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From the given quantities, \(0.10\text{ mol}\) of hydrocarbon yields \(0.40\text{ mol}\) of \(\text{CO}_2\), which corresponds to \(0.40\text{ mol}\) of carbon atoms. The same amount yields \(0.50\text{ mol}\) of \(\text{H}_2\text{O}\), which corresponds to \(1.00\text{ mol}\) of hydrogen atoms. Therefore, \(1\text{ mol}\) of the hydrocarbon contains \(4\text{ mol}\) of carbon and \(10\text{ mol}\) of hydrogen. This gives a molecular formula of \(\text{C}_4\text{H}_{10}\). The empirical formula is the simplest whole-number ratio, which is \(\text{C}_2\text{H}_5\).

Award [1] for the correct answer B. Deduce the mole ratio of C to H atoms (4:10) to find the molecular formula (C4H10) and simplify to find the empirical formula (C2H5).

\(\text{NH}_4^+\) has a tetrahedral shape with 4 bonding pairs and 0 lone pairs, giving a bond angle of \(109.5^\circ\). \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair, with a trigonal pyramidal shape and a bond angle of approximately \(107^\circ\). \(\text{NH}_2^-\right) has 2 bonding pairs and 2 lone pairs, with a bent shape and a bond angle of approximately \)104.5^\circ\). \(\text{BF}_3\) has a trigonal planar shape with a bond angle of \(120^\circ\). Therefore, \(\text{NH}_2^-\right) has the smallest bond angle due to increased lone pair-lone pair repulsion.

Award [1] for correct answer C. Correctly identify the electron domain geometry and the effect of lone pairs on bond angles for each species.

Let the initial rate be \(\text{Rate}_1 = k[\text{A}][\text{B}]^2\). If \([\text{A}]\) becomes \(0.5[\text{A}]\) and \([\text{B}]\) becomes \(2[\text{B}]\), the new rate is \(\text{Rate}_2 = k(0.5[\text{A}])(2[\text{B}])^2 = k \cdot 0.5[\text{A}] \cdot 4[\text{B}]^2 = 2k[\text{A}][\text{B}]^2 = 2\text{Rate}_1\). Therefore, the rate is doubled.

Award [1] for B. Use the rate law to substitute the changes in concentration and calculate the overall multiplier.

A conjugate acid-base pair consists of two species that differ by a single proton \((\text{H}^+)\). In this reaction, \(\text{H}_2\text{PO}_4^-\right) acts as an acid by donating a proton to form its conjugate base, \)\text{HPO}_4^{2-}\). Therefore, \(\text{H}_2\text{PO}_4^-\right) and \)\text{HPO}_4^{2-}\) represent a conjugate acid-base pair.

Award [1] for C. Identify conjugate partners by finding species that differ by exactly one H+ ion.

In \(\text{SF}_4\), S is +4. In \(\text{Na}_2\text{S}_2\text{O}_3\), S has an average oxidation state of +2. In \(\text{H}_2\text{SO}_3\), S is +4. In \(\text{SO}_2\text{Cl}_2\), S is +6 (O is -2, Cl is -1, so \(x + 2(-2) + 2(-1) = 0\), which gives \(x = +6\)). Therefore, sulfur has the highest oxidation state in \(\text{SO}_2\text{Cl}_2\).

Award [1] for D. Calculate the oxidation states of sulfur in each option to identify the highest state (+6).

The electron configuration of a neutral Fe atom is [Ar] 3d^6 4s^2. When forming the \(\text{Fe}^{2+}\) ion, the two 4s electrons are lost first, giving the configuration [Ar] 3d^6. According to Hund's rule, the five 3d orbitals are occupied as follows: one orbital contains a paired set of electrons, and the other four orbitals each contain a single unpaired electron. Thus, there are 4 unpaired electrons.

Award [1] for B. State the correct electron configuration for Fe2+ (leaving 3d6) and apply Hund's rule to find the number of unpaired electrons.

The longest carbon chain contains 4 carbon atoms with an aldehyde group at carbon-1 (taking priority). This gives the parent suffix '-al' (butanal). A hydroxyl group (-\(\text{OH}\)) is attached to carbon-3. Thus, the correct name is 3-hydroxybutanal.

Award [1] for A. Identify the principal functional group (aldehyde) to establish carbon numbering and name the substituent group with the correct position prefix.

Using the ideal gas relationship \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\), we have \(\frac{P V}{T} = \frac{(0.5P) V_2}{2T}\). Solving for \(V_2\) gives \(V_2 = \frac{2 T \cdot P V}{0.5 P \cdot T} = 4V\).

Award [1] for D. Set up the combined gas law or ideal gas equation and solve for the new volume using the proportional changes.

To determine the molecular geometry, we first count the valence shell electron pairs on the central atom:

- In \(\text{NH}_4^+\), nitrogen has 4 bonding pairs and 0 lone pairs. The geometry is tetrahedral.
- In \(\text{XeF}_4\), xenon has 8 valence electrons, plus 4 from fluorine, giving 12 electrons (6 electron domains). There are 4 bonding pairs and 2 lone pairs. To minimize repulsion, the 2 lone pairs occupy axial positions, resulting in a square planar molecular geometry.
- In \(\text{SF}_4\), sulfur has 6 valence electrons, plus 4 from fluorine, giving 10 electrons (5 electron domains). There are 4 bonding pairs and 1 lone pair, resulting in a see-saw molecular geometry.
- In \(\text{PCl}_4^+\), phosphorus has 4 bonding pairs and 0 lone pairs. The geometry is tetrahedral.

Award 1 mark for the correct option (B).

A conjugate acid-base pair consists of two species that differ by exactly one hydrogen ion (proton, \(\text{H}^+\)).

- \(\text{NH}_4^+\) (acid) can lose a proton to form \(\text{NH}_3\) (conjugate base): \(\text{NH}_4^+ \rightleftharpoons \text{NH}_3 + \text{H}^+\). These differ by exactly one proton.
- In options A, B, and D, the species differ by two protons (e.g., \(\text{H}_2\text{SO}_4\) and \(\text{SO}_4^{2-}\)), which does not constitute a conjugate acid-base pair.

