WAEC SSCE - Chemistry - 1990

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The compound of copper which is used in electroplating dyeing, printing, wood preservation and as a fungicide is copper (ll) tetraoxosulphate (VI) pentahydrate, also known as copper sulphate. Copper sulphate is a blue crystalline solid that contains copper, sulphur, and oxygen. It is commonly used in electroplating to coat objects with a thin layer of copper. It is also used as a mordant in dyeing and printing fabrics, to preserve wood, and as a fungicide in agriculture. Other compounds of copper listed in the options, such as copper (ll) hydroxide, copper (ll) trioxonitrate (V) pentahydrate, copper (ll) oxide, and copper (ll) trioxocarbonate (lV) are also used for various purposes. However, copper sulphate is the specific compound used in the applications mentioned in the question.

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In the given chemical reaction, the oxidation number of sulfur changes from 4 in SO₂ to 6 in H₂SO₄. Oxidation number is the charge an atom would have if the shared electrons were assigned to the more electronegative atom in a compound. In SO₂, oxygen has an oxidation number of -2, and since there are two oxygen atoms, the total charge from oxygen is -4. Therefore, to balance the charge of SO₂, sulfur has an oxidation number of +4. In H₂SO₄, oxygen still has an oxidation number of -2, and since there are four oxygen atoms, the total charge from oxygen is -8. Hydrogen has an oxidation number of +1, and there are two hydrogen atoms, for a total charge of +2. To balance the charge of H₂SO₄, sulfur must have an oxidation number of +6. Therefore, the oxidation number of sulfur changes from +4 to +6 in this reaction.

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The alloy that does not contain copper is magnalium. Bronze and brass are copper alloys. Bronze is an alloy of copper and tin, while brass is an alloy of copper and zinc. Duralumin is an alloy of aluminum, copper, and magnesium. Type metal is an alloy of lead, tin, and antimony. Magnalium is an aluminum alloy that contains magnesium but not copper. Therefore, the answer is magnalium as it is the only option that does not contain copper.

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The heat accompanying the reaction represented by the equation H2O(l) → H2O(g) is described as the heat of vaporization. This is because the process involves a change in the state of matter from liquid to gas, which requires energy to overcome the intermolecular forces of attraction between the water molecules. The energy required to vaporize a given amount of a substance is called the heat of vaporization. In this case, it is the amount of heat energy required to vaporize one mole of water at a constant temperature and pressure.

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Ethyne (also known as acetylene) is the compound that will undergo additional reactions. Ethyne has a triple bond between the two carbon atoms, which is a reactive functional group that can easily participate in addition reactions with other compounds. Additionally, the sp hybridization of the carbon atoms in the triple bond makes them more electronegative and reactive towards electrophiles. In contrast, butane, pentane, and tetrachloromethane are saturated hydrocarbons that do not have any reactive functional groups and are not expected to undergo additional reactions. Ethanol has an -OH functional group, but it is also saturated and does not undergo additional reactions under normal conditions.

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The empirical formula of a compound gives the simplest whole number ratio of the atoms present in it. To determine the molecular formula of a compound from its empirical formula and molar mass, we need to find the actual number of atoms of each element in the molecule. In this case, the empirical formula of the compound is CH₂O, which has a molar mass of 30 (12 for carbon, 2 for hydrogen, and 16 for oxygen). However, the molar mass of the compound is given as 180 g/mol, which is six times larger than the empirical formula. To find the molecular formula, we need to multiply the empirical formula by a whole number n that gives the actual number of atoms of each element in the molecule. Since the molar mass of the compound is six times larger than its empirical formula, we can assume that the molecular formula is six times larger than the empirical formula. Therefore, we can write: Empirical formula mass = 30 g/mol Molecular formula mass = 6 x empirical formula mass = 6 x 30 g/mol = 180 g/mol The molecular formula will have 6 times the number of atoms as the empirical formula. Thus, the molecular formula of the compound is C₆H₁₂O₆

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Trioxonitrate (V) acid (HNO₃) is a strong oxidizing agent that can react with many metals, including those that can react with acids to form hydrogen gas. However, it does not produce hydrogen gas because it oxidizes the metal instead of producing hydrogen gas. The nitrate ion in the HNO₃ oxidizes the metal to form an oxide layer on the surface of the metal, rendering it passive. This oxide layer prevents further reaction between the metal and the acid, and as a result, hydrogen gas is not produced. Therefore, trioxonitrate (V) acid is not used for preparing hydrogen from metals.

