Correct Answer: D. S

Explanation

(a)(i) Electronegatively, ionisation energy, electron affinity.

(ii)

(i) P and S are isotopes of the same element.

(ii) R contains 18 neutrons

(iii) T is chemically unreactive

(iv) Q readily

(b)

(I) P and S are isotopes of the same elements

(ii) R contains 18 neutrons

(iii) T is chemically unreactive

(iv) Q readily forms an ion with two positive charges

(v) R attains an octet structure by accepting one electron

(vi) Q forms an ionic bond with R

(vii) Q belongs to the s - block in the periodic table

(c)(i) To convert CuCO\(_3\) to Cu, heat strongly ; CuCO\(_3\) -> CuO + CO\(_2\) . Pass hydrogen over heated residue of CuO. H\(_2\) + CuO ---> Cu + H\(_2\)O

or pass NH\(_3\) over heated copper II oxide. 2CuO + 3NH\(_3\) —› 3Cu + 3H\(_2\)O + N\(_2\)

OR add dilute mineral acid to CuCO\(_3\) to form soluble salt: CuCO\(_3\) + 2HCI -> CuCl\(_2\) + H\(_2\) O + CO\(_2\) . Electrolysis using platinum electrodes or any inert electrode. Equation at the cathode Cu\(^{2+}\) + 2e\(^-\) \(\to\) Cu

OR add mineral acid to CuCO\(_3\) to form soluble salt CuCO + 2HCI -> CuCl\(_2\) + H\(_2\)O+ CO\(_2\) then add reactive metal e.g. Zinc or Aluminium CuCl\(_2\) + Zn -> Cu + ZnCl\(_2\)

(ii) To convert MgO to Mg\(_2\)SO\(_4\) , acid dilute H\(_2\)SO\(_4\) to excess MgO, filter off excess MgO. The filtrate is MgSO\(_4\) , MgO + H\(_2\)SO\(_4\) -> MgSO\(_4\) + H\(_2\)O. Evaporate / heat to dryness to obtain MgSO\(_4\).

Correct Answer: B. degree of disorder in the system

Explanation

(a)(i) Variable oxidation states (oxidation states of +2 or +3) formation of coloured ions (green or brown) Catalytic ability e.g (in the Haber process) formation of complex ions e.g K\(_4\)Fe (CN)\(_6\). Ability to be magnetised strongly (Ferromagnetism / paramagnetism).

(ii) C + O\(_2\) --> CO\(_2\) CO + C \(\to\) CO Fe\(_2\)O\(_3\) + 3CO \(\to\) 2Fe + 3CO\(_2\)

or Fe\(_2\)O\(_3\) + 3C \(\to\) 2Fe + 3CO CaCO\(_3\) \(\to\) CaO + CO\(_2\) CaO + SiO\(_2\) \(\to\) CaCO\(_3\)

(iii) The reaction of iron to produce rust (Fe\(_3\)O\(_2\) x H\(_2\)O) can be prevented by galvanising, painting, electroplating, greasing or oiling, cathodic protection or anodic sacrificing.

(iv) The oxidation number of iron in iron rust is +3.

(b)(i) Au, Sn, Zn, Mg, Ca, Na in order of reactivity. Mg, Ca, and Na are extracted by electrolysis

(ii) Zinc is said to be amphoteric because it can react with acids and with bases to form salts.

(c)(i) Oxidation is loss of electrons

(ii) If or 96500C is equivalent to 1 mole of electrons

4225 C = \(\frac{4825}{96500}\) moles = 0.05 mole

(iii) 1 mole of Cu\(^{2+}\) is discharged by 2F. 1 mole of Cu\(^{2+}\) contains 6.02 x 10\(^{23}\) ions

6.02 x 10\(^2\) copper ions is 6.02 x 10\(^{23}\) discharged by 2F.

Hence ; 0.25 ions will be discharged by 0.25

F = 7.53 x 10\(^{22}\)

Correct Answer: C. NH₄

Explanation

(a)(i) Allotropes of sulphur prismatic / monoclinic; Rhombic sulphur. Also plastic/amorphous sulphurs.

(ii) When H\(_2\)S is bubbled through acidified KMnO\(_4\), the solution turns from purple to colourless / is decolourised with yellow deposit of sulphur. H\(_2\)S is a reducing agent. It reduces manganese from oxidation state of + 7 to +2 and itself oxidized to sulphur.

(iii) The products are sulphur and water

(b)(i) The raw materials for the manufacture of H\(_2\)SO\(_4\) are sulphur air / oxygen.

(ii) The equation of the reaction that requires a catalyst in the contact process is 2SO\(_2\) + O\(_2\) -> 2SO\(_3\). The catalyst used is V\(_2\)O\(_5\) (vanadium (V) oxide)

(iii) with copper turnings a blue solution and a gas (with irritating smell) are observed. Products are CuSO\(_4\), SO\(_2\) and water - with a cube of sugar, a black mass of carbon is obtained.

(c)(i) Uses of Na\(_2\)CO\(_3\)

- As a water softener

- In glass / paper manufacture

- As analytical reagent, or specific reaction.

(ii) Na\(_2\)CO\(_3\).

Exposure

10H\(_2\)O \(\to\) Na\(_2\)CO\(_3\) . H\(_2\)O + 9H\(_2\)O. This equation illustrates efflorescence in air

in air

(iii) Solubility calculation. Mass of salt per 20cm of saturated solution = 1.75g 1000cm\(^3\) of saturated solution contains

\(\frac{1000 x 1.75g}{20}\) = 87.5g

molar mass of Na\(_2\)CO\(_3\) = (2 x 23) + 12 + (3 x 16) = 106g

Solubility of Na\(_2\)CO\(_3\) = 87.5

= 0.825 mol /dm\(^{-3}\)

Correct Answer: C. dilute trioxonitrate (V) acid

Correct Answer: B. Destructive distillation

Explanation

(a)(i) Esterification is the reaction between alkanols and alkanoic acids to form alkanoates in the presence of mineral acids

(ii) Saponification is the reaction between caustic soda/sodium hydroxide/alkalis and fats or oils to produce salts or fatty acids/soap.

(b)(i) The general molecular formula of alkynes is C\(_n\)H\(_{2n-2}\)

(ii) The molecular formula of ethylethanoate is C\(_4\)H\(_8\)O\(_2\) and its empirical formula is C\(_2\)H\(_4\)O.

(iii) The structure of 1, 1, 2, 2 - tetrabromoethane is

2C\(_2\)H\(_5\)OH + 2Na \(\to\) 2C\(_2\)H\(_5\)ONa + H\(_2\)

(c)(i) Stage (i) is fractional distillation;

Stage (ii) is cracking;

Stage (iii) is chlorination or alogenation or addition:

Stage (iv) is polymerisation

(ii) X is ethene / C\(_2\)H\(_4\). Y is chlorine / CI

(iii) IUPAC name of the structure below is 1, 2, dichlorethane

(iv) The reaction conditions for the polymerisation of ethene (stage iv) are high temperature (150°C to 300°C). High pressure (any pressure 500 to 1500 atmosphere) presence of a trace of oxygen or peroxides.

(d)(i) Palm wine Froths or foams because of the process of fermentation. The CO\(_2\) bubbles produced, cause the mixture to foam.

(ii) Palm wine tastes sour after some days because it has been oxidized to ethanoic by bacteria Atmospheric oxygen i.e aerial oxidation.

Correct Answer: A. contains more ions