Award 1 mark for the correct option (C).

1. Assume a \(100\text{ g}\) sample of the hydrocarbon:
- Mass of \(\text{C} = 85.7\text{ g}\)
- Mass of \(\text{H} = 14.3\text{ g}\)

2. Calculate the number of moles of each element:
- Moles of \(\text{C} = \frac{85.7\text{ g}}{12.01\text{ g mol}^{-1}} \approx 7.14\text{ mol}\)
- Moles of \(\text{H} = \frac{14.3\text{ g}}{1.01\text{ g mol}^{-1}} \approx 14.16\text{ mol}\)

3. Determine the simplest whole-number ratio:
- Ratio of \(\text{C} : \text{H} = \frac{7.14}{7.14} : \frac{14.16}{7.14} \approx 1 : 2\)

Therefore, the empirical formula is \(\text{CH}_2\).

Award 1 mark for the correct option (B).

When the temperature is increased, the average kinetic energy of the molecules increases. On the Maxwell-Boltzmann distribution:
- The curve flattens out, meaning the peak shifts to the right (higher energy) and becomes lower.
- The activation energy (\(E_a\)) remains constant, but a significantly greater fraction of molecules have kinetic energy greater than or equal to \(E_a\). This leads to a higher rate of successful collisions, and thus an increased reaction rate.

Award 1 mark for the correct option (B).

Let us analyze the structure \(\text{HO-CH}_2\text{-CH}_2\text{-CO-OCH}_3\):
- The \(\text{HO-}\) group attached to a saturated carbon atom is a hydroxyl group, which classifies the compound as an **alcohol**.
- The \(\text{-CO-OCH}_3\) group (or \(\text{-COO-}\) linked to an alkyl group) is an **ester** group.

Therefore, the functional groups present are alcohol and ester.

Award 1 mark for the correct option (C).

Iron (\(\text{Fe}\)) has an atomic number of 26.
1. The ground-state electron configuration of neutral \(\text{Fe}\) is \([\text{Ar}] 4\text{s}^2 3\text{d}^6\).
2. When transition metals form ions, they lose electrons from the highest principal energy level orbital first, which is the \(4\text{s}\) orbital, followed by the \(3\text{d}\) orbitals.
3. To form the \(\text{Fe}^{3+}\) ion, three electrons must be removed: two from the \(4\text{s}\) orbital and one from the \(3\text{d}\) orbital.
4. This yields the electron configuration \([\text{Ar}] 3\text{d}^5\).

Award 1 mark for the correct option (C).

1. For dichromate, \(\text{Cr}_2\text{O}_7^{2-}\):
- Let the oxidation state of Cr be \(x\).
- Oxygen is typically \(-2\).
- \(2x + 7(-2) = -2 \Rightarrow 2x - 14 = -2 \Rightarrow 2x = +12 \Rightarrow x = +6\).

2. For permanganate, \(\text{MnO}_4^-\):
- Let the oxidation state of Mn be \(y\).
- Oxygen is typically \(-2\).
- \(y + 4(-2) = -1 \Rightarrow y - 8 = -1 \Rightarrow y = +7\).

Therefore, the oxidation states are \(\text{Cr} = +6\) and \(\text{Mn} = +7\).

Award 1 mark for the correct option (B).

Using the combined gas law for a fixed mass of gas:
\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]

Rearrange to solve for the final volume \(V_2\):
\[V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}\]

Substitute the given values:
\[V_2 = \frac{100\text{ kPa} \times 2.0\text{ dm}^3 \times 600\text{ K}}{200\text{ kPa} \times 300\text{ K}}\]
\[V_2 = \frac{120000}{60000} = 2.0\text{ dm}^3\]

Alternatively, doubling the pressure halves the volume (to \(1.0\text{ dm}^3\)), and doubling the absolute temperature doubles the volume (returning it to \(2.0\text{ dm}^3\)).

Award 1 mark for the correct option (B).

To determine the geometry, we look at the central atom's valence electrons and bonding domains:
- For \(\text{NO}_2^-\), nitrogen is the central atom (5 valence electrons + 1 from negative charge = 6 electrons). It forms two bonding domains with two oxygen atoms and has one lone pair. With 3 electron domains, its electron domain geometry is trigonal planar. The presence of one lone pair gives it a bent molecular geometry.
- For \(\text{H}_2\text{O}\), oxygen has 4 electron domains (2 bonding, 2 lone pairs), resulting in tetrahedral electron domain geometry and bent molecular geometry.
- For \(\text{CO}_2\), carbon has 2 electron domains, resulting in linear geometries.
- For \(\text{OF}_2\), oxygen has 4 electron domains, resulting in tetrahedral electron domain geometry and bent molecular geometry.

Award [1] for the correct option A.
Award [0] for any other option.

A Brønsted-Lowry base is defined as a proton (\(\text{H}^+\)) acceptor:
- In the forward reaction, \(\text{CO}_3^{2-}\) accepts a proton from \(\text{H}_2\text{PO}_4^-\) to become \(\text{HCO}_3^-\). Thus, \(\text{CO}_3^{2-}\) is the base.
- In the reverse reaction, \(\text{HPO}_4^{2-}\) accepts a proton from \(\text{HCO}_3^-\cap\) to become \(\text{H}_2\text{PO}_4^-\). Thus, \(\text{HPO}_4^{2-}\) is the base.

Award [1] for the correct option A.
Award [0] for any other option.

Cobalt has an atomic number of 27. The ground-state electron configuration of a neutral cobalt atom is \([\text{Ar}] 4\text{s}^2 3\text{d}^7\). When transition metals form cations, they lose electrons from the outermost \(\text{s}\) subshell (4s) before the \(\text{d}\) subshell (3d). Therefore, losing 2 electrons to form \(\text{Co}^{2+}\) results in the configuration \([\text{Ar}] 3\text{d}^7\).

