2025 Edexcel IGCSE Chemistry Practice Paper with Answers

To calculate the enthalpy change (\(\Delta H\)):
1. Calculate the energy needed to break the bonds in the reactants:
Energy needed = \(1 \times \text{H}-\text{H} + 1 \times \text{Cl}-\text{Cl} = 436 + 243 = 679\text{ kJ/mol}\).

2. Calculate the energy released when making the bonds in the products:
Energy released = \(2 \times \text{H}-\text{Cl} = 2 \times 431 = 862\text{ kJ/mol}\).

3. Calculate the overall enthalpy change:
\(\Delta H = \text{Energy needed} - \text{Energy released}\)
\(\Delta H = 679 - 862 = -183\text{ kJ/mol}\).

Award 1 mark for the correct option (A).

- Reject B (+183 kJ/mol) as it has the incorrect sign for an exothermic reaction.
- Reject C and D as they result from incorrect bond energy combinations (e.g. failing to multiply the H-Cl bond by 2).

Both lithium and sodium react with water to produce hydrogen gas (bubbles are seen), both float on the surface because they are less dense than water, and both form alkaline metal hydroxide solutions that turn phenolphthalein pink. However, the heat produced by the reaction of sodium with water is enough to melt the sodium metal into a shiny silver sphere because of its lower melting point compared to lithium. Lithium has a higher melting point and reacts more slowly, so it does not melt.

Award 1 mark for the correct option (B).

- Option A is incorrect because both lithium and sodium produce hydrogen gas bubbles.
- Option C is incorrect because both produce metal hydroxides which turn phenolphthalein pink.
- Option D is incorrect because both metals have densities less than 1.0 g/cm³ and float on water.

1. Find the percentage of oxygen in the compound:
\(\% \text{ O} = 100.0\% - 70.0\% = 30.0\%\)

2. Divide the percentage of each element by its relative atomic mass to find the molar ratio:
\(\text{Moles of Fe} = \frac{70.0}{56} = 1.25\)
\(\text{Moles of O} = \frac{30.0}{16} = 1.875\)

3. Divide by the smallest value to obtain the simplest whole-number ratio:
\(\text{Fe} = \frac{1.25}{1.25} = 1\)
\(\text{O} = \frac{1.875}{1.25} = 1.5\)

Multiply by 2 to convert to integers:
\(\text{Fe} = 2\), \(\text{O} = 3\).

Therefore, the empirical formula is \(\text{Fe}_2\text{O}_3\).

Award 1 mark for the correct option (B).

- Option A (FeO) results from an incorrect 1:1 atomic ratio.
- Option C (Fe3O4) is a different oxide but does not match the 70.0% composition.
- Option D (Fe3O2) represents an inverted mole ratio.

To find the monomer of an addition polymer:
1. Identify the repeating unit of the polymer chain. The repeating unit consists of two carbon atoms along the backbone:
\(-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\)

2. Reconstruct the monomer by replacing the single carbon-carbon bond in the backbone of the repeating unit with a double bond:
\(\text{CH}(\text{CH}_3)=\text{CH}(\text{CH}_3)\)

3. Determine the IUPAC name. The longest continuous carbon chain contains 4 carbon atoms (but-), and the double bond starts at carbon 2. Thus, the monomer is but-2-ene.

Award 1 mark for the correct option (C).

- Option A (propene) has only 3 carbons in the monomer.
- Option B (but-1-ene) is an isomer but would produce a repeating unit with a different substituent distribution (\(-\text{CH}_2-\text{CH}(\text{CH}_2\text{CH}_3)-\)).
- Option D (methylpropene) would yield a polymer with two methyl groups on the same carbon atom (\(-\text{CH}_2-\text{C}(\text{CH}_3)_2-\)).

1. Use the equation \(Q = mc\Delta T\) where \(m = 100\text{ g}\), \(c = 4.18\text{ J/g/}^\circ\text{C}\), and \(\Delta T = 21.0\ ^\circ\text{C}\).
2. \(Q = 100 \times 4.18 \times 21.0 = 8778\text{ J}\).
3. Convert J to kJ by dividing by 1000: \(8778 / 1000 = 8.778\text{ kJ}\), which rounds to \(8.78\text{ kJ}\) (to 3 significant figures).

M1: Substitution of correct values into formula \(Q = 100 \times 4.18 \times 21.0\) (1 mark).
M2: Calculation of energy in J (8778 J) (1 mark).
M3: Conversion to kJ (8.78 kJ or 8.8 kJ) (0.5 mark).

1. Butane is a hydrocarbon because it contains only carbon and hydrogen atoms, and it is saturated because it contains only single carbon-carbon covalent bonds.
2. The balanced chemical equation for the complete combustion of butane is: \(2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}\).

M1: Explanation of 'saturated hydrocarbon' (only single C-C bonds and contains only carbon and hydrogen) (1 mark).
M2: Correct formulae for reactants and products in equation: \(\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\) (1 mark).
M3: Correct balancing of the equation: \(2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}\) (0.5 mark).

1. Calculate mass of water lost: \(4.92\text{ g} - 2.40\text{ g} = 2.52\text{ g}\).
2. Calculate moles of anhydrous \(\text{MgSO}_4\): \(2.40 / 120 = 0.02\text{ mol}\).
3. Calculate moles of water lost: \(2.52 / 18 = 0.14\text{ mol}\).
4. Determine the molar ratio of water to anhydrous salt: \(x = 0.14 / 0.02 = 7\).

M1: Determination of the mass of water lost (2.52 g) (0.5 mark).
M2: Calculation of the moles of both \(\text{MgSO}_4\) (0.02 mol) and \(\text{H}_2\text{O}\) (0.14 mol) (1 mark).
M3: Division by the smallest value to find ratio \(x = 7\) (1 mark).

1. A covalent bond is the electrostatic attraction between a shared pair of electrons and the nuclei of the bonded atoms.
2. Simple molecular substances have low boiling points because they have weak intermolecular forces between the molecules, which require very little energy to overcome.

M1: Definition of covalent bond: electrostatic attraction between shared pair of electrons and nuclei of the bonded atoms (1 mark).
M2: Low boiling point explanation: mention of weak intermolecular forces / weak forces between molecules (1 mark).
M3: Statement that little thermal energy is needed to break these weak intermolecular forces (0.5 mark).

