WAEC SSCE - Chemistry - 1994

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Glucose can be obtained from starch by hydrolysis. Hydrolysis is a chemical reaction in which water is used to break down a compound into smaller molecules. In the case of starch, hydrolysis breaks down the long chains of glucose molecules into individual glucose molecules. This reaction is often catalyzed by enzymes such as amylase, which are found in saliva and the pancreas. The resulting glucose molecules can then be used by the body as a source of energy.

Bayanin Amsa

The compound that does not contain the same type of bonds as the others is hydrogen chloride. Hydrogen chloride is a covalent compound, while the others are all ionic compounds. Covalent bonds are formed by the sharing of electrons between atoms, while ionic bonds are formed by the transfer of electrons from one atom to another. In hydrogen chloride, the hydrogen atom and chlorine atom share electrons to form a covalent bond. In contrast, in sodium chloride, magnesium chloride, potassium chloride, and lithium chloride, the metal atom donates electrons to the nonmetal atom to form an ionic bond. Therefore, all the other compounds contain ionic bonds, while hydrogen chloride contains a covalent bond.

Bayanin Amsa

The electronic structure of a noble gas has a completely filled outermost shell. Among the options, only 1s²2s²2p⁶3s²3p⁶, has a completely filled outermost shell. This electronic configuration corresponds to the noble gas argon (Ar). Therefore, is the correct answer.

Bayanin Amsa

Bayanin Amsa

The volume of a gas is directly proportional to its temperature, provided the pressure and the amount of gas remain constant. This relationship is known as Charles's law. According to Charles's law, when the temperature of a gas is increased, its volume will also increase, and when the temperature of a gas is decreased, its volume will also decrease. In this case, the initial volume of the gas is 27.3cm³ at 0^oC. If the temperature is increased to 10^oC while the pressure remains constant, we can use Charles's law to calculate the new volume of the gas. Charles's law states that the ratio of the initial volume to the initial temperature is equal to the ratio of the final volume to the final temperature. In other words: (V1 / T1) = (V2 / T2) Where: V1 = initial volume T1 = initial temperature V2 = final volume (unknown) T2 = final temperature Substituting the given values, we have: (27.3 / 273) = (V2 / 283) Solving for V2, we get: V2 = (27.3 / 273) x 283 V2 ≈ 28.3cm³ Therefore, the volume of the gas at 10^oC, pressure remaining constant, is approximately 28.3cm³. Hence, the correct option is (ii) 28.3cm³.

Bayanin Amsa

Bayanin Amsa

The question doesn't provide options to choose from, but based on the given equation, hydrogen sulfide (H2S) is a reducing agent. A reducing agent is a substance that causes another substance to be reduced by donating electrons to it. In the given equation, hydrogen sulfide (H2S) is donating electrons to sulfuric acid (H2SO4), which causes sulfuric acid to be reduced, and hydrogen sulfide is oxidized in the process. Therefore, hydrogen sulfide is a reducing agent in the given equation.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The type of process involved in the reverse reaction is hydrolysis. Hydrolysis is a chemical reaction in which water is used to break down a compound. In the given reaction, an alkanol and an alkanoic acid react in the presence of concentrated sulfuric acid to form an alkanoate and water. The reverse reaction involves the breakdown of the alkanoate using water to form the alkanol and alkanoic acid. This process is called hydrolysis.

Bayanin Amsa

The oxidation number of phosphorus in PO3/4 is +5. To determine the oxidation number of phosphorus, we can use the fact that the sum of the oxidation numbers of all atoms in a neutral compound is zero. Since oxygen has an oxidation number of -2, we can set up the equation: x + 4(-2) = 0 where x is the oxidation number of phosphorus. Solving for x gives: x = +5 Therefore, the oxidation number of phosphorus in PO3/4 is +5.

Bayanin Amsa

Bayanin Amsa

Tetraoxosulphate (IV) acid, also known as sulfuric acid, is considered a strong acid because it completely ionizes in water. When sulfuric acid dissolves in water, it breaks down completely into hydrogen ions (H+) and sulfate ions (SO4^2-). This means that all the sulfuric acid molecules present in the solution donate their hydrogen ions to the water, resulting in a high concentration of H+ ions in the solution. These H+ ions can react with other substances, making sulfuric acid a powerful and reactive acid. Therefore, - "ionized in water" - is the correct answer.

