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The structural formula \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) contains the ester functional group (\(\text{-COO-}\)). Thus, the molecule is ethyl ethanoate, an ester. Ketones have a carbonyl group (\(\text{-CO-}\)), ethers have an ether group (\(\text{-O-}\)), and carboxylic acids have a carboxyl group (\(\text{-COOH}\)).

Award [1] for the correct choice B. Incorrect choices are: A (ketone), C (ether), D (carboxylic acid).

A heterogeneous mixture has a non-uniform composition and consists of two or more distinct phases. A mixture of sand and water does not dissolve or mix uniformly, forming separate solid and liquid phases. Sodium chloride dissolved in water, brass (a copper-zinc alloy), and air (nitrogen and oxygen) are homogeneous mixtures (solutions) with uniform compositions throughout.

Award [1] for the correct choice C. Other options represent homogeneous mixtures.

Oxygen in \(\text{H}_2\text{O}\) has four electron domains (two bonding pairs and two lone pairs), which gives a tetrahedral electron domain geometry. Due to the greater repulsion of the two lone pairs, the bond angle is compressed from the ideal tetrahedral angle of \(109.5^\circ\) to approximately \(104.5^\circ\), resulting in a bent molecular geometry.

Award [1] for the correct choice C. Other choices are incorrect because linear geometry has a \(180^\circ\) angle (A), bent geometry with \(120^\circ\) is typical for species with one lone pair (B), and trigonal planar is not the geometry of water (D).

According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain equal numbers of particles. The mole ratio of \(\text{H}_2\text{S}\) to \(\text{SO}_2\) is \(2:2\) (or \(1:1\)). Therefore, the volume ratio is also \(1:1\), and \(15.0\text{ dm}^3\) of \(\text{H}_2\text{S}\) will yield exactly \(15.0\text{ dm}^3\) of \(\text{SO}_2\).

Award [1] for the correct choice C. A is half the correct volume, B is incorrect, D is twice the correct volume.

Using the relation: \(\text{mass} = \text{amount in moles} \times \text{molar mass}\), we get \(\text{mass} = 0.50\text{ mol} \times 100.1\text{ g mol}^{-1} = 50.05\text{ g}\).

Award [1] for the correct choice A. B represents 1 mole, C represents 2 moles, and D represents 0.25 moles.

The atomic number (Z) represents the number of protons. Therefore, there are \(28\) protons. Since the atom is neutral, the number of electrons is equal to the number of protons, which is \(28\). The mass number (A) represents the sum of protons and neutrons. Thus, the number of neutrons is \(A - Z = 60 - 28 = 32\).

Award [1] for the correct choice A. Other options incorrectly assign the subatomic particle counts.

As you go down Group 17, atomic radius increases. Since the outermost electrons are further from the nucleus and experience more shielding, they are less strongly attracted to the nucleus, making them easier to remove. Therefore, first ionization energy decreases down the group. Ionic radius and atomic number increase down the group, and melting points increase due to stronger London dispersion forces between larger molecules.

Award [1] for the correct choice B. A, C, and D describe properties that increase down the group.

The heat absorbed by the solution is given by the formula \(q = m \cdot c \cdot \Delta T\). The mass \(m\) is the total mass of the mixture (solute + solvent), which is \(5.00\text{ g} + 100.0\text{ g} = 105.0\text{ g}\). Hence, \(q = 105.0 \times c \times \Delta T\).

Award [1] for the correct choice C. A only uses the solute mass, B only uses the solvent mass, and D divides by the temperature change.

During boiling, which is a phase transition from liquid to gas at constant pressure, the temperature of the substance remains constant. Since temperature is a measure of the average kinetic energy of the particles, the average kinetic energy does not change. The heat energy supplied during boiling is used to overcome the intermolecular forces (hydrogen bonds) between water molecules, which increases their potential energy as they move further apart. No intramolecular covalent bonds are broken during boiling. The gas produced occupies a much larger volume than the liquid, so the density of the substance decreases.

Award [1] for B.
- Option A is incorrect because temperature remains constant during boiling, so average kinetic energy is constant.
- Option C is incorrect because boiling is a physical change and does not involve breaking covalent bonds.
- Option D is incorrect because gas has a much lower density than liquid, so density decreases.

Let us analyze the molecular structure of methyl 3-hydroxybutanoate: \(\text{CH}_3\text{CH(OH)CH}_2\text{COOCH}_3\).
1. The \(-\text{OH}\) group attached to the second carbon atom represents an alcohol functional group.
2. The \(-\text{COOCH}_3\) group at the end of the chain represents an ester functional group.
Therefore, the compound contains both an alcohol and an ester functional group.

Award [1] for A.
- Reject B because there is no ketone functional group present (the carbonyl carbon is part of the ester group).
- Reject C because there is no ether (\(\text{R-O-R}\)) functional group.
- Reject D because there is no carboxylic acid (\(-\text{COOH}\)) functional group.

To find the number of nitrogen atoms, we calculate the number of moles of nitrogen atoms in each sample:
- **For Option A:** Moles of \(\text{N}_2 = \frac{14\text{ g}}{28\text{ g mol}^{-1}} = 0.5\text{ mol}\). Each mole of \(\text{N}_2\) contains 2 moles of N atoms, so \(n(\text{N}) = 0.5 \times 2 = 1.0\text{ mol}\).
- **For Option B:** Moles of \(\text{NH}_3 = \frac{17\text{ g}}{17\text{ g mol}^{-1}} = 1.0\text{ mol}\). Each mole of \(\text{NH}_3\) contains 1 mole of N atoms, so \(n(\text{N}) = 1.0 \times 1 = 1.0\text{ mol}\).
- **For Option C:** Moles of \(\text{HNO}_3 = \frac{63\text{ g}}{63\text{ g mol}^{-1}} = 1.0\text{ mol}\). Each mole of \(\text{HNO}_3\) contains 1 mole of N atoms, so \(n(\text{N}) = 1.0 \times 1 = 1.0\text{ mol}\).
- **For Option D:** Moles of \(\text{NH}_4\text{NO}_3 = \frac{60\text{ g}}{80\text{ g mol}^{-1}} = 0.75\text{ mol}\). Each mole of ammonium nitrate contains 2 moles of N atoms, so \(n(\text{N}) = 0.75 \times 2 = 1.5\text{ mol}\).

