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An atom containing 8 valence electrons is typically classified as a noble gas. Noble gases are found in Group 0 of the periodic table and are known for their stability due to having full valence shells, which means they possess eight electrons in their outermost shell.

In the blast furnace during iron extraction, limestone acts as a flux, reacting with impurities like silicon dioxide (sand) to form a molten slag that can be easily separated from the molten iron. A flux is a substance added to a smelting process to react with and remove impurities from the molten metal. In the blast furnace, limestone (calcium carbonate, CaCO\(_3\)) decomposes at high temperatures into calcium oxide (CaO) and carbon dioxide (CO\(_2\)). The calcium oxide (CaO) then reacts with the silicon dioxide (SiO\(_2\)), which is a common impurity in iron ore, to form calcium silicate (CaSiO\(_3\)), which is the molten slag.

A monomer is a small, basic building block molecule that can bond with other similar molecules to form a larger, more complex molecule called a polymer. The monomer of polythene (also known as polyethylene) is ethylene (ethene), a gaseous hydrocarbon with the chemical formula C\(_2\)H\(_4\).

The oxidation of ethanol involves two steps: first, ethanol is oxidized to acetaldehyde (ethanal), and then acetaldehyde is further oxidized to acetic acid (ethanoic acid). Ethanol (CH\(_3\)CH\(_2\)OH) is oxidized to acetaldehyde (CH\(_3\)CHO). This process involves the removal of a hydrogen atom from the carbon atom that is bonded to the alcohol group (-OH). Acetaldehyde (CH\(_3\)CHO) is then further oxidized to acetic acid (CH\(_3\)COOH). This involves the addition of an oxygen atom to the carbon atom that is bonded to the aldehyde group (-CHO). Strong oxidizing agents like acidified potassium dichromate (K\(_2\)Cr\(_2\)O\(_7\)) or alkaline potassium permanganate (KMnO\(_4\)) are commonly used to oxidize ethanol.

Tetraoxosulphate(VI) acid, also known as sulfuric acid (H₂SO₄), is considered a heavy chemical because it is produced and used on a large industrial scale and is a key raw material in many industries. It serves as a crucial raw material in various industries, including fertilizer production, manufacturing of explosives, dyes, and other chemicals.

C\(_5\)H\(_{12}\) has three (3) isomers which are; - pentane, - 2- methyl butane and - 2,2-dimethyl propane.

The oxidation number of iron in its free, or elemental, state is 0.

Water pipes can be made from a variety of materials, including copper, steel, concrete, PVC, and plastic. However, Polyvinyl chloride is a common replacement for metal piping. PVC pipes are resistant to corrosion and chlorinated disinfectants.

Firstly, ensure you balance the chemical reaction x Al + yCl\(_2\) → z AlCl\(_3\) 2Al + 3Cl\(_2\) → 2 AlCl\(_3\) Therefore from the balanced equation, x =2, y=3 and z = 2

(a) An electrochemical cell is a device that can generate electrical energy from the chemical reactions occurring in it, or use the electrical energy supplied to it to facilitate chemical reactions in it. (b) Electrochemical cell Electrolytic cell Chemical reaction produces electric current Electricity is used to produce chemical reaction The positive electrode is cathode Cathode is the negative electrode Anode is the negative electrode Anode is the positive electrode It does not require any external source of energy. It does require an external source of energy, thus it is non-spontaneous Electrodes are in separate compartments Electrodes are in the same compartment (c) When powdered zinc is added to a copper (II) tetraoxosulphate (IV) solution, a reddish-brown solid of copper metal will form on the zinc, and the blue solution will become colourless as zinc displaces copper. d(i) The reaction can be represented by the following equation: Zn(s) + CuSO\(_4\)(aq) → ZnSO\(_4\)(aq) + Cu(s). (ii) Oxidation-Reduction (Redox) reaction (e) Ammonia, Chlorine, Hydrogen Sulphide (f) Factors that favor cracking include: high temperatures, high pressures, the presence of catalysts, and the nature of the material's composition and structure. (g) Noble gases are chemically inert because their outermost electron shells are completely filled, making them stable and unlikely to react with other elements. (h) Properties of homologous series include the following: - All members of a homologous series have the same general formula. - All members of a homologous series have the same functional group. e.g Alcohols contain the (-OH) functional group. - The molecular formula of successive members of a homologous series differs by CH\(_2\). - Members of a homologous series have similar physical properties, such as melting point, boiling point, and density. - Members of a homologous series have similar chemical properties because they have the same functional group. (i) Global warming is the increase in Earth's average temperature caused by greenhouse gases. Greenhouse gases trap heat in the atmosphere, which warms the planet. Carbon dioxide, methane, and nitrous oxide are greenhouse gases that trap heat in the atmosphere. (j) Global warming is primarily caused by - burning of fossil fuels - increased concentration of greenhouse gases, like carbon dioxide (CO\(_2\))