Award [1] for the correct option B.
Award [0] for any other option.

First, count the number of atoms in one formula unit of \(\text{(NH}_4\text{)}_2\text{SO}_4\):
- Nitrogen: \(2 \times 1 = 2\)
- Hydrogen: \(2 \times 4 = 8\)
- Sulfur: \(1 \times 1 = 1\)
- Oxygen: \(4 \times 1 = 4\)
Total atoms per formula unit = \(2 + 8 + 1 + 4 = 15\).
For \(0.10\text{ mol}\) of \(\text{(NH}_4\text{)}_2\text{SO}_4\), the total amount of atoms is \(0.10\text{ mol} \times 15 = 1.5\text{ mol}\).
Using the Avogadro constant \(L\), the number of atoms is \(1.5 L\).

Award [1] for the correct option C.
Award [0] for any other option.

Let the initial rate be \(\text{Rate}_1 = k [\text{NO}]^2 [\text{H}_2]\).
If \([\text{NO}]\) is halved, it becomes \(0.5[\text{NO}]\). If \([\text{H}_2]\) is tripled, it becomes \(3[\text{H}_2]\).
Substituting these into the rate expression:
\(\text{Rate}_2 = k (0.5 [\text{NO}])^2 (3 [\text{H}_2]) = k (0.25 [\text{NO}]^2) (3 [\text{H}_2]) = 0.75 k [\text{NO}]^2 [\text{H}_2] = 0.75 \times \text{Rate}_1\).
Thus, the rate changes by a factor of 0.75.

Award [1] for the correct option A.
Award [0] for any other option.

Analyzing the structural formula \(\text{CH}_3\text{CH(OH)CH}_2\text{COOCH}_3\):
- The \(\text{-OH}\) group attached to a carbon atom represents a hydroxyl group (characteristic of an alcohol).
- The \(\text{-COO-}\) linkage (specifically \(\text{-COOCH}_3\)) represents an ester group.
Therefore, the two functional groups present are hydroxyl (alcohol) and ester.

Award [1] for the correct option A.
Award [0] for any other option.

We use the combined gas law: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\).
First, convert the temperatures from Celsius to Kelvin:
- \(T_1 = 27 + 273 = 300\text{ K}\)
- \(T_2 = 327 + 273 = 600\text{ K}\)
Now rearrange the equation to solve for \(V_2\):
\(V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}\)
\(V_2 = 2.0 \times \frac{100}{200} \times \frac{600}{300} = 2.0 \times 0.5 \times 2 = 2.0\text{ dm}^3\).

Award [1] for the correct option B.
Award [0] for any other option.

Determine the oxidation state of nitrogen in the reactant and product:
- In \(\text{HNO}_3\): Hydrogen is \(+1\) and oxygen is \(-2\). Let \(x\) be the oxidation state of nitrogen. Then, \(+1 + x + 3(-2) = 0 \Rightarrow x - 5 = 0 \Rightarrow x = +5\).
- In \(\text{NO}\): Oxygen is \(-2\). Let \(y\) be the oxidation state of nitrogen. Then, \(y + (-2) = 0 \Rightarrow y = +2\).
Therefore, the oxidation state of nitrogen changes from \(+5\) to \(+2\).

Award [1] for the correct option A.
Award [0] for any other option.

Chlorine trifluoride (\(\text{ClF}_3\)) has 5 electron domains around the central chlorine atom (3 bonding pairs and 2 lone pairs), giving a trigonal bipyramidal electron domain geometry. To minimize lone pair-lone pair repulsion, the lone pairs occupy equatorial positions, resulting in a T-shaped molecular geometry. \(\text{BF}_3\) and \(\text{SO}_3\) are trigonal planar, while \(\text{NF}_3\) is trigonal pyramidal.

[1 mark] Award for selecting correct option A. No partial marks.

The second reaction is the reverse of the first reaction, and its coefficients are divided by 2 (or multiplied by \(\frac{1}{2}\)). Thus, its equilibrium constant is the reciprocal of the first reaction's constant raised to the power of \(\frac{1}{2}\): \(K_2 = \left(\frac{1}{K_1}\right)^{1/2} = \frac{1}{\sqrt{K_1}}\).

[1 mark] Award for selecting option C. No partial marks.

A conjugate acid-base pair consists of two species that differ by a single proton (\(\text{H}^+\)). In this reaction, \(\text{H}_2\text{PO}_4^-\) acts as an acid by donating a proton to become its conjugate base, \(\text{HPO}_4^{2-}\). Therefore, \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) represent a conjugate acid-base pair.

[1 mark] Award for selecting option A.

Assume a 100 g sample of the oxide: mass of \(\text{N} = 30.4\text{ g}\), mass of \(\text{O} = 100 - 30.4 = 69.6\text{ g}\). Number of moles of \(\text{N} = \frac{30.4}{14.01} \approx 2.17\text{ mol}\). Number of moles of \(\text{O} = \frac{69.6}{16.00} \approx 4.35\text{ mol}\). Divide by the smaller value to find the ratio: \(\text{N} = \frac{2.17}{2.17} = 1\), \(\text{O} = \frac{4.35}{2.17} \approx 2\). The empirical formula is \(\text{NO}_2\).

[1 mark] Award for selecting option B.

We can find the target reaction by combining the equations using Hess's Law: target = Reaction(1) - 0.5 * Reaction(2). Enthalpy change: \(\Delta H = -394 - (0.5 \times -566) = -394 + 283 = -111\text{ kJ}\).

[1 mark] Award for selecting option A.

The oxidizing agent is the species that is reduced (gains electrons, decrease in oxidation state). Here, the oxidation state of Mn in \(\text{MnO}_4^-\) is +7, which decreases to +2 in \(\text{Mn}^{2+}\). Therefore, \(\text{MnO}_4^-\) is reduced and acts as the oxidizing agent.

[1 mark] Award for selecting option B.