Therefore, 60 g of \(\text{NH}_4\text{NO}_3\) has the greatest number of nitrogen atoms (1.5 moles of N atoms).

Award [1] for D.
- Reject A, B, and C because they each contain exactly 1.0 mol of N atoms, which is less than the 1.5 mol of N atoms in D.

To determine the geometry of the nitrite ion, \(\text{NO}_2^-\):
1. The total number of valence electrons is: \(5\text{ (from N)} + 2 \times 6\text{ (from O)} + 1\text{ (negative charge)} = 18\text{ electrons}\).
2. The Lewis structure features a central nitrogen atom bonded to two oxygen atoms, with one lone pair of electrons remaining on the nitrogen atom (i.e., \(\text{O}=\text{N}-\text{O}^-\)).
3. This gives 3 electron domains around the nitrogen atom (2 bonding domains + 1 non-bonding domain).
4. The electron domain geometry is trigonal planar.
5. The molecular geometry is bent (V-shaped).
6. The repulsion from the non-bonding lone pair compresses the bond angle to slightly less than the ideal \(120^\circ\) (typically around \(115^\circ\)).
Therefore, the geometry is bent and the angle is \(< 120^\circ\).

Award [1] for B.
- Reject A because the presence of the lone pair prevents the molecule from being linear.
- Reject C because the electron domain geometry is trigonal planar (3 domains), not tetrahedral (4 domains), so the reference angle is \(120^\circ\) rather than \(109.5^\circ\).
- Reject D because 'trigonal planar' describes the electron domain geometry, not the molecular geometry (which only considers atom positions).

According to the balanced equation:
\(2\text{ mol of Al}\) reacts with \(3\text{ mol of Cl}_2\) to produce \(2\text{ mol of AlCl}_3\).

1. Let's find the limiting reactant:
- If all \(2.0\text{ mol of Al}\) reacted, we would need: \(2.0 \times \frac{3}{2} = 3.0\text{ mol of Cl}_2\).
- Since only \(2.0\text{ mol of Cl}_2\) is available, \(\text{Cl}_2\) is the limiting reactant.

2. Now calculate the maximum yield of \(\text{AlCl}_3\) based on the limiting reactant, \(\text{Cl}_2\):
- \(\text{Moles of AlCl}_3 = \text{moles of Cl}_2 \times \frac{2}{3} = 2.0 \times \frac{2}{3} = 1.33\text{ mol}\).
Therefore, the limiting reactant is \(\text{Cl}_2\) and the maximum yield is approximately \(1.3\text{ mol}\).

Award [1] for C.
- Reject A and D because \(\text{Al}\) is in excess (only \(1.3\text{ mol}\) of \(\text{Al}\) is needed to react with all \(2.0\text{ mol}\) of \(\text{Cl}_2\)).
- Reject B because a yield of \(2.0\text{ mol}\) assumes that \(\text{Al}\) is the limiting reactant or that the mole ratio is \(1:1\).

The heat energy absorbed, \(q\), in Joules (J) is calculated using the formula:
\(q = m \cdot c \cdot \Delta T\)

Where:
- \(m = 50.0\text{ g}\) (mass of the water)
- \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\) (specific heat capacity)
- \(\Delta T = T_{\text{final}} - T_{\text{initial}} = 60.0^\circ\text{C} - 20.0^\circ\text{C} = 40.0^\circ\text{C} = 40.0\text{ K}\) (temperature change)

Substituting these values gives:
\(q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 40.0\text{ K}\) (in Joules)

To convert the energy from Joules (J) to kilojoules (kJ), we divide by 1000:
\(q\text{ (in kJ)} = \frac{50.0 \times 4.18 \times 40.0}{1000}\)

Award [1] for B.
- Reject A because it gives the heat in Joules instead of kilojoules.
- Reject C because a change in temperature (\(\Delta T\)) is the same in Celsius and Kelvin; 273 should not be added to the temperature difference.
- Reject D because the terms are incorrectly arranged.

When temperature is increased:
1. The reactant molecules gain kinetic energy and move faster. This leads to an increase in the frequency of collisions between molecules.
2. More importantly, the Maxwell-Boltzmann distribution shifts to show that a significantly larger fraction of molecules have kinetic energy greater than or equal to the activation energy (\(E \ge E_a\)).
Both factors contribute to the increased reaction rate, although the increase in the fraction of molecules with \(E \ge E_a\) has a much larger effect. Temperature does not change the activation energy of a reaction (only a catalyst can change the activation energy).

Award [1] for B.
- Reject A because temperature has no effect on activation energy (only a catalyst lowers it).
- Reject C because although collision frequency increases, it is not the only factor; the increase in the fraction of molecules with \(E \ge E_a\) is actually the major contributor to the increased rate.
- Reject D because activation energy does not increase, and if it did, the rate would decrease.

