WAEC SSCE - Chemistry - 2011

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The element that can conveniently be placed in two groups in the periodic table is hydrogen. Hydrogen is the first element in the periodic table and has an atomic number of 1. It is located in group 1 of the periodic table because it has only one valence electron and shares many chemical properties with the alkali metals, such as lithium and sodium. However, hydrogen also shares some chemical properties with the halogens, such as fluorine and chlorine, which are located in group 17 of the periodic table. Like the halogens, hydrogen can form a single covalent bond by sharing one electron with another atom, and it can also form negative ions called hydrides, which have similar properties to halides. Therefore, hydrogen can be placed in two groups in the periodic table: group 1 and group 17. However, its properties are more similar to the alkali metals than the halogens, so it is generally classified as a nonmetal or a metalloid rather than a halogen. In conclusion, hydrogen is the element that can conveniently be placed in two groups in the periodic table: group 1 and group 17.

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A displacement reaction is a type of chemical reaction where an element or a group of atoms of a molecule is replaced by another element or group of atoms. The correct statement about a displacement reaction is that a more electropositive element displaces a less electropositive one. This is because an element with higher electropositivity has a greater tendency to lose electrons and form positive ions, making it more reactive. Therefore, it can easily displace a less electropositive element from its compound by forming a new compound with it. The position of elements in the reactivity series plays a crucial role in determining the outcome of a displacement reaction. The higher an element is in the reactivity series, the more likely it is to displace a lower element. Finally, displacement reactions can occur at any point during the reaction, and they do not require the reaction to be at equilibrium.

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The number of atoms in one mole of a substance is equal to Avogadro's number. Avogadro's number is a constant, denoted as 'N\(_A\)', which represents the number of particles (atoms, molecules, ions, etc.) in one mole of a substance. The value of Avogadro's number is approximately 6.022 x 10\(^{23}\) particles per mole. Therefore, if we know the molar mass of a substance (in grams per mole), we can calculate the number of atoms (or molecules) in one mole of that substance by multiplying the molar mass by Avogadro's number. This relationship is commonly used in chemistry to relate the mass of a substance to the number of particles it contains.

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Out of the four options given, graphite is the substance that is not an ore of metal. Graphite is a form of carbon and is not considered as an ore of metal. Bauxite is an ore of aluminum, cuprite is an ore of copper, and cassiterite is an ore of tin. Ores are naturally occurring minerals or rocks that contain metal or metal compounds in sufficient amounts to make it economically feasible to extract the metal.

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To prevent the corrosion of iron, a more reactive metal can be used as a sacrificial anode. This metal will undergo oxidation instead of iron, and hence, protect it from rusting. Out of the given options, magnesium is the most reactive metal, followed by zinc and aluminium. Therefore, magnesium can be used as a sacrificial anode to protect iron from corrosion. Copper, lead, and silver are less reactive and cannot be used as sacrificial anodes for preventing the corrosion of iron.

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To solve this problem, we need to use the ideal gas law, which relates the volume, pressure, temperature, and number of moles of a gas. At standard temperature and pressure (STP), which is defined as 0°C and 1 atmosphere of pressure, one mole of any gas occupies 22.4 dm³ of volume. First, we need to calculate the number of moles of H2S present in 17g of the gas. To do this, we divide the mass of the gas by its molar mass. Molar mass of H2S = (2 × atomic mass of H) + atomic mass of S = (2 × 1.00 g/mol) + 32.0 g/mol = 34.0 g/mol Number of moles of H2S = mass of H2S ÷ molar mass of H2S = 17g ÷ 34.0 g/mol = 0.5 mol Now, we can use the ideal gas law to calculate the volume of 0.5 mol of H2S at STP. PV = nRT where P = 1 atm, V is the volume we want to find, n = 0.5 mol, R is the gas constant (0.0821 L atm/K mol), and T = 273 K. Solving for V, we get: V = nRT/P = (0.5 mol) × (0.0821 L atm/K mol) × (273 K) ÷ (1 atm) = 11.2 L or dm³ Therefore, the volume occupied by 17g of H2S at STP is 11.2 dm³. Answer: 11.2 dm³.