A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_{\text{a}}\)). Since the rate constant \(k\) depends exponentially on the activation energy through the Arrhenius equation \(k = Ae^{-E_{\text{a}}/RT}\), lowering \(E_{\text{a}}\) increases the value of \(k\) at a constant temperature.

[1 mark] Award for selecting option B.

Amino acids like glycine (\(\text{H}_2\text{NCH}_2\text{COOH}\)) contain an amino group (\(-\text{NH}_2\)) and a carboxyl group (\(-\text{COOH}\)). Option B is an amide, option C is a primary amine, and option D is an amino alcohol containing a hydroxyl group and an amino group.

[1 mark] Award for selecting option A.

1. Calculate the moles of each element in a 100 g sample: Moles of carbon = \(85.6 \text{ g} / 12.01 \text{ g mol}^{-1} = 7.13 \text{ mol}\). Moles of hydrogen = \(14.4 \text{ g} / 1.01 \text{ g mol}^{-1} = 14.26 \text{ mol}\). 2. Find the simplest whole number ratio: Carbon : Hydrogen = \(7.13 / 7.13 : 14.26 / 7.13 = 1 : 2\). Therefore, the empirical formula is \(CH_2\).

Award [1] for the correct answer B. Award [0] for any other response.

The carbon-oxygen bond length depends on the bond order. In \(CO\), there is a triple bond (bond order 3). In \(CO_2\), there are double bonds (bond order 2). In \(CO_3^{2-}\), resonance gives a bond order of 1.33. In \(CH_3OH\), there is a single bond (bond order 1). Higher bond order results in stronger and shorter bonds, so \(CO\) has the shortest carbon-oxygen bond length.

Award [1] for the correct answer A. Award [0] for any other response.

According to the Bronsted-Lowry theory, a conjugate base is formed when an acid loses a proton (\(H^+\)). Removing \(H^+\) from \(HPO_4^{2-}\) yields the phosphate ion, \(PO_4^{3-}\).

Award [1] for the correct answer B. Award [0] for any other response.

The ground-state electron configuration of a neutral iron atom is \([Ar] 4s^2 3d^6\). When forming the \(Fe^{2+}\) transition metal cation, electrons are lost from the outer \(4s\) orbital before the \(3d\) orbitals. Therefore, two electrons are removed from the \(4s\) subshell, leaving \([Ar] 3d^6\).

Award [1] for the correct answer A. Award [0] for any other response.

Step 2 is the rate-determining step, so the rate expression is initially written as \(\text{Rate} = k_2[N_2O_2][O_2]\). Since \(N_2O_2\) is an unstable intermediate, its concentration is determined from the fast pre-equilibrium Step 1: \(K_c = \frac{[N_2O_2]}{[NO]^2}\), which gives \([N_2O_2] = K_c[NO]^2\). Substituting this back into the rate-determining step rate expression gives: \(\text{Rate} = k_2 K_c [NO]^2 [O_2] = k [NO]^2 [O_2]\).

Award [1] for the correct answer C. Award [0] for any other response.

1. Calculate heat absorbed by water: \(q = m c \Delta T = 50.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 25.0 \text{ K}\) (since \(\Delta T = 45.0 - 20.0 = 25.0\)). 2. Convert Joules to kilojoules by dividing by 1000: \(q \text{ (kJ)} = \frac{50.0 \times 4.18 \times 25.0}{1000}\). 3. Moles of methanol used: \(n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.40}{32.05}\). 4. Enthalpy of combustion (which is exothermic, hence negative sign) is given by \(\Delta H_c = -\frac{q}{n} = -\frac{50.0 \times 4.18 \times 25.0}{1000 \times (0.40 / 32.05)}\).

Award [1] for the correct answer A. Award [0] for any other response.

For \(K_2Cr_2O_7\): potassium has an oxidation state of \(+1\) and oxygen has \(-2\). Thus, \(2(+1) + 2(\text{Cr}) + 7(-2) = 0 \Rightarrow 2 + 2(\text{Cr}) - 14 = 0 \Rightarrow 2(\text{Cr}) = 12 \Rightarrow \text{Cr} = +6\). For \([Fe(H_2O)_6]^{3+}\): water is a neutral ligand (charge of 0). The total charge of the complex is \(+3\), so the oxidation state of iron is \(+3\).

Award [1] for the correct answer B. Award [0] for any other response.

The carboxyl group (\(-COOH\)) is characteristic of carboxylic acids. The hydroxyl group (\(-OH\)) is present in alcohols/phenols. The methoxy group (\(-OCH_3\)) contains an oxygen atom bonded to two carbon atoms (one from the benzene ring and one from the methyl group), which classifies it as an ether group. Therefore, the groups present are carboxyl, hydroxyl, and ether.

Award [1] for the correct answer B. Award [0] for any other response.

(a) Total volume of mixture = \(100.0 \text{ cm}^3\), so mass \(m = 100.0 \text{ g}\). Temperature change \(\Delta T = 26.8 - 20.2 = 6.6 \text{ K}\). Heat energy \(q = m c \Delta T = 100.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 6.6 \text{ K} = 2758.8 \text{ J} = 2.76 \text{ kJ}\). (b) Moles of \(\text{HCl} = 0.0500 \text{ dm}^3 \times 1.00 \text{ mol dm}^{-3} = 0.0500 \text{ mol}\), moles of \(\text{NaOH} = 0.0500 \text{ mol}\). Both reactants are in a 1:1 stoichiometric ratio, so they are both fully consumed. Moles of water formed = \(0.0500 \text{ mol}\). Enthalpy of neutralization \(\Delta H_{\text{neu}} = -q / n = -2.7588 \text{ kJ} / 0.0500 \text{ mol} = -55.2 \text{ kJ mol}^{-1}\). (c) Temperature change is derived from two readings, so absolute uncertainty = \(0.1 + 0.1 = \pm 0.2^\circ\text{C}\). Percentage uncertainty = \((0.2 / 6.6) \times 100\% = 3.03\%\). (d) Modification: Use a tight-fitting plastic lid on the polystyrene cup (or double-nest the cups). Explanation: This minimizes heat loss to the surroundings, resulting in a higher measured maximum temperature, a larger calculated \(\Delta T\), and therefore a more exothermic (more negative) value for \(\Delta H_{\text{neu}}\).