1. Moles of \(\text{NaOH} = \text{volume} \times \text{concentration} = 0.0250\text{ dm}^3 \times 0.150\text{ mol/dm}^3 = 0.00375\text{ mol}\).
2. From the stoichiometry of the equation, the mole ratio of \(\text{NaOH} : \text{H}_2\text{SO}_4\) is 2:1. Moles of \(\text{H}_2\text{SO}_4\) needed = \(0.00375 / 2 = 0.001875\text{ mol}\).
3. Volume of \(\text{H}_2\text{SO}_4\) needed = \(\text{moles} / \text{concentration} = 0.001875 / 0.100 = 0.01875\text{ dm}^3\).
4. Convert to \(\text{cm}^3\): \(0.01875 \times 1000 = 18.75\text{ cm}^3\) (accept 18.8).

M1: Correct calculation of moles of \(\text{NaOH}\) (0.00375 mol) (1 mark).
M2: Correct calculation of moles of \(\text{H}_2\text{SO}_4\) using stoichiometry (0.001875 mol) (0.5 mark).
M3: Correct calculation of volume of \(\text{H}_2\text{SO}_4\) in \(\text{cm}^3\) (18.75 cm\(^3\)) (1 mark).

1. Two observations: the potassium melts into a ball, floats on the surface, moves rapidly, produces a lilac flame, fizzes/bubbles, or disappears.
2. The chemical reaction is: \(2\text{K}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{KOH}(aq) + \text{H}_2(g)\).
3. The ionic equation, omitting spectator ions (though there are no spectator ions directly participating on both sides since the reactants are elements/compounds), is represented as: \(2\text{K}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{K}^+(aq) + 2\text{OH}^-(aq) + \text{H}_2(g)\).

M1: Any two correct observations (e.g., lilac flame, moves rapidly, melts, fizzes) (1 mark - 0.5 each).
M2: Correct chemical species in the ionic equation (1 mark).
M3: Correct state symbols for all species: (s), (l), (aq), (aq), (g) (0.5 mark).

1. The repeat unit is drawn by changing the double C=C bond to a single C-C bond, leaving free single bonds extending out from each carbon atom, surrounded by brackets with 'n' outside: \(-(\text{CH}_2-\text{CHCl})_n-\).
2. Addition polymers are non-biodegradable / inert, which means they do not decompose naturally and persist in landfill sites for hundreds of years, taking up space.

M1: Drawing of repeat unit with single C-C bond, correct atoms attached (2 hydrogens on one carbon, 1 hydrogen and 1 chlorine on the other) (1 mark).
M2: Extension bonds extending outside brackets with subscript 'n' (0.5 mark).
M3: Environmental problem: they are non-biodegradable / inert / do not decompose (1 mark).

1. Zinc atoms lose 2 electrons to form zinc ions (\(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\)), which is oxidation (loss of electrons).
2. Copper(II) ions gain 2 electrons to form copper metal (\(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\)), which is reduction (gain of electrons). Since both oxidation and reduction occur simultaneously, it is a redox reaction.
3. The ionic equation is: \(\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)\).

M1: Explanation of oxidation: Zinc atoms lose electrons (0.5 mark).
M2: Explanation of reduction: Copper ions gain electrons (0.5 mark).
M3: Correct ionic equation: \(\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}\) (1.5 marks; 1 mark for correct species, 0.5 mark for balancing/states).

The heat energy change is calculated using the formula \( Q = m c \Delta T \), where m is the mass of water (100 g), c is the specific heat capacity of water (4.18 J/g/\( ^\circ \text{C} \)), and \( \Delta T \) is the temperature change (24.0 \( ^\circ \text{C} \)). \( Q = 100 \times 4.18 \times 24.0 = 10032 \text{ J} \). To convert joules to kilojoules, divide by 1000: \( 10032 / 1000 = 10.032 \text{ kJ} \). To three significant figures, this is 10.0 kJ.

1 mark for substituting correct values into the formula: \( Q = 100 \times 4.18 \times 24.0 \). 1 mark for evaluating the calculation as 10032 J or 10.032 kJ. 0.5 marks for rounding correctly to three significant figures with appropriate units (10.0 kJ).

Divide the percentage of each element by its relative atomic mass to find the molar ratio: Moles of C = \( 40.0 / 12 = 3.33 \). Moles of H = \( 6.7 / 1 = 6.7 \). Moles of O = \( 53.3 / 16 = 3.33 \). Divide each value by the smallest number of moles (3.33): C = \( 3.33 / 3.33 = 1 \); H = \( 6.7 / 3.33 = 2 \); O = \( 3.33 / 3.33 = 1 \). The empirical formula is \( \text{CH}_2\text{O} \).

1 mark for dividing percentages by relative atomic masses to find moles (C = 3.33, H = 6.7, O = 3.33). 1 mark for dividing by the smallest value to obtain the simplest whole-number ratio (1:2:1). 0.5 marks for writing the correct empirical formula (\( \text{CH}_2\text{O} \)).

The \( R_f \) value is calculated by dividing the distance travelled by the solute by the distance travelled by the solvent front: \( R_f = 5.2 / 8.0 = 0.65 \). If a different solvent is used, the \( R_f \) value will change because the dye will have a different solubility in the new solvent compared to its affinity for the stationary phase.

1 mark for calculating the correct \( R_f \) value of 0.65. 1 mark for stating that the \( R_f \) value will change (or be different). 0.5 marks for explaining that solubility of the dye in the solvent (or relative partition between mobile and stationary phases) changes.

The reaction between ethane and chlorine in the presence of UV light is a substitution reaction. The organic product of this mono-substitution reaction is chloroethane. The role of the UV radiation is to provide energy to break the covalent bond in the chlorine molecule (homolytic fission of Cl-Cl bond) to form chlorine free radicals.

0.5 marks for stating the reaction type is substitution. 1 mark for identifying the product as chloroethane. 1 mark for explaining that UV light provides the activation energy or breaks the Cl-Cl bond / creates free radicals.

Sodium chloride has a giant ionic lattice structure. There are strong electrostatic forces of attraction between the oppositely charged sodium ions (\( \text{Na}^+ \)) and chloride ions (\( \text{Cl}^- \)). Because these forces act in all directions and are very strong, a large amount of thermal energy is required to overcome these ionic bonds and melt the substance.

0.5 marks for mentioning a giant ionic lattice structure. 1 mark for describing the strong electrostatic attraction between oppositely charged ions (or sodium and chloride ions). 1 mark for stating that a large amount of energy is required to overcome/break these strong bonds.

When potassium reacts with water, observations include: potassium floats/moves on the surface, melts into a ball, burns with a lilac flame, bubbles of gas are produced, and potassium disappears. The balanced chemical equation with state symbols is: \( 2\text{K(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{KOH(aq)} + \text{H}_2\text{(g)} \).

1 mark for any two correct observations (0.5 marks each). 1 mark for the correct balanced chemical equation: \( 2\text{K} + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2 \). 0.5 marks for correct state symbols: (s), (l), (aq), (g).

