1 mark: Particles have more kinetic energy / move faster.
1 mark: More frequent collisions / more collisions per second / more successful collisions per second.
(Do not accept 'more collisions' on its own without a time reference like 'frequent' or 'per second').

Greenhouse gases include carbon dioxide and methane. Human activities such as burning fossil fuels (for electricity and transport) and deforestation (which reduces the uptake of carbon dioxide by trees) increase carbon dioxide levels. Intensive livestock farming (cattle release methane during digestion) and anaerobic decay in landfill sites increase methane levels.

Award 1 mark for each correct activity up to a maximum of 2:
- Burning fossil fuels / coal / oil / gas (1)
- Deforestation / cutting down trees (1)
- Cattle farming / livestock farming (1)
- Decomposition of waste in landfills (1)

Reject: 'breathing' or 'volcanoes' (natural causes).

To calculate the relative formula mass of \(\text{Ca(OH)}_2\):
\(M_r = \text{mass of Ca} + 2 \times (\text{mass of O} + \text{mass of H})\)
\(M_r = 40 + 2 \times (16 + 1)\)
\(M_r = 40 + 2 \times 17\)
\(M_r = 40 + 34 = 74\)

1 mark: Correct working showing the addition of the relative atomic masses, e.g. \(40 + 2 \times (16 + 1)\) or \(40 + 32 + 2\).
1 mark: Correct final calculation of 74.

To test for sulfate ions, dilute hydrochloric acid is first added to the sample (to remove any carbonate impurities that could form a false precipitate). Then, barium chloride solution is added. If sulfate ions are present, they react with the barium ions to form a white precipitate of insoluble barium sulfate.

1 mark: Add dilute hydrochloric acid followed by barium chloride solution (accept barium nitrate instead of barium chloride, and dilute nitric acid instead of hydrochloric acid).
1 mark: White precipitate.
Note: Reagent(s) and correct observation must both be present to score both marks.

Chlorine is more reactive than bromine, so it displaces bromine from potassium bromide. The reactants are chlorine and potassium bromide, and the products are potassium chloride and bromine.

\(\text{chlorine} + \text{potassium bromide} \rightarrow \text{potassium chloride} + \text{bromine}\)

1 mark: Correct reactants on the left side: chlorine + potassium bromide.
1 mark: Correct products on the right side: potassium chloride + bromine.
(Accept reactants/products in any order on their respective sides. Reject chemical symbols/formulas; a word equation is requested).

The reaction is endothermic because the temperature of the reaction mixture decreases. In an endothermic reaction, heat energy is absorbed/taken in from the surroundings, resulting in a temperature drop.

1 mark: Endothermic.
1 mark: Heat energy is taken in / absorbed from the surroundings.

During incomplete combustion, carbon in the hydrocarbon fuel is not fully oxidized. It can form carbon monoxide (\(\text{CO}\)), which is a toxic, colorless gas, and carbon (\(\text{C}\)) in the form of particulate soot. Some carbon dioxide (\(\text{CO}_2\)) may also be produced.

Award 1 mark for each correct name up to a maximum of 2:
- Carbon monoxide (1)
- Carbon / soot (1)
- Carbon dioxide (1)

Reject: 'hydrogen', 'water' (these do not contain carbon).

To calculate the mean rate of reaction between 10 seconds and 40 seconds: 1. Find the change in volume of gas: \(45\text{ cm}^3 - 15\text{ cm}^3 = 30\text{ cm}^3\). 2. Find the time interval: \(40\text{ s} - 10\text{ s} = 30\text{ s}\). 3. Calculate the mean rate of reaction: \(\text{mean rate} = \frac{\text{change in volume}}{\text{time taken}} = \frac{30\text{ cm}^3}{30\text{ s}} = 1.0\text{ cm}^3\text{/s}\).

1. Award 1 mark for finding either the change in volume (\(30\text{ cm}^3\)) or the change in time (\(30\text{ s}\)). 2. Award 1 mark for dividing the change in volume by the change in time (\(\frac{30}{30}\)). 3. Award 1 mark for the correct final answer of \(1.0\) (or \(1\)).