Part (a): [3 marks] - Award [1] for calculating mass of 100.0 g, [1] for finding temperature change of 6.6 K (or 6.6 degrees C), [1] for correct calculation of q as 2.76 kJ. Part (b): [3 marks] - Award [1] for calculating moles of reactants (0.0500 mol) and identifying either as limiting, [1] for dividing q by moles, [1] for correct value of -55.2 kJ/mol with the negative sign. Part (c): [3 marks] - Award [1] for absolute uncertainty of +/- 0.2 C, [2] for percentage uncertainty of 3.03% (accept 3%). Part (d): [2 marks] - Award [1] for suggesting a lid/double cup/insulating wrap, [1] for explaining that calculated enthalpy becomes more negative/exothermic.

(a) The initial rate is obtained by drawing a tangent line to the curve at time \(t = 0 \text{ s}\) and calculating the gradient of this tangent (change in volume divided by change in time). (b) Average rate = \((31.0 - 0.0) / 40 = 0.775 \text{ cm}^3 \text{ s}^{-1}\). (c) As the reaction proceeds, \(\text{H}_2\text{O}_2\) is consumed, decreasing its concentration. This reduces the frequency of collisions between reactant particles, lowering the rate of reaction. (d) Using a single pellet decreases the surface area of the catalyst. This reduces the number of active sites available, lowering the frequency of successful collisions and thus decreasing the initial rate. (e) Percentage uncertainty = \((0.5 / 50.0) \times 100\% = 1.0\%\). (f) A delay in inserting the stopper after adding the catalyst leads to gas escaping, which systematically underestimates the volume of gas collected.

Part (a): [2 marks] - Award [1] for stating tangent at t=0, [1] for finding gradient. Part (b): [2 marks] - Award [1] for setup (31.0 / 40), [1] for 0.775 cm3/s. Part (c): [2 marks] - Award [1] for reactant concentration decreasing, [1] for lower frequency of collisions. Part (d): [2 marks] - Award [1] for identifying smaller surface area, [1] for fewer active sites / lower collision frequency. Part (e): [2 marks] - Award [1] for correct expression, [1] for 1.0%. Part (f): [2 marks] - Award [1] for identifying gas loss during stopper insertion / gas leakage, [1] for its effect on reducing recorded volume.

(a) Mass of vaporized liquid = \(82.885 - 82.415 = 0.470 \text{ g}\). (b) \(T = 99.5 + 273.15 = 372.65 \text{ K}\) (accept \(373 \text{ K}\)). \(P = 100.8 \times 10^3 = 100800 \text{ Pa}\). (c) \(PV = nRT \Rightarrow n = PV / RT\). Convert volume to \(\text{m}^3\): \(250.0 \text{ cm}^3 = 2.500 \times 10^{-4} \text{ m}^3\). \(n = (100800 \times 2.500 \times 10^{-4}) / (8.31 \times 372.65) = 25.2 / 3096.7 = 8.138 \times 10^{-3} \text{ mol}\). (d) Molar mass \(M = m / n = 0.470 / 8.138 \times 10^{-3} = 57.8 \text{ g mol}^{-1}\). (e) If some liquid remains unvaporized, the measured mass of condensed liquid at the end will include both the actual vaporized gas mass and the unvaporized liquid, making the experimental mass \(m\) falsely high. Since \(M = m / n\), this increases the calculated molar mass. (f) Intermolecular forces exist between the real gas molecules, or the volume of the gas molecules is not negligible compared to the flask volume.

Part (a): [1 mark] - Award [1] for 0.470 g. Part (b): [2 marks] - Award [1] for 372.65 K (or 373 K), [1] for 100800 Pa. Part (c): [3 marks] - Award [1] for volume conversion to 2.500 x 10^-4 m3, [1] for rearranging ideal gas equation, [1] for 8.14 x 10^-3 mol. Part (d): [2 marks] - Award [1] for M = m/n calculation, [1] for 57.8 g/mol (accept 57.7 - 58.0 g/mol depending on rounding). Part (e): [2 marks] - Award [1] for identifying that calculated mass is too high, [1] for explaining that this results in a higher calculated molar mass. Part (f): [1 mark] - Award [1] for stating either that real gas molecules have volume or that there are intermolecular forces between them.

(i) Comparing Exp 1 and Exp 2: [B] is constant, [A] doubles, and rate doubles, so the reaction is first order with respect to A. Comparing Exp 1 and Exp 3: [A] is constant, [B] doubles, and rate quadruples (\(4.8 \times 10^{-3} / 1.2 \times 10^{-3} = 4\)), so the reaction is second order with respect to B. Overall rate expression: \(\text{Rate} = k[\text{A}][\text{B}]^2\). (ii) Substitute values from Exp 1: \(1.2 \times 10^{-3} = k(0.10)(0.10)^2 \Rightarrow k = 1.2\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\). (iii) Increasing temperature increases the average kinetic energy of the molecules. This increases the collision frequency slightly, but more significantly, it greatly increases the fraction of molecules that possess energy equal to or greater than the activation energy (\(E \ge E_a\)), causing a larger proportion of collisions to be successful. Hence, the rate constant \(k\) and the reaction rate increase. (iv) The Maxwell-Boltzmann curve at 308 K has a lower peak shifted to the right compared to 298 K. The area under the curve to the right of the activation energy line (\(E_a\)) is larger at 308 K, representing more particles with sufficient energy to react.