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To determine the number of moles of AgNO₃ in a solution, we need to use the formula: moles = concentration x volume Here, the concentration is given as 0.01 M (M represents molarity, which is the number of moles of a solute per liter of solution), and the volume is given as 500 cm³ (cubic centimeters). However, we need to convert the volume to liters before we can use the formula. There are 1000 cm³ in 1 liter, so: 500 cm³ = 500/1000 L = 0.5 L Now we can use the formula: moles = 0.01 M x 0.5 L = 0.005 moles Therefore, the answer is 0.005 moles of AgNO₃.

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To double the pressure and volume of a gas, we need to use the combined gas law, which states that: (P1 × V1) / T1 = (P2 × V2) / T2 where P is the pressure, V is the volume, and T is the temperature of the gas. Let's assume that the initial temperature, pressure, and volume of the gas are T1 = 273 K, P1 = 1 atm, and V1 = 1 L, respectively. If we want to double both the pressure and volume of the gas, then the final values will be P2 = 2 atm and V2 = 2 L. Substituting these values into the combined gas law, we get: (1 atm × 1 L) / 273 K = (2 atm × 2 L) / T2 Solving for T2, we get: T2 = (2 atm × 2 L × 273 K) / (1 atm × 1 L) = 1092 K Therefore, the gas must be raised to a temperature of 1092 K in order to double both its volume and pressure.

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The catalytic hydrogenation of oils involves adding hydrogen gas to unsaturated fats, which are commonly found in vegetable oils, to create a more saturated fat. This process is commonly used to produce margarine and other types of spreads. During this process, the double bonds in the unsaturated fats are broken, and hydrogen atoms are added to the carbon atoms to create single bonds. This results in the formation of more saturated fats, which are typically solid at room temperature and have a longer shelf life. Therefore, the catalytic hydrogenation of oils results in the production of more saturated fats, which can be used to create products such as margarine, rather than soaps, detergents or alkanes. Butter is a natural product made from milk fat and is not typically produced through catalytic hydrogenation.

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The statement that explains why tetraoxosulphate (IV) acid is regarded as a strong acid is that the acid is completely ionized in aqueous solution. This means that when tetraoxosulphate (IV) acid dissolves in water, all the acid molecules dissociate into ions. In the case of tetraoxosulphate (IV) acid, the hydrogen ion (H+) and tetraoxosulphate (IV) ion (SO4²-) are formed when it dissolves in water. The completeness of ionization of a strong acid in aqueous solution is due to the stability of the resulting ions, which are highly charged and polar. This means that the ions are strongly attracted to the water molecules, making it difficult for them to recombine and reform the original acid molecule. As a result, the concentration of H+ ions in the solution is high, leading to a low pH value. In contrast, weak acids are only partially ionized in aqueous solution, meaning that only a fraction of the acid molecules dissociate into ions. This is due to the fact that the resulting ions are not as stable as those produced by strong acids. As a result, the concentration of H+ ions in a weak acid solution is lower, leading to a higher pH value. Therefore, tetraoxosulphate (IV) acid is regarded as a strong acid because it completely ionizes in aqueous solution, leading to a high concentration of H+ ions and a low pH value.

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Aluminum does not corrode easily on exposure to air while iron does, even though aluminum is above iron in the electrochemical series. This is because aluminum forms a thin layer of inert oxide (Al2O3) on its surface when exposed to moist air, which acts as a protective layer and prevents further corrosion. On the other hand, iron does not form a stable oxide layer and hence corrodes easily. Therefore, the correct answer is "forms a thin layer of inert oxide in moist air".

