WAEC SSCE - Chemistry - 2011 (Essay)

Question 1 Report

(a)Define the term solubility.

(b) The table below gives the solubility of salt Z at various mperatures

Temperature (°C)	0	10	20	30	40	50	60
Solubility (mol dm\(^{-3}\))	0.13	0.21	0.31	0.45	0.63	0.85	1.10

(i) Plot a graph of solubility against temperature.

(ii) From the graph determine the solubility of salt Z at 35°C.

(iii) If 100cm\(^{-3}\) of the saturated solution is cooled from 55°C to 35°C, calculate the mass of salt Z that would crystallize out. [Molar mass of salt Z = 100 g]

(c)(i) Write a balanced equation to illustrate the reaction of AI\(_2\)O\(_3\) with dilute I. HCI; II. NaOH.

(ii) What is the name given to an oxide that exhibits both acidic and basic properties?

(iii) Give one metallic oxide which exhibite these properties.

(d)(i) Determine the oxidation number of: I. Al in [Al (H\(_2\)O)\(_6\)]\(^{3+}\); II. H in NaH.

(ii) Give the IUPAC name of each of the following substances; I. CuSO\(_4\).5H\(_2\)O; II. CaCO\(_3\); Ill. KMnO\(_4\).

Answer

(a) Solubility is the maximum amount of a solute that dissolves in 1 dm³ of solution at a specified temperature to form a saturated solution.

(b)(i) The solubility curve of salt Z is plotted below. Temperature is on the horizontal axis and solubility is on the vertical axis.

(ii) From the graph, at 35°C:

Solubility of Z = 0.54 mol dm⁻³.

(iii) From the graph:

Solubility at 55°C = 0.97 mol dm⁻³
Solubility at 35°C = 0.54 mol dm⁻³

Decrease in solubility:

\[0.97-0.54=0.43\ \text{mol dm}^{-3}\]

Volume of saturated solution = \(100\text{ cm}^3=0.100\text{ dm}^3\).

Moles of salt crystallized:

\[0.43\times0.100=0.043\ \text{mol}\]

\[\text{Mass crystallized}=0.043\times100=\boxed{4.3\ \text{g}}\]

(c)(i)

I. Reaction with dilute hydrochloric acid:

\[\mathrm{Al_2O_3(s)+6HCl(aq)\rightarrow2AlCl_3(aq)+3H_2O(l)}\]

II. Reaction with sodium hydroxide solution:

\[\mathrm{Al_2O_3(s)+2NaOH(aq)+3H_2O(l)\rightarrow2NaAl(OH)_4(aq)}\]

(c)(ii) Such an oxide is called an amphoteric oxide.

(c)(iii) Zinc oxide, \(\mathrm{ZnO}\), is an amphoteric metallic oxide.

(d)(i)

I. In \(\mathrm{[Al(H_2O)_6]^{3+}}\), water is neutral. Therefore, oxidation number of Al = +3.

II. In \(\mathrm{NaH}\), Na is \(+1\). Therefore, oxidation number of H = −1.

(d)(ii)

Formula	IUPAC name
\(\mathrm{CuSO_4\cdot5H_2O}\)	Copper(II) tetraoxosulfate(VI) pentahydrate
\(\mathrm{CaCO_3}\)	Calcium trioxocarbonate(IV)
\(\mathrm{KMnO_4}\)	Potassium tetraoxomanganate(VII)

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Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

Answer Details

(a) Solubility is the maximum amount of a solute that dissolves in 1 dm³ of solution at a specified temperature to form a saturated solution.

(b)(i) The solubility curve of salt Z is plotted below. Temperature is on the horizontal axis and solubility is on the vertical axis.

(ii) From the graph, at 35°C:

Solubility of Z = 0.54 mol dm⁻³.

(iii) From the graph:

Solubility at 55°C = 0.97 mol dm⁻³
Solubility at 35°C = 0.54 mol dm⁻³

Decrease in solubility:

\[0.97-0.54=0.43\ \text{mol dm}^{-3}\]

Volume of saturated solution = \(100\text{ cm}^3=0.100\text{ dm}^3\).