A redox reaction involves changes in the oxidation states of elements:
- In **Option A** (acid-carbonate reaction): \(\text{Ca}\) remains \(+2\), \(\text{C}\) remains \(+4\), \(\text{O}\) remains \(-2\), \(\text{H}\) remains \(+1\), and \(\text{Cl}\) remains \(-1\). No oxidation states change.
- In **Option B** (precipitation reaction): \(\text{Ag}\) remains \(+1\), \(\text{N}\) remains \(+5\), \(\text{O}\) remains \(-2\), \(\text{Na}\) remains \(+1\), and \(\text{Cl}\) remains \(-1\). No oxidation states change.
- In **Option C** (single displacement reaction): Zinc metal, \(\text{Zn}\), goes from an oxidation state of \(0\) to \(+2\) (it is oxidized). Copper in copper(II) sulfate, \(\text{Cu}^{2+}\), goes from an oxidation state of \(+2\) to \(0\) in copper metal (it is reduced). Thus, this is a redox reaction.
- In **Option D** (acid-base neutralization): \(\text{Na}\) remains \(+1\), \(\text{O}\) remains \(-2\), \(\text{H}\) remains \(+1\), and \(\text{N}\) remains \(+5\). No oxidation states change.

Award [1] for C.
- Reject A, B, and D because there are no changes in the oxidation states of any elements in these reactions.

The molecule given is \(\text{CH}_3\text{CH(OH)CH}_2\text{COOCH}_3\).
- The \(-\text{OH}\) group bonded to a carbon atom represents an alcohol (hydroxyl group).
- The \(-\text{COO}-\) group bonded to two carbon groups represents an ester group.
Therefore, the functional groups present are an alcohol and an ester.

Award 1 mark for the correct answer B.
- Option A is incorrect because there is no ketone group present.
- Option C is incorrect because there is no ether group present.
- Option D is incorrect because there is no carboxylic acid group present.

A homogeneous mixture has a uniform composition throughout its mass.
- Brass is a solid solution (an alloy) of copper and zinc, which is homogeneous.
- A suspension of sand in water, oil and water, and sulfur powder mixed with iron filings are all heterogeneous mixtures because their components remain distinct and are not uniformly distributed.

Award 1 mark for the correct answer B.
- Reject other choices as they represent heterogeneous mixtures.

To determine the molecular geometry, we look at the number of bonding domains and lone pairs around the central atom:
- \(\text{BF}_3\): Boron has 3 valence electrons, forming 3 bonding pairs and 0 lone pairs. Shape is trigonal planar.
- \(\text{SO}_3\): Sulfur has 6 valence electrons, forming 3 double bonds with 3 oxygen atoms and 0 lone pairs. Shape is trigonal planar.
- \(\text{H}_3\text{O}^+\): Oxygen has 6 valence electrons, minus 1 electron for the positive charge, leaving 5. It forms 3 bonding pairs with hydrogen and has 1 lone pair. The geometry is trigonal pyramidal.
- \(\text{CH}_4\): Carbon has 4 valence electrons, forming 4 bonding pairs and 0 lone pairs. Shape is tetrahedral.

Award 1 mark for the correct answer C.
- Option A, B: trigonal planar.
- Option D: tetrahedral.

According to Avogadro's law, the ratio of gas volumes is equal to the ratio of their moles at constant temperature and pressure.
From the balanced equation:
\(2\text{ dm}^3\) of \(\text{SO}_2\text{(g)}\) reacts with \(1\text{ dm}^3\) of \(\text{O}_2\text{(g)}\) to produce \(2\text{ dm}^3\) of \(\text{SO}_3\text{(g)}\).

If all \(15\text{ dm}^3\) of \(\text{O}_2\) reacts, the volume of \(\text{SO}_2\) required is:
\(15 \times 2 = 30\text{ dm}^3\).
Since we have \(40\text{ dm}^3\) of \(\text{SO}_2\), \(\text{SO}_2\) is in excess and \(\text{O}_2\) is the limiting reactant.

The volume of \(\text{SO}_3\) produced is based on the limiting reactant:
\(\text{Volume of SO}_3 = 2 \times \text{Volume of O}_2 = 2 \times 15 = 30\text{ dm}^3\).

Award 1 mark for the correct answer A.
- Option B is incorrect (fails to identify limiting reactant).
- Option C is incorrect (sum of reactant volumes).
- Option D is incorrect (incorrect stoichiometry).

Assume a \(100\text{ g}\) sample of the oxide:
- Mass of \(\text{N} = 30.4\text{ g}\)
- Mass of \(\text{O} = 100 - 30.4 = 69.6\text{ g}\)

Calculate the amount in moles of each element:
- \(n(\text{N}) = \frac{30.4}{14.0} \approx 2.17\text{ mol}\)
- \(n(\text{O}) = \frac{69.6}{16.0} \approx 4.35\text{ mol}\)

Divide by the smallest value to obtain the simplest whole-number ratio:
- Ratio of \(\text{N} : \text{O} = \frac{2.17}{2.17} : \frac{4.35}{2.17} \approx 1 : 2\)

Therefore, the empirical formula is \(\text{NO}_2\).

Award 1 mark for the correct answer B.
- Option A is incorrect (1:1 ratio).
- Option C is incorrect (2:1 ratio).
- Option D is incorrect (2:5 ratio).

For Phosphorus (\(\text{P}\)):
- Atomic number (\(Z\)) is 15, so there are 15 protons.
- Mass number (\(A\)) is 31, so the number of neutrons is \(31 - 15 = 16\).
- The charge of the phosphide ion is \(3-\), meaning it has 3 more electrons than protons: \(15 + 3 = 18\) electrons.

