2023 Cambridge IGCSE Chemistry (0620) Practice Paper with Answers

Methane (\(\text{CH}_4\)) is a key greenhouse gas that is released during the decay of organic material in anaerobic conditions, such as those found in landfill sites.

Award [1] mark for Methane (or correct chemical formula \(\text{CH}_4\)).

Iron (\(\text{Fe\)}) is a transition element used as a catalyst to increase the rate of reaction in the Haber process where nitrogen and hydrogen react to form ammonia.

Award [1] mark for Iron (or correct chemical symbol \(\text{Fe\)}).

Argon (\(\text{Ar}\)) is a Group VIII/0 noble gas. Because it has a stable full outer shell of electrons, it is chemically inert and is used to prevent the hot metal filament in light bulbs from burning.

Award [1] mark for Argon (or correct chemical symbol \(\text{Ar}\)).

Anhydrous copper(II) sulfate (\(\text{CuSO}_4\)) is white. When water is added, it forms hydrated copper(II) sulfate (\(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\)), which is blue. This serves as a chemical test for the presence of water.

Award [1] mark for Copper(II) sulfate (or copper sulfate / \(\text{CuSO}_4\)). Do not accept hydrated copper(II) sulfate.

Sulfur dioxide (\(\text{SO}_2\)) is formed by the combustion of sulfur impurities in fossil fuels. It dissolves in atmospheric water to produce acid rain.

Award [1] mark for Sulfur dioxide (or correct chemical formula \(\text{SO}_2\)).

Sodium chloride (\(\text{NaCl}\)) has a giant ionic lattice. In the solid state, ions are locked in place and cannot move. When molten or in aqueous solution, the ionic bonds are broken/weakened and the ions are free to move and carry electric charge.

Award [1] mark for Sodium chloride (or correct chemical formula \(\text{NaCl}\)).

Isotopes of the same element have identical chemical properties because chemical reactions depend on the arrangement of electrons in the outer shell. Since both \({}^{35}_{17}\text{Cl}\) and \({}^{37}_{17}\text{Cl}\) have the same atomic number of 17, they have the same number of electrons (17) and the same electronic configuration (2,8,7).

M1: State that they have the same number of electrons / same electronic configuration / same number of valency electrons (1 mark). M2: Explain that chemical properties or chemical reactions are determined by the number of electrons / outer-shell electrons (1 mark).

To calculate the relative atomic mass (\(A_r\)) of element \(Y\):
1. Multiply the mass number of each isotope by its percentage abundance:
- For \(^{39}\text{Y}\): \(39 \times 92.20 = 3595.8\)
- For \(^{40}\text{Y}\): \(40 \times 0.12 = 4.8\)
- For \(^{41}\text{Y}\): \(41 \times 7.68 = 314.88\)
2. Sum these values together:
\(3595.8 + 4.8 + 314.88 = 3915.48\)
3. Divide the sum by 100 to find the weighted average:
\(A_r = \frac{3915.48}{100} = 39.1548\)
4. Round to two decimal places as requested:
\(A_r = 39.15\)

- [1 mark] for correctly setting up the relative atomic mass calculation: \(\frac{(39 \times 92.20) + (40 \times 0.12) + (41 \times 7.68)}{100}\)
- [1 mark] for calculating the intermediate sum as \(3915.48\) (or intermediate division result \(39.1548\))
- [1 mark] for the final answer of \(39.15\) (correctly rounded to 2 decimal places)
- Note: award 3 marks for a correct final answer of 39.15 with no working shown. Award 2 marks for 39.1548.

(a)(i) The atomic number (subscript) represents the number of protons: 13. The mass number (superscript) minus the atomic number gives the number of neutrons: 27 - 13 = 14.
(a)(ii) Isotopes are atoms of the same element containing the same number of protons but different numbers of neutrons (or having the same atomic number but different nucleon numbers).
(b)(i) Pure alumina has an extremely high melting point (approx. 2000°C). Dissolving it in molten cryolite reduces the operating temperature to around 950°C, significantly reducing energy requirements and costs. It also increases the electrical conductivity of the electrolyte.
(b)(ii) At the cathode (negative electrode), aluminium ions gain electrons (reduction) to form liquid aluminium: \(\text{Al}^{3+}(\text{l}) + 3\text{e}^- \rightarrow \text{Al}(\text{l})\).
(c)(i) At the high operating temperatures of the cell, the oxygen gas liberated at the anode reacts with the graphite (carbon) electrodes, forming carbon dioxide gas: \(\text{C}(\text{s}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g})\). This causes the anodes to gradually burn away, so they must be replaced.
(c)(ii) An oxygen molecule contains a double covalent bond. This involves sharing two pairs of electrons (4 electrons in total) between the two oxygen atoms. Each oxygen atom is left with 4 non-bonding valence electrons to complete its octet (outer shell).