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The bond formed when two electrons that are shared between two atoms are donated by only one of the atoms is called a dative bond. In this type of bond, also known as a coordinate covalent bond, one atom donates a pair of electrons to be shared with another atom, which accepts the electrons. The atom that donates the electrons is called the donor or Lewis base, while the atom that accepts the electrons is called the acceptor or Lewis acid. Dative bonds are commonly found in coordination compounds, where a metal ion accepts electron pairs from surrounding ligands to form a complex.

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The reaction equation given is X_(g) + Y_(g) ⇌ XY_(g); ∆H = +220KJ mol^-1. The ∆H value is positive which indicates that the reaction is endothermic, meaning that heat is absorbed when the reaction takes place. According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change in temperature, pressure, or concentration of reactants/products, the system will respond in a way that tends to counteract that change. In this case, if the temperature of the system is increased, the equilibrium position will shift in the direction that absorbs heat. Since the reaction is endothermic, the forward reaction will absorb heat, so the system will favor the forward reaction to absorb the excess heat. Therefore, the forward reaction would be favored if the temperature of the system is increased. Hence, the correct option is "forward reaction would be favored."

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An amphoteric oxide is one that can react as both an acid and a base. This means that it can react with both acids and bases to form salts. The only oxide listed that can do this is aluminum oxide (Al2O3). Therefore, the answer is AI₂O₃. MgO and Na₂O are basic oxides, which means they react with acids to form salts. Fe₂O₃ is an acidic oxide, which means it reacts with bases to form salts.

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The compound that determines the octane rating of petrol is 2,2,4-trimethylpentane. Octane rating is a measure of a fuel's ability to resist "knocking" during combustion, caused by the air/fuel mixture detonating spontaneously in the engine. The higher the octane rating, the more resistant the fuel is to knocking. The octane rating of a fuel is determined by comparing its performance to that of a mixture of 2,2,4-trimethylpentane and n-heptane, which are assigned octane ratings of 100 and 0, respectively. Therefore, 2,2,4-trimethylpentane is used as a standard reference for octane rating, and is one of the compounds that determines the octane rating of petrol.

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The equation P = \(\frac{K}{V}\) illustrates the relationship between pressure and volume of a gas at constant temperature. This is known as Boyle's law, which states that the pressure of a gas is inversely proportional to its volume when temperature is kept constant. In other words, as the volume of a gas decreases, its pressure will increase, and vice versa, as long as the temperature is held constant.

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The electrode potential is positive when electrons flow from the metal electrode, M_(s), to the hydrogen electrode, H_2(g). The electrochemical cell notation M(s) / M2+ (aq) // H+ (aq) / H2(g) represents a cell with a metal electrode M(s) in contact with a solution containing M2+ ions and a hydrogen electrode in contact with a solution containing H+ ions. The double vertical lines (||) represent the salt bridge or porous membrane used to separate the two solutions while allowing the flow of ions to maintain electrical neutrality. The electrode potential of a cell is a measure of the tendency of electrons to flow from one electrode to another. In this case, the electrode potential is positive when electrons flow from the metal electrode, M(s), to the hydrogen electrode, H2(g). This means that the metal electrode has a higher tendency to lose electrons and become oxidized, while the hydrogen electrode has a lower tendency to gain electrons and become reduced. When electrons flow from the metal electrode to the hydrogen electrode, a current is generated and the flow of electrons is accompanied by the flow of positive ions from the metal solution to the hydrogen solution, and vice versa. The flow of current is related to the electrode potential and the resistance of the circuit, and can be used to do work such as powering a light bulb or charging a battery. In conclusion, the electrode potential is positive when electrons flow from the metal electrode, M(s), to the hydrogen electrode, H2(g), in an electrochemical cell. This is because the metal electrode has a higher tendency to lose electrons and become oxidized, while the hydrogen electrode has a lower tendency to gain electrons and become reduced.