(i) M1: Deduce first order for A with explanation [1 mark]. M2: Deduce second order for B with explanation [1 mark]. M3: Correct rate expression [1 mark]. (ii) M4: Calculation of k value as 1.2 [1 mark]. M5: Correct units of \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\) [1 mark]. (iii) M6: Higher temperature increases kinetic energy and collision frequency [1 mark]. M7: More significant effect is the increase in the fraction of successful collisions with \(E \ge E_a\) [1 mark]. M8: Explicitly links this to an increase in the rate constant, k [1 mark]. (iv) M9: Axes correctly labeled (y: Number of particles / probability density, x: Kinetic energy) [1 mark]. M10: Curve for 308 K drawn flatter, peak to the right, and crossing the 298 K curve only once [1 mark]. M11: Activation energy marked, showing larger area under the curve for higher temperature [1 mark].

(i) A Brønsted-Lowry acid is a proton (\(\text{H}^+\)) donor. (ii) \(\text{CH}_3\text{CH}_2\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}\). Acid-conjugate base pair: \(\text{CH}_3\text{CH}_2\text{COOH}\) / \(\text{CH}_3\text{CH}_2\text{COO}^-\); Base-conjugate acid pair: \(\text{H}_2\text{O}\) / \(\text{H}_3\text{O}^+\). (iii) Assuming \([\text{H}^+] \approx [\text{CH}_3\text{CH}_2\text{COO}^-]\) and \([\text{CH}_3\text{CH}_2\text{COOH}]_{\text{eq}} \approx 0.150\text{ mol dm}^{-3}\): \(K_a = \frac{[\text{H}^+]^2}{[\text{CH}_3\text{CH}_2\text{COOH}]} \Rightarrow 1.35 \times 10^{-5} = \frac{[\text{H}^+]^2}{0.150}\). \([\text{H}^+]^2 = 2.025 \times 10^{-6} \Rightarrow [\text{H}^+] = 1.423 \times 10^{-3}\text{ mol dm}^{-3}\). \(\text{pH} = -\log_{10}(1.423 \times 10^{-3}) = 2.85\). (iv) When \(\text{H}^+\) (from strong acid) is added, the large reservoir of conjugate base \(\text{CH}_3\text{CH}_2\text{COO}^-\)\reacts with it: \(\text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{COOH(aq)}\). When \(\text{OH}^-\) (from strong base) is added, the reservoir of weak acid \(\text{CH}_3\text{CH}_2\text{COOH}\) reacts with it: \(\text{CH}_3\text{CH}_2\text{COOH(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{COO}^-\text{(aq)} + \text{H}_2\text{O(l)}\). Thus, the concentration ratio of weak acid to conjugate base remains relatively constant, minimizing pH changes.

(i) M1: Acid as a proton donor [1 mark]. (ii) M2: Correct balanced equation with reversible arrows [1 mark]. M3: Both conjugate pairs correctly identified [1 mark]. (iii) M4: Setting up the correct expression for \(K_a\) [1 mark]. M5: Correct calculation of \([\text{H}^+] = 1.42 \times 10^{-3}\text{ mol dm}^{-3}\) [1 mark]. M6: pH calculation yielding 2.85 (accept 2.8-2.9) [1 mark]. (iv) M7: Mentions high concentrations of both \(\text{CH}_3\text{CH}_2\text{COOH}\) and \(\text{CH}_3\text{CH}_2\text{COO}^-\) [1 mark]. M8: Equation for addition of acid [1 mark]. M9: Explanation of acid removal [1 mark]. M10: Equation for addition of base [1 mark]. M11: Explanation of base removal and why pH remains stable [1 mark].

(i) 1-bromobutane: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}\); 2-bromo-2-methylpropane: \((\text{CH}_3)_3\text{CBr}\). (ii) The reaction of 1-bromobutane with \(\text{OH}^-\)\is an \(\text{S}_\text{N}2\) mechanism. Curly arrow from lone pair on oxygen in \(\text{OH}^-\)\to the carbon bonded to bromine. Simultaneous curly arrow from the C-Br bond to the bromine atom. Draw the transition state with partial bonds shown by dashed lines, showing the negative charge on the whole transition state. Arrow showing bromide ion leaving to form butan-1-ol and \(\text{Br}^-\). (iii) Mechanism classification: \(\text{S}_\text{N}2\) (substitution nucleophilic bimolecular). 2-bromo-2-methylpropane is a tertiary halogenoalkane; it reacts via the \(\text{S}_\text{N}1\) mechanism. Reasons: 1. Steric hindrance: The bulky methyl groups prevent the nucleophile from attacking the carbon atom from the back side. 2. Stability of carbocation: The tertiary carbocation intermediate formed, \((\text{CH}_3)_3\text{C}^+\), is highly stable due to the positive inductive effect of three electron-donating methyl groups. (iv) The alcohol formed is butan-1-ol (a primary alcohol). Heating a primary alcohol under reflux with excess acidified potassium dichromate(VI) oxidizes it completely to butanoic acid, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\).

(i) M1: Both structures drawn correctly [1 mark]. M2: Both names correct [1 mark]. (ii) M3: Curly arrow from hydroxide lone pair to electron-deficient carbon [1 mark]. M4: Transition state shown correctly with partial bonds and charge [1 mark]. M5: Curly arrow from C-Br bond to Br, and correct products shown [1 mark]. (iii) M6: Identify 1-bromobutane as \(\text{S}_\text{N}2\) [1 mark]. M7: Explain steric hindrance in tertiary halogenoalkanes [1 mark]. M8: Explain inductive effect / stability of tertiary carbocation in \(\text{S}_\text{N}1\) [1 mark]. (iv) M9: Identify primary alcohol as butan-1-ol [1 mark]. M10: Deduce complete oxidation product as butanoic acid [1 mark]. M11: Structural formula of butanoic acid provided correctly [1 mark].