(a)(i) [1 mark] Both values correct: 13 protons AND 14 neutrons.
(a)(ii) [1 mark] Correct definition: atoms of the same element with the same number of protons / same atomic number but different numbers of neutrons / different mass numbers.
(b)(i) [2 marks] Any two from:
- Lowers the operating temperature / melting point [1]
- Saves energy / reduces electricity costs [1]
- Acts as a solvent / improves electrical conductivity of the electrolyte [1]
(b)(ii) [1 mark] Correct ionic equation with state symbols: \(\text{Al}^{3+}(\text{l}) + 3\text{e}^- \rightarrow \text{Al}(\text{l})\). Accept state symbols of (l) or (aq). Reject if state symbols are omitted or incorrect (e.g., solid aluminium).
(c)(i) [1 mark] Carbon reacts with oxygen gas to form carbon dioxide (which escapes as a gas) / anodes burn away.
(c)(ii) [1 mark] Correctly states 4 shared electrons (or double bond) between the oxygen atoms AND 4 non-bonding outer electrons on each oxygen atom.

(a) Carbon cannot reduce aluminium oxide because aluminium is more reactive than carbon (higher in the reactivity series). (b) Pure aluminium oxide has a very high melting point (around 2045 °C). Dissolving it in molten cryolite lowers the melting point of the mixture to around 950 °C, which saves a large amount of energy and reduces operating costs. (c) At the cathode, aluminium ions gain three electrons to form aluminium atoms: \(Al^{3+} + 3e^{-} \rightarrow Al\). (d) Oxygen gas is produced at the anode. At the high operating temperature, this oxygen reacts with the carbon anodes to form carbon dioxide gas, which causes the anodes to burn away over time: \(C(s) + O_2(g) \rightarrow CO_2(g)\).

(a) [1 mark] for stating that aluminium is more reactive than carbon (or carbon is less reactive than aluminium). (b) [1 mark] for stating that cryolite lowers the melting point of the electrolyte / reduces energy costs / acts as a solvent. (c) [1 mark] for correct balanced ionic half-equation: \(Al^{3+} + 3e^{-} \rightarrow Al\) (accept correct multiples, ignore state symbols). (d) [1 mark] for stating that the carbon/graphite anodes react with oxygen to form carbon dioxide / burn away.

(a)(i) The number of neutrons is calculated by subtracting the atomic number (13) from the mass number (27): 27 - 13 = 14 neutrons. (a)(ii) An aluminium atom has 3 electrons in its outer shell (electronic configuration 2,8,3). To obtain a stable full outer shell, it loses these 3 valence electrons to form an \(\text{Al}^{3+}\) cation. (b) Aluminium oxide is electrolysed in a molten state (dissolved in molten cryolite) at approximately 950 °C. At this high temperature, both the aluminium reactant ions and the produced aluminium metal are liquids. Therefore, the state symbols for both species must be (l): \(\text{Al}^{3+}\text{(l)} + 3\text{e}^- \rightarrow \text{Al(l)}\). (c)(i) An oxygen atom is in Group VI and has 6 outer-shell electrons. It needs 2 more electrons to achieve a stable octet. Two oxygen atoms share two pairs of electrons to form a double covalent bond. (c)(ii) Each oxygen atom starts with 6 outer-shell electrons and shares 2, leaving 4 unshared (non-bonding) outer-shell electrons (or 2 lone pairs) on each oxygen atom.