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The electron configuration 1s²2s²2p⁶ belongs to the noble gas neon (Ne). Elements in the noble gas family have a full outer shell of electrons, which makes them stable and unreactive. Combining power, also known as valency, is a measure of how many chemical bonds an element can form with other atoms. Noble gases have a combining power of 0 because they already have a full outer shell of electrons and do not need to gain or lose any electrons to achieve stability. Therefore, the answer is 0.

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The electron configuration 2,8,8 represents an atom with 18 electrons, which corresponds to the noble gas argon (Ar). To determine which ion has the same electron configuration, we need to look at the number of electrons that each ion has lost or gained compared to the neutral atom. Na+ has lost one electron, which means it now has 10 electrons, with a configuration of 2,8. This is not the same as 2,8,8, so Na+ is not the correct ion. Mg2+ has lost two electrons, which means it now has 10 electrons, with a configuration of 2,8. This is also not the same as 2,8,8, so Mg2+ is not the correct ion. F- has gained one electron, which means it now has 10 electrons, with a configuration of 2,8. This is not the same as 2,8,8, so F- is not the correct ion. Cl- has gained one electron, which means it now has 18 electrons, with a configuration of 2,8,8. Therefore, the correct answer is Cl-.

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The given equation represents a nuclear reaction, where the nuclei of two hydrogen isotopes, deuterium (D) and tritium (H), combine to form a helium nucleus and a neutron. This process is called nuclear fusion. Nuclear fusion is a type of nuclear reaction in which two atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy. In this reaction, the two hydrogen isotopes, deuterium (D) and tritium (H), combine to form helium-4 (He-4) and a neutron (n), with the release of a large amount of energy. So, the type of reaction represented by the given equation is Nuclear Fusion.

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The energy change that occurs when an electron is added to an isolated gaseous atom is called "electron affinity." This process involves adding an electron to a neutral atom, which then becomes a negatively charged ion. Electron affinity is a measure of how much energy is released or absorbed when this happens. If energy is released, the electron affinity is negative, meaning the atom is more likely to attract electrons. If energy is absorbed, the electron affinity is positive, meaning the atom is less likely to attract electrons. The electron affinity is related to the ionization energy, which is the energy required to remove an electron from an atom.

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Aluminium is extracted from bauxite using the process of electrolysis. Bauxite is a naturally occurring mineral that contains a mixture of aluminium, iron, titanium, and silica. The first step in the extraction of aluminium from bauxite involves the purification of bauxite by removing any impurities. This is done by treating the bauxite with sodium hydroxide solution to produce a solution of sodium aluminate. Next, the purified solution of sodium aluminate is subjected to electrolysis in a cell called Hall-Heroult cell. In this cell, the purified solution of sodium aluminate is mixed with molten cryolite, which acts as a solvent to lower the melting point of the aluminium oxide. The mixture is then electrolyzed at high temperatures to separate the aluminium from the other elements present in the solution. During the electrolysis process, aluminium ions are reduced at the cathode, forming liquid aluminium metal, while oxygen is produced at the anode. The liquid aluminium is then siphoned off from the bottom of the cell and sent for further processing to obtain pure aluminium metal. Therefore, the correct answer is bauxite.

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Avogadro's constant is the number of particles in one mole of a substance, which is approximately equal to 6.02 x 10^23 particles. Therefore, if we have 1.20 x 10^24 particles of magnesium, we can calculate the number of moles by dividing the number of particles by Avogadro's constant: n = (1.20 x 10^24 particles) / (6.02 x 10^23 particles/mole) n = 2.0 moles Therefore, the amount of magnesium that would contain 1.20 x 10^24 particles is 2.0 moles. So, the correct option is "2.0 moles".