(i) Lewis structure of \(\text{SF}_4\): Sulfur is the central atom with 4 bonding pairs (to F atoms) and 1 lone pair (total of 10 valence electrons on S). Each F has 3 lone pairs. Lewis structure of \(\text{XeF}_4\): Xenon is the central atom with 4 bonding pairs (to F atoms) and 2 lone pairs (total of 12 valence electrons on Xe). Each F has 3 lone pairs. (ii) \(\text{SF}_4\): 5 electron domains. Electron domain geometry: trigonal bipyramidal. Molecular geometry: see-saw. \(\text{XeF}_4\): 6 electron domains. Electron domain geometry: octahedral. Molecular geometry: square planar. (iii) In \(\text{SF}_4\), the see-saw molecular geometry is asymmetrical. The polar S-F bond dipoles do not cancel each other out, resulting in a net molecular dipole. In \(\text{XeF}_4\), the square planar molecular geometry is highly symmetrical. The four polar Xe-F bonds are directed towards the corners of a square, and their dipoles cancel out. The two lone pairs are opposite each other, meaning their effects also cancel. (iv) The S-F bond in \(\text{SF}_4\) is a single covalent bond, whereas the bonds in \(\text{SO}_2\) are double covalent bonds (due to resonance, bond order of 1.5, or simply double bonds). Double bonds have greater electron density between the nuclei, leading to stronger electrostatic attraction. Therefore, the S=O bond in \(\text{SO}_2\) is stronger and shorter than the S-F single bond in \(\text{SF}_4\).

(i) M1: Correct Lewis structure for \(\text{SF}_4\) with 10 valence electrons on S [1 mark]. M2: Correct Lewis structure for \(\text{XeF}_4\) with 12 valence electrons on Xe [1 mark]. (ii) M3: 5 domains, trigonal bipyramidal, and see-saw for \(\text{SF}_4\) [1 mark]. M4: 6 domains, octahedral, and square planar for \(\text{XeF}_4\) [1 mark]. (iii) M5: Explains polarity of \(\text{SF}_4\) due to asymmetry/non-cancelling dipoles [1 mark]. M6: Explains non-polarity of \(\text{XeF}_4\) due to symmetrical square planar shape [1 mark]. M7: Identifies the role of opposing lone pairs in \(\text{XeF}_4\) [1 mark]. (iv) M8: Identifies S-F as a single bond and S-O as a double / resonance bond [1 mark]. M9: States double bonds have higher bond order / greater electron density between nuclei [1 mark]. M10: Correctly links higher bond order to stronger bonds (higher bond enthalpy) [1 mark]. M11: Correctly links higher bond order to shorter bonds [1 mark].

(i) Find moles of each element in 100 g: Moles of C = \(85.7 / 12.01 = 7.136\text{ mol}\). Moles of H = \(14.3 / 1.01 = 14.158\text{ mol}\). Divide by the smallest: C = \(7.136 / 7.136 = 1\), H = \(14.158 / 7.136 = 1.98 \approx 2\). Empirical formula is \(\text{CH}_2\). (ii) Empirical formula mass of \(\text{CH}_2 = 12.01 + (2 \times 1.01) = 14.03\text{ g mol}^{-1}\). Ratio = \(56.1 / 14.03 \approx 4\). Molecular formula is \(\text{C}_4\text{H}_8\). (iii) Balanced combustion equation: \(\text{C}_4\text{H}_8\text{(g)} + 6\text{O}_2\text{(g)} \rightarrow 4\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}\). (iv) Determine limiting reactant: Moles of X (\(\text{C}_4\text{H}_8\)) = \(5.61 / 56.1 = 0.100\text{ mol}\). Moles of \(\text{O}_2\) = \(24.0 / 32.00 = 0.750\text{ mol}\). According to the equation, 1 mole of X requires 6 moles of \(\text{O}_2\). Thus, \(0.100\text{ mol}\) of X requires \(0.100 \times 6 = 0.600\text{ mol}\) of \(\text{O}_2\). Since we have \(0.750\text{ mol}\) of \(\text{O}_2\), \(\text{O}_2\) is in excess, and X (\(\text{C}_4\text{H}_8\)) is the limiting reactant. Moles of \(\text{CO}_2\) produced = \(4 \times 0.100 = 0.400\text{ mol}\). Volume of \(\text{CO}_2\) at STP = \(0.400\text{ mol} \times 22.7\text{ dm}^3\text{ mol}^{-1} = 9.08\text{ dm}^3\).

(i) M1: Calculating moles of C and H [1 mark]. M2: Deducing the 1:2 ratio and empirical formula \(\text{CH}_2\) [1 mark]. (ii) M3: Correct calculation of empirical formula mass and molecular formula \(\text{C}_4\text{H}_8\) [1 mark]. (iii) M4: Correct formulas of reactants and products [1 mark]. M5: Correctly balanced coefficients [1 mark]. (iv) M6: Calculation of moles of X (0.100 mol) [1 mark]. M7: Calculation of moles of \(\text{O}_2\) (0.750 mol) [1 mark]. M8: Showing that X is the limiting reactant [1 mark]. M9: Determining that 0.400 mol of \(\text{CO}_2\) is produced [1 mark]. M10: Calculation of gas volume using 22.7 (or 22.4) yielding 9.08 (or 8.96) \(\text{dm}^3\) [1 mark]. M11: Correct units and significant figures [1 mark].

(i) Oxidation is an increase in the oxidation state of an element. (ii) Write half-equations: Oxidation: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\). Reduction: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\). Multiply the oxidation half-equation by 6 and combine: \(\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 14\text{H}^+\text{(aq)} + 6\text{Fe}^{2+}\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 6\text{Fe}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)}\). (iii) Diagram requirements: Zinc electrode in \(\text{Zn}^{2+}\) solution (anode, negative). Copper electrode in \(\text{Cu}^{2+}\) solution (cathode, positive). External wire connecting electrodes with electrons flowing from Zn (anode) to Cu (cathode). Salt bridge connecting the two solutions, with anions moving toward the Zn half-cell (anode) and cations moving toward the Cu half-cell (cathode). (iv) \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = E^\theta(\text{Cu}^{2+}/\text{Cu}) - E^\theta(\text{Zn}^{2+}/\text{Zn}) = +0.34\text{ V} - (-0.76\text{ V}) = +1.10\text{ V}\).