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Chromatography can be used to detect the yellow dye in the sample of orange juice suspected to be contaminated. Chromatography is a technique used to separate mixtures of substances into their individual components. It works by passing a mixture through a stationary phase, which is typically a solid or a liquid supported on a solid, and a mobile phase, which is typically a liquid or a gas. The components in the mixture separate out based on their interactions with the stationary and mobile phases, and can be identified by their distinct positions on the stationary phase. In this case, the orange juice sample suspected to be contaminated with the yellow dye can be passed through a chromatographic column, and the dye can be separated from the other components in the sample based on its interactions with the stationary and mobile phases. Once the dye is separated, it can be identified by its characteristic position on the stationary phase or by using other analytical techniques. Therefore, the correct answer is "Chromatography".

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A catalyst increases the rate of a chemical reaction by decreasing the activation energy required for the reaction to occur. In simpler terms, the catalyst provides an alternative pathway for the reaction to take place, which requires less energy than the original pathway. This lower energy pathway allows more reactant molecules to collide with enough energy to overcome the activation energy barrier, which speeds up the rate of the reaction. Therefore, a catalyst does not change the initial or final energy levels of the reaction, but it lowers the energy barrier that must be overcome for the reaction to take place, making the reaction occur more quickly.

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Sewage is a pollutant that is biodegradable. Biodegradable means that the substance can be broken down naturally by microorganisms into simpler substances, such as water and carbon dioxide. Sewage contains organic matter that can be easily decomposed by bacteria, which is why it is considered biodegradable. This is also why wastewater treatment plants use bacteria to break down organic matter in sewage before releasing it into the environment. Plastics, metal scrap, lead compounds, and hydrogen sulfide are not biodegradable pollutants. Plastics can take hundreds of years to decompose, and even then, they only break down into smaller plastic particles that can cause harm to the environment and wildlife. Metal scrap and lead compounds do not break down naturally and can accumulate in the environment over time, leading to toxicity issues. Hydrogen sulfide is a gas that can be toxic to living organisms, but it does not biodegrade as it is a simple molecule that does not contain carbon.

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The compound that crystallizes without water of crystallization is NaCI (sodium chloride). When some compounds crystallize, they form a specific number of water molecules as part of their crystal structure. These water molecules are called water of crystallization. For example, CuSO₄⋅5H₂O means that each CuSO₄ unit is bonded to 5 water molecules. However, NaCI does not form water of crystallization when it crystallizes. Instead, the sodium ions (Na⁺) and chloride ions (CI^-) bond together in a lattice structure without any water molecules. This is why NaCI is known as an anhydrous compound. Therefore, the compound that crystallizes without water of crystallization is NaCI.

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The salt that will have a pH greater than 7 in aqueous solution is K₂CO₃. When a salt is dissolved in water, it may undergo hydrolysis, which is a chemical reaction between the ions of the salt and the water molecules. Depending on the nature of the ions, the hydrolysis of a salt can result in an acidic, basic or neutral solution. K₂CO₃ is a salt of a strong base (KOH) and a weak acid (H₂CO₃). When it is dissolved in water, the carbonate ions react with water molecules to produce hydroxide ions (OH^-) and bicarbonate ions (HCO₃^-). HCO₃^- + H₂O ⇌ H₃O⁺ + OH^- The presence of hydroxide ions in solution makes it basic. Therefore, the pH of a solution of K₂CO₃ in water will be greater than 7. In contrast, NaCl, ACI₃, FeCI₃, and Na₂SO₄ do not contain hydroxide ions or any other ions that can significantly affect the pH of the solution. They will therefore result in neutral solutions with a pH of 7. Therefore, the answer is K₂CO₃, because it is the only salt of the options given that will produce a basic solution with a pH greater than 7.

