Question Library

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.

(a) Gases that are soluble in water are: - Ammonia - Hydrogen chloride - oxygen, - carbon dioxide, - Sulphur dioxide, - chlorine, and - hydrogen sulfide. (b) Apparatus that could be used to measure volume of solutions accurately: - volumetric flask, - graduated cylinder, - burette, or - volumetric pipette. (c) In a laboratory, fume cupboards are essential safety equipment used to capture and remove airborne hazardous or poisonous substances like gases, vapours, aerosols, and dust, protecting users from exposure to harmful chemicals. (d) Personal protective equipment (PPE) used in the laboratory. - Lab coats - safety glasses - respiratory protector - Hand gloves - Face shield

(a) Volume of pipette used = 25.00 cm\(^3\) Burette readings (cm\(^3\)) Rough 1st titre 2nd titre Final burette reading 20.90 40.90 21.00 Initial burette reading 0.00 20.10 0.00 Volume of acid, A used 20.90 20.80 21.00 Volume of A used = \(\frac{Rough + 1st titre + 2nd titre}{3}\) = \(\frac{20.90 + 20.80 + 21.00}{3}\) = 20.90 cm\(^3\) (b) 2HNO\(_3\)\(_{aq}\) + Na\(_2\)CO\(_3\).xH\(_2\)O\(_{aq}\) → 2NaNO\(_3\) + CO\(_2\) + (X+1)H\(_2\)0(l) c(i) concentration of E in moldm\(^{-3}\) Since we do not know the concentration of D in moldm\(^{-3}\), but in g per cm\(^3\), it will be difficult using the fomula \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\). Thus, we must first convert the Conc. of D to moldm\(^{-3}\). We were told from the question that D contains 6.30 g of HNO\(_3\) in 500 cm\(_3\), So, if D 6.30 g of HNO\(_3\) was dissolved in 500 cm\(_3\) then, X g of HNO\(_3\) will be dissolved in 1000 cm\(_3\) X = \(\frac{{6.30}\times{1000}}{500}\) = 12.6 gdm\(^{-3}\) Recall that Molar concentration (moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\) Molar mass od HNO\(_3\) = 1 + 14 + (16 X 3) = 63 g/mol concentration of D(moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\) C\(_D\) = \(\frac{12.6}{63}\) = 0.20 moldm\(^{-3}\) Now we can use the formula \(\frac{C_D V_D}{C_E V_E}\) = \(\frac{N_D}{N_E}\). Given that C\(_D\) = 0.20 moldm\(^{-3}\) N\(_D\) = 2 V\(_D\) = 20.90cm\(^3\) N\(_E\) = 1 C\(_E\) = ? V\(_E\) = 25.0cm\(^3\) C\(_B\) = \(\frac{C_D V_D N_E}{N_D V_E}\) C\(_B\) = \(\frac{ 0.2 X 20.90 X 1}{25 X 2}\) = 0.0836 moldm\(^{-3}\) = 0.084 moldm\(^{-3}\) (ii) To get the value of x in Na\(_2\)CO\(_3\).xH\(_2\)O First, from the question, if 8.20 g of Na\(_2\)CO\(_3\).xH\(_2\)O was dissolved in 250 cm\(^3\) of distilled water. then, Xg of Na\(_2\)CO\(_3\).xH\(_2\)O will be dissolved in 1000 cm\(^3\) of distilled water. So, X = \(\frac{{8.20}\times{1000}}{250}\) = 32.8 gdm\(^{-3}\) Mass concentration of E = 32.8 gdm\(^{-3}\) Recall that concentration of E(moldm\(^{-3}\)) = \(\frac{Mass concentration(gdm^{-3})}{Molar mass}\) Then, Molar mass of E ( Na\(_2\)CO\(_3\).xH\(_2\)O ) = \(\frac{Mass concentration}{Molar concentration}\) = \(\frac{32.8}{0.0836}\) = 392.34 g/mol But Molar mass of Na\(_2\)CO\(_3\).xH\(_2\)O = Na\(_2\)CO\(_3\) + x(H\(_2\)O) 392.34 = (23 x 2) + 12 + (16 x 3) + 18x 392.34 = 46 + 12 + 48 + 18x 392.34 = 106 + 18x 18x = 392.34 - 106 18x = 286.34 x = \(\frac{286.34}{18}\) x = 15.9 x ≈ 16