To find the relative formula mass of \(\text{Na}_2\text{CO}_3\): 1. Multiply the relative atomic mass of each element by its quantity in the formula: Sodium (Na): \(2 \times 23 = 46\), Carbon (C): \(1 \times 12 = 12\), Oxygen (O): \(3 \times 16 = 48\). 2. Add these values together: \(M_r = 46 + 12 + 48 = 106\).

1. Award 1 mark for showing correct individual contributions for at least one element (e.g. \(2 \times 23 = 46\) or \(3 \times 16 = 48\)). 2. Award 1 mark for showing a complete sum: \(46 + 12 + 48\) (allow arithmetic error). 3. Award 1 mark for the correct final answer of 106.

To find the volume of nitrogen in the sample: 1. Express the percentage as a fraction or decimal: \(78\% = \frac{78}{100} = 0.78\). 2. Multiply this by the total volume of air: \(250\text{ cm}^3 \times 0.78 = 195\text{ cm}^3\).

1. Award 1 mark for converting the percentage to a fraction or decimal: \(\frac{78}{100}\) or \(0.78\). 2. Award 1 mark for multiplying the fraction or decimal by the total volume: \(250 \times 0.78\) (or equivalent). 3. Award 1 mark for the correct final answer of 195.

To find the \(R_f\) value: 1. Use the equation: \(R_f = \frac{\text{distance moved by the spot}}{\text{distance moved by the solvent front}}\). 2. Substitute the values: \(R_f = \frac{5.2}{8.0}\). 3. Calculate the value: \(R_f = 0.65\). This is already to 2 significant figures.

1. Award 1 mark for correct substitution into the equation: \(\frac{5.2}{8.0}\). 2. Award 1 mark for evaluation to \(0.65\). 3. Award 1 mark for giving the final answer to 2 significant figures (\(0.65\)).

To find the mass of helium: 1. Substitute the density and volume values into the equation: \(\text{mass} = 0.178\text{ g/dm}^3 \times 3.5\text{ dm}^3\). 2. Calculate the mass: \(\text{mass} = 0.623\text{ g}\). 3. Round to 2 decimal places: \(0.623\text{ g} \approx 0.62\text{ g}\).

1. Award 1 mark for correct substitution into the equation: \(0.178 \times 3.5\). 2. Award 1 mark for correct evaluation of the product: \(0.623\). 3. Award 1 mark for rounding the final answer to 2 decimal places: \(0.62\).

To find the activation energy: 1. Recall that the activation energy is the difference between the peak energy and the energy of the reactants: \(\text{Activation energy} = \text{peak energy} - \text{energy of reactants}\). 2. Substitute the values: \(\text{Activation energy} = 280\text{ kJ/mol} - 150\text{ kJ/mol}\). 3. Calculate the value: \(130\text{ kJ/mol}\).

1. Award 1 mark for identifying the correct peak energy (\(280\)) and reactant energy (\(150\)). 2. Award 1 mark for subtracting the reactant energy from the peak energy: \(280 - 150\). 3. Award 1 mark for the correct final answer of 130.

To calculate the percentage by mass of iron: 1. Find the total mass of the iron oxide sample: \(5.6\text{ g} + 2.4\text{ g} = 8.0\text{ g}\). 2. Divide the mass of iron by the total mass: \(\frac{5.6}{8.0} = 0.70\). 3. Multiply by 100 to get the percentage: \(0.70 \times 100\% = 70\%\).

1. Award 1 mark for finding the correct total mass: \(5.6 + 2.4 = 8.0\) (g). 2. Award 1 mark for dividing the mass of iron by the total mass: \(\frac{5.6}{8.0}\) (or \(\frac{5.6}{\text{their total mass}}\)). 3. Award 1 mark for the correct final percentage of 70 (%).