a(i)I. Deliquescence is the phenomenon where a solid substance absorbs moisture from the air and dissolves, turning into a liquid solution. II. Efflorescence refers to the phenomenon or product of a substance losing water of crystallization, often resulting in a powdery or crusty deposit on a surface. (ii)Examples of Deliquescence include; Calcium chloride (CaCl\(_2\)); Potassium hydroxide (KOH);Sodium hydroxide (NaOH); Magnesium chloride (MgCl\(_2\)); Zinc chloride (ZnCl\(_2\)) Example of Efflorescence include; washing soda, Glauber's salt, and copper sulphate, gypsum b(i) Ionization energy is the minimum energy needed to remove an electron from a gaseous atom or ion, essentially measuring how strongly an atom holds onto its electrons. (ii) The second ionization energy of sodium (Na) is higher than its first ionization energy because the second electron is removed from a positively charged ion (Na+), which experiences a stronger electrostatic attraction to the nucleus than the neutral atom. c(i) Charles' law states that the volume of a given mass of gas is directly proportional to the temperature provided pressure remains constant. (ii) A graph illustrating Charles' law: Volume on the y-axis and temperature on the -axis with a line intercepting the y-axis and drawn to the negative x-axis at - 273\(^0\)C. (iii) Given: V\(_1\) = 150 cm\(^3\) T\(_1\) = 21 + 273 = 294K V\(_2\) = ? T\(_2\) = 45 + 273 = 318K Using \(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) V\(_2\) = \(\frac{V_1}{T_1}\times{T_2}\) V\(_2\) = \(\frac{150}{294}\times{318}\) V\(_2\) = 162.24 cm\(^3\) d(i) I. Na₂O + H₂O → 2NaOH. II. P\(_4\)O\(_6\) + 6H\(_2\)O → 4H\(_3\)PO\(_3\) III. Cl₂O₇ + H₂O → 2HClO₄. IV. MgO + H₂O → Mg(OH)₂. d(ii) Increasing acidity is arranged from NaOH < Mg(OH)₂. < H\(_3\)PO\(_3\) < HClO₄.