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The balanced chemical equation for the decomposition of hydrogen peroxide is: 2H₂O₂(l) → 2H₂O(l) + O₂(g) From the equation, it is clear that 2 moles of hydrogen peroxide will produce 1 mole of oxygen gas. Also, 1 mole of any gas at s.t.p. occupies a volume of 22.4 dm³. Therefore, to produce 22.4 dm³ of oxygen gas, we need 2 moles of hydrogen peroxide. This is because 2 moles of hydrogen peroxide will produce 1 mole of oxygen gas. Now, we can use the molar mass of hydrogen peroxide to find the mass of hydrogen peroxide needed. The molar mass of hydrogen peroxide (H₂O₂) is: 2(1.01) + 2(16.00) = 34.02 g/mol So, 2 moles of hydrogen peroxide will have a mass of: 2 mol x 34.02 g/mol = 68.04 g Therefore, 68.04 g of hydrogen peroxide is required to produce 22.4 dm³ of oxygen gas at s.t.p. Thus, the answer is (E) 68 g.

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(a)(i) Oxidation is the loss of electron(s) by a substance.

(a)(ii)

Oxidation half-reaction:

\[\mathrm{Fe^{2+}_{(aq)} \rightarrow Fe^{3+}_{(aq)} + e^-}\]

Reduction half-reaction:

\[\mathrm{MnO_4^-_{(aq)} + 8H^+_{(aq)} + 5e^- \rightarrow Mn^{2+}_{(aq)} + 4H_2O_{(l)}}\]

(b)(i)

Electrolytic cell	Electrochemical cell
Electrical energy is converted into chemical energy. An external source of electricity drives a non-spontaneous reaction.	Chemical energy is converted into electrical energy. A spontaneous reaction produces electricity.
The anode is positive and the cathode is negative.	The anode is negative and the cathode is positive.

(b)(ii) Cell for the electrolysis of molten magnesium chloride:

At the cathode:

\[\mathrm{Mg^{2+}_{(l)} + 2e^- \rightarrow Mg_{(l)}}\]

At the anode:

\[\mathrm{2Cl^-_{(l)} \rightarrow Cl_{2(g)} + 2e^-}\]

(b)(iii) A platinum anode is not economical for this electrolysis because platinum is very expensive.

(c) Both copper(II) ions and lead(II) ions require two electrons for deposition:

\[\mathrm{Cu^{2+}+2e^-\rightarrow Cu}\]\[\mathrm{Pb^{2+}+2e^-\rightarrow Pb}\]

Hence, for the same quantity of electricity, the masses deposited are proportional to their molar masses:

\[\frac{m_{\mathrm{Pb}}}{m_{\mathrm{Cu}}}=\frac{207}{63.5}\]\[m_{\mathrm{Pb}}=1.35\times\frac{207}{63.5}=4.40\ \mathrm{g}\]

Mass of lead deposited = \(4.40\ \mathrm{g}\).

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(a)(i) Kekulé structures of benzene, C₆H₆

(a)(ii) The two structures illustrate resonance. The actual benzene molecule is a resonance hybrid in which the c0-electrons are delocalized over all six carbon atoms. Thus, all the carbon-carbon bonds are identical and are intermediate between a single and a double bond.

(b)(i) Uses of benzene

1. It is used in the manufacture of chemicals such as disinfectants.
2. It is used in the manufacture of perfumes.

(b)(ii) Industrial source of benzene

Benzene is obtained from coal tar, produced during the destructive distillation of coal.

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(a)

\[K_c=\frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2[\mathrm{O_2}]}\]

(b) The forward reaction is exothermic; hence the products are at a lower energy level than the reactants. The catalysed pathway has a lower activation energy.

(c)

1. Increase in temperature: The equilibrium shifts to the left, towards \(\mathrm{SO_2}\) and \(\mathrm{O_2}\), because the forward reaction is exothermic and added heat favours the endothermic reverse reaction. The yield of \(\mathrm{SO_3}\) decreases.
2. Increase in pressure: The equilibrium shifts to the right, producing more \(\mathrm{SO_3}\), because there are fewer moles of gaseous products (2 moles) than gaseous reactants (3 moles).
3. Removal of \(\mathrm{SO_3}\): The equilibrium shifts to the right to replace the removed \(\mathrm{SO_3}\). More \(\mathrm{SO_3}\) is formed.
4. Presence of \(\mathrm{V_2O_5}\): \(\mathrm{V_2O_5}\) acts as a catalyst. It increases the rates of both forward and reverse reactions equally, so it does not change the position of equilibrium or the equilibrium yield. It only enables equilibrium to be reached faster.