Moles of salt crystallized:

\[0.43\times0.100=0.043\ \text{mol}\]

\[\text{Mass crystallized}=0.043\times100=\boxed{4.3\ \text{g}}\]

(c)(i)

I. Reaction with dilute hydrochloric acid:

\[\mathrm{Al_2O_3(s)+6HCl(aq)\rightarrow2AlCl_3(aq)+3H_2O(l)}\]

II. Reaction with sodium hydroxide solution:

\[\mathrm{Al_2O_3(s)+2NaOH(aq)+3H_2O(l)\rightarrow2NaAl(OH)_4(aq)}\]

(c)(ii) Such an oxide is called an amphoteric oxide.

(c)(iii) Zinc oxide, \(\mathrm{ZnO}\), is an amphoteric metallic oxide.

(d)(i)

I. In \(\mathrm{[Al(H_2O)_6]^{3+}}\), water is neutral. Therefore, oxidation number of Al = +3.

II. In \(\mathrm{NaH}\), Na is \(+1\). Therefore, oxidation number of H = −1.

(d)(ii)

Formula	IUPAC name
\(\mathrm{CuSO_4\cdot5H_2O}\)	Copper(II) tetraoxosulfate(VI) pentahydrate
\(\mathrm{CaCO_3}\)	Calcium trioxocarbonate(IV)
\(\mathrm{KMnO_4}\)	Potassium tetraoxomanganate(VII)

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Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

Solubility Of Substances (Ghana Only)

Kinetic Theory Of Matter And Gas Laws

Acids, Bases And Salts (Ghana Only)

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Question 2 Report

(a)(i) Draw the energy profile diagram for the reaction; H\(_{2(g)}\) + I\(_{2(g)}\) \(\to\) 2Hl\(_{(g)}\); \(\Delta\)H = –13 KJ mol\(^{-1}\)

(ii) If the concentration of HI\(_{(g)}\) increases from 0.000 to 0.002 mol dm\(^{-3}\) in 80 seconds, what is the rate of t reaction?

(b)(i)Give one use of each of the following compounds: I. NaHCO\(_{3}\); II. CaSO\(_{4}\); III. CaCO\(_{3}\).

(ii) State a drying agent that can be used for each of the following gases: I. SO\(_{2}\); II. HCI; Ill. NH\(_{3}\)

(c)(i) Write an equation for the complete combustion of carbon in oxygen.

(ii) Calculate the number moles of carbon (IV) oxide produced from the complete combustion of 2.5 g of carbon. [ C = 12.0, O = 16]

(iii) Mention one use of I. carbon (II) oxide; II. carbon (IV) oxide.

(d) An industrial raw material has the following composition by mass:

Iron = 28.1%; Chlorine = 35.7%; Water cf crystallization = 36.2%.

Calculate the formula for the material. [H = 1.00, O = 16.0, CI = 35.5, Fe = 56.0]

(e) Give one example of a (i) metal that is liquid at room temperature,

(ii) non-metal that is liquid room temperature.

Answer

Answer Details

(a)(i) The reaction is exothermic because \(\Delta H=-13\ \text{kJ mol}^{-1}\). The products are therefore at a lower energy level than the reactants.

Energy profile diagram for \(\mathrm{H_2(g)+I_2(g)\rightarrow 2HI(g)}\).

(ii)

\[\text{Rate of reaction}=\frac{\text{change in concentration of HI}}{\text{time}}\]

\[=\frac{0.002-0.000}{80}=2.5\times10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}\]

Rate of reaction = \(2.5\times10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}\).

(b)(i)

\(\mathrm{NaHCO_3}\): manufacture of baking powder.
\(\mathrm{CaSO_4}\): manufacture of plaster of Paris.
\(\mathrm{CaCO_3}\): manufacture of cement.

(ii)

\(\mathrm{SO_2}\): concentrated \(\mathrm{H_2SO_4}\).
\(\mathrm{HCl}\): concentrated \(\mathrm{H_2SO_4}\).
\(\mathrm{NH_3}\): calcium oxide, \(\mathrm{CaO}\).

(c)(i)

\[\mathrm{C_{(s)}+O_{2(g)}\rightarrow CO_{2(g)}}\]

(ii)

From the equation, 1 mol of carbon produces 1 mol of \(\mathrm{CO_2}\).

\[n(\mathrm C)=\frac{\text{mass}}{\text{molar mass}}=\frac{2.5}{12.0}=0.2083\ \text{mol}\]

\[n(\mathrm{CO_2})=0.2083\ \text{mol}\approx\boxed{0.208\ \text{mol}}\]

(iii)

Carbon(II) oxide, \(\mathrm{CO}\): reducing agent in the extraction of metals.
Carbon(IV) oxide, \(\mathrm{CO_2}\): used in fire extinguishers.