a(i) Production of Lead(II)Chloride by: I. direct combination : Pb + Cl\(_2\) → PbCl\(_2\) II. double decomposition: Pb(NO₃)₂(aq) + 2NaCl(aq) → PbCl₂(s) + 2NaNO₃(aq) III. displacement reaction: Pb + CuCl₂ → PbCl₂ + Cu. (ii) Condition that would increase the rate of reaction for: I. direct decomposition: increase the temperature, concentration of reactants, or introduce a catalyst. II.double decomposition: increase the concentration of reactants, and/or increase the surface area of solid reactants. III. displacement reaction: increase the temperature. b(i) Conditions that would increase the yield of SO\(_3\)\(_(g)\) - Increase in temperature, - Increase in pressure, - Removing the product formed (ii) 2SO\(_2\)(g) + O\(_2\)(g) ⇌ 2SO\(_3\)\(_(g)\) Reacting ratio: 2 moles : 1 mole Total Reacting Volume: 60cm\(^3\) : 40cm\(^3\) Actual reacted Volume: 60cm\(^3\) : 30cm\(^3\) → (Following the ratio from the balanced chemical equation) Residual Volume: (Total reacting volume - Actual reacted volume) For SO\(_2\): 60 - 60 = 0 ⇒ SO\(_2\) was completely used up in the reaction, hence it is called the limiting reactant. For O\(_2\) : 40 - 30 = 10cm\(^3\) → Volume of the residual gas is 10cm\(^3\). (iii) Change in the oxidation number of sulphur in the reaction: 2SO\(_2\)(g) + O\(_2\)(g) ⇌ 2SO\(_3\)\(_(g)\) i.e From SO\(_2\)(g) → SO\(_3\)\(_(g)\) S + O\(_2\)= 0 → S + O\(_3\) = 0 S + (-2)2 = 0 → S + (-2)3 = 0 S - 4 =0 → S - 6 = 0 S = + 4 → S = + 6 Change in oxidation state is from +4 to +6. (c) cyclopropane n-propane CH\(_2\) H H H / \ I I I CH\(_2\) — CH\(_2\) H —C — C — C — H ( C\(_3\)H\(_6\)) I I I H H H ( C\(_3\)H\(_8\)) For any two or more compounds to be isomers, their molecular formula must be the same while their structural formular would be different from each other. In this case, the molecular formula of the two compounds are not the same. i.e C\(_3\)H\(_6\) and C\(_3\)H\(_8\) are not the same, hence cyclopropane cannot be said to be an isomer of n-propane. d(i) Compound X is Propanol while Compound Y is Propylbutanoate (ii) The functional group present in X is - OH (hydroxyl) (iii) Propanol + Butanoic acid → Propyl butanoate + Water CH\(_3\)CH\(_2\)CH\(_2\)OH + CH\(_3\)CH\(_2\)CH\(_2\)COOH → CH\(_3\)CH\(_2\)CH\(_2\)COOCH\(_2\)CH\(_2\)CH\(_3\) + H\(_2\)O

a(i) 2NH\(_3\) + 3CuO → Cu + N\(_2\) + 3H\(_2\)O (ii) 4NH\(_3\)(g) + 5O\(_2\)(g) → 4NO(g) + 6H\(_2\)O(g) (iii) Cl₂(g) + 2 KI(aq) → 2 KCl(aq) + I₂(s) (iv) Fe + H\(_2\)SO\(_4\) → FeSO\(_4\) + H\(_2\) (v) Pb(NO\(_3\))\(_2\) + 2HCl → PbCl\(_2\) +2HNO\(_3\) b(i) The process involved in the extraction of tin from tin ore Tin extraction from its ore, primarily cassiterite (SnO₂), involves crushing, concentration (using techniques like gravity separation or flotation), smelting (heating with carbon to produce metallic tin), and finally, refining to achieve high purity. The tin ore is first crushed and ground to a fine powder.This step aims to increase the tin content in the ore. Techniques like gravity separation, flotation, or magnetic separation are used to separate the tin-rich minerals from the gangue (waste rock). The concentrated ore is then heated with carbon in a furnace to a high temperature (around 1300-1400^\circC or 2375-2550^\circF) to reduce the tin oxide to metallic tin. The resulting impure tin is refined to achieve the desired purity for industrial applications. (ii) Uses of hydrogen include the following: - Hydrogen can be used in fuel cells to power vehicles. - Hydrogen can be used to generate electricity. - It is also used to produce "green steel". - Hydrogen is used to extract tungsten from ore and copper from copper oxide. - Hydrogen is used to synthesize ammonia. - Hydrogen is used to produce nitric acid and nitrogenous fertilizers. - Hydrogen is used to produce hydrogen chloride. - It is used to produce water gas. - Filling balloons. (iii) Physical properties of hydrogen include; - colourless, - odourless, - tasteless, - highly flammable, - It has a low density and - low boiling and melting points. c(i) Na\(_2\)O - basic oxide (ii) ZnO - amphoteric oxide (iii)SO\(_2\) - acidic oxide (iv) CO\(_2\) - acidic oxide (v) N\(_2\)O - neutral oxide