F = (NH\(_4\))\(_2\)CO\(_3\) TEST OBSERVATION INFERENCE (a) F + t.t + dil. HCl Effervescence of colourless, odourless gas. Evolved gas turned lime water milky CO\(_2\) from CO\(_3\)\(^{2-}\) or HCO\(_3\) confirmed. b(i) 1st portion + t.t + NaOH in drops and then in excess + heat to warm No visible reaction, No ppt was also formed Effervescence of a colourless gas with an irritating smell. Gas turned red litmus to blue,also produced white fumes with conc. HCl NH\(_3\) from NH\(_4\)\(^+\) confirmed. (ii) 2nd portion + t.t + few drops of potassium hexacyano ferrate(II) solution No ppt formed and no visible reaction was observed Fe\(^{2+}\)/ Fe\(^{3+}\) absent

a(i) The chemical formula for alumina,(aluminium oxide) is Al\(_2\)O\(_3\) (ii) In the extraction of aluminium,the compound that is used; I. to obtain alumina from Bauxite - Sodium hydroxide, NaOH II. as solvent for alumina - Cryolite, Na\(_3\)AlF\(_6\) (iii) The electrodes used in the extraction of aluminium are - cathode (negative electrode) graphite, - anode (positive electrode) carbon rods or slabs. (iv)The extraction of aluminium is expensive due to the high energy requirements of the electrolysis process, which involves purifying bauxite, melting aluminium oxide, and using electricity to break down electrolytes. (v) In the extraction of aluminium: I. the substance liberated at the anode is oxygen gas (O₂) II. the equation for the reaction that leads to decrease in the size of the anode is: C + O\(_2\) → CO\(_2\) b(i) Ions that cause hardness of water are Ca\(^{2+}\) and Mg\(^{2+}\) (ii) Advantages of hard water - strengthening bones and teeth, - potentially aiding digestion, and - reducing the risk of cardiovascular disease. - improves the taste of water due to the presence of the ions. Disadvantages of hard water - skin and hair dryness, - reduced soap effectiveness, - potential appliance damage, and plumbing issues. c(i) Destructive distillation of coal is a process of heating coal in the absence of air (or with limited oxygen) to break down its complex molecules into simpler, useful products like coke, coal gas, coal tar, and ammoniacal liquor. (ii) products obtained from the destructive distillation of coal are: - coke - coal gas - coal tar and - ammoniacal liquor d(i) Gases which can be used to demonstrate the fountain experiment are; - Ammonia, NH\(_3\) - Hydrogen chloride gas, HCl (ii) Ammonia (NH\(_3\)) and hydrogen chloride (HCl) are used in the fountain experiment to demonstrate the high solubility of gases in water. When water is introduced into a flask containing either gas, the gas dissolves rapidly, creating a pressure drop that forces water upwards, forming a "fountain".

a(i) Salt solution is acidic - Reaction is exothermic. - White precipitate is likely to be a trioxocarbonate(IV) salt of Al, Ca and Zn. - Prepared solution contained soluble salt of Al, Ca and Zn. b(i)PbSO\(_4\) (ii) NaHCO\(_3\) (iii) NaHCO\(_3\) ; Cu(NO\(_3\))\(_2\). (iv) NaHCO\(_3\) (v) NH\(_4\)Cl (c) Precautions to be taken during titration include: - Wear a lab coat to protect your clothing from spills. - Ensure to rinse the burette with an acid solution - Also ensure the funnel is removed immediately after filling the acid into the burette. - Avoid error due to parallax while taking the reading to determine the end point. - Ensure all glassware is clean and free of bubbles.

a(i) Salt solution is acidic - Reaction is exothermic. - White precipitate is likely to be a trioxocarbonate(IV) salt of Al, Ca and Zn. - Prepared solution contained soluble salt of Al, Ca and Zn. b(i)PbSO\(_4\) (ii) NaHCO\(_3\) (iii) NaHCO\(_3\) ; Cu(NO\(_3\))\(_2\). (iv) NaHCO\(_3\) (v) NH\(_4\)Cl (c) Precautions to be taken during titration include: - Wear a lab coat to protect your clothing from spills. - Ensure to rinse the burette with an acid solution - Also ensure the funnel is removed immediately after filling the acid into the burette. - Avoid error due to parallax while taking the reading to determine the end point. - Ensure all glassware is clean and free of bubbles.