(a)(i) [1 mark] 14. (a)(ii) [1 mark] Loses 3 electrons (reject: shares/gains electrons). (b) [1 mark] \(\text{Al}^{3+}\text{(l)} + 3\text{e}^- \rightarrow \text{Al(l)}\) (1 mark for correct balanced equation with correct state symbols; reject (s) or (aq) for state symbols; allow e instead of e-). (c)(i) [1 mark] 2 (shared pairs) / double covalent bond. (c)(ii) [1 mark] 4 (unshared outer-shell electrons) / 2 lone pairs.

(a)(i) Order of decreasing reactivity: zinc, X, copper.
(ii) Ionic equation: \( \text{X(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{X}^{2+}(\text{aq}) + \text{Cu(s)} \)

(b)(i) Color change: from colorless to orange / yellow / brown.
(ii) Ionic equation: \( \text{Cl}_2(\text{g}) + 2\text{Br}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{Br}_2(\text{aq}) \) (or aq/l for bromine).

(a)(i) [1 mark] zinc, X, copper (or Zn, X, Cu).
(a)(ii) [2 marks]
- Correct species and balancing: \( \text{X} + \text{Cu}^{2+} \rightarrow \text{X}^{2+} + \text{Cu} \) [1]
- Correct state symbols: (s), (aq), (aq), (s) [1]
(b)(i) [1 mark] colorless to orange / yellow / brown (reject 'clear' to orange).
(b)(ii) [2 marks]
- Correct species and balancing: \( \text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2 \) [1]
- Correct state symbols: (g), (aq), (aq), (aq) or (l) [1]

Part (a)(i): The starting solution containing potassium iodide is colorless. Since chlorine is more reactive than iodine, it displaces the iodide ions to form aqueous iodine, which colors the solution brown (or orange-brown).

Part (a)(ii): In an ionic equation, spectator ions (potassium, \( \text{K}^+ \)) are omitted. Chlorine molecules react with iodide ions to form chloride ions and iodine molecules. The balanced equation with correct state symbols is: \( \text{Cl}_2(\text{g}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{I}_2(\text{aq}) \).

Part (b): Bromine is lower down Group VII than chlorine, meaning it is less reactive and cannot displace chloride ions. At the atomic level, a bromine atom is larger and has more electron shells than a chlorine atom. This increased distance and shielding weaken the electrostatic attraction between the positive nucleus and an incoming electron, making it harder for bromine to gain an electron to form a halide ion.

(a)(i)
[1] (colorless to) brown / orange-brown / yellow-brown. (Reject red, blue, or purple)

(a)(ii)
[1] Correct species and balancing: \( \text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2 \)
[1] Correct state symbols: \( \text{Cl}_2(\text{g}) \) [accept \( \text{aq} \)], \( \text{I}^-(\text{aq}) \), \( \text{Cl}^-(\text{aq}) \), \( \text{I}_2(\text{aq}) \)

(b)
[1] Bromine is less reactive than chlorine (or chlorine is more reactive than bromine)
[1] Bromine atoms are larger / have more shells / have more shielding than chlorine atoms
[1] Weaker attraction from the nucleus to gain an electron (or less ease of gaining an electron)

(a) The gas produced is carbon dioxide. To test for carbon dioxide, bubble it through limewater. If carbon dioxide is present, the limewater turns cloudy (or milky).

(b) According to collision theory, increasing the concentration of the acid means there are more particles in a given volume (higher particle density). This increases the frequency of collisions (more collisions per second) between the reactant particles, leading to a higher rate of reaction.

(c) Step 1: Calculate the moles of hydrochloric acid used.
$$\text{Moles of HCl} = \text{concentration} \times \text{volume in dm}^3 = 0.500\text{ mol/dm}^3 \times 0.0400\text{ dm}^3 = 0.0200\text{ mol}$$

Step 2: Determine the moles of carbon dioxide produced using the stoichiometric ratio.
From the equation, $2\text{ moles of HCl}$ produce $1\text{ mole of CO}_2$.
$$\text{Moles of CO}_2 = \frac{0.0200\text{ mol}}{2} = 0.0100\text{ mol}$$

Step 3: Calculate the volume of carbon dioxide at r.t.p. in $\text{cm}^3$.
$$\text{Volume of CO}_2 = 0.0100\text{ mol} \times 24000\text{ cm}^3/\text{mol} = 240\text{ cm}^3$$