In carbon dioxide (\( \text{CO}_2 \)), the carbon atom forms two double covalent bonds, one with each oxygen atom. In each double bond, four electrons are shared (two pairs) - two electrons come from the carbon atom and two from the oxygen atom. This allows all three atoms to achieve a stable outer shell of eight electrons.

1 mark for stating that there are two double covalent bonds. 1 mark for explaining that each double bond consists of two shared pairs of electrons (four electrons shared per bond). 0.5 marks for stating that this sharing allows both carbon and oxygen atoms to achieve a full outer shell (or eight valence electrons).

First, dissolve both solid lead(II) nitrate and solid sodium sulfate in separate containers of distilled water to make aqueous solutions. Mix the two solutions together to form a precipitate of lead(II) sulfate. Filter the mixture to collect the insoluble lead(II) sulfate residue on the filter paper. Wash the residue with distilled water to remove any soluble impurities. Finally, dry the pure precipitate in a warm oven or leave it to dry between pieces of filter paper.

1 mark for dissolving both solids in water and mixing the solutions together. 0.5 marks for filtering the mixture to obtain the precipitate/residue. 1 mark for washing the residue with distilled water and drying it (using filter paper or a warm oven).

1. The reaction is a substitution reaction because a hydrogen atom in propane is replaced by a bromine atom.
2. The balanced chemical equation is: \(\text{C}_3\text{H}_8 + \text{Br}_2 \rightarrow \text{C}_3\text{H}_7\text{Br} + \text{HBr}\).
3. UV radiation provides the activation energy required to break the covalent bond in the bromine molecule (homolytic fission) to form reactive bromine radicals.

M1: State 'substitution' (1 mark)
M2: Correct chemical equation (1 mark) (allow structural or molecular formulae: \(\text{C}_3\text{H}_8 + \text{Br}_2 \rightarrow \text{C}_3\text{H}_7\text{Br} + \text{HBr}\))
M3: State that UV radiation provides energy to break the Br-Br bond / initiate the reaction (0.5 mark)

1. Two unique observations for potassium compared to lithium are: potassium melts into a sphere due to its lower melting point, and it catches fire with a lilac flame (or reacts with an explosion/crackle).
2. The balanced equation with state symbols is: \(2\text{K(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{KOH(aq)} + \text{H}_2\text{(g)}\).

M1: State two unique observations (lilac flame / melts into a ball / explodes) (1.5 marks - 0.75 marks each)
M2: Correctly balanced chemical equation with correct state symbols (1 mark)

1. Calculate heat energy change (\(q\)):
\(q = m \times c \times \Delta T = 150.0 \text{ g} \times 4.18 \text{ J/g/}^{\circ}\text{C} \times 24.0 \text{ }^{\circ}\text{C} = 15048 \text{ J} = 15.05 \text{ kJ}\).
2. Calculate the number of moles of alcohol burned:
\(n = \frac{\text{mass}}{M_r} = \frac{2.00}{46.0} = 0.0435 \text{ mol}\).
3. Calculate the enthalpy change of combustion (\(\Delta H\)):
\(\Delta H = -\frac{q}{n} = -\frac{15.048 \text{ kJ}}{0.0435 \text{ mol}} = -346 \text{ kJ/mol}\) (or \(-346.1 \text{ kJ/mol}\)).

M1: Calculate \(q = 15.05 \text{ kJ}\) (or \(15.0 \text{ kJ}\)) (1 mark)
M2: Calculate moles of alcohol = 0.0435 mol (0.5 mark)
M3: Calculate \(\Delta H\) with negative sign as \(-346 \text{ kJ/mol}\) (allow \(-345\) to \(-347\)) (1 mark) (deduct 0.5 mark if negative sign is missing)

Carbon dioxide has a simple molecular structure. There are weak intermolecular forces between the \(\text{CO}_2\) molecules, which require very little thermal energy to overcome, so it has a low boiling point and is a gas at room temperature. In contrast, silicon dioxide has a giant covalent structure. It consists of a large network of atoms held together by many strong covalent bonds. Breaking these strong covalent bonds requires a very large amount of thermal energy, giving it a high melting point.

M1: Identify that \(\text{CO}_2\) has a simple molecular structure with weak intermolecular forces requiring little energy to overcome (1 mark)
M2: Identify that \(\text{SiO}_2\) has a giant covalent structure with strong covalent bonds (1 mark)
M3: Explain that breaking the strong covalent bonds in \(\text{SiO}_2\) requires a large amount of energy (0.5 mark)

1. Difference: Both isotopes have the same number of protons (31), but \(^{71}\text{Ga}\) has 40 neutrons (\(71 - 31\)) while \(^{69}\text{Ga}\) has 38 neutrons (\(69 - 31\)). Therefore, \(^{71}\text{Ga}\) has 2 more neutrons.
2. Calculation of abundance:
Let the abundance of \(^{69}\text{Ga}\) be \(x\%\). Then the abundance of \(^{71}\text{Ga}\) is \((100 - x)\%\).
\(69.72 = \frac{69x + 71(100 - x)}{100}\)
\(6972 = 69x + 7100 - 71x\)
\(-128 = -2x\)
\(x = 64\%\).

M1: State that \(^{71}\text{Ga}\) has 2 more neutrons than \(^{69}\text{Ga}\) (or correctly state neutron counts 40 and 38) (0.5 mark)
M2: Set up a correct algebraic equation for the weighted average (1 mark)
M3: Obtain the correct percentage of 64% (1 mark)

1. Find the mass of water lost:
\(4.93 \text{ g} - 2.41 \text{ g} = 2.52 \text{ g}\) of \(\text{H}_2\text{O}\).
2. Calculate the moles of anhydrous \(\text{MgSO}_4\) and \(\text{H}_2\text{O}\):
\(n(\text{MgSO}_4) = \frac{2.41}{120} = 0.0201 \text{ mol}\).
\(n(\text{H}_2\text{O}) = \frac{2.52}{18} = 0.140 \text{ mol}\).
3. Find the mole ratio:
\(x = \frac{0.140}{0.0201} = 6.97 \approx 7\).
Therefore, \(x = 7\).

M1: Determine mass of water lost = 2.52 g (0.5 mark)
M2: Calculate moles of anhydrous salt (0.0201 mol) and water (0.140 mol) (1 mark)
M3: Determine the simplest whole-number ratio, leading to \(x = 7\) (1 mark)

1. Filtration: Pour the mixture through filter paper in a funnel. The insoluble sand remains on the filter paper as residue, while the sodium chloride solution passes through as the filtrate.
2. Obtain pure dry sand: Wash the sand residue with distilled water to remove any remaining salt solution, then leave it to dry in a warm oven or press between filter papers.
3. Obtain dry sodium chloride: Heat the filtrate (sodium chloride solution) to evaporate the water, leaving behind solid sodium chloride crystals, which can then be dried.