(d)(i) In the contact process, sulphur(VI) oxide is absorbed in concentrated tetraoxosulphate(VI) acid to form oleum, which is then diluted:

\[\mathrm{SO_3(g)+H_2SO_4(l)\rightarrow H_2S_2O_7(l)}\]

\[\mathrm{H_2S_2O_7(l)+H_2O(l)\rightarrow 2H_2SO_4(aq)}\]

(d)(ii) Uses of tetraoxosulphate(VI) acid include:

• manufacture of fertilizers, such as ammonium sulphate and superphosphate;
• manufacture of detergents, paints and pigments.

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(a)(i) Addition polymerization. It is the process in which many small unsaturated molecules (monomers containing C=C double bonds) join together to form a single large molecule (the polymer) without the loss of any small molecule.

(a)(ii) Type of compounds that undergo it: unsaturated organic compounds, that is alkenes (and other molecules containing a carbon-to-carbon double bond).

(a)(iii) Two factors affecting the strength of polymers:

• the chain length (relative molecular mass) of the polymer;
• the degree of branching / cross-linking between the chains (and the strength of intermolecular forces between chains).

(b) Reaction pathways of ethanol. Reading the diagram: ethanol (C₂H₅OH) reacts with propanoic acid (CH₃CH₂COOH) and an H⁺ catalyst to give X at the top; step A converts ethanol to CH₃COOH; step B converts ethanol to C₂H₄.

(b)(i) Product X. Ethanol and propanoic acid undergo esterification, so X is ethyl propanoate.

\[ \text{CH}_3\text{CH}_2\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{H}^+} \text{CH}_3\text{CH}_2\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]

Structural formula of X: CH₃–CH₂–C(=O)–O–CH₂–CH₃

(b)(ii) Reagent for step A (ethanol \(\rightarrow\) ethanoic acid): acidified potassium dichromate(VI), K₂Cr₂O₇/H⁺ (acidified KMnO₄ is also acceptable), used as an oxidising agent with heat under reflux.

(b)(iii) Type of reaction in step B (ethanol \(\rightarrow\) ethene): a dehydration (elimination) reaction, in which a molecule of water is removed.

\[ \text{C}_2\text{H}_5\text{OH} \xrightarrow[180^{\circ}\text{C}]{\text{conc. H}_2\text{SO}_4} \text{C}_2\text{H}_4 + \text{H}_2\text{O} \]

(c) Complete combustion of ethanol.

\[ \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \]

From the equation, 1 mole of ethanol requires 3 moles of oxygen. At s.t.p. the volume of oxygen needed to burn one mole of ethanol completely is:

\[ V(\text{O}_2) = 3 \times 22.4 = 67.2\ \text{dm}^3 \]

(For a given mass \(m\) of ethanol, molar mass \(=46\), the volume is \(\dfrac{m}{46}\times 3\times 22.4\ \text{dm}^3\).)

(d)(i) Two substances produced when coal is heated in the absence of air: coke (solid residue) and coal gas; coal tar and ammoniacal liquor are also produced.

(d)(ii) Name of the process: destructive distillation of coal.

(d)(iii) Importance of the non-volatile residue (coke) to the iron and steel industry: coke is used in the blast furnace both as the fuel that supplies heat and as the reducing agent that reduces iron(III) oxide to iron:

\[ \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \]

(the carbon monoxide reducer is itself generated from the burning coke).

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(a) Why carbon(IV) oxide is used to extinguish fire

• It is denser (heavier) than air, so it settles over the burning material and forms a blanket that cuts off the supply of oxygen (air) needed for burning.
• It does not support combustion and is itself non-combustible, so once it covers the flame the fire goes out.

(b) Water-softening action of a cation-exchange resin

A cation-exchange resin carries loosely held sodium ions, \(Na^+\) (or \(H^+\)) on its surface. When hard water containing the \(Ca^{2+}\) and \(Mg^{2+}\) ions that cause hardness passes through the resin, the resin exchanges its \(Na^+\) ions for the \(Ca^{2+}\) and \(Mg^{2+}\) ions, which are held by the resin: \[2NaR + Ca^{2+} \to CaR_2 + 2Na^+\] The water leaving the resin now contains only \(Na^+\) ions, which do not cause hardness, so the water is softened. The exhausted resin is regenerated by running concentrated brine (sodium chloride solution) through it.