Award 1 mark for the correct answer B.
- Option A represents a neutral P-31 atom.
- Option C swaps the proton and neutron counts.
- Option D is incorrect as it uses the mass number for the neutron count.

1. Find the mass of the reacting mixture:
\(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3\).
Since density is \(1.00\text{ g cm}^{-3}\), \(m = 100.0\text{ g}\).

2. Calculate heat evolved, \(q\), in Joules:
\(q = m \cdot c \cdot \Delta T = 100.0 \times c \times \Delta T\text{ J}\).
In kJ:
\(q = \frac{100.0 \times c \times \Delta T}{1000}\text{ kJ}\).

3. Find the number of moles of water formed:
\(n(\text{H}^+) = n(\text{OH}^-) = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\).

4. Enthalpy of neutralization, \(\Delta H\):
Since the temperature rose, the reaction is exothermic, so \(\Delta H\) is negative:
\(\Delta H = -\frac{q}{n} = - \frac{100.0 \times c \times \Delta T}{0.0500 \times 1000}\text{ kJ mol}^{-1}\).

Award 1 mark for the correct answer A.
- Option B incorrectly uses only the volume of one reactant (50.0 g) as the mass of the mixture.
- Option C neglects the conversion from J to kJ and the division by moles.
- Option D multiplies by 1000 instead of dividing.

- The oxidation state of copper changes from \(0\) in \(\text{Cu(s)}\) to \(+2\) in \(\text{Cu(NO}_3)_2\). Since copper is oxidized, it acts as the reducing agent.
- Nitrogen in \(\text{HNO}_3\) has an oxidation state of \(+5\) (since \(\text{H} = +1\) and \(\text{O} = -2\)). In \(\text{NO(g)}\), nitrogen has an oxidation state of \(+2\). Since the oxidation state of nitrogen decreases (is reduced), \(\text{HNO}_3\) is the oxidizing agent.
- The change in oxidation state of nitrogen is from \(+5\) to \(+2\).

Award 1 mark for the correct answer B.
- Option A is incorrect because copper is the reducing agent.
- Option C is incorrect because the oxidation state of nitrogen in \(\text{NO}\) is \(+2\), not \(+4\).
- Option D is incorrect because copper is the reducing agent and the oxidation states are reversed.

Ethyl propanoate contains an ester functional group (\(-COO-\)). Its molecular formula is \(C_5H_{10}O_2\). Pentanoic acid is a carboxylic acid with five carbon atoms, also having the molecular formula \(C_5H_{10}O_2\). Since they share the same molecular formula but have different structural layouts and functional groups, they are functional group isomers.

Award 1 mark for the correct option A.
- Correctly identifies the ester functional group.
- Correctly determines that both compounds have the same molecular formula \(C_5H_{10}O_2\) and are therefore isomers.

1. Determine the number of atoms in one formula unit of \((NH_4)_3PO_4\):
\((3 \times \text{N}) + (12 \times \text{H}) + (1 \times \text{P}) + (4 \times \text{O}) = 3 + 12 + 1 + 4 = 20\text{ atoms}\).
2. Calculate the total moles of atoms in \(0.10\text{ mol}\) of the compound:
\(0.10\text{ mol} \times 20 = 2.0\text{ mol of atoms}\).
3. Convert moles of atoms to the total number of atoms:
\(2.0\text{ mol} \times 6.0 \times 10^{23}\text{ mol}^{-1} = 1.2 \times 10^{24}\text{ atoms}\).

Award 1 mark for the correct option C.
- 1 mark: Calculating that one mole of \((NH_4)_3PO_4\) contains 20 moles of atoms, and multiplying by the mole amount to get 2.0 moles of atoms, yielding \(1.2 \times 10^{24}\).

\(H_2O\) has 4 electron domains (tetrahedral arrangement) around the central oxygen atom: 2 bonding pairs and 2 lone pairs. The strong repulsion from the two lone pairs reduces the bond angle from the tetrahedral angle of \(109.5^\circ\) to approximately \(104.5^\circ\), resulting in a bent shape.

- \(NH_3\) has 1 lone pair and 3 bonding pairs, resulting in a trigonal pyramidal shape with a bond angle of \(107^\circ\).
- \(SO_2\) has 3 electron domains, resulting in a bent shape with a bond angle of approximately \(117^\circ\).
- \(CO_2\) has 2 electron domains (linear) with a bond angle of \(180^\circ\).

Award 1 mark for the correct option A.
- Distinguishes water as having a bent geometry with a bond angle of \(104.5^\circ\) due to its two lone pairs.

1. Determine oxidation states of reactants and products:
- In \(MnO_2\), oxygen is \(-2\), so manganese is \(+4\).
- Elemental \(Al\) is \(0\).
- Elemental \(Mn\) is \(0\).
- In \(Al_2O_3\), oxygen is \(-2\), so aluminum is \(+3\).
2. The reducing agent is the species that undergoes oxidation (loses electrons). Since the oxidation state of aluminum increases from \(0\) to \(+3\), \(Al\) is oxidized and is therefore the reducing agent.
3. The oxidation state of manganese decreases (is reduced) from \(+4\) in \(MnO_2\) to \(0\) in \(Mn\).

Award 1 mark for the correct option A.
- Correctly identifies \(Al (s)\) as the reducing agent because its oxidation state increases.
- Correctly determines the oxidation state of manganese changes from +4 to 0.

A conjugate acid-base pair consists of two species that differ by only a single proton (\(H^+\)). In this reaction, \(H_2PO_4^-\text{ (aq)}\) acts as an acid by donating a proton to form its conjugate base, \(HPO_4^{2-}\text{ (aq)}\). Therefore, \(H_2PO_4^-\text{ (aq)}\) and \(HPO_4^{2-}\text{ (aq)}\) are a conjugate acid-base pair.