Bayanin Amsa

The first step to solving this problem is to determine the number of moles of carbon-12 present in 1.2g of the isotope. We do this by dividing the mass of the isotope by its molar mass. The molar mass of carbon-12 is 12 g/mol. Number of moles of carbon-12 = 1.2 g / 12 g/mol = 0.1 mol Since each mole of carbon-12 contains Avogadro's number of atoms (6.02 x 10^23), we can determine the number of atoms in 0.1 mol of carbon-12 as: Number of atoms = 0.1 mol x 6.02 x 10^23 atoms/mol = 6.02 x 10^22 atoms Therefore, the answer is 6.02 x 10^22.

Bayanin Amsa

When water is dropped on calcium carbide, the gaseous product formed is acetylene (C2H2), which is an alkyne. Calcium carbide (CaC2) reacts with water (H2O) to produce acetylene gas and calcium hydroxide (Ca(OH)2) according to the following equation: CaC2 + 2H2O → C2H2 + Ca(OH)2 Acetylene gas is a colorless, highly flammable gas with a distinctive odor. It is commonly used in welding and cutting torches, as well as in the production of various organic chemicals.

Bayanin Amsa

The measure of the degree of disorder or randomness in a chemical system is known as entropy. Entropy is a thermodynamic quantity that describes the number of ways in which the energy of a system can be distributed among its particles. In simpler terms, it is a measure of how much the energy of a system is spread out or dispersed. The greater the entropy of a system, the more disordered or random it is. Entropy is usually denoted by the symbol S and has units of joules per kelvin (J/K).

Bayanin Amsa

Calcium is usually extracted by the electrolysis of its chloride. Electrolysis is a process that involves the use of electrical energy to break down a compound into its constituent elements. In the case of calcium, the most common compound that it is extracted from is calcium chloride (CaCl2). Calcium chloride is an ionic compound, which means that it consists of positively charged calcium ions and negatively charged chloride ions. During electrolysis, an electric current is passed through a molten or dissolved compound, causing the ions to move towards the electrodes. At the negative electrode (cathode), the positively charged calcium ions gain electrons and are reduced to form calcium atoms. At the positive electrode (anode), the negatively charged chloride ions lose electrons and are oxidized to form chlorine gas. The overall reaction during the electrolysis of calcium chloride is: CaCl2(l) → Ca(l) + Cl2(g) This process is used to extract calcium on an industrial scale, as it is an efficient method for producing large quantities of calcium.

Bayanin Amsa

Bayanin Amsa

Halogens are a group of chemical elements that include fluorine, chlorine, bromine, iodine, and astatine. These elements share similar chemical properties and can form salts with metals. Silicon, on the other hand, is not a halogen. It is a metalloid element that is commonly found in the Earth's crust. Therefore, is the correct answer as silicon is not a halogen.

Bayanin Amsa

When naphthalene changes from the solid state directly to the gaseous state upon heating, it undergoes sublimation. Sublimation is the process in which a solid directly changes into a gas without going through the liquid state. This occurs when the vapor pressure of the solid is greater than the atmospheric pressure. In the case of naphthalene, it has a high vapor pressure at room temperature and atmospheric pressure, which allows it to undergo sublimation when heated.

Bayanin Amsa

Grinding the metal into powder would increase the rate of reaction of chips of a metal M in the given equation. The rate of reaction depends on the surface area of the reactants in contact with each other. When the metal is in the form of chips, the surface area in contact with the hydrochloric acid is relatively small, and the rate of reaction is slow. However, by grinding the metal into powder, the surface area of the metal in contact with the acid increases significantly, allowing more metal to react at a faster rate. Therefore, grinding the metal into powder increases the rate of reaction. The other options would not increase the rate of reaction of chips of a metal M in the given equation. Diluting the hydrochloric acid or adding distilled water would decrease the concentration of the acid, which would slow down the reaction rate. Decreasing the temperature would decrease the kinetic energy of the molecules, leading to a slower rate of reaction. Increasing the pressure would not have a significant effect on the rate of the reaction in this case.