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The compound that would react with ethanoic acid to give a sweet smelling liquid is an alkanol. An alkanol is an organic compound that has a hydroxyl (-OH) group attached to an alkyl group. When an alkanol such as ethanol (C2H5OH) reacts with ethanoic acid (CH3COOH), the reaction is an esterification reaction, which produces an ester and water. The reaction is catalyzed by the presence of an acid, such as sulfuric acid (H2SO4). The general equation for this reaction is: alcohol + carboxylic acid ⇌ ester + water In this case, the reaction between ethanol and ethanoic acid produces the ester ethyl acetate, which has a sweet, fruity odor. This sweet smelling liquid is used in the production of perfumes, artificial flavors, and solvents. Therefore, the compound that would react with ethanoic acid to give a sweet smelling liquid is an alkanol, which has a hydroxyl group (-OH) attached to an alkyl group.

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A gas that is collected by upward delivery is likely to be lighter than air. This is because upward delivery involves collecting the gas in an inverted container with its open end downwards in a trough filled with water. The gas being collected is less dense than air, and thus it rises upwards and displaces the air in the container, eventually filling it up. The principle of this method is based on the fact that gases with lower density rise upwards, and gases with higher density sink downwards. Therefore, the correct option is that the gas is lighter than air.

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An inert element is an element that has a stable electron configuration, which means it is not likely to lose, gain or share electrons. It is also known as a noble gas. The correct electron configuration for an inert element is a full outermost energy level, which is represented as an electron configuration with eight valence electrons. Therefore, the electron configuration that correctly represents an inert element is 1s²2s²2p⁶. This configuration represents the noble gas, neon (Ne), which has a full outermost energy level and does not easily react with other elements.

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Phenolphthalein is a commonly used acid-base indicator. It is colourless in acidic solutions and pink in basic solutions. Therefore, at the end-point of an acid-base titration, when the solution has become basic, the phenolphthalein indicator will turn pink. Thus, the answer is pink.

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This is an application of Charles's Law, which states that at constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature. Mathematically, V/T = k, where V is the volume of the gas, T is its absolute temperature, and k is a constant. So, if the initial volume of the gas is 220 cm ³ at a temperature of 300 K, we can use the above equation to find the value of k. V/T = k 220/300 = k k = 0.7333 (rounded to four decimal places) Now, if the volume of the gas changes to 250 cm ³, we can use the same equation to find its new temperature, T. V/T = k 250/T = 0.7333 T = 250/0.7333 T = 341 K (rounded to the nearest whole number) Therefore, the temperature of the gas when its volume changes from 220 cm ³ to 250 cm ³ is 341 K. Hence, the correct answer is.

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Hydrogen has three isotopes - protium, deuterium, and tritium. Protium is the most common and has one proton and no neutrons, deuterium has one proton and one neutron, and tritium has one proton and two neutrons. Therefore, the correct answer is (2) three isotopes.

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A molar solution is a solution that contains one mole of solute per liter of solution. This means that the correct statement about a molar solution is: "contains one mole of the solute in 1 dm³ of solution". Option A, "is a supersaturated solution", is incorrect because a supersaturated solution contains more solute than it can normally dissolve at a given temperature. Option B, "cannot dissolve more of the solute at that temperature", is incorrect because a saturated solution is the one that cannot dissolve more solute at a given temperature, not a molar solution. Option C, "contains any amount of solute in a given volume of solution", is incorrect because it does not describe the specific concentration of a molar solution, which has a defined amount of solute per liter of solution.

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In the given chemical equation, the electrons (ne^-) appear on the left-hand side of the equation as a reactant. This indicates that the equation is showing a reduction half-reaction, where electrons are gained. To balance the electrons gained in the reduction half-reaction, we need to identify the number of electrons that are being transferred. In this case, we can see that 6 electrons are being gained by the reactants (Cr₂O₇^2- and 14H⁺) to form 2 moles of Cr³⁺ and 7 moles of H₂O. Therefore, the value of n in the equation is 6.