Part (a):
• Bubble gas through limewater [1]
• Limewater turns cloudy / milky / white precipitate forms [1]
(Reject: 'clear' instead of 'colorless' for initial limewater, but ignore initial state if not wrong)

Part (b):
• More particles per unit volume / particles are closer together [1]
• Increased frequency of collisions / more collisions per second / greater rate of successful collisions [1]
(Reject: 'more collisions' without a time reference)

Part (c):
• Calculate moles of HCl = 0.0200 mol OR moles of CO2 = 0.0100 mol [1]
• Correct final volume = 240 cm³ [1]
(Allow: 0.24 dm³ for 1 mark if unit is explicitly converted/written as dm³; correct numerical answer of 240 with no working gains 2 marks)

(a) The gas produced is carbon dioxide. The standard test is to bubble the gas through limewater. The positive observation is that the limewater turns cloudy or milky due to the formation of insoluble calcium carbonate.

(b) By increasing the concentration of hydrochloric acid, there are more acid particles (H⁺ ions) in a given volume of solution. This increases the density of the particles, leading to a higher frequency of collisions (more collisions per unit time) between the acid particles and the calcium carbonate chips, which increases the reaction rate.

(c) Step 1: Find the number of moles of HCl used:
$$\text{Moles of HCl} = \text{Concentration} \times \text{Volume (in dm}^3\text{)}$$
$$\text{Moles of HCl} = 0.500\text{ mol/dm}^3 \times 0.0400\text{ dm}^3 = 0.0200\text{ mol}$$

Step 2: Use the stoichiometric ratio from the balanced equation to find the moles of CO2:
The ratio of HCl to CO2 is 2:1.
$$\text{Moles of CO}_2 = \frac{0.0200}{2} = 0.0100\text{ mol}$$

Step 3: Convert moles of CO2 to volume in cm³ at r.t.p.:
$$\text{Volume} = 0.0100\text{ mol} \times 24000\text{ cm}^3/\text{mol} = 240\text{ cm}^3$$

Part (a):
- Bubble gas through limewater [1]
- Limewater turns cloudy / milky / chalky [1] (Reject: cloudy white solution without mentioning limewater)

Part (b):
- More particles per unit volume / particles closer together [1]
- Greater frequency of collisions / more collisions per unit time / more collisions per second [1] (Reject: 'more collisions' on its own without time reference)

Part (c):
- Correctly calculates moles of HCl (0.0200 mol) OR moles of CO2 (0.0100 mol) [1]
- Correctly calculates volume as 240 cm³ [1] (Allow 1 mark for 0.24 dm³)

Step 1: Calculate the moles of \(\text{NaOH}\) used in the titration. \(\text{Moles of NaOH} = \text{concentration} \times \text{volume} = 0.500\text{ mol/dm}^3 \times \frac{20.0}{1000}\text{ dm}^3 = 0.0100\text{ mol}\). Step 2: Determine the moles of excess \(\text{HCl}\). Since the reaction ratio is 1:1, \(\text{moles of excess HCl} = 0.0100\text{ mol}\). Step 3: Calculate the total initial moles of \(\text{HCl}\) added. \(\text{Initial moles of HCl} = 1.00\text{ mol/dm}^3 \times \frac{50.0}{1000}\text{ dm}^3 = 0.0500\text{ mol}\). Step 4: Calculate the moles of \(\text{HCl}\) that reacted with the calcium carbonate. \(\text{Reacted HCl} = 0.0500\text{ mol} - 0.0100\text{ mol} = 0.0400\text{ mol}\). Step 5: Calculate the moles of \(\text{CaCO}_3\) that reacted. From the balanced equation, 1 mole of \(\text{CaCO}_3\) reacts with 2 moles of \(\text{HCl}\). Therefore, \(\text{moles of CaCO}_3\) = \(\frac{0.0400\text{ mol}}{2} = 0.0200\text{ mol}\). Step 6: Calculate the mass and percentage purity of \(\text{CaCO}_3\). \(M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100\). \(\text{Mass of pure CaCO}_3\) = \(0.0200\text{ mol} \times 100\text{ g/mol} = 2.00\text{ g}\). \(\text{Percentage purity} = \frac{2.00\text{ g}}{2.50\text{ g}} \times 100\% = 80.0\%\).