Bayanin Amsa

Bayanin Amsa

Destructive distillation of coal means heating coal in the absence of air. This process breaks down the complex organic compounds in coal into simpler molecules such as gases, liquids, and solids. These products are then collected and can be used for a variety of purposes such as the production of coal gas, tar, and coke. This process is called destructive distillation because the coal is being broken down or "destroyed" into simpler substances through the process of heating.

Bayanin Amsa

When an aqueous solution of sodium hydroxide (NaOH) is electrolyzed using graphite electrodes, the following occurs: At the cathode (negative electrode), hydrogen gas is produced through the reduction of water molecules. This is because water has a lower reduction potential than Na+ ions in the solution. 2H2O(l) + 2e- -> H2(g) + 2OH-(aq) At the anode (positive electrode), the hydroxide ions (OH-) are oxidized to form oxygen gas and water molecules. This is because OH- ions have a higher oxidation potential than water molecules. 4OH-(aq) -> 2H2O(l) + O2(g) + 4e- Therefore, the correct option is (c) Oxygen gas is produced at the anode.

Bayanin Amsa

The positive charge of an element in the Periodic Table is determined by the number of protons in its atom. This is because the positive charge of an atom is equal to the number of protons in its nucleus, which is also known as its atomic number. The atomic number is the defining characteristic of an element and determines its position in the Periodic Table. Each element has a unique atomic number which corresponds to the number of protons in its nucleus. Therefore, the number of protons in an atom determines its positive charge and its position in the Periodic Table.

Bayanin Amsa

The correct answer is (c) C_nH_2n+2. When alkenes undergo catalytic hydrogenation, each double bond in the molecule is converted into a single bond, and two hydrogen atoms are added to the molecule. This process is a type of addition reaction, and the general formula of the resulting compounds is C_nH_2n+2. Therefore, is the correct answer.

Bayanin Amsa

Bayanin Amsa

The property of magnesium oxide that is not correct is "low binding energy." Magnesium oxide is an ionic compound made up of magnesium and oxygen ions. It has a high melting point and possesses a crystal lattice structure due to the presence of ionic bonds between the ions. Magnesium oxide is also known to dissolve in polar solvents, such as water, due to its ionic nature. However, it is not known for having low binding energy, as it has a strong ionic bond between the magnesium and oxygen ions. Therefore, the property of magnesium oxide that is not correct is "low binding energy."

Bayanin Amsa

The energy of motion possessed by the reacting species in a chemical reaction is known as kinetic energy. Kinetic energy is the energy of motion. When a substance is in motion, it possesses kinetic energy. In a chemical reaction, the reactants are in motion and collide with each other, causing the chemical reaction to occur. This motion is due to the kinetic energy of the particles. The kinetic energy of the particles in a substance depends on its temperature. As the temperature of a substance increases, the kinetic energy of the particles also increases, causing them to move faster and collide more frequently. This, in turn, increases the rate of the chemical reaction. Therefore, the reacting species in a chemical reaction possess kinetic energy, which is the energy of motion.

Bayanin Amsa

Alkanols, also known as alcohols, have unexpectedly high boiling points relative to their molar masses because of intermolecular hydrogen bonding. Hydrogen bonding is a type of intermolecular force that occurs when a hydrogen atom bonded to an electronegative atom, such as oxygen or nitrogen, is attracted to another electronegative atom in a nearby molecule. In the case of alkanols, the hydrogen atom attached to the oxygen atom in one molecule can form a hydrogen bond with the oxygen atom of another molecule. This results in a network of intermolecular hydrogen bonds that are much stronger than the weak van der Waals' forces that exist between non-polar molecules of similar size and shape. Due to the presence of these strong intermolecular forces, more thermal energy is required to overcome these forces and cause the molecules to separate, resulting in higher boiling points of alkanols relative to other organic compounds of similar molar mass.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The statement "In any chemical reaction, the total mass of the products is always equal to that of the reactants" is the statement of the law of conservation of matter. This law, also known as the law of conservation of mass, states that in a closed system, the total mass of the system remains constant during a chemical reaction. This means that the mass of the reactants before the reaction is the same as the mass of the products after the reaction. In other words, matter cannot be created or destroyed in a chemical reaction; it can only be transformed from one form to another. This law is a fundamental principle of chemistry and is essential in understanding the behavior of matter in chemical reactions. It was first proposed by Antoine Lavoisier, a French chemist in the late 18th century, and has since become one of the most important principles in chemistry.