Award 1 mark for the correct option C.
- Identifies the conjugate acid-base pair based on a difference of exactly one proton (\(H^+\)).

Increasing temperature increases the average kinetic energy of the reacting particles. This results in:
1. A higher velocity of the particles, which increases the frequency of collisions (more collisions per unit of time).
2. A significantly larger fraction of colliding particles that possess kinetic energy equal to or greater than the activation energy (\(E_a\)), leading to a higher proportion of successful collisions.
Note that activation energy itself is a constant for a given reaction pathway and does not change with temperature.

Award 1 mark for the correct option B.
- Correctly identifies that higher temperature increases both the frequency of collisions and the fraction of particles meeting the activation energy requirement.

(a) A homogeneous mixture has a uniform composition throughout and exists in a single phase (e.g., salt water, clean air, brass). A heterogeneous mixture has a non-uniform composition and contains separate phases (e.g., sand in water, oil and vinegar dressing, muddy water).

(b) (i) Arrangement: changes from a highly ordered, regular lattice structure to a random, highly dispersed arrangement where particles are far apart. Movement: changes from vibrating about fixed positions to rapid, random, straight-line motion.
(ii) London (dispersion) forces / instantaneous dipole-induced dipole forces.

(c) (i) Using the ideal gas equation: \( PV = nRT \)
Convert temperature to Kelvin: \( T = 25.0 + 273.15 = 298.15\text{ K} \) (or \( 298\text{ K} \)).
\( n = \frac{PV}{RT} = \frac{100\text{ kPa} \times 2.50\text{ dm}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 298.15\text{ K}} = \frac{250}{2477.6} \approx 0.101\text{ mol} \) (or \( 0.101\text{ mol} \) using \( 298\text{ K} \)).
(ii) Any two of:
- The volume of gas particles is negligible compared to the volume of the container.
- There are no intermolecular attractive forces between the particles.
- Collisions between gas particles, and between particles and container walls, are perfectly elastic.
- Gas particles are in continuous, rapid, random motion.

(a) Award [1 mark] for defining homogeneous mixture (uniform composition/single phase).
Award [1 mark] for defining heterogeneous mixture (non-uniform composition/multiple phases).
Award [1 mark] for one correct example of each.

(b) (i) Award [1 mark] for describing arrangement (regular/ordered to random/dispersed).
Award [1 mark] for describing movement (vibrations to rapid/random straight line motion).
(ii) Award [1 mark] for London (dispersion) forces / dispersion forces. Do NOT accept covalent bonds.

(c) (i) Award [1 mark] for correct substitution of values into \( n = \frac{PV}{RT} \) (temperature must be in K, i.e., 298 K).
Award [1 mark] for final answer \( 0.101\text{ mol} \) (accept range 0.100 - 0.101).
(ii) Award [1 mark] for each correct assumption stated, up to [2 marks] max.

(a) (i) Structural isomers are molecules with the same molecular formula but different arrangements of atoms / different bonding connectivity.
(ii) Carboxylic acid: Butanoic acid structure, \( \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} \) (or 2-methylpropanoic acid).
Ester: e.g., Ethyl ethanoate, \( \text{CH}_3\text{COOCH}_2\text{CH}_3 \) (or methyl propanoate, \( \text{CH}_3\text{CH}_2\text{COOCH}_3 \)).
(iii) Carboxylic acid: butanoic acid (or 2-methylpropanoic acid).
Ester: ethyl ethanoate (or methyl propanoate, propyl methanoate).

(b) (i) The carboxylic acid contains an \( -\text{O}-\text{H} \) bond, allowing it to form strong intermolecular hydrogen bonds. The ester lacks this bond and can only form weaker dipole-dipole and London dispersion forces. More thermal energy is required to break the hydrogen bonds in the carboxylic acid.
(ii) Class: Esters. Reactants (for ethyl ethanoate): ethanol and ethanoic acid (or the corresponding alcohol and carboxylic acid based on the ester drawn in (a)(ii)).

(c) A homologous series is a series of organic compounds with the same functional group / general formula, where each successive member differs by a \( -\text{CH}_2- \) unit.

(a) (i) Award [1 mark] for stating same molecular formula but different structures / arrangements of atoms.
(ii) Award [1 mark] for a correct drawing of a carboxylic acid with formula \( \text{C}_4\text{H}_8\text{O}_2 \).
Award [1 mark] for a correct drawing of an ester with formula \( \text{C}_4\text{H}_8\text{O}_2 \).
(iii) Award [1 mark] for the correct IUPAC name of the carboxylic acid drawn in (ii).
Award [1 mark] for the correct IUPAC name of the ester drawn in (ii).

(b) (i) Award [1 mark] for identifying that carboxylic acids form hydrogen bonds. Award [1 mark] for stating that esters cannot form hydrogen bonds (only dipole-dipole/London dispersion forces) and therefore require less energy to separate.
(ii) Award [1 mark] for identifying the class as esters. Award [1 mark] for naming the correct alcohol and carboxylic acid that correspond to the ester drawn in (a)(ii).

(c) Award [1 mark] for stating that successive members differ by a \( -\text{CH}_2- \) group / share the same functional group.

(a) (i) \( \text{PF}_3 \) Lewis structure: P is the central atom, single bonded to three F atoms. P has one lone pair. Each F atom has three lone pairs (8 valence electrons around each atom).
\( \text{BF}_3 \) Lewis structure: B is the central atom, single bonded to three F atoms. B has NO lone pairs. Each F atom has three lone pairs (B has an incomplete octet of 6 valence electrons).
(ii) \( \text{PF}_3 \): Geometry is trigonal pyramidal; bond angle is \( < 109.5^\circ \) (accept any value in the range \( 95^\circ - 107^\circ \)).
\( \text{BF}_3 \): Geometry is trigonal planar; bond angle is \( 120^\circ \).