Award 1 mark for each of the following points: 1. Correct calculation of moles of NaOH (0.0100 mol). 2. Identification that moles of excess HCl is equal to moles of NaOH (0.0100 mol). 3. Correct calculation of initial moles of HCl added (0.0500 mol). 4. Correct calculation of moles of HCl reacted with CaCO3 (0.0400 mol). 5. Correct division of reacted HCl moles by 2 to find moles of CaCO3 (0.0200 mol). 6. Correct calculation of percentage purity (80.0%). Accept 80% as a correct final answer.

Part (a):
Dynamic equilibrium has two key characteristics:
1. The rate of the forward reaction is equal to the rate of the backward reaction.
2. The concentration of all reactants and products remains constant (must occur in a closed system).

Part (b):
(i) The forward reaction is exothermic (\( \Delta H = -165\text{ kJ/mol} \)). According to Le Chatelier's principle, increasing the temperature will cause the equilibrium to shift in the endothermic direction to absorb heat. Therefore, the position of equilibrium shifts to the left / backward direction.
(ii) There are 5 moles of gas on the reactant side (\( 1\text{ mol of CO}_2 + 4\text{ mol of H}_2 \)) and 3 moles of gas on the product side (\( 1\text{ mol of CH}_4 + 2\text{ mol of H}_2\text{O} \)). Increasing the pressure shifts the equilibrium to the side with fewer gas moles to reduce the pressure. Thus, the position of equilibrium shifts to the right / forward direction.

Part (c):
(i) Short-wavelength radiation from the sun passes through the atmosphere and warms the Earth's surface. The Earth then emits longer-wavelength infrared (thermal) radiation. Greenhouse gases absorb this thermal radiation and re-emit it in all directions, including back to Earth, raising atmospheric temperatures.
(ii) Water vapour (\( \text{H}_2\text{O} \)) or nitrous oxide (\( \text{N}_2\text{O} \)) are other common greenhouse gases in the atmosphere.

(a) [Total: 2 marks]
- Rate of forward reaction equals rate of reverse/backward reaction [1]
- Concentrations of reactants and products remain constant (in a closed system) [1]

(b)(i) [Total: 2 marks]
- Position of equilibrium shifts to the left / reactant side / backward direction [1]
- Forward reaction is exothermic / backward reaction is endothermic AND system shifts to absorb heat / lower temperature [1]

(b)(ii) [Total: 1 mark]
- Position of equilibrium shifts to the right / product side / forward direction [1] (Accept: because there are fewer moles/molecules of gas on the right (3 moles) than the left (5 moles))

(c)(i) [Total: 2 marks]
- Absorb infrared / thermal radiation emitted or reflected from Earth's surface [1]
- Re-radiate / re-emit this thermal energy back towards the Earth / trap the heat in the atmosphere [1]

(c)(ii) [Total: 1 mark]
- Water vapour / nitrous oxide / nitrogen dioxide / oxides of nitrogen [1]
(Reject: sulfur dioxide, carbon monoxide, nitrogen, oxygen)

**(a)** For fermentation to happen efficiently, yeast must be present as it provides the enzymes (zymase) needed to catalyze the reaction. The temperature must be maintained warm, ideally between \(25\,^{\circ}\text{C}\) and \(35\,^{\circ}\text{C}\). If it is too cold, the reaction is too slow; if it is too hot, the yeast enzymes denature. Oxygen must also be excluded (anaerobic conditions) to prevent the ethanol from oxidizing to ethanoic acid.

**(b) (i)** Ethene gas reacts with steam in an addition reaction:
\(\text{C}_2\text{H}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightarrow \text{C}_2\text{H}_5\text{OH}(\text{g})\)

**(b) (ii)** This industrial process requires a phosphoric acid (\(\text{H}_3\text{PO}_4\)) catalyst absorbed on silica, and a high temperature of approximately \(300\,^{\circ}\text{C}\) (pressures of about \(60\,\text{atm}\) are also used).