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(a) Electronic configuration of the element with atomic number 15 (phosphorus)

Filling the sub-levels in order of increasing energy:

That is, \(2, 8, 5\) in the main shells. The three 3p electrons occupy the three p-orbitals singly (Hund's rule): \(3p_x^{1}\,3p_y^{1}\,3p_z^{1}\).

(b) Complex formed between ammonia and copper(II) ions

Formula: \([Cu(NH_3)_4]^{2+}\) (the tetraamminecopper(II) ion).

Colour: deep (royal) blue.

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(a) Fraction that distills over first

The fraction that distills over first is nitrogen.

Reason: Fractional distillation separates the components according to their boiling points, the component with the lower boiling point (the more volatile one) boiling off and distilling over first. Nitrogen has a lower boiling point (about \(-196\,^{\circ}C\)) than oxygen (about \(-183\,^{\circ}C\)). Therefore, as the liquefied air warms up, nitrogen reaches its boiling point first, vaporizes and distills over before oxygen.

(b) Another industrial application of fractional distillation

Fractional distillation is used in the refining of petroleum (crude oil), where crude oil is separated into useful fractions such as petrol, kerosene, diesel and lubricating oil according to their different boiling ranges. (It is also used to separate ethanol from a fermented ethanol-water mixture.)

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(a) Balancing the nuclear equation

Balancing the mass numbers: \(238 = 234 + A \Rightarrow A = 4\).

Balancing the atomic numbers: \(92 = 90 + Z \Rightarrow Z = 2\).

So \(Y\) has mass number 4 and atomic number 2, i.e. \(^{4}_{2}He\). Y is an alpha particle (a helium nucleus), and the process is alpha decay:

(b) Cathode-ray experiment: observations and deductions

(A further acceptable pair: the rays turn a small paddle wheel placed in their path, showing the particles have mass and momentum.)

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(a) Increasing order of electronegativity of the first three halogens

The first three members of the halogen family are fluorine, chlorine and bromine. In increasing order of electronegativity:

Reason: Electronegativity is the tendency of an atom to attract the shared (bonding) pair of electrons towards itself. Down the group the atomic size (number of shells) increases, so the bonding electrons are farther from the nucleus and are more shielded from it. The effective nuclear attraction for the shared electrons therefore decreases down the group. Since fluorine is the smallest atom its nucleus attracts electrons most strongly, while bromine, being the largest of the three, attracts them least. Hence electronegativity increases from bromine to chlorine to fluorine.

(b) Reaction of chlorine with starch-iodide paper

Observation: the starch-iodide paper turns blue-black.

Explanation: Chlorine is a stronger oxidizing agent (more electronegative) than iodine, so it displaces/oxidizes the iodide ions in the paper to liberate free iodine:

The liberated iodine then reacts with the starch present in the paper to form the characteristic blue-black starch-iodine complex, confirming the presence of chlorine.

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(a) Type of solid structure

• (i) Diamond: giant covalent (atomic/macromolecular) structure. Each carbon atom is covalently bonded tetrahedrally to four other carbon atoms.
• (ii) Iodine: simple molecular (molecular crystal) structure. Discrete \(I_2\) molecules are held together in the lattice by weak van der Waals forces.
• (iii) Sodium chloride: giant ionic (ionic crystal) structure. \(Na^+\) and \(Cl^-\) ions are held in a regular lattice by strong electrostatic (ionic) forces.

(b)

• (i) One alloy of tin: solder (an alloy of tin and lead). Other acceptable answers are bronze (tin and copper) or pewter (tin, antimony and copper).
• (ii) A common reducing agent that is a compound of tin: tin(II) chloride, \(SnCl_2\). The \(Sn^{2+}\) ion is readily oxidized to \(Sn^{4+}\), so it acts as a reducing agent.

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