(b) \( \text{PF}_3 \) is polar. The phosphorus atom has a lone pair of electrons which causes an asymmetrical molecular geometry (trigonal pyramidal). Therefore, the individual polar \( \text{P}-\text{F} \) bond dipoles do not cancel out, resulting in a net dipole moment.

(c) The fourth bond is a coordinate (dative covalent) bond. Both of the shared electrons in this bond are donated by the fluoride ion (\( \text{F}^- \)), which acts as a Lewis base. The electron-deficient boron atom (with an empty orbital) accepts this electron pair, acting as a Lewis acid.

(a) (i) Award [1 mark] for correct Lewis structure of \( \text{PF}_3 \) showing all bonding pairs and lone pairs (including 3 lone pairs on each F and 1 on P).
Award [1 mark] for correct Lewis structure of \( \text{BF}_3 \) showing no lone pairs on B and 3 lone pairs on each F.
(ii) Award [1 mark] for trigonal pyramidal for \( \text{PF}_3 \).
Award [1 mark] for predicting angle \( < 109.5^\circ \) (or a specific value between \( 95^\circ \) and \( 107^\circ \)) for \( \text{PF}_3 \).
Award [1 mark] for trigonal planar for \( \text{BF}_3 \).
Award [1 mark] for predicting \( 120^\circ \) for \( \text{BF}_3 \).

(b) Award [1 mark] for identifying \( \text{PF}_3 \) as polar.
Award [1 mark] for explaining that the shape is asymmetrical / the dipoles do not cancel out / the presence of the lone pair on the central atom causes asymmetry.

(c) Award [1 mark] for identifying the bond as a coordinate / dative covalent bond.
Award [1 mark] for explaining that both shared electrons are donated by the fluoride ion (\( \text{F}^- \)).

(a) (i) Molar mass of \( \text{Fe}_2\text{O}_3 = 2 \times 55.85 + 3 \times 16.00 = 159.70\text{ g mol}^{-1} \).
Amount of \( \text{Fe}_2\text{O}_3 = \frac{20.0\text{ g}}{159.70\text{ g mol}^{-1}} = 0.125\text{ mol} \).
(ii) Amount of \( \text{CO} = \frac{11.2\text{ dm}^3}{22.7\text{ dm}^3\text{ mol}^{-1}} = 0.493\text{ mol} \).

(b) (i) According to the stoichiometry, 1 mole of \( \text{Fe}_2\text{O}_3 \) reacts with 3 moles of \( \text{CO} \).
For \( 0.125\text{ mol} \) of \( \text{Fe}_2\text{O}_3 \), the amount of \( \text{CO} \) required is \( 3 \times 0.125 = 0.375\text{ mol} \).
Since \( 0.493\text{ mol} \) of \( \text{CO} \) is present, which is greater than \( 0.375\text{ mol} \), \( \text{CO} \) is in excess and \( \text{Fe}_2\text{O}_3 \) is the limiting reactant.
(ii) The ratio of \( \text{Fe}_2\text{O}_3 \) to \( \text{Fe} \) is 1:2.
Moles of \( \text{Fe} \) produced = \( 2 \times 0.125\text{ mol} = 0.250\text{ mol} \).
Mass of \( \text{Fe} = 0.250\text{ mol} \times 55.85\text{ g mol}^{-1} = 14.0\text{ g} \) (or \( 13.96\text{ g} \)).

(c) (i) Percentage yield = \( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% = \frac{11.5\text{ g}}{13.96\text{ g}} \times 100\% = 82.4\% \) (or \( \frac{11.5}{14.0} \times 100\% = 82.1\% \)).
(ii) The reaction may not have gone to completion, some iron could be lost during separation/purification, or side reactions may have occurred.

(a) (i) Award [1 mark] for calculating the correct molar mass of \( \text{Fe}_2\text{O}_3 \) (\( 159.70\text{ g mol}^{-1} \)). Award [1 mark] for the amount of \( 0.125\text{ mol} \) (accept range 0.125 - 0.126).
(ii) Award [1 mark] for \( 0.493\text{ mol} \) (accept range 0.493 - 0.494).

(b) (i) Award [1 mark] for showing a valid comparison (e.g., comparing required vs. available moles, or mole ratios of reactants).
Award [1 mark] for concluding that \( \text{Fe}_2\text{O}_3 \) is the limiting reactant.
(ii) Award [1 mark] for moles of iron = \( 0.250\text{ mol} \) (ECF from (b)(i)).
Award [1 mark] for mass of iron = \( 14.0\text{ g} \) or \( 13.96\text{ g} \) (accept range 13.9 - 14.0).

(c) (i) Award [1 mark] for correct setup of percentage yield. Award [1 mark] for final answer of \( 82.4\% \) (or \( 82.1\% \)) (ECF from (b)(ii)).
(ii) Award [1 mark] for any reasonable suggestion (e.g., incomplete reaction, product lost during transfer, side reactions).

(a) (i) In \( \text{ZnSO}_4 \), the oxidation state of Zn is \( +2 \) and oxygen is \( -2 \). Since the compound is neutral: \( (+2) + \text{S} + 4(-2) = 0 \implies \text{S} = +6 \) (or \( \text{VI} \)).
(ii) The reaction at the anode is: \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^- \). This is oxidation.