M1: Filter the mixture to obtain sand as residue and NaCl solution as filtrate (1 mark)
M2: Wash the sand residue with distilled water and dry it (0.5 mark)
M3: Heat/evaporate the filtrate to remove water and obtain dry sodium chloride (1 mark)

1. A catalyst provides an alternative reaction pathway with a lower activation energy.
2. Consequently, a larger proportion of colliding particles have energy greater than or equal to the activation energy, leading to a higher frequency of successful collisions and thus a faster reaction rate.
3. At the end of the reaction, the catalyst is chemically unchanged (and its mass is conserved).

M1: State that a catalyst provides an alternative pathway with a lower activation energy (1 mark)
M2: Explain that this results in more frequent successful collisions (1 mark)
M3: State that the catalyst is chemically unchanged / mass is conserved at the end (0.5 mark)

To calculate the relative atomic mass:
\(A_r = \frac{(63 \times 69.2) + (65 \times 30.8)}{100}\)
\(A_r = \frac{4359.6 + 2002.0}{100}\)
\(A_r = \frac{6361.6}{100} = 63.616\)
Rounding to 1 decimal place gives 63.6.

- 1 mark for correct mathematical substitution: \(\frac{(63 \times 69.2) + (65 \times 30.8)}{100}\)
- 1 mark for calculating the intermediate sum/value: \(63.616\)
- 0.5 marks for rounding correctly to 1 decimal place: \(63.6\)

Carbon dioxide consists of small molecules. It has a simple molecular structure. Between these molecules, there are weak intermolecular forces of attraction. These forces require very little thermal energy to break, resulting in a low boiling point.

- 1 mark for identifying that carbon dioxide has a simple molecular structure.
- 1 mark for identifying that there are weak intermolecular forces (of attraction) between molecules (do not accept weak covalent bonds).
- 0.5 marks for stating that very little energy is needed to overcome these forces.

A magnesium atom has an electronic configuration of 2.8.2. It loses 2 outer shell electrons to form a magnesium ion (\(\text{Mg}^{2+}\)) with a configuration of 2.8. An oxygen atom has an electronic configuration of 2.6. It gains these 2 electrons to form an oxide ion (\(\text{O}^{2-}\)) with a configuration of 2.8.

- 1 mark for stating magnesium loses two electrons (to form a \(2+\) ion / achieve a full outer shell).
- 1 mark for stating oxygen gains two electrons (to form a \(2-\) ion / achieve a full outer shell).
- 0.5 marks for describing the transfer of electrons from magnesium to oxygen to form ions with the stable configuration 2.8.

Iron is higher in the reactivity series than copper, so a displacement reaction occurs. The blue colour of the copper(II) sulfate solution fades and the solution becomes pale green as iron(II) sulfate forms. A pink-brown or brown solid (copper metal) is deposited on the surface of the iron nail.

- 1 mark for stating at least one correct observation: blue solution turns pale green/fades OR a brown/pink-brown solid forms on the nail.
- 1 mark for explaining that iron is more reactive than copper / iron displaces copper.
- 0.5 marks for completing the explanation by stating that copper ions are reduced to copper metal / iron oxidized to iron(II) ions.

The balanced chemical equation for the incomplete combustion is:
\(2\text{C}_3\text{H}_8 + 7\text{O}_2 \rightarrow 6\text{CO} + 8\text{H}_2\text{O}\) (or fractional: \(\text{C}_3\text{H}_8 + 3.5\text{O}_2 \rightarrow 3\text{CO} + 4\text{H}_2\text{O}\)).
Carbon monoxide is toxic because it binds strongly and irreversibly to haemoglobin in red blood cells, which prevents oxygen from binding and being transported around the body.

- 1 mark for a fully balanced chemical equation with correct formulas.
- 1 mark for stating that carbon monoxide binds to haemoglobin in red blood cells.
- 0.5 marks for explaining that this reduces the capacity of the blood to carry/transport oxygen.

First, find the moles of each element:
Moles of S = \(\frac{40.0}{32} = 1.25\)
Moles of O = \(\frac{60.0}{16} = 3.75\)

Divide by the smallest number of moles (1.25):
S = \(\frac{1.25}{1.25} = 1\)
O = \(\frac{3.75}{1.25} = 3\)

The empirical formula is \(\text{SO}_3\).

- 1 mark for calculating correct moles of sulfur (1.25) and oxygen (3.75).
- 1 mark for dividing by the smallest value to find the simplest ratio (1 : 3).
- 0.5 marks for writing the correct final empirical formula: \(\text{SO}_3\).

First, calculate the temperature change:
\(\Delta T = 31.8 - 20.2 = 11.6\text{ }^\circ\text{C}\)

The mass of the solution is:
\(m = 50.0\text{ cm}^3 \times 1.00\text{ g/cm}^3 = 50.0\text{ g}\)

Now, use the heat energy formula:
\(Q = m \cdot c \cdot \Delta T\)
\(Q = 50.0\text{ g} \times 4.18\text{ J/g/}^\circ\text{C} \times 11.6\text{ }^\circ\text{C}\)
\(Q = 2424.4\text{ J}\)

- 1 mark for calculating \(\Delta T = 11.6\text{ }^\circ\text{C}\).
- 1 mark for correct substitution into the formula: \(Q = 50.0 \times 4.18 \times 11.6\).
- 0.5 marks for the correct final value of \(2424.4\text{ J}\) (accept \(2424\text{ J}\) or \(2.42\text{ kJ}\)).

During the reaction, sodium melts into a shiny metal ball because the reaction is highly exothermic. It floats and moves rapidly across the surface of the water, fizzing/effervescing as hydrogen gas is produced. The phenolphthalein indicator turns pink because sodium hydroxide, an alkaline solution, is formed.

- 1 mark for any two correct physical observations (e.g., melts into a ball, floats, moves around, fizzes/bubbles, disappears/gets smaller) [0.5 marks for each observation].
- 1 mark for stating that the phenolphthalein indicator turns pink.
- 0.5 marks for explaining that the color change is due to the formation of an alkaline solution / hydroxide ions.

1. Calculate the mass of water lost: 5.56 g - 3.04 g = 2.52 g. 2. Calculate the moles of anhydrous \(\text{FeSO}_4\): \(M_r(\text{FeSO}_4) = 56 + 32 + (4 \times 16) = 152\). \(\text{Moles} = 3.04 / 152 = 0.02\text{ mol}\). 3. Calculate the moles of \(\text{H}_2\text{O}\): \(M_r(\text{H}_2\text{O}) = 18\). \(\text{Moles} = 2.52 / 18 = 0.14\text{ mol}\). 4. Determine the simplest molar ratio: \(x = 0.14 / 0.02 = 7\).