(d)

Using 100 g of the material:

Constituent	Mass (g)	Number of moles	Mole ratio
Fe	28.1	\(\frac{28.1}{56.0}=0.502\)	\(\frac{0.502}{0.502}=1\)
Cl	35.7	\(\frac{35.7}{35.5}=1.006\)	\(\frac{1.006}{0.502}=2\)
\(\mathrm{H_2O}\)	36.2	\(\frac{36.2}{18.0}=2.011\)	\(\frac{2.011}{0.502}=4\)

Therefore, the formula of the material is \(\boxed{\mathrm{FeCl_2\cdot4H_2O}}\).

(e)

(i) Mercury, \(\mathrm{Hg}\).

(ii) Bromine, \(\mathrm{Br_2}\).

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Chemistry Of Carbon Compounds (Ghana Only)

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Question 3 Report

(a)The table below gives the volume/pressure data for a particular sample of a gas-at a given temperature.

Volumedm\(^3\) (V)	4.00	2.00	1.00
Pressure/atm (P)	1.00	2.00	4.00

(i) Deduce a mathematical relationship between volume (V) and pressure (P).

(ii) Name law that can be deduced from the data.

(iii) Calculate the pressure of the gas when the volume is 3.20 dm\(^3\)

(b)(i) What is the role of a salt bridge in an electrochemical cell?

(ii) What type of ions must flow into the cathode? Give a reason for your answer.

(iii)A standard galvanic cell constructed with \( \mathrm{Ag^+_{(aq)}} \) \( \mathrm{Ag_{(s)}} \) and \( \mathrm{Zn^{2+}_{(aq)}} \) couple is discharged until 3.3 g of Ag forms.

I. Write the overall cell reaction and standard cell potential?

II. How many moles of electrons flowed through the circuit during the discharge?

III. How many coulombs of charges flowed through the circuit?

\( \mathrm{Ag^+_{(aq)} + e^- \rightleftharpoons Ag_{(s)}} \); E° = + 0.80V

\( \mathrm{Zn^{2+}_{(aq)} + 2^- \rightleftharpoons Zn_{(s)}} \), E° = – 0.76 V [Ag = 108]

(c)(i) Define each of the following terms: I. Activation energy; II. Exothermic reaction.

(ii) Give one example of an endothermic process.

(iii) What is the significance of activated complex in a chemical reaction?

Answer

(a) Volume-pressure data

Volume/dm\(^3\) (V)	4.00	2.00	1.00
Pressure/atm (P)	1.00	2.00	4.00

(a)(i) Relationship between V and P

The product PV is constant (4.00, 4.00, 4.00), so:

\[ PV = \text{constant} \qquad \Rightarrow \qquad V \propto \frac{1}{P} \]

(a)(ii) Law: Boyle's law.

(a)(iii) Pressure when V = 3.20 dm\(^3\)

\[ PV = 4.00 \Rightarrow P = \frac{4.00}{3.20} = 1.25\ \text{atm} \]

(b)(i) Role of a salt bridge: it completes the electrical circuit by allowing ions to flow between the two half-cells, and it keeps each solution electrically neutral (preventing a build-up of charge).

(b)(ii) Ions that flow into the cathode: cations (positive ions). Reason: reduction (gain of electrons) takes place at the cathode, so positive ions migrate there to collect electrons.

(b)(iii) Galvanic cell: Ag\(^+\)/Ag and Zn\(^{2+}\)/Zn; 3.3 g of Ag deposited

I. Overall reaction and standard cell potential

Silver has the higher (more positive) potential, so it is reduced (cathode) and zinc is oxidised (anode):

\[ \text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag} \] \[ E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (+0.80) - (-0.76) = +1.56\ \text{V} \]

II. Moles of electrons that flowed

\[ \text{Moles of Ag} = \frac{3.3}{108} = 0.0306\ \text{mol} \]

Since Ag\(^+\) + e\(^-\) \(\rightarrow\) Ag needs one electron per silver atom, moles of electrons = 0.0306 mol.

