SECTION "A" (20 marks)

Here are the solutions to the questions: SECTION "A" (20 marks) 1. What is the valence of M in the formula (M₂(SO₄)₃? The sulfate ion (SO_4) has a charge of -2. Since there are 3 sulfate ions, the total negative charge is 3 × (-2) = -6. For the compound to be neutral, the total positive charge from M must be +6. As there are 2 M atoms, each M atom must have a charge of +6 / 2 = +3. Therefore, the valence of M is 3. C. 3. 2. Which of the following acids is a dibasic acid? A dibasic acid can donate two protons (H^+ ions) per molecule. A. Oxalic acid (H_2C_2O_4) is dibasic. B. Nitric acid (HNO_3) is monobasic. C. Phosphoric acid (H_3PO_4) is tribasic. D. Hydrochloric acid (HCl) is monobasic. A. Oxalic acid. 3. 21.0 g of potassium nitrate was strongly heated in a crucible dish until there was no further change. Which of the following is the mass of the residue? Potassium nitrate (KNO_3) decomposes upon strong heating to form potassium nitrite (KNO_2) and oxygen gas: 2KNO_3(s) → 2KNO_2(s) + O_2(g) The residue is potassium nitrite. Molar mass of KNO_3 = 39 (K) + 14 (N) + 3 × 16 (O) = 101 g/mol. Molar mass of KNO_2 = 39 (K) + 14 (N) + 2 × 16 (O) = 85 g/mol. From the stoichiometry, 2 moles of KNO_3 produce 2 moles of KNO_2. This means 101 g of KNO_3 produces 85 g of KNO_2. Mass of KNO_2 produced from 21.0 g of KNO_3: Mass of KNO_2 = 21.0 g KNO_3 × 85 g KNO_2101 g KNO_3 A. (21.0 × 85)/(101) 4. Separation by separating funnel is made possible based on A separating funnel is used to separate immiscible liquids (liquids that do not mix) that have different densities. The denser liquid settles at the bottom and can be drained off. C. Difference in densities of substance. 5. An element has atomic number of 15, the electronic configuration of the ion of the element is An element with atomic number 15 is Phosphorus (P). Its electronic configuration is 2:8:5. To achieve a stable octet, phosphorus typically gains 3 electrons to form a P^3- ion. The electronic configuration of the P^3- ion will be 2:8:8. D. 2:8:8. 6. Which of the following hydrocarbon is unsaturated? Unsaturated hydrocarbons contain carbon-carbon double or triple bonds. A. C_3H_3: This formula suggests a highly unsaturated compound (e.g., a radical or a compound with multiple double/triple bonds). B. C_4H_8: This fits the general formula for alkenes (C_nH_2n), which are unsaturated. C. C_2H_6: This is ethane, an alkane (C_nH_2n+2), which is saturated. D. C_5H_12: This is pentane, an alkane (C_nH_2n+2), which is saturated. Among the given options, C_4H_8 is clearly an unsaturated hydrocarbon (butene). B. C_4H_8 7. An element has got 24 atomic masses and 12 neutron numbers. To which group and period in the periodic table does the element belong? Number of protons = Atomic mass - Neutron number = 24 - 12 = 12. The element with atomic number 12 is Magnesium (Mg). Electronic configuration of Mg: 2:8:2. It has 2 valence electrons, so it belongs to Group II. It has 3 electron shells, so it belongs to Period 3. C. Group II period 3. 8. 2.4 g of magnesium powder reacted with 0.1 m hydrochloric acid. If 50 cm³ of the solution of hydrochloric acid reacted the temperature rise was 8.0°C, calculate the enthalpy that occurs when one mole of magnesium completely react with hydrochloric acid. The heat absorbed by the solution (q) is calculated using q = mc T. Assume the density of the solution is 1 g/cm³, so mass (m) = 50 g. Specific heat capacity (c) = 4.2 J/g/K. Temperature change ( T) = 8.0 K. q = 50 g × 4.2 J/g/K × 8.0 K = 1680 J The question asks for the enthalpy change per mole of magnesium. To match the options, we assume that the 2.4 g of magnesium was the amount that reacted, and the heat released was 1680 J. Moles of magnesium = 2.4 g / 24 g/mol = 0.1 mol. Enthalpy change ( H) per mole of magnesium = Heat releasedMoles of Mg H = 1680 J0.1 mol = 16800 J/mol = 16.8 kJ/mol B. 16.8KJ/mol. 9. How many electrons are present in sodium ion? Sodium (Na) has an atomic number of 11, meaning a neutral sodium atom has 11 protons and 11 electrons. A sodium ion (Na^+) is formed when a sodium atom loses one electron. Number of electrons in Na^+ = 11 - 1 = 10 electrons. C. 10. 10. Nitrogen is considered almost inert (unreacted) in chemistry. This is because nitrogen Nitrogen gas (N_2) consists of two nitrogen atoms joined by a very strong triple covalent bond (N). A large amount of energy is required to break this bond, making nitrogen unreactive under normal conditions. B. Has triple covalent bonds which are very strong. 11. During electrolysis of copper (I) chloride using copper electrodes, 0.4 amperes was passed through the electrolyte for one and a half hours. The mass of copper deposited at the cathode is First, calculate the total charge (Q) passed: Q = It I = 0.4 A t = 1.5 hours × 3600 s/hour = 5400 s Q = 0.4 A × 5400 s = 2160 C The question states "copper (I) chloride", implying Cu^+ ions. However, if we assume copper (II) chloride (Cu^2+) (which is more common and leads to one of the options): At the cathode, Cu^2+(aq) + 2e^- → Cu(s). This means 2 moles of electrons (2 Faraday constants) are required to deposit 1 mole of copper. 1 Faraday constant (F) ≈ 96485 C/mol. Moles of electrons passed = Q / F = 2160 C / 96485 C/mol ≈ 0.02238 mol. Moles of copper deposited = Moles of electrons / 2 = 0.02238 mol / 2 ≈ 0.01119 mol. Molar mass of Cu = 63.5 g/mol. Mass of copper deposited = Moles × Molar mass = 0.01119 mol × 63.5 g/mol ≈ 0.710 g. C. 0.71 g. 12. In the laboratory preparation of chlorine gas by S.4 students, concentrated hydrochloric acid was heated with substance Q and chlorine gas was evolved. Substance Q is likely to be Chlorine gas can be prepared by heating concentrated hydrochloric acid with an oxidizing agent. Manganese(IV) oxide (MnO_2) and Lead(IV) oxide (PbO_2) are common choices that require heating. A. Potassium permanganate (KMnO_4) reacts at room temperature. B. Lead (IV) oxide (PbO_2) reacts with concentrated HCl upon heating to produce chlorine gas. C. Potassium chlorate (KClO_3) is an oxidizing agent but less commonly used for this specific reaction. D. Hydrogen peroxide is not typically used for this purpose. B. Lead (IV) oxide. 13. 15 cm³ of 0.1 m a basic acid reacted with 10 cm³ of 0.15 m sodium hydroxide solution. What is the basicity of the acid? Let the acid be H_xA and the base be NaOH. The reaction is: H_xA + xNaOH → Na_xA + xH_2O Moles of NaOH = Molarity × Volume = 0.15 mol/dm^3 × (10/1000) dm^3 = 0.0015 mol. Moles of H_xA = Molarity × Volume = 0.1 mol/dm^3 × (15/1000) dm^3 = 0.0015 mol. From the stoichiometry, 1 mole of acid reacts with x moles of base. So, Moles of acid1 = Moles of basex (0.0015)/(1) = (0.0015)/(x) This implies x = 1. Therefore, the acid is monobasic, and its basicity is 1. A. 1. 14. Burning magnesium was lowered into a gas jar of carbon dioxide gas. Which one of the following was observed? Magnesium burns in carbon dioxide according to the reaction: 2Mg(s) + CO_2(g) → 2MgO(s) + C(s) The observations would be the formation of white magnesium oxide (ash) and black specks of carbon. A. White ash and black specks. 15. Which one of the particles does not affect the final atomic mass of an element? Atomic mass (mass number) is determined by the total number of protons and neutrons. Alpha particle emission decreases the mass number by 4. Beta particle emission (an electron) results from a neutron converting to a proton, so the mass number remains unchanged (the mass of an electron is negligible). Protons have significant mass. Therefore, a beta particle does not significantly affect the atomic mass (mass number). B. Beta particle. 16. The following metals are extracted by electrolysis EXCEPT. Electrolysis is used to extract highly reactive metals (those above hydrogen in the reactivity series) from their molten compounds. A. Calcium (Ca) is a highly reactive metal, extracted by electrolysis. B. Silver (Ag) is a less reactive metal, typically extracted by chemical reduction or displacement, not electrolysis of its molten compounds. C. Potassium (K) is a highly reactive metal, extracted by electrolysis. D. Aluminium (Al) is a highly reactive metal, extracted by electrolysis (Hall-Héroult process). B. Silver. 17. A gaseous hydrocarbon W contains 82.15% carbon. What could be the empirical formula of W? Assume 100 g of the compound. Mass of Carbon (C) = 82.15 g. Mass of Hydrogen (H) = 100 g - 82.15 g = 17.85 g. Moles of C = 82.15 g / 12 g/mol ≈ 6.846 mol. Moles of H = 17.85 g / 1 g/mol = 17.85 mol. Divide by the smallest number of moles (6.846): C: 6.846 / 6.846 = 1. H: 17.85 / 6.846 ≈ 2.607. To get whole numbers, we can recognize 2.607 is close to 2.5 (or 5/2). Multiplying by 2 gives: C: 1 × 2 = 2. H: 2.607 × 2 ≈ 5.214 ≈ 5. The empirical formula is C_2H_5. Let's check the percentage of carbon for C_2H_5: Molar mass of C_2H_5 = (2 × 12) + (5 × 1) = 24 + 5 = 29 g/mol. Percentage of C = (24 / 29) × 100\% ≈ 82.76\%, which is very close to 82.15%. A. C_2H_5 18. Which of the following properties of Sulphur dioxide gas is useful to textile industries? Sulphur dioxide (SO_2) is known for its bleaching property, particularly for delicate fabrics like wool and silk, where chlorine bleach would be too harsh. It bleaches by reduction. C. Bleaching property. 19. Ethanol