(a) Volume of pipette used (Volume of B used) = 25.0 cm\(^3\) Burette Readings (cm\(^3\)) Rough 1st titre 2nd titre Final burette readings 18.20 18.50 38.40 Initial burette readings 0.00 0.40 20.20 Volume of A used (KMnO\(_4\)) 18.20 18.10 18.20 Volume of A used = \(\frac{Rough + 1st titre + 2nd titre}{3}\) = \(\frac{18.20 + 18.10 + 18.20}{3}\) = 18.166 V\(_A\) = 18.17 cm\(^3\) (i) Concentration of B in moldm\(^{-3}\) From the balanced equation : MnO\(_4\)\(^-\)\(_{(aq)}\) + 5Fe\(^{2+}\)\(_{(aq)}\) + 8H\(^+\)\(_{(aq)}\) → Mn\(^{2+}\)\(_{(aq)}\) + Fe\(^{3+}\)\(_{(aq)}\) + 4H\(_2\)O\(_{(l)}\) C\(_A\) = 0.020 moldm\(^{-3}\) N\(_A\) = 1 V\(_A\) = 18.17cm\(^3\) N\(_B\) = 5 C\(_B\) = ? V\(_B\) = 25.0cm\(^3\) Using \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\) C\(_B\) = \(\frac{C_A V_A N_B}{N_A V_B}\) C\(_B\) = \(\frac{ 0.02 X 18.17 X 5}{25 X 1}\) C\(_B\) = 0.07268 C\(_B\) = 0.070 moldm\(^{-3}\) (ii) Concentration of B in gdm\(^{-3}\) If 3.8 g of Fe granules was dissolved in 250 cm\(^3\) of B Xg of Fe granules will be dissolved in 1000 cm\(^3\) of B X = \(\frac{ 3.8 X 1000}{250}\) = 15.2 gdm\(^{-3}\) Concentration of B in gdm\(^{-3}\) will be 15.2 gdm\(^{-3}\) (iii) Mass of Fe\(^{2+}\) in 250cm\(^3\) of B. To get the mass, we'll use the formula n = \(\frac{mass}{Molar mass}\) However, since do not know the number of moles, n , hence we'll need to obtain n, using n = \(\frac{CV}{1000}\) since we have C and V So, n = \(\frac{{0.070}\times{250}}{1000}\) = 0.025 mole Now, we can apply n = \(\frac{mass}{Molar mass}\) to obtain the mass, given that the Molar mass of Fe was given to be 56 n = \(\frac{mass}{Molar mass}\) m = n x M = 0.018 x 56 = 1.02 g Therefore, 1.02 g of Fe\(^{2+}\) in 250 cm\(^3\) of B. (iv) Percentage of Fe\(^{2+}\) in the granules = \(\frac{Mass of Fe^{2+}}{Total mass of Fe granules}\times 100\) = \(\frac{1.02}{3.8}\times 100\) = 26.89% = 26.9%

Test Observation Inference (a) C + 10cm\(^3\) of H\(_2\)O C partially dissolves C is a mixture of a soluble and insoluble salt. b(i) 2cm\(^3\) of filtrate + heat Colourless and odourless gas in form of bubbles CO\(_3\)\(^{2-}\) suspected (ii) residue + t.t + dil. HCl No visible reaction NO\(_3\)\(^-\), SO\(_4\)\(^{2-}\) suspected (iii) Solution from b(ii) + NaOH in drops and then in excess White ppt which is dissolves in excess NaOH Zn\(^{2+}\) , Pb\(^{2+}\) , Al\(^{3+}\) suspected (iv) Solution from b(ii) + NH\(_3\) in drops and then in excess White ppt insoluble in excess NH\(_3\) Pb\(^{2+}\) confirmed.