Bayanin Amsa

Universal indicator solution is a mixture of different indicators that changes color over a range of pH values. The best method to separate the components of universal indicator solution is chromatography. Chromatography is a method used to separate different components of a mixture based on their physical and chemical properties. In this case, the components of universal indicator solution can be separated based on their different solubility and adsorption properties. A small amount of the solution is placed on a chromatography paper and allowed to separate through the paper by capillary action. The different components of the mixture will travel at different rates based on their solubility in the solvent and their interaction with the paper. As a result, the different components can be separated and identified by their distinct color changes as they travel through the paper. Filtration, evaporation, crystallization, and fractional distillation are all separation methods that are based on the physical properties of the components of a mixture, such as their boiling points or solubility in a particular solvent. These methods are not suitable for separating the components of universal indicator solution, which are all indicators with similar physical properties.

Bayanin Amsa

(a) Oxidation (in terms of electron transfer)

Oxidation is the loss of electrons by an atom, ion or molecule.

(b) \( Cu_{(s)} + 2Ag^+_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)} \)

(i) The species that is reduced is Ag⁺, because it gains electrons (its oxidation state falls from +1 to 0).

(ii) Half-cell equations:

Oxidation: \[ Cu \to Cu^{2+} + 2e^- \]

Reduction: \[ 2Ag^+ + 2e^- \to 2Ag \]

Bayanin Amsa

(a) Chemical differences between metals and non-metals

• Metals form basic oxides (and hydroxides), whereas non-metals form acidic oxides.
• Metals are electropositive and lose electrons to form positive ions (acting as reducing agents), whereas non-metals are electronegative and gain electrons to form negative ions (acting as oxidising agents).

(b) Two general methods of extracting metals from their ores

• Reduction of the metal oxide with carbon (coke), used for moderately reactive metals such as iron and zinc.
• Electrolysis of the molten ore or compound, used for very reactive metals such as sodium and aluminium.

Bayanin Amsa

(a) (i) Heat of combustion is the enthalpy change when one mole of a substance is completely burnt in excess oxygen under standard conditions.

(a) (ii) The container used is a calorimeter.

(b) (i) The energy profile for the combustion reaction is shown below.

(b) (ii)

The thermochemical equation is:

\[\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}\qquad \Delta H = -3935\ \mathrm{kJ\ mol^{-1}}\]

\[n(\mathrm C)=\frac{60}{12}=5\ \mathrm{mol}\]

\[\Delta H = 5\times(-3935)=-19675\ \mathrm{kJ}\]

Therefore, the heat change is \(-19675\ \mathrm{kJ}\), that is, 19675 kJ of heat is evolved.

(b) (iii) Strong acids and strong bases ionize completely in water. Thus, every neutralization between them has the same net ionic equation:

\[\mathrm{H^+(aq)+OH^-(aq)\rightarrow H_2O(l)}\]

Since the same reaction occurs in each case, the enthalpy of neutralization is constant.

(c) (i) Rusting is a slow combustion process because iron combines slowly with oxygen, in the presence of water, to form iron oxide. Like combustion, it is an oxidation process and is exothermic.

(c) (ii) Iron filings have a much larger surface area than iron nails for the same mass. More of the iron is exposed to air and moisture, so rusting occurs faster.

(c) (iii) Zinc is more reactive than iron and provides sacrificial protection. If the zinc coating is damaged, zinc oxidizes in preference to iron. Tin is less reactive than iron; therefore, if a tin coating is damaged, iron becomes the anode and rusts more rapidly.