(i) M1: Oxidation defined as increase in oxidation state [1 mark]. (ii) M2: Correct chromium half-equation [1 mark]. M3: Correct iron half-equation [1 mark]. M4: Correctly combined overall equation [1 mark]. (iii) M5: Identifies Zn as the anode and Cu as the cathode [1 mark]. M6: Shows electron flow from Zn to Cu in external wire [1 mark]. M7: Shows salt bridge with correct direction of ion migration (cations to Cu, anions to Zn) [1 mark]. M8: Solutions labeled with correct electrolytes (Zn2+ and Cu2+) [1 mark]. (iv) M9: Formula for cell potential [1 mark]. M10: Calculation yielding +1.10 V [1 mark]. M11: Inclusion of correct positive sign and unit (V) [1 mark].

(i) Total volume of solution = \(50.0 + 50.0 = 100.0\text{ cm}^3\). Mass of solution (assuming density \(1.00\text{ g cm}^{-3}\)) = \(100.0\text{ g}\). Temperature change (\(\Delta T\)) = \(27.8 - 21.3 = 6.5\text{ K}\). Heat released (\(q\)) = \(m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.5\text{ K} = 2717\text{ J} = 2.717\text{ kJ}\). Moles of \(\text{H}^+\) reacting = \(1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). Enthalpy of neutralization (\(\Delta H\)) = \(-q / n = -2.717\text{ kJ} / 0.0500\text{ mol} = -54.34 \approx -54.3\text{ kJ mol}^{-1}\). Assumptions: specific heat capacity is that of pure water, density of solution is that of water, and no heat is lost to the surroundings or absorbed by the calorimeter. (ii) Systematic errors: 1. Heat loss to the surroundings: causes measured \(\Delta T\) to be lower than theoretical, making the calculated \(\Delta H\) less exothermic (less negative). 2. Heat absorbed by the calorimeter/thermometer: also reduces measured \(\Delta T\), leading to a less exothermic calculated value. (iii) Target equation for formation of propene: \(3\text{C(s)} + 3\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_6\text{(g)}\). Using combustion values: \(\Delta H_f^\theta = \sum \Delta H_c^\theta(\text{reactants}) - \sum \Delta H_c^\theta(\text{products}) = [3 \times \Delta H_c^\theta(\text{C}) + 3 \times \Delta H_c^\theta(\text{H}_2)] - \Delta H_c^\theta(\text{C}_3\text{H}_6)\). \(\Delta H_f^\theta = [3(-394) + 3(-286)] - (-2058) = [-1182 - 858] + 2058 = -2040 + 2058 = +18\text{ kJ mol}^{-1}\).

(i) M1: Calculate mass of solution (100 g) and temperature change (6.5 K) [1 mark]. M2: Calculate heat released (2.717 kJ) [1 mark]. M3: Calculate moles of acid/base reacting (0.0500 mol) [1 mark]. M4: Enthalpy of neutralization as -54.3 kJ/mol with negative sign [1 mark]. M5: State one valid assumption [1 mark]. (ii) M6: First systematic error (e.g. heat loss) and its effect (less exothermic) [1 mark]. M7: Second systematic error (e.g. heat capacity of cup ignored) and its effect [1 mark]. (iii) M8: State the equation of formation or correct Hess cycle [1 mark]. M9: Set up the correct algebraic expression for calculation [1 mark]. M10: Correct calculation leading to +18 [1 mark]. M11: Correct units of kJ/mol and positive sign [1 mark].

(i) Copper atom: \(1s^2 2s^2 2s^6 3s^2 3p^6 3d^{10} 4s^1\); Copper(II) ion: \(1s^2 2s^2 2s^6 3s^2 3p^6 3d^9\). (ii) The transition element copper has a fully filled d-subshell (\(3d^{10}\)) and a half-filled s-subshell (\(4s^1\)). This configuration is lower in energy and more stable than the expected \(3d^9 4s^2\) because a symmetrical, completely filled d-subshell offers extra stability due to decreased electron-electron repulsion. (iii) Across Period 3, first ionization energy generally increases. This is because the nuclear charge (number of protons) increases, while shielding from inner-shell electrons remains relatively constant. Therefore, the outer-shell electrons experience a stronger effective nuclear charge and are held more tightly. The drop between P and S occurs because phosphorus has the configuration \([\text{Ne}]3s^2 3p_x^1 3p_y^1 3p_z^1\) (three singly-occupied 3p orbitals), whereas sulfur has \([\text{Ne}]3s^2 3p_x^2 3p_y^1 3p_z^1\). In sulfur, one 3p orbital contains a pair of electrons. The mutual repulsion between these paired electrons in the same orbital makes it easier to remove one of them, resulting in a lower first ionization energy than phosphorus. (iv) When atoms absorb energy, electrons are promoted to higher energy levels (excited state). These excited electrons are unstable and drop back down to lower energy levels (ground state). As they do, they emit energy in the form of electromagnetic radiation (photons). The energy of the photon matches the difference between the two energy levels (\(E = h\nu\)), resulting in discrete, sharp lines in the emission spectrum.

(i) M1: Full electron configuration of Cu [1 mark]. M2: Full electron configuration of Cu2+ (electrons removed from 4s first) [1 mark]. (ii) M3: Identifies \(3d^{10} 4s^1\) configuration [1 mark]. M4: Explains extra stability associated with a symmetrical, completely filled d-subshell [1 mark]. (iii) M5: General increase across Period 3 linked to increasing nuclear charge [1 mark]. M6: Constant shielding leading to higher effective nuclear charge [1 mark]. M7: Identifies the drop between P and S [1 mark]. M8: Explains electron pairing in S [1 mark]. M9: Refers to inter-electron repulsion in the paired 3p orbital of S lowering the required energy [1 mark]. (iv) M10: Mentions promotion of electrons to higher levels followed by relaxation to lower levels [1 mark]. M11: Explains emission of photons of discrete energy/frequency proportional to the gap between levels [1 mark].