- Award 1 mark for calculating the amount of anhydrous salt as 0.02 mol. - Award 1 mark for calculating the amount of water lost as 0.14 mol. - Award 0.5 marks for the final correct integer value of 7.

1. Energy required to break reactant bonds: \(436 + 158 = 594\text{ kJ/mol}\). 2. Energy released when product bonds are formed: \(2 \times 562 = 1124\text{ kJ/mol}\). 3. Enthalpy change: \(\Delta H = 594 - 1124 = -530\text{ kJ/mol}\).

- Award 1 mark for calculating energy absorbed to break bonds as 594 kJ/mol. - Award 1 mark for calculating energy released when forming product bonds as 1124 kJ/mol. - Award 0.5 marks for the correct enthalpy change value of -530 (the negative sign must be present for full marks).

Add bromine water (or aqueous bromine) to both gases. Propane is an alkane (saturated) and will not react under standard conditions, meaning the bromine water remains orange/yellow/brown. Propene is an alkene (unsaturated) and will undergo an addition reaction, decolourising the bromine water from orange to colourless.

- Award 0.5 marks for naming bromine water, bromine solution, or aqueous bromine as the reagent (reject bromine liquid or bromine gas alone). - Award 1 mark for stating that propane stays orange/yellow/brown (or has no reaction). - Award 1 mark for stating that propene decolourises the bromine water / turns colourless (reject 'turns clear').

Step 1: Calculate the number of moles of iron and oxygen atoms. \(\text{Moles of Fe} = 11.2 / 56 = 0.20\text{ mol}\). \(\text{Moles of O} = 4.8 / 16 = 0.30\text{ mol}\). Step 2: Find the simplest ratio of the atoms by dividing by the smaller value. \(\text{Fe} = 0.20 / 0.20 = 1\), \(\text{O} = 0.30 / 0.20 = 1.5\). Step 3: Multiply by 2 to obtain the simplest whole-number ratio of 2:3. Therefore, the empirical formula is \(\text{Fe}_2\text{O}_3\).

M1: Calculating moles of iron (0.20 mol) and oxygen (0.30 mol) (1 mark). M2: Dividing by the smallest to find the ratio of 1 : 1.5 (or direct 2 : 3 ratio) (1 mark). M3: Writing the correct empirical formula Fe2O3 (1 mark).

Step 1: Calculate the temperature change. \(\Delta T = 44.5 - 18.5 = 26.0\,^{\circ}\text{C}\). Step 2: Use the formula \(Q = m c \Delta T\) to calculate the heat energy in Joules. \(Q = 150\text{ g} \times 4.18\text{ J/g/}^{\circ}\text{C} \times 26.0\,^{\circ}\text{C} = 16302\text{ J}\). Step 3: Convert Joules to kilojoules. \(Q = 16302 / 1000 = 16.302\text{ kJ}\). Rounding to 3 significant figures gives \(16.3\text{ kJ}\).

M1: Calculation of temperature change \(\Delta T = 26.0\,^{\circ}\text{C}\) (1 mark). M2: Correct calculation of heat energy in J as \(16302\text{ J}\) (1 mark). M3: Conversion to kJ and rounding to 3 significant figures to give \(16.3\text{ kJ}\) (1 mark).

Step 1: Calculate the moles of sulfuric acid used. \(\text{Moles of H}_2\text{SO}_4 = (20.0 / 1000) \times 0.150 = 0.0030\text{ mol}\). Step 2: Determine the moles of NaOH that react. From the balanced equation, 2 moles of NaOH react with 1 mole of H2SO4. Therefore, \(\text{Moles of NaOH} = 2 \times 0.0030 = 0.0060\text{ mol}\). Step 3: Calculate the concentration of the NaOH solution. \(\text{Concentration} = 0.0060 / (25.0 / 1000) = 0.24\text{ mol/dm}^3\).

M1: Calculation of moles of acid = 0.0030 mol (1 mark). M2: Calculation of moles of NaOH = 0.0060 mol (1 mark). M3: Calculation of concentration of NaOH = 0.24 mol/dm^3 (1 mark).

Step 1: Calculate the moles of magnesium reacted. \(\text{Moles of Mg} = 6.00 / 24 = 0.25\text{ mol}\). Step 2: Determine the theoretical moles and mass of magnesium oxide. The ratio of Mg to MgO is 1:1, so theoretical moles of MgO = 0.25 mol. Theoretical mass of MgO = \(0.25 \times (24 + 16) = 0.25 \times 40 = 10.0\text{ g}\). Step 3: Calculate the percentage yield. \(\text{Percentage yield} = (8.50 / 10.0) \times 100 = 85.0\%\).

M1: Calculation of moles of Mg = 0.25 mol (1 mark). M2: Calculation of theoretical mass of MgO = 10.0 g (1 mark). M3: Calculation of percentage yield = 85.0% or 85% (1 mark).

Step 1: Calculate the moles of Mg reacting. \(\text{Moles of Mg} = 4.80 / 24 = 0.20\text{ mol}\). Step 2: Use the stoichiometric ratio from the balanced equation to find the moles of H2 gas. The ratio of Mg to H2 is 1:1, so moles of H2 produced = 0.20 mol. Step 3: Calculate the volume of H2 gas. \(\text{Volume of H}_2 = 0.20 \times 24 = 4.8\text{ dm}^3\).

M1: Calculation of moles of Mg = 0.20 mol (1 mark). M2: Using the 1:1 ratio to state/use moles of H2 = 0.20 mol (1 mark). M3: Calculation of H2 volume = 4.8 dm^3 (1 mark).

Step 1: Calculate the energy required to break bonds. Bonds broken = \(4 \times (\text{C-H}) + 1 \times (\text{Cl-Cl}) = (4 \times 413) + 243 = 1895\text{ kJ/mol}\) (or net bonds broken = 656 kJ/mol). Step 2: Calculate the energy released when making bonds. Bonds made = \(3 \times (\text{C-H}) + 1 \times (\text{C-Cl}) + 1 \times (\text{H-Cl}) = (3 \times 413) + 346 + 432 = 2017\text{ kJ/mol}\) (or net bonds made = 778 kJ/mol). Step 3: Calculate the enthalpy change. \(\Delta H = 1895 - 2017 = -122\text{ kJ/mol}\).

M1: Calculation of total energy for bonds broken = 1895 kJ/mol (or 656 kJ/mol net) (1 mark). M2: Calculation of total energy for bonds made = 2017 kJ/mol (or 778 kJ/mol net) (1 mark). M3: Calculation of enthalpy change of -122 kJ/mol (including minus sign) (1 mark).