III. Quantity of charge (coulombs)

\[ Q = n \times F = 0.0306 \times 96500 = 2.95 \times 10^{3}\ \text{C} \ (\approx 2949\ \text{C}) \]

(c)(i) Definitions

I. Activation energy: the minimum energy that the reacting particles must acquire before they can react to form products.
II. Exothermic reaction: a reaction in which heat energy is given out (released) to the surroundings, so \(\Delta H\) is negative.

(c)(ii) One example of an endothermic process: the dissolving of ammonium nitrate (or ammonium chloride) in water; photosynthesis is also acceptable.

(c)(iii) Significance of the activated complex: it is the short-lived, high-energy species formed at the peak of the energy barrier, in which old bonds are partly broken and new bonds are partly formed; it is the state through which reactants must pass to become products.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

Answer Details

(a) Volume-pressure data

Volume/dm\(^3\) (V)	4.00	2.00	1.00
Pressure/atm (P)	1.00	2.00	4.00

(a)(i) Relationship between V and P

The product PV is constant (4.00, 4.00, 4.00), so:

\[ PV = \text{constant} \qquad \Rightarrow \qquad V \propto \frac{1}{P} \]

(a)(ii) Law: Boyle's law.

(a)(iii) Pressure when V = 3.20 dm\(^3\)

\[ PV = 4.00 \Rightarrow P = \frac{4.00}{3.20} = 1.25\ \text{atm} \]

(b)(i) Role of a salt bridge: it completes the electrical circuit by allowing ions to flow between the two half-cells, and it keeps each solution electrically neutral (preventing a build-up of charge).

(b)(ii) Ions that flow into the cathode: cations (positive ions). Reason: reduction (gain of electrons) takes place at the cathode, so positive ions migrate there to collect electrons.

(b)(iii) Galvanic cell: Ag\(^+\)/Ag and Zn\(^{2+}\)/Zn; 3.3 g of Ag deposited

I. Overall reaction and standard cell potential

Silver has the higher (more positive) potential, so it is reduced (cathode) and zinc is oxidised (anode):

\[ \text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag} \] \[ E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (+0.80) - (-0.76) = +1.56\ \text{V} \]

II. Moles of electrons that flowed

\[ \text{Moles of Ag} = \frac{3.3}{108} = 0.0306\ \text{mol} \]

Since Ag\(^+\) + e\(^-\) \(\rightarrow\) Ag needs one electron per silver atom, moles of electrons = 0.0306 mol.

III. Quantity of charge (coulombs)

\[ Q = n \times F = 0.0306 \times 96500 = 2.95 \times 10^{3}\ \text{C} \ (\approx 2949\ \text{C}) \]

(c)(i) Definitions

I. Activation energy: the minimum energy that the reacting particles must acquire before they can react to form products.
II. Exothermic reaction: a reaction in which heat energy is given out (released) to the surroundings, so \(\Delta H\) is negative.

(c)(ii) One example of an endothermic process: the dissolving of ammonium nitrate (or ammonium chloride) in water; photosynthesis is also acceptable.

(c)(iii) Significance of the activated complex: it is the short-lived, high-energy species formed at the peak of the energy barrier, in which old bonds are partly broken and new bonds are partly formed; it is the state through which reactants must pass to become products.

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Kinetic Theory Of Matter And Gas Laws

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Question 4 Report

The following table gives the atomic numbers of elements V, W, X, Y and Z.

Element	V	W	X	Y	Z
Atomic number	11	16	18	19	24

(a)Which of the elements: (i) belong(s) to group 1?

(ii) is/are riotle gas(es)?

(iii) form(s) coloured compound(s)?

(iv) form(s) an anion?

(v) react(s) with water to liberate hydrogen?

(vi) react(s) with water to form alkaline solution?

(b) What is the:

(i) charge on the ion formed in

(a)(iv) above?

(ii) group of the element(s) in (a)(i) above?

(c)(i) Write the formula of the compound formed between element V and element W.

(ii) Sate the type of bond formed in (c)(i) above. Explain your answer.

(d)(i) What is a covalent compound?

(ii) Give two factors that influence covalent bonding.

(iii) State the type of bond that exists in each of the following substances: MgO, \( \mathrm{NH_3} \) and Fe.

(iv) What are intermolecular forces?