**(c) (i)** Acidified potassium manganate(VII) is a strong oxidizing agent. When it oxidizes ethanol to ethanoic acid, the manganese(VII) ion (purple) is reduced to manganese(II) ions (colorless). Thus, the color change is from purple to colorless.

**(c) (ii)** Ethanoic acid is a two-carbon carboxylic acid with the structural formula \(\text{CH}_3\text{COOH}\).

**(d)** When ethanoic acid reacts with magnesium, it behaves as a weak acid. Magnesium displaces hydrogen to form magnesium ethanoate. Since magnesium forms a \(\text{Mg}^{2+}\) ion and ethanoate is \(\text{CH}_3\text{COO}^-\), the formula of the salt is \((\text{CH}_3\text{COO})_2\text{Mg}\).

**(a)** Any two from:
- Yeast / zymase [1]
- Warm temperature / temperature in the range \(25\text{--}35\,^{\circ}\text{C}\) [1]
- Absence of oxygen / anaerobic conditions (if not already given in prompt) [1]
*(Max 2 marks)*

**(b) (i)**
- \(\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}\) [1] (Accept structural representations)

**(b) (ii)**
- Catalyst: Phosphoric acid / \(\text{H}_3\text{PO}_4\) [1] (Reject other acids)
- Temperature: \(300\,^{\circ}\text{C}\) (accept any value from \(250\text{--}350\,^{\circ}\text{C}\)) [1]

**(c) (i)**
- Purple to colorless / decolored [1] (Reject: clear instead of colorless)

**(c) (ii)**
- \(\text{CH}_3\text{COOH}\) [1] (Accept displayed formula showing all bonds)

**(d)**
- \((\text{CH}_3\text{COO})_2\text{Mg}\) or \(\text{Mg}(\text{CH}_3\text{COO})_2\) [1]

Step-by-step explanation:

(a) (i) In terms of electron transfer, oxidation is defined as the loss of electrons (OIL: Oxidation Is Loss).
(ii) In terms of oxidation numbers, reduction is defined as a decrease in the oxidation number of an element.

(b) (i) According to the Brønsted-Lowry theory, an acid is defined as a proton (\(\text{H}^+\)) donor.
(ii) The strength of an acid refers to its degree of dissociation in water. A strong acid (such as hydrochloric acid) dissociates completely into its ions: \(\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-\). A weak acid (such as ethanoic acid) only partially dissociates into ions, establishing an equilibrium: \(\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+\).
(iii) When ethanoic acid reacts with sodium hydroxide, a neutralisation reaction occurs: \(\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}\). The salt formed is sodium ethanoate, which has the chemical formula \(\text{CH}_3\text{COONa}\).

Part (a)
* (i) Loss of electrons [1] (Reject: gain of oxygen / loss of hydrogen as question asks in terms of electron transfer)
* (ii) Decrease in oxidation number / decrease in oxidation state [1]

Part (b)
* (i) Proton donor / \(\text{H}^+\) donor [1]
* (ii) Strong acid: completely/fully ionises or dissociates (in aqueous solution) [1]; Weak acid: partially/incompletely ionises or dissociates (in aqueous solution) [1]
* (iii) \(\text{CH}_3\text{COONa}\) [1] (Accept: \(\text{CH}_3\text{COO}^-\text{Na}^+\); Reject: sodium ethanoate / name of the salt as the formula was requested)

To answer this question:

(a) Carboxylic acids are formed by the oxidation of primary alcohols with the same number of carbon atoms. Propanoic acid contains three carbon atoms, so the corresponding starting alcohol is propan-1-ol (propanol).

(b) The molecular formula shows the actual number of atoms of each element in a molecule. For propanoic acid, this is \(\text{C}_3\text{H}_6\text{O}_2\).

(c) When a carboxylic acid reacts with a metal carbonate, it forms a metal carboxylate salt, carbon dioxide, and water. Propanoic acid reacts with calcium carbonate to form the salt calcium propanoate.

(a) propan-1-ol [1] (accept propanol)
(b) \(\text{C}_3\text{H}_6\text{O}_2\) [1] (accept \(\text{CH}_3\text{CH}_2\text{COOH}\) or \(\text{C}_2\text{H}_5\text{COOH}\))
(c) calcium propanoate [1] (reject calcium propionate)