To calculate the emissions after a \(15\%\) reduction: Method A: 1. Calculate the mass reduced: \(120,000 \times 0.15 = 18,000\text{ tonnes}\). 2. Subtract this from the original mass: \(120,000 - 18,000 = 102,000\text{ tonnes}\). Method B: 1. Calculate the remaining percentage: \(100\% - 15\% = 85\%\). 2. Multiply the original mass by this percentage: \(120,000 \times 0.85 = 102,000\text{ tonnes}\).

1. Award 1 mark for finding the reduction: \(120,000 \times 0.15 = 18,000\) OR for finding the remaining percentage: \(85\%\). 2. Award 1 mark for subtracting the reduction: \(120,000 - 18,000\) OR for multiplying by the remaining fraction: \(120,000 \times 0.85\). 3. Award 1 mark for the correct final answer of 102,000.

To investigate the rate of reaction between magnesium and hydrochloric acid:
1. Measure a set volume of the first concentration of hydrochloric acid using a measuring cylinder and pour it into a conical flask.
2. Set up the gas collection apparatus (either a gas syringe or an inverted measuring cylinder filled with water in a trough).
3. Cut a piece of magnesium ribbon to a standard length (e.g., 3 cm) and clean it with emery paper to remove any oxide layer.
4. Add the magnesium ribbon to the flask, quickly replace the bung connected to the gas syringe, and start the stopwatch.
5. Record the volume of gas produced at regular time intervals, such as every 10 or 20 seconds, until the reaction stops or a set time is reached.
6. Repeat the entire procedure using different concentrations of hydrochloric acid.
7. Keep control variables constant: temperature of the acid, total volume of the acid, and the length, surface area, and mass of the magnesium ribbon.

Level 1 (1-2 marks): The response identifies a basic method or setup, such as mixing magnesium and acid and observing bubbles, or makes a simple reference to changing concentration. The plan lacks structure and key details are missing.
Level 2 (3-4 marks): The response describes a workable method to measure the volume of gas produced (e.g., using a gas syringe or measuring cylinder) and mentions repeating the test with different concentrations. Some control variables (such as volume of acid or mass of magnesium) are identified.
Level 3 (5-6 marks): The response provides a detailed, logical, and repeatable method. It clearly explains how to measure the rate (volume of gas at timed intervals or time taken to produce a fixed volume), identifies multiple key control variables (temperature, mass/length of magnesium, volume of acid) to ensure a fair test, and describes how to vary the independent variable (concentration).

The experiment is designed to show displacement reactions among the halogens:
1. Use a pipette to place a few drops of each halide solution (potassium chloride, potassium bromide, potassium iodide) into separate wells of a dimple tile or into separate test tubes.
2. Add chlorine water to a sample of each of the three halide solutions and record any color changes.
3. Add bromine water to a separate sample of each of the three halide solutions and record any color changes.
4. Add iodine water to a separate sample of each of the three halide solutions and record any color changes.
5. Analyze the results: a reaction (color change) indicates that the free halogen is more reactive than the halide ion in solution and has displaced it.
- Chlorine displaces bromide (solution turns orange) and iodide (solution turns brown).
- Bromine displaces iodide (solution turns brown) but not chloride (no change).
- Iodine cannot displace chloride or bromide (no change).
This demonstrates the reactivity order: chlorine > bromine > iodine.

Level 1 (1-2 marks): The response describes a simple procedure of mixing some solutions, or identifies a single observation/displacement reaction. The link to reactivity is weak or absent.
Level 2 (3-4 marks): The response describes a structured method to mix the halogens with the halide solutions. At least two correct observations of displacement are described (e.g., chlorine reacting with bromide and iodide). A partial conclusion about the order of reactivity is made.
Level 3 (5-6 marks): The response describes a systematic, comprehensive method covering all key mixtures. All correct observations are described clearly (including cases where no reaction occurs). The response provides a clear, logical explanation of how these displacement reactions prove the trend in reactivity: chlorine is the most reactive (displaces both), bromine is in the middle (displaces only iodine), and iodine is the least reactive (displaces neither).