Answer

First identify each element from its atomic number, then place it in the Periodic Table.

Element	Atomic number	Identity	Group / block
V	11	Sodium, Na	Group 1
W	16	Sulfur, S	Group 6
X	18	Argon, Ar	Group 8 (noble gas)
Y	19	Potassium, K	Group 1
Z	24	Chromium, Cr	Transition (d-block) metal

(a)

(i) Group 1: V and Y (sodium and potassium). Each has a single electron in its outermost shell.
(ii) Noble gas: X (argon). Its electron configuration 2,8,8 is a complete, stable octet.
(iii) Coloured compound(s): Z (chromium). Transition metals have partly filled d-orbitals, so d-d electron transitions absorb visible light and give coloured compounds.
(iv) Forms an anion: W (sulfur). With 6 outer electrons it gains 2 electrons to complete its octet, forming S^2-.
(v) Reacts with water to liberate hydrogen: V and Y (sodium and potassium).
(vi) Reacts with water to form an alkaline solution: V and Y. They form NaOH and KOH respectively, which are alkalis.

The reaction of a Group 1 metal with water is illustrated by:

\[2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)\]

(b)

(i) The ion in (a)(iv) is the sulfide ion S^2-; its charge is 2- (a double negative charge).
(ii) The elements in (a)(i) belong to Group 1.

(c)

(i) V is Na (forms Na⁺) and W is S (forms S^2-). Balancing the charges gives the formula Na₂S.
(ii) The bond is ionic (electrovalent). Sodium, a Group 1 metal, loses one electron to become Na⁺; sulfur, a Group 6 non-metal, gains two electrons to become S^2-. The complete transfer of electrons from the metal to the non-metal, and the strong electrostatic attraction between the oppositely charged ions, is the mark of ionic bonding.

(d)

(i) A covalent compound is a compound in which the atoms are held together by the sharing of one or more pairs of electrons, usually between non-metallic atoms.
(ii) Two factors that influence covalent bonding:
1. High ionisation energy of the combining atoms (they hold their electrons too tightly to transfer them, so they share instead).
2. A small difference in electronegativity between the atoms (electrons are shared rather than fully transferred).
(iii) Type of bond in each substance:

Substance Bond type

MgO Ionic (electrovalent)

NH₃ Covalent

Fe Metallic
(iv) Intermolecular forces are the relatively weak forces of attraction that exist between separate molecules of a substance, as distinct from the strong bonds that hold atoms together within a molecule.

Substance	Bond type
MgO	Ionic (electrovalent)
NH₃	Covalent
Fe	Metallic

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Answer Details

First identify each element from its atomic number, then place it in the Periodic Table.

Element	Atomic number	Identity	Group / block
V	11	Sodium, Na	Group 1
W	16	Sulfur, S	Group 6
X	18	Argon, Ar	Group 8 (noble gas)
Y	19	Potassium, K	Group 1
Z	24	Chromium, Cr	Transition (d-block) metal

(a)

(i) Group 1: V and Y (sodium and potassium). Each has a single electron in its outermost shell.
(ii) Noble gas: X (argon). Its electron configuration 2,8,8 is a complete, stable octet.
(iii) Coloured compound(s): Z (chromium). Transition metals have partly filled d-orbitals, so d-d electron transitions absorb visible light and give coloured compounds.
(iv) Forms an anion: W (sulfur). With 6 outer electrons it gains 2 electrons to complete its octet, forming S^2-.
(v) Reacts with water to liberate hydrogen: V and Y (sodium and potassium).
(vi) Reacts with water to form an alkaline solution: V and Y. They form NaOH and KOH respectively, which are alkalis.

The reaction of a Group 1 metal with water is illustrated by:

\[2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)\]

(b)

(i) The ion in (a)(iv) is the sulfide ion S^2-; its charge is 2- (a double negative charge).
(ii) The elements in (a)(i) belong to Group 1.

(c)

(i) V is Na (forms Na⁺) and W is S (forms S^2-). Balancing the charges gives the formula Na₂S.
(ii) The bond is ionic (electrovalent). Sodium, a Group 1 metal, loses one electron to become Na⁺; sulfur, a Group 6 non-metal, gains two electrons to become S^2-. The complete transfer of electrons from the metal to the non-metal, and the strong electrostatic attraction between the oppositely charged ions, is the mark of ionic bonding.