Bayanin Amsa

(a)(i) Laboratory preparation of a dry sample of chlorine

Heat manganese(IV) oxide with concentrated hydrochloric acid. The chlorine produced is passed through saturated brine (or aqueous potassium hydrogencarbonate) to remove hydrogen chloride, then through concentrated sulphuric acid to dry it. It is collected in an upright gas jar by downward displacement of air, since chlorine is denser than air.

\[\mathrm{MnO_2(s)+4HCl(conc)\rightarrow MnCl_2(aq)+Cl_2(g)+2H_2O(l)}\]

(ii) Test for chlorine

Hold moist starch-iodide paper in the gas. It turns blue-black because chlorine liberates iodine from iodide ions.

\[\mathrm{Cl_2+2I^-\rightarrow 2Cl^-+I_2}\]

(b) Equations for the reactions of chlorine

(i) With iron(II) chloride solution:

\[\mathrm{Cl_2+2FeCl_2\rightarrow 2FeCl_3}\]

(ii) With potassium iodide solution:

\[\mathrm{Cl_2+2KI\rightarrow 2KCl+I_2}\]

(iii) With hot concentrated sodium hydroxide solution:

\[\mathrm{3Cl_2+6NaOH\rightarrow 5NaCl+NaClO_3+3H_2O}\]

(c)

(i) A white precipitate of lead(II) chloride is formed when hydrogen chloride gas is bubbled into lead(II) trioxonitrate(V) solution.

\[\mathrm{Pb(NO_3)_2(aq)+2HCl(aq)\rightarrow PbCl_2(s)+2HNO_3(aq)}\]

(ii) The white precipitate dissolves on heating to boiling. On cooling, white crystals of lead(II) chloride reappear.

(d)

A white precipitate or milky suspension of bismuth oxychloride, \(\mathrm{BiOCl}\), forms when water is added.

Adding water dilutes the hydrochloric acid and shifts the equilibrium to the left. Soluble \(\mathrm{BiCl_3}\) is thereby converted into insoluble white \(\mathrm{BiOCl}\):

\[\mathrm{BiOCl(s)+2HCl(aq)\rightleftharpoons BiCl_3(aq)+H_2O(l)}\]

Bayanin Amsa

(a)(i) Maximum electrons in the M shell (principal quantum number n = 3): \( 2n^2 = 2 \times 3^2 = \mathbf{18} \).

(ii) Formation of sodium fluoride

Sodium (2, 8, 1) loses its one outer electron to become Na⁺ (2, 8); fluorine (2, 7) gains that electron to become F^- (2, 8). Both attain the stable octet:

\[ Na \to Na^+ + e^- \qquad F + e^- \to F^- \qquad Na^+ + F^- \to NaF \]

A dot-and-cross diagram showing the single electron transferring from Na to F is expected.

(iii) Properties of NaF (ionic compound): high melting and boiling points; conducts electricity when molten or dissolved in water; a hard, crystalline solid that is soluble in water.

(b)

(i) \[ 4Na + O_2 \to 2Na_2O \]

(ii) \[ Na_2O + H_2O \to 2NaOH \]

(c) Comparing the conductance of 1 mol dm^-3 NaOH and ethanoic acid

Set up a simple conductivity cell: two clean carbon (or platinum) electrodes connected through a battery and a bulb (or ammeter). Dip the electrodes into a fixed volume of 1 mol dm^-3 NaOH and note the brightness of the bulb (or the ammeter reading). Rinse the electrodes and repeat with the same volume of 1 mol dm^-3 ethanoic acid. The bulb glows brightly (large current) with NaOH, which is a strong electrolyte (fully ionised), but only dimly (small current) with ethanoic acid, which is a weak electrolyte (only partially ionised). Hence NaOH conducts better.

Bayanin Amsa

(a) Avogadro's law

Equal volumes of all gases, measured at the same temperature and pressure, contain the same number of molecules.

(b) States of matter

• (i) Particles that can move readily and flow (but are still fairly close together): liquid.
• (ii) Particles held firmly together by strong forces of cohesion: solid.
• (iii) Particles in rapid, random motion and far apart: gas.