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The mass of copper that would be formed can be calculated using Faraday's law of electrolysis. According to the law, the amount of substance produced at an electrode is directly proportional to the amount of electric charge passed through the electrolyte. Mathematically, the amount of substance produced (in moles) is given by: moles of substance = (electric charge passed) / (Faraday constant * ionic charge) In this case, copper ions (Cu2+) are being reduced at the cathode, so the ionic charge is 2+. The Faraday constant is 96500 C mol-1, which means that one mole of electrons corresponds to a charge of 96500 C. Now, we need to determine the amount of electric charge that is passed through the solution of CuSO4. We are given that a current of 10.0 A is passed through the solution for 1 hour. This means that the total electric charge passed is: electric charge passed = current x time electric charge passed = 10.0 A x 3600 s electric charge passed = 36000 C Substituting the values of electric charge, Faraday constant, and ionic charge into the above equation, we get: moles of Cu = (36000 C) / (96500 C mol-1 x 2) moles of Cu = 0.1869 mol Finally, we can calculate the mass of copper (Cu) that is formed using its molar mass: mass of Cu = moles of Cu x molar mass of Cu mass of Cu = 0.1869 mol x 63.5 g/mol mass of Cu = 11.86 g (rounded to two decimal places) Therefore, the mass of copper that would be formed when a current of 10.0 A is passed through a solution of CuSO4 for 1 hour is 11.86 g. Hence, the correct answer is.

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When bromine is added to ethene at room temperature, the compound formed is 1,2-dibromoethane. This reaction is an example of an addition reaction in which the double bond of the ethene molecule breaks, and the two bromine atoms are added to the carbon atoms of the ethene molecule. The product formed in this reaction is 1,2-dibromoethane because the two bromine atoms are added to adjacent carbon atoms of the ethene molecule.

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When NH₄CI is dissolved in water, it undergoes an endothermic process. This is because the dissolution of NH₄CI in water requires energy in the form of heat to break the bonds between the NH₄CI molecules and allow the NH₄CI to mix with the water. This energy is absorbed from the surrounding environment, resulting in a decrease in temperature and the container feeling cold to the touch. Therefore, the correct option is "the process is endothermic".

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Enthalpy of neutralization is the heat released or absorbed when an acid and a base react together to form a salt and water. The enthalpy of neutralization depends on the strength of the acid and the base involved in the reaction. The correct statement about enthalpy of neutralization is that it is constant for a strong acid and a strong base. This is because, in a strong acid-strong base neutralization reaction, the acid and the base completely ionize in solution, forming H⁺ and OH^- ions respectively. The reaction between H⁺ and OH^- ions is highly exothermic and releases a constant amount of heat. This constant value of heat released per mole of acid-base reaction is known as the enthalpy of neutralization. Therefore, the correct option is: "enthalpy of neutralization is constant for a strong acid and a strong base."

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Chromatography would show that black ink is a mixture of chemical compounds. Chromatography is a separation technique that separates different components of a mixture based on their ability to move through a medium, such as paper or a column, at different rates. Since black ink is not a pure substance, but rather a mixture of different chemical compounds, chromatography would be able to separate the different compounds and show that they are present in the ink. By comparing the separation patterns of the black ink to those of known compounds, the different components of the ink can be identified.

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The combination of elements with atomic numbers 20 and 8 (A and Y, respectively) would result in the formation of an ionic compound, AY. Element 20 (A) is calcium, a metal that easily loses two electrons to form a stable Ca²⁺ ion. Element 8 (Y) is oxygen, a non-metal that readily accepts two electrons to form a stable O^2- ion. When calcium (A) reacts with oxygen (Y), calcium loses two electrons to form Ca²⁺, and oxygen gains two electrons to form O^2-. The oppositely charged ions, Ca²⁺ and O^2-, are then held together by electrostatic forces of attraction to form the ionic compound, AY. Therefore, the correct option is: "an ionic compound, AY is formed."