(b) (i) Anode half-reaction: \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^- \)
Cathode half-reaction: \( \text{Ag}^+\text{(aq)} + \text{e}^- \rightarrow \text{Ag(s)} \)
Multiplying the cathode half-reaction by 2 and adding both gives the overall balanced ionic equation:
\( \text{Zn(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{Ag(s)} \).
(ii) Electrons flow through the external wire from the zinc electrode (anode) to the silver electrode (cathode).

(c) (i) The salt bridge completes the electrical circuit and maintains electrical neutrality in both half-cells by allowing anions to migrate toward the anode half-cell and cations to migrate toward the cathode half-cell.
(ii) The mass of the silver electrode increases because silver ions (\( \text{Ag}^+ \)) in the solution are reduced to solid silver metal (\( \text{Ag(s)} \)), which is deposited on the surface of the silver electrode.

(a) (i) Award [1 mark] for \( +6 \) or \( \text{VI} \).
(ii) Award [1 mark] for the correct half-equation: \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^- \) (state symbols not required). Award [1 mark] for identifying it as oxidation.

(b) (i) Award [1 mark] for correct species. Award [1 mark] for correct balancing.
(ii) Award [1 mark] for stating that electrons flow from the zinc electrode (anode) to the silver electrode (cathode).

(c) (i) Award [1 mark] for completing the electrical circuit. Award [1 mark] for maintaining electrical neutrality / preventing charge build-up by migration of ions.
(ii) Award [1 mark] for stating that the mass increases. Award [1 mark] for explaining that \( \text{Ag}^+ \) ions are reduced / deposit as solid silver on the electrode.

**(a)**
* Mass of anhydrous \(\text{CaCl}_2\) = \(14.56\text{ g} - 9.21\text{ g} = 5.35\text{ g}\)
* Uncertainty in mass: \(\pm (0.01\text{ g} + 0.01\text{ g}) = \pm 0.02\text{ g}\)
* Mass of \(\text{CaCl}_2 = 5.35 \pm 0.02\text{ g}\)

**(b)**
* Assumption 1: The density of the solution is equal to pure water (\(1.00\text{ g cm}^{-3}\)), meaning the mass of the solution is assumed to be exactly \(50.0\text{ g}\) (or the mass of the solute does not affect the mass of the solution used in the calculation).
* Assumption 2: The specific heat capacity of the solution is equal to that of pure water (\(4.18\text{ J g}^{-1}\text{ K}^{-1}\)).

**(c)**
* Temperature change, \(\Delta T = 38.4\text{ }^\circ\text{C} - 21.2\text{ }^\circ\text{C} = 17.2\text{ }^\circ\text{C}\) (or \(17.2\text{ K}\))
* Using \(m = 50.0\text{ g}\):
\(q = m \cdot c \cdot \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 17.2\text{ K} = 3594.8\text{ J} = 3.59\text{ kJ}\)
* *Alternative (using total mass of solution \(m = 50.0\text{ g} + 5.35\text{ g} = 55.35\text{ g}\)):
\(q = 55.35\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 17.2\text{ K} = 3980\text{ J} = 3.98\text{ kJ}\)*

**(d)**
* Amount in moles, \(n\) of \(\text{CaCl}_2\) = \(\frac{5.35\text{ g}}{110.98\text{ g mol}^{-1}} = 0.0482\text{ mol}\)

**(e)**
* Enthalpy of solution, \(\Delta H_{\text{sol}} = -\frac{q}{n}\)
* Because the temperature increased, the reaction is exothermic and \(\Delta H_{\text{sol}}\) must be negative.
* Using \(q = 3.5948\text{ kJ}\):
\(\Delta H_{\text{sol}} = -\frac{3.5948\text{ kJ}}{0.048207\text{ mol}} = -74.6\text{ kJ mol}^{-1}\)
* *Alternative (using \(q = 3.980\text{ kJ}\)):
\(\Delta H_{\text{sol}} = -\frac{3.980\text{ kJ}}{0.048207\text{ mol}} = -82.6\text{ kJ mol}^{-1}\)*

**(f)**
* Absolute uncertainty in temperature change \(\Delta T = 0.1\text{ }^\circ\text{C} + 0.1\text{ }^\circ\text{C} = 0.2\text{ }^\circ\text{C}\)
* Percentage uncertainty in \(\Delta T = \frac{0.2}{17.2} \times 100\% = 1.16\% \approx 1.2\%\)
* Percentage uncertainty in Volume = \(\frac{0.5}{50.0} \times 100\% = 1.0\%\)
* Comparing the two, the temperature change measurement (\(1.16\%\)) contributes more to the overall experimental uncertainty than the volume of water (\(1.0\%\)).

**(g)**
* Percentage error calculation (using standard value of \(-74.6\text{ kJ mol}^{-1}\)):
\(\text{Percentage error} = \left| \frac{-74.6 - (-81.3)}{-81.3} \right| \times 100\% = 8.24\% \approx 8.2\%\)
*(Alternative percentage error if using \(-82.6\text{ kJ mol}^{-1}\): \(\left| \frac{-82.6 - (-81.3)}{-81.3} \right| \times 100\% = 1.6\%\))*
* Systematic error: Heat loss to the surroundings / calorimeter lid was open / some heat was absorbed by the polystyrene cup.
* Modification: Use a double-walled polystyrene cup with a tight-fitting lid to minimize heat transfer OR record the temperature at regular intervals and plot a temperature-time cooling curve to extrapolate the true maximum temperature at the moment of mixing.