Step 1: Calculate the mass of water of crystallization lost. \(\text{Mass of water} = 5.88 - 4.44 = 1.44\text{ g}\). Step 2: Calculate the moles of anhydrous calcium chloride and moles of water. \(\text{Moles of CaCl}_2 = 4.44 / 111 = 0.040\text{ mol}\). \(\text{Moles of H}_2\text{O} = 1.44 / 18 = 0.080\text{ mol}\). Step 3: Determine the mole ratio. \(\text{Ratio} = 0.040 : 0.080 = 1 : 2\). Therefore, \(x = 2\).

M1: Calculation of water mass = 1.44 g (1 mark). M2: Calculation of moles of calcium chloride (0.040 mol) and moles of water (0.080 mol) (1 mark). M3: Simplification of ratio to show x = 2 (1 mark).

Propanoic acid (\(CH_3CH_2COOH\)) and methanol (\(CH_3OH\)) react in an esterification reaction to produce methyl propanoate and water (\(H_2O\)). The structural formula of methyl propanoate is \(CH_3CH_2COOCH_3\). By counting the atoms, we find it has 4 carbon atoms, 8 hydrogen atoms, and 2 oxygen atoms. Therefore, its molecular formula is \(C_4H_8O_2\).

Award 1 mark for the correct answer (A). Reject all other options.

1. Calculate the energy required to break bonds in the reactants: Breaking 4 \(C-H\) bonds: \(4 \times 413 = 1652\) kJ/mol. Breaking 2 \(O-H\) bonds: \(2 \times 463 = 926\) kJ/mol. Total energy input = \(1652 + 926 = 2578\) kJ/mol. 2. Calculate the energy released making bonds in the products: Making 1 \(C\equiv O\) bond: 1077 kJ/mol. Making 3 \(H-H\) bonds: \(3 \times 436 = 1308\) kJ/mol. Total energy output = \(1077 + 1308 = 2385\) kJ/mol. 3. Calculate enthalpy change: \(\Delta H = \text{energy input} - \text{energy output} = 2578 - 2385 = +193\) kJ/mol.

Award 1 mark for the correct answer (B). Reject A (incorrect sign), C or D (calculation errors).

Alkanes are hydrocarbons because they are compounds containing hydrogen and carbon atoms only. They are described as saturated because all the bonds between carbon atoms are single covalent bonds, meaning they contain the maximum possible number of hydrogen atoms per carbon.

M1: (hydrocarbon) contains ONLY hydrogen and carbon (atoms/elements) (1)
M2: (saturated) contains only single bonds / no double bonds between carbon atoms (1)

When potassium is placed in water, it reacts vigorously. It floats and moves rapidly across the surface, melts into a shiny ball, fizzes due to hydrogen gas release, burns with a lilac flame, and eventually disappears as it is used up.

Any two observations from:
- floats / moves on the water surface (1)
- melts into a ball/sphere (1)
- fizzes / bubbles / effervescence (1)
- burns with a lilac flame (1)
- disappears / gets smaller (1)

Using the formula: \(q = m \times c \times \Delta T\) where \(m = 100\text{ g}\) (mass of water), \(c = 4.18\text{ J/g/}^{\circ}\text{C}\), and \(\Delta T = 24\text{ }^{\circ}\text{C}\). Thus: \(q = 100 \times 4.18 \times 24 = 10032\text{ J}\) (or \(10\text{ kJ}\)).

M1: Correct substitution of values: \(100 \times 4.18 \times 24\) (1)
M2: Correct evaluation to 10032 (J) / 10.0 (kJ) (1)
(Accept 10000 J or 10030 J due to significant figures)

Carbon dioxide consists of simple molecules. While the covalent bonds within the molecules are strong, the forces of attraction between the molecules (intermolecular forces) are weak and require very little energy to overcome, resulting in a low boiling point.

M1: Weak forces of attraction between molecules / weak intermolecular forces (1)
M2: Little thermal energy is needed to overcome / break these forces (1)

Chlorine has an atomic number of 17, which corresponds to the number of protons. The mass number is the sum of protons and neutrons, so the number of neutrons is \(37 - 17 = 20\).

M1: 17 protons (1)
M2: 20 neutrons (1)

Calculate the relative formula mass (\(M_r\)) of \(\text{NH}_4\text{NO}_3\): \(M_r = (14 \times 2) + (1 \times 4) + (16 \times 3) = 28 + 4 + 48 = 80\). The total mass of nitrogen in the formula is \(2 \times 14 = 28\). The percentage by mass of nitrogen is: \(\frac{28}{80} \times 100 = 35\%\).

M1: Correct calculation of \(M_r = 80\) (or showing total nitrogen mass is 28) (1)
M2: Correct percentage: \(35\%\) (1)

Increasing the temperature favors the endothermic reaction to absorb the added thermal energy. Since the forward reaction is exothermic (negative enthalpy change), the reverse reaction is endothermic. Thus, the equilibrium position shifts to the left, decreasing the yield of ammonia.

M1: Shifts to the left / towards the reactants (1)
M2: (Because) the forward reaction is exothermic / reverse reaction is endothermic, and increasing temperature favors the endothermic direction (1)

The reaction of an alcohol and a carboxylic acid to form an ester is catalyzed by concentrated sulfuric acid. The word equation is: \(\text{ethanol} + \text{ethanoic acid} \rightleftharpoons \text{ethyl ethanoate} + \text{water}\).

M1: Concentrated sulfuric acid / \(\text{H}_2\text{SO}_4\) (1)
M2: ethanol + ethanoic acid \(\rightleftharpoons\) (or single arrow) ethyl ethanoate + water (1)

First, calculate the temperature change (\(\Delta T\)):\
\(\Delta T = 38.6 - 20.2 = 18.4\text{ °C}\)\
\
Next, use the formula \(q = m c \Delta T\) to find the heat energy change in joules:\
\(q = 75.0\text{ g} \times 4.18\text{ J/g/°C} \times 18.4\text{ °C} = 5768.4\text{ J}\)\
\
Convert joules to kilojoules by dividing by 1000:\
\(q = 5768.4 / 1000 = 5.77\text{ kJ}\) (to 3 significant figures).

- M1: Correctly calculates temperature change: \(\Delta T = 18.4\text{ °C}\) (1 mark)\
- M2: Correct substitution into formula: \(75.0 \times 4.18 \times 18.4 = 5768.4\text{ J}\) (1 mark)\
- M3: Correct conversion to kJ and final value: \(5.77\text{ kJ}\) (0.4 marks)

In a substitution reaction with alkanes, one hydrogen atom is replaced by a halogen atom under UV light.\
Equation:\
\(C_4H_{10} + Br_2 \rightarrow C_4H_9Br + HBr\)\
The organic product is bromobutane.