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Iron is the element that would produce colored ions in aqueous solution. This is because iron has variable oxidation states, which means that it can lose different numbers of electrons to form different ions. When iron ions are in solution, they can absorb specific wavelengths of light and reflect others, resulting in the appearance of color. The color of the ion depends on the oxidation state of iron and the ligands (molecules or ions that surround the metal ion) attached to it. For example, Fe2+ ions can form a complex with the ligand SCN- to give a blood-red color, while Fe3+ ions can form a complex with the ligand EDTA to give a blue color. In contrast, calcium, magnesium, and sodium ions do not usually produce colored ions in aqueous solution.

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In metallic colids, the forces of attraction are between the mobile valence electrons and the positively charged nuclei. In a metallic solid, the outermost valence electrons of metal atoms are not tightly bound to their respective atoms, but instead, they move freely throughout the entire crystal structure. These electrons are referred to as delocalized electrons. The positive metal ions are arranged in a lattice structure and are surrounded by these delocalized electrons. These electrons are attracted to the positively charged nuclei, which keeps the metal ions together in the lattice structure. Therefore, the forces of attraction in metallic colids are between the mobile valence electrons and the positively charged nuclei.

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

(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:

(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.

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(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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(a)(i) A functional group is an atom or group of atoms in an organic compound that determines its characteristic chemical reactions.

(a)(ii)

1. In CH₃CH₂CH(CH₃)OH, the functional group is the hydroxyl group, OH.
2. In CH₃CH₂CH₂COOH, the functional group is the carboxyl group, COOH.

(b)(i) The completed equations are:

1. \[\mathrm{CH_3COOCH_2CH_2CH_3 + H_2O \xrightarrow[\text{heat}]{H^+} CH_3COOH + CH_3CH_2CH_2OH}\]
2. \[\mathrm{CH_3CH_2CH_2CH_2COOCH_3 + NaOH \xrightarrow{\text{heat}} CH_3CH_2CH_2CH_2COONa + CH_3OH}\]

(b)(ii) The displayed structures of the products are shown below.

(c)(i) Butan-2-ol is prepared from butene by acid-catalysed hydration:

\[\mathrm{CH_2{=}CHCH_2CH_3 + H_2O \xrightarrow{H^+\,/\,H_2SO_4} CH_3CH(OH)CH_2CH_3}\]

(d)(i)

1. CH₂=CH₂  CH₃CH₂OH: steam and concentrated H₂SO₄ or H₃PO₄ catalyst.
2. CH₃CH₂OH  CH₃COOH: acidified potassium dichromate(VI), K₂Cr₂O₇/H₂SO₄, heat under reflux. Acidified KMnO₄ is also suitable.
3. CH₃COOH  CH₃COOCH₂CH₃: ethanol and concentrated H₂SO₄, warmed.

(e)(i) For CH₃CH(OH)CH=CHCOOH:

1. With cold aqueous NaHCO₃, brisk effervescence occurs as carbon dioxide is evolved.
   \[\mathrm{RCOOH + NaHCO_3 \rightarrow RCOONa + H_2O + CO_2}\]
2. With hot iodine solution in aqueous sodium hydroxide, a yellow precipitate of iodoform, CHI₃, is formed.
3. Bromine water is decolourised, showing the presence of the carbon-carbon double bond.

Awọn alaye Idahun

(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.

(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)

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

(ii)

1. \(\mathrm{SO_2}\): concentrated \(\mathrm{H_2SO_4}\).
2. \(\mathrm{HCl}\): concentrated \(\mathrm{H_2SO_4}\).
3. \(\mathrm{NH_3}\): 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)

1. Carbon(II) oxide, \(\mathrm{CO}\): reducing agent in the extraction of metals.
2. 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, \(\mathrm{Br_2}\).

Awọn alaye Idahun

None