**(a) [2 marks]**
* \(5.35\text{ g}\) [1]
* \(\pm 0.02\text{ g}\) [1]

**(b) [2 marks]**
* Density of solution is equal to pure water / is \(1.00\text{ g cm}^{-3}\) (or mass of solution is equal to mass of water) [1]
* Specific heat capacity of solution is equal to that of pure water / is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\) [1]

**(c) [2 marks]**
* Correct temperature change: \(\Delta T = 17.2\text{ K}\) (or \(17.2\text{ }^\circ\text{C}\)) [1]
* Correct calculation of \(q\): \(3.59\text{ kJ}\) (or \(3.59 \times 10^3\text{ J}\)) OR \(3.98\text{ kJ}\) (if using total mass \(55.35\text{ g}\)) [1]

**(d) [1 mark]**
* \(0.0482\text{ mol}\) (accept values in the range \(0.0481\) to \(0.0483\)) [1]

**(e) [2 marks]**
* Negative sign (indicating an exothermic reaction) [1]
* Magnitude: \(74.6\text{ kJ mol}^{-1}\) (or \(82.6\text{ kJ mol}^{-1}\) if total mass was used in part c) [1] *(Accept error carried forward from previous steps)*

**(f) [3 marks]**
* Percentage uncertainty in \(\Delta T = 1.16\%\) (accept \(1.2\%\)) [1] *(Note: must use absolute uncertainty of \(0.2\text{ }^\circ\text{C}\) in temperature change)*
* Percentage uncertainty in Volume = \(1.0\%\) [1]
* Identifies temperature change (\(\Delta T\)) as contributing more to uncertainty, supported by the calculated percentages [1]

**(g) [3 marks]**
* Percentage error: \(8.2\%\) (accept \(8.24\%\)) OR \(1.6\%\) (if alternative value was calculated in part e) [1]
* Systematic error: Heat lost to surroundings / polystyrene cup absorbed heat [1]
* Modification: Use better insulation (e.g., lid / vacuum flask / double-cup calorimeter) OR record temperature over time and perform a cooling curve extrapolation back to the time of mixing [1]

(a) Reagent: ethanoic anhydride (or acetyl chloride) [1]; Catalyst: concentrated sulfuric acid, \(\text{H}_2\text{SO}_4\) (or concentrated phosphoric acid, \(\text{H}_3\text{PO}_4\)) [1]. (b) Dissolve the crude aspirin in the minimum volume of hot solvent [1]. Allow the solution to cool slowly so that pure aspirin crystals precipitate out while soluble impurities remain in the solvent, then filter and wash the crystals [1]. (c) The crude product would have a lower melting point [1] and a wider melting range/interval [1] than pure aspirin.

Part (a): Award [1] for ethanoic anhydride or acetyl chloride (accept acetic anhydride). Award [1] for concentrated sulfuric acid or concentrated phosphoric acid. Part (b): Award [1] for dissolving crude product in the minimum volume of hot solvent. Award [1] for cooling the solution to recrystallize the pure product and filtering/washing. Part (c): Award [1] for lower melting point. Award [1] for broader/wider melting range.

(a) \(\text{Mg(OH)}_2(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + 2\text{H}_2\text{O}(\text{l})\) [1]. (b) Amount of \(\text{HCl}\): \(n(\text{HCl}) = 0.1500\text{ dm}^3 \times 0.0800\text{ mol dm}^{-3} = 0.0120\text{ mol}\) [1]. Amount of \(\text{Mg(OH)}_2\): \(n(\text{Mg(OH)}_2) = \frac{0.0120\text{ mol}}{2} = 0.00600\text{ mol}\) [1]. Molar mass of \(\text{Mg(OH)}_2 = 58.33\text{ g mol}^{-1}\), so mass = \(0.00600\text{ mol} \times 58.33\text{ g mol}^{-1} = 0.350\text{ g}\) [1]. (c) Ranitidine is an \(\text{H}_2\)-receptor antagonist that blocks histamine receptors in the parietal cells, reducing acid production [1]. Omeprazole is a proton pump inhibitor that directly inhibits the \(\text{H}^+/\text{K}^+\)-ATPase enzyme system (proton pump) [1], preventing the secretion of \(\text{H}^+\) ions into the stomach [1].

Part (a): Award [1] for correct balanced equation. Part (b): Award [1] for \(n(\text{HCl}) = 0.0120\text{ mol}\). Award [1] for \(n(\text{Mg(OH)}_2) = 0.00600\text{ mol}\). Award [1] for mass of \(0.350\text{ g}\) (accept 0.35 g). Part (c): Award [1] for ranitidine blocks \(\text{H}_2\)/histamine receptors. Award [1] for omeprazole inhibits proton pump / \(\text{H}^+/\text{K}^+\)-ATPase. Award [1] for preventing secretion of \(\text{H}^+\) ions.

(a) Both drugs act as neuraminidase inhibitors [1]. They bind to the active site of the neuraminidase enzyme, preventing it from releasing newly formed viral particles from the host cell [1]. (b) (i) Carboxamide / ether / alkenyl group [1]. (ii) Hydroxyl / carboxylic acid / guanidino group [1]. (iii) Ester / primary amino group [1]. (c) Overuse or improper disposal can lead to the selection and development of antiviral-resistant strains of the virus [1], making these drugs ineffective in treating future outbreaks [1].

Part (a): Award [1] for neuraminidase inhibitor. Award [1] for preventing release/spread of viral particles. Part (b): (i) Award [1] for carboxamide (amide) OR ether OR alkenyl group (alkene). (ii) Award [1] for hydroxyl (alcohol) OR carboxylic acid OR guanidino. (iii) Award [1] for ester OR primary amino. Part (c): Award [1] for development of drug resistance. Award [1] for making medications ineffective / threat of uncontrollable pandemics.