- M1: Correct balanced chemical equation with correct molecular formulas: \(C_4H_{10} + Br_2 \rightarrow C_4H_9Br + HBr\) (1.4 marks)\
- M2: Correct name of organic product: bromobutane (1 mark)

The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, increasing the temperature favors the endothermic direction to absorb heat. Therefore, the equilibrium position shifts to the left, which decreases the yield of ammonia.

- M1: State that the yield of ammonia decreases (1 mark)\
- M2: State that the forward reaction is exothermic / reverse reaction is endothermic (1 mark)\
- M3: Explain that equilibrium shifts to the left / in the endothermic direction (0.4 marks)

Write the balanced chemical equation:\
\(HNO_3 + KOH \rightarrow KNO_3 + H_2O\)\
\
Calculate the moles of nitric acid:\
\(\text{Moles of } HNO_3 = \text{volume in dm}^3 \times \text{concentration} = 0.0200\text{ dm}^3 \times 0.0500\text{ mol/dm}^3 = 0.00100\text{ mol}\)\
\
Since the reaction ratio is 1:1, moles of \(KOH\) required = 0.00100 mol.\
\
Calculate the volume of \(KOH\) solution:\
\(\text{Volume of } KOH = \frac{\text{moles}}{\text{concentration}} = \frac{0.00100\text{ mol}}{0.0250\text{ mol/dm}^3} = 0.0400\text{ dm}^3\)\
\
Convert to cm³:\
\(0.0400 \times 1000 = 40.0\text{ cm}^3\)

- M1: Calculate moles of nitric acid: \(0.00100\text{ mol}\) (1 mark)\
- M2: Relate moles to KOH (1:1 ratio) and calculate volume in dm³ or cm³: \(40.0\text{ cm}^3\) (1.4 marks)\
[Allow full marks for correct answer with no working]

Nitrogen consists of simple molecules. While the triple covalent bonds holding the nitrogen atoms together inside each molecule are very strong, the forces of attraction between different nitrogen molecules (intermolecular forces) are very weak. Boiling only requires overcoming these weak intermolecular forces, not breaking the strong covalent bonds, so very little energy is needed.

- M1: Identify that nitrogen has a simple molecular structure (1 mark)\
- M2: State that the intermolecular forces (forces between molecules) are weak (1 mark)\
- M3: Conclude that little thermal energy is needed to overcome these weak forces (0.4 marks)\
[Reject references to breaking covalent bonds during boiling]

To test for sulfate ions, dilute hydrochloric acid is first added to remove any contaminating carbonate or sulfite ions. Then, barium chloride solution is added. If sulfate ions are present, a white precipitate of insoluble barium sulfate (\(BaSO_4\)) forms.

- M1: Add dilute hydrochloric acid (or nitric acid) and barium chloride solution (or barium nitrate solution) (1.4 marks)\
- M2: Observe a white precipitate (1 mark)\
[Reject sulfuric acid as the acid, reject other incorrect reagents]

1. Divide the percentage of each element by its relative atomic mass:\
Carbon: \(40.0 / 12 = 3.33\)\
Hydrogen: \(6.7 / 1 = 6.7\)\
Oxygen: \(53.3 / 16 = 3.33\)\
\
2. Divide each result by the smallest value (3.33):\
Carbon: \(3.33 / 3.33 = 1\)\
Hydrogen: \(6.7 / 3.33 \approx 2\)\
Oxygen: \(3.33 / 3.33 = 1\)\
\
3. The simplest whole-number ratio is 1:2:1, so the empirical formula is \(CH_2O\).

- M1: Divide percentages by respective \(A_r\) values to get ratio moles: \(C = 3.33\), \(H = 6.7\), \(O = 3.33\) (1.4 marks)\
- M2: Divide by the smallest to show the 1:2:1 ratio and deduce \(CH_2O\) (1 mark)

Fermentation of glucose to produce ethanol requires yeast (which provides enzymes). The conditions must be:\
1. A warm temperature (usually between 30 °C and 40 °C) to ensure enzymes work at their optimum rate without being denatured.\
2. Anaerobic conditions (absence of oxygen) to prevent the ethanol from oxidizing to ethanoic acid and to ensure the yeast respires anaerobically.

- M1: Mention a suitable temperature range: 30–40 °C (or 'warm temperature') (1.2 marks)\
- M2: Mention anaerobic conditions / absence of oxygen (1.2 marks)

First, calculate the heat energy change per mole of reactant: \(2.4\text{ kJ} / 0.025\text{ mol} = 96\text{ kJ/mol}\). Second, since the reaction is exothermic (temperature increases and heat is released), the sign of the enthalpy change must be negative. Therefore, \(\Delta H = -96\text{ kJ/mol}\).

M1: Calculation of numerical value of 96 (kJ/mol) by dividing 2.4 by 0.025 (1 mark). M2: Giving the final answer with a negative sign as -96 (kJ/mol) (1 mark).

1. Alkanes react with halogens in the presence of ultraviolet light via a substitution reaction, where one hydrogen atom is replaced by a halogen atom. 2. The molecular equation for this reaction is: \(\text{C}_2\text{H}_6 + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{HCl}\).

M1: Substitution (1 mark). M2: Correct chemical equation with molecular formulae: \(\text{C}_2\text{H}_6 + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{HCl}\) (1 mark). Accept reactants and products in any order. Ignore state symbols.

To test for ammonium ions, add sodium hydroxide solution to the sample and warm gently. This produces ammonia gas. Test the gas with damp red litmus paper, which will turn blue because ammonia is an alkaline gas.

M1: Add sodium hydroxide (solution) and warm / heat (1 mark). M2: Gas produced turns damp red litmus paper blue (1 mark). Reject direct placement of litmus paper in the liquid.

Addition polymers consist of long chains of carbon atoms held together by strong, non-polar carbon-carbon (\(\text{C}-\text{C}\)) single covalent bonds. Because these bonds are very strong and non-polar, they require a high amount of energy to break, making the polymer chemically unreactive (inert) and preventing microorganisms or decomposers from breaking them down.

M1: State that they contain strong carbon-carbon (\(\text{C}-\text{C}\)) single bonds / covalent bonds (1 mark). M2: State that these bonds are non-polar / require a lot of energy to break, so decomposers/microorganisms cannot break them down (1 mark).