Bayanin Amsa

The equation shown on the diagram is the action of hot concentrated tetraoxosulphate(VI) acid on ethanedioic (oxalic) acid:

\[ \underset{\text{(COOH)}_2}{\text{HOOC-COOH}} \;\xrightarrow[\;\Delta\;]{\text{conc. } H_2SO_4}\; CO_2 + CO + H_2O \]

(a) Type of process involved

The process is dehydration. The concentrated tetraoxosulphate(VI) acid acts as a dehydrating agent: it removes the elements of water (\(H_2O\)) from each mole of ethanedioic acid, leaving behind the two oxides of carbon, \(CO_2\) and \(CO\). Notice that the atoms balance exactly, \((COOH)_2 = C_2H_2O_4\) splits into \(CO_2 + CO + H_2O\), so no oxidation state change is imposed by the acid; it simply strips out water.

(b) Basicity of ethanedioic acid

The basicity of an acid is the number of replaceable (ionisable) hydrogen atoms it releases per molecule in aqueous solution. Ethanedioic acid, \((COOH)_2\), contains two carboxyl (-COOH) groups, each supplying one ionisable hydrogen:

\[ (COOH)_2 \rightleftharpoons 2H^+ + (COO)_2^{2-} \]

Therefore its basicity is 2 (it is a dibasic / diprotic acid).

(c) Three differences in the chemical properties of the two oxides of carbon (\(CO_2\) and \(CO\))

A further valid distinction: \(CO_2\) turns lime water milky (forming \(CaCO_3\)), whereas \(CO\) does not; and \(CO\) is highly poisonous because it combines with haemoglobin, while \(CO_2\) is not poisonous in that way.

Bayanin Amsa

(a)(i) The loss in mass of a hydrated crystal (loss of water of crystallisation) to the atmosphere is called efflorescence.

(ii) Another example: sodium trioxocarbonate(IV) decahydrate, Na₂CO₃·10H₂O (washing soda).

(b) Finding X in MgSO₄·XH₂O

Molar mass of the hydrate \( = \dfrac{\text{mass}}{\text{moles}} = \dfrac{123}{0.50} = 246\ \text{g mol}^{-1} \)

Molar mass of anhydrous MgSO₄ \( = 24 + 32 + (4\times16) = 120\ \text{g mol}^{-1} \)

Mass of water \( = 246 - 120 = 126\ \text{g} \)

\[ X = \frac{126}{18} = \mathbf{7} \]

The compound is MgSO₄·7H₂O.

Bayanin Amsa

Thermal decomposition: \( 2KClO_3 \to 2KCl + 3O_2 \)

(a)(i) Substance to increase the rate of the reaction (a catalyst): manganese(IV) oxide, MnO₂.

(ii) Substance to absorb the oxygen produced: alkaline pyrogallol solution (pyrogallol in KOH).

(b) An aqueous solution of silver trioxonitrate(V) gives a white precipitate (AgCl) with KCl because KCl contains free chloride ions, Cl^-:

\[ Ag^+ + Cl^- \to AgCl\downarrow \]

KClO₃ gives no precipitate because its chlorine is locked in the chlorate ion, ClO₃^- (not free Cl^-), and silver chlorate, AgClO₃, is soluble in water. Only free chloride ions form the insoluble AgCl.

Bayanin Amsa

(a) Examples of naturally occurring substances

• (i) Conforms to the general formula C_x(H₂O)_y (a carbohydrate): glucose (or starch, cellulose, sucrose).
• (ii) Contains the carboxyl group (-COOH) as its functional group: a fatty acid such as stearic acid (or citric acid found in citrus fruits).

(b) Names of the processes

• (i) Obtaining paraffin oil (kerosene) from crude oil: fractional distillation.
• (ii) Obtaining benzene from ethyne: polymerisation (cyclic trimerisation), \(3C_2H_2 \to C_6H_6\).

Bayanin Amsa

Reading the shortened Periodic Table. The columns are the groups, labelled I, II (on the far left), then the wide transition-metal block, then III, IV, V, VI, VII, 0 (on the far right). The rows 1, 2, 3, 4 are the periods. From the diagram the five elements sit as follows:

• A: far top-right, in Group 0 (period 1) - a noble gas.
• B: on the right, in the Group VII region (period 3) - a non-metal.
• C: far left, in Group I (period 3) - an alkali metal.
• D: on the left, in Group II (period 4) - an alkaline earth metal.
• E: in the central transition block (period 4), between Groups II and III.