(d)

(i) A covalent compound is a compound in which the atoms are held together by the sharing of one or more pairs of electrons, usually between non-metallic atoms.
(ii) Two factors that influence covalent bonding:
1. High ionisation energy of the combining atoms (they hold their electrons too tightly to transfer them, so they share instead).
2. A small difference in electronegativity between the atoms (electrons are shared rather than fully transferred).
(iii) Type of bond in each substance:

Substance Bond type

MgO Ionic (electrovalent)

NH₃ Covalent

Fe Metallic
(iv) Intermolecular forces are the relatively weak forces of attraction that exist between separate molecules of a substance, as distinct from the strong bonds that hold atoms together within a molecule.

Substance	Bond type
MgO	Ionic (electrovalent)
NH₃	Covalent
Fe	Metallic

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Atomic Structure And Bonding

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Question 5 Report

(a) A compound X reacts with excess HNO\(_{3(aq)}\) to give carbon (IV) oxide and another compound Y. A solution of Y reacts with NaOH\(_{(aq)}\) to form a white precipitate which is insoluble in excess NaOH\(_{(aq)}\). Identify X and Y.

(b) (i) Write a balanced equation to illustrate the reducing property of ammonia in its reaction with CuO.

(ii) Explain why it is not advisable to heat ammonium dioxonitrate (III) directly.

(iii) Give two uses nitrogen.

(c) Give the reason why (i) dilute H\(_{2}\)SO\(_{4}\) is not suitable for the preparation of CO\(_{2(g)}\) from CaCO\(_{3(s)}\) (ii) concentrated H\(_{2}\)SO\(_{4}\) cannot be used to dry ammonia gas.

(d) State two: (i) physical properties; (ii) chemical properties of metals.

(e) What is the oxidation number of: (i) chlorine in I. Cl\(_{2}\). II. ClO\(_{-(3)}\)

(ii) vanadium in V\(_{2}\)O\(_{5}\)

(f)(i) Explain the term half-life. (ii) Two radioactive elements, P and Q have half-life of 1200 seconds and 3600 seconds respectively.

I. Which of the elements is more stable? II. Give a reason for your answer.

Answer

(a) X gives carbon(IV) oxide with acid, so X is a trioxocarbonate(IV); Y forms a white hydroxide precipitate insoluble in excess NaOH. Taking the metal as calcium: X is calcium trioxocarbonate(IV), CaCO₃, and Y is calcium trioxonitrate(V), Ca(NO₃)₂. \(CaCO_3 + 2HNO_3 \rightarrow Ca(NO_3)_2 + H_2O + CO_2\); \(Ca(NO_3)_2 + 2NaOH \rightarrow Ca(OH)_2 + 2NaNO_3\).

(b)(i) \(2NH_3 + 3CuO \rightarrow 3Cu + N_2 + 3H_2O\) (ammonia reduces copper(II) oxide to copper).

(ii) Ammonium dioxonitrate(III), NH₄NO₂, decomposes explosively when heated directly, so direct heating is dangerous. \(NH_4NO_2 \rightarrow N_2 + 2H_2O\)

(iii) Two uses of nitrogen: manufacture of ammonia; providing an inert atmosphere (also food packaging/preservation, or liquid nitrogen as a refrigerant).

(c)(i) Dilute H₂SO₄ is unsuitable because insoluble calcium tetraoxosulphate(VI) forms and coats the marble (CaCO₃), stopping further reaction.

(ii) Concentrated H₂SO₄ cannot dry ammonia because it reacts with the (basic) ammonia to form ammonium tetraoxosulphate(VI). \(2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4\)

(d)(i) Two physical properties of metals: they are good conductors of heat and electricity; they are malleable and ductile (also lustrous, sonorous, high density).
(ii) Two chemical properties: they form basic oxides; they act as reducing agents (are electropositive) and displace hydrogen from dilute acids.

(e)(i) Chlorine: in Cl₂ it is 0; in ClO₃^- it is +5.
(ii) Vanadium in V₂O₅: \(2V + 5(-2) = 0 \Rightarrow V = +5\).

(f)(i) Half-life is the time taken for half of the atoms (nuclei) of a radioactive element to decay.
(ii) I. Q is more stable. II. Because Q has the longer half-life (3600 s), it decays more slowly, so it is more stable.

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