1. Calculate the heat energy absorbed (\(Q\)) using \(Q = m \cdot c \cdot \Delta T\):
\(Q = 50.0\text{ g} \times 4.18\text{ J/g/}^\circ\text{C} \times 5.6\text{ }^\circ\text{C} = 1170.4\text{ J} = 1.1704\text{ kJ}\)

2. Calculate the amount in moles of \(\text{NH}_4\text{NO}_3\) dissolved:
\(\text{Moles} = \frac{\text{mass}}{M_{\text{r}}} = \frac{4.00\text{ g}}{80.0\text{ g/mol}} = 0.0500\text{ mol}\)

3. Calculate the molar enthalpy change (\(\Delta H\)):
\(\Delta H = \frac{Q}{\text{moles}} = \frac{1.1704\text{ kJ}}{0.0500\text{ mol}} = 23.408\text{ kJ/mol}\)

Since the temperature decreased, the process is endothermic, so the sign must be positive (\(+\)).

Rounding to 3 significant figures gives \(+23.4\text{ kJ/mol}\).

- **1 mark** for calculating heat energy absorbed (\(1170.4\text{ J}\) or \(1.1704\text{ kJ}\)).
- **1 mark** for calculating moles of \(\text{NH}_4\text{NO}_3\) (\(0.0500\text{ mol}\)).
- **1 mark** for dividing the energy by the number of moles to find the magnitude of the molar enthalpy change (\(23.4\text{ kJ/mol}\)).
- **0.5 mark** for the correct positive sign (\(+\)).

*Accept \(23.4\text{ kJ/mol}\) without sign for 3 marks. Reject negative sign.*

1. Calculate the mass of oxygen in the copper oxide:
\(\text{Mass of oxygen} = 14.3\text{ g} - 12.7\text{ g} = 1.6\text{ g}\)

2. Calculate the moles of copper and oxygen:
\(\text{Moles of Cu} = \frac{12.7}{63.5} = 0.20\text{ mol}\)
\(\text{Moles of O} = \frac{1.6}{16.0} = 0.10\text{ mol}\)

3. Find the simplest whole-number ratio:
\(\text{Cu} : \text{O} = \frac{0.20}{0.10} : \frac{0.10}{0.10} = 2 : 1\)

Therefore, the empirical formula of the copper oxide is \(\text{Cu}_2\text{O}\).

- **1 mark** for calculating the mass of oxygen (\(1.6\text{ g}\)).
- **1 mark** for calculating the correct moles of both copper (\(0.20\text{ mol}\)) and oxygen (\(0.10\text{ mol}\)).
- **1 mark** for obtaining the simplest molar ratio of \(2 : 1\).
- **0.5 mark** for writing the correct empirical formula (\(\text{Cu}_2\text{O}\)).

1. Calculate the amount in moles of \(\text{H}_2\text{SO}_4\) used:
\(\text{Moles of H}_2\text{SO}_4 = \text{concentration} \times \text{volume} = 0.100\text{ mol/dm}^3 \times \frac{18.50}{1000}\text{ dm}^3 = 0.00185\text{ mol}\)

2. Determine the moles of \(\text{NaOH}\) that reacted:
From the balanced equation, \(2\text{ mol}\) of \(\text{NaOH}\) react with \(1\text{ mol}\) of \(\text{H}_2\text{SO}_4\).
\(\text{Moles of NaOH} = 2 \times 0.00185\text{ mol} = 0.00370\text{ mol}\)

3. Calculate the concentration of the \(\text{NaOH}\) solution:
\(\text{Volume of NaOH} = \frac{25.0}{1000} = 0.0250\text{ dm}^3\)
\(\text{Concentration of NaOH} = \frac{\text{moles}}{\text{volume}} = \frac{0.00370\text{ mol}}{0.0250\text{ dm}^3} = 0.148\text{ mol/dm}^3\)

The final answer to 3 significant figures is \(0.148\text{ mol/dm}^3\).

- **1 mark** for calculating the moles of \(\text{H}_2\text{SO}_4\) (\(0.00185\text{ mol}\)).
- **1 mark** for multiplying by 2 to find the moles of \(\text{NaOH}\) (\(0.00370\text{ mol\)}).
- **1 mark** for dividing the moles of \(\text{NaOH}\) by the volume in \(\text{dm}^3\) (\(\frac{0.00370}{0.0250}\)).
- **0.5 mark** for giving the correct final value to 3 significant figures (\(0.148\text{ mol/dm}^3\)).

1. Calculate the moles of propane in \(880\text{ g}\):
\(\text{Moles of C}_3\text{H}_8 = \frac{880\text{ g}}{44.0\text{ g/mol}} = 20.0\text{ mol}\)

2. Use the molar ratio from the equation to find the moles of \(\text{CO}_2\) produced:
The ratio of \(\text{C}_3\text{H}_8\) to \(\text{CO}_2\) is \(1 : 3\).
\(\text{Moles of CO}_2 = 3 \times 20.0\text{ mol} = 60.0\text{ mol}\)

3. Calculate the volume of \(\text{CO}_2\) at rtp:
\(\text{Volume of CO}_2 = 60.0\text{ mol} \times 24.0\text{ dm}^3\text{/mol} = 1440\text{ dm}^3\)

- **1 mark** for calculating the moles of propane (\(20.0\text{ mol}\)).
- **1 mark** for using the 1:3 ratio to find the moles of \(\text{CO}_2\) (\(60.0\text{ mol}\)).
- **1 mark** for multiplying the moles of \(\text{CO}_2\) by \(24.0\text{ dm}^3\text{/mol}\) (\(1440\)).
- **0.5 mark** for the correct unit (\(\text{dm}^3\)).

1. Calculate the mass of carbon dioxide gas lost:
\(\text{Mass lost} = 210.50\text{ g} - 209.87\text{ g} = 0.63\text{ g}\)

2. Calculate the moles of carbon dioxide produced:
\(\text{Moles of CO}_2 = \frac{0.63\text{ g}}{44.0\text{ g/mol}} = 0.014318\text{ mol}\)

3. Calculate the mean rate of reaction in \(\text{mol/s}\):
\(\text{Rate} = \frac{0.014318\text{ mol}}{45.0\text{ s}} = 0.00031818...\text{ mol/s}\)

Expressing the final answer to 3 significant figures: \(3.18 \times 10^{-4}\text{ mol/s}\) (or \(0.000318\text{ mol/s}\)).

- **1 mark** for calculating the mass of carbon dioxide lost (\(0.63\text{ g}\)).
- **1 mark** for calculating either the moles of \(\text{CO}_2\) (\(0.0143\text{ mol}\)) or the rate in \(\text{g/s}\) (\(\frac{0.63}{45} = 0.014\text{ g/s}\)).
- **1 mark** for dividing the moles of \(\text{CO}_2\) by time, or dividing mass rate by \(44.0\).
- **0.5 mark** for the correct final answer to 3 significant figures (\(3.18 \times 10^{-4}\text{ mol/s}\) or \(0.000318\text{ mol/s}\)).