(a)(i) Transition metal. Transition metals occupy the block wedged between Group II and Group III. The only element sitting in that central block is E.

(a)(ii) Alkaline earth metal. The alkaline earth metals are the Group II elements. The element placed in the second column (Group II) is D.

(a)(iii) Least reactive. The least reactive elements are the noble gases of Group 0, which have stable, filled outer electron shells and rarely form compounds. The Group 0 element is A.

(a)(iv) Most electronegative. Electronegativity (the pull an atom exerts on a shared electron pair) increases across a period towards Group VII and decreases down a group; the halogens of Group VII are the most electronegative reactive elements (noble gases are excluded as they scarcely bond). The Group VII element here is B.

(b)(i) Bond type in D's oxide. Element D is a Group II metal, so it readily loses its two outer electrons to form a

D²⁺ cation, while oxygen (Group VI) gains two electrons to form the O^2- anion. A metal reacting with a non-metal by electron transfer gives an ionic (electrovalent) bond.

(b)(ii) Formula of the compound. Balancing the charges of the ions:

One D²⁺ exactly balances one O^2-, so the formula of the oxide is DO.

Bayanin Amsa

(a)(i) Hydrocarbons are compounds that contain only carbon and hydrogen.

(ii) Molecular formula

Carbon = 92.3%, so hydrogen = 100 - 92.3 = 7.7%.

C: \( \dfrac{92.3}{12} = 7.69 \)    H: \( \dfrac{7.7}{1} = 7.70 \)   ratio \(1:1\), so empirical formula = CH (mass 13).

Molar mass \( = 2 \times \text{vapour density} = 2 \times 39 = 78 \)

\( n = \dfrac{78}{13} = 6 \), so molecular formula \( = \mathbf{C_6H_6} \) (benzene).

(b)(i) Ethanol from cassava

Cassava is rich in starch. Grate the tubers and mix with water; hydrolyse the starch to glucose (using malt/diastase enzyme or by boiling with dilute acid then neutralising). Cool, add yeast and allow to ferment anaerobically for a few days at about 35°C; the enzyme zymase in yeast converts glucose to ethanol and carbon dioxide. Finally distil (fractional distillation) to concentrate the ethanol.

\[ C_6H_{12}O_6 \xrightarrow{\text{yeast}} 2C_2H_5OH + 2CO_2 \]

(ii) Reagents for oxidising ethanol to ethanoic acid: acidified potassium dichromate(VI) (K₂Cr₂O₇/H₂SO₄); acidified potassium tetraoxomanganate(VII) (KMnO₄/H₂SO₄).

(c) Names of processes

• (i) Conversion of alkanols to alkanoates: esterification.
• (ii) Breakdown of proteins to amino acids: hydrolysis.
• (iii) Conversion of oils to fats: hydrogenation.
• (iv) Alkaline hydrolysis of fats and oils: saponification.

Bayanin Amsa

(a) Action of strong heat

(i) Potassium trioxonitrate(V) (potassium nitrate) decomposes to the nitrite and oxygen:

\[ 2KNO_3 \to 2KNO_2 + O_2 \]

(ii) Sucrose chars (is dehydrated) to carbon and water vapour:

\[ C_{12}H_{22}O_{11} \to 12C + 11H_2O \]

(b) Granulated zinc with dilute tetraoxosulphate(VI) acid

• Effervescence occurs as bubbles of a colourless gas (hydrogen) are given off and the zinc gradually dissolves/gets smaller.
• The mixture warms up (the reaction is exothermic), leaving a colourless solution of zinc sulphate.

\[ Zn + H_2SO_4 \to ZnSO_4 + H_2 \]

Bayanin Amsa

(a) Entropy change

• (i) Condensation of steam (gas → liquid): entropy decreases.
• (ii) Melting of wax (solid → liquid): entropy increases.
• (iii) Dissolution of sugar in water: entropy increases.
• (iv) Adsorption of a gas on charcoal: entropy decreases (gas particles become ordered on the surface).

(b) Free energy change, ΔG

• (i) ΔG = zero: the reaction is at equilibrium.
• (ii) ΔG negative: the reaction is spontaneous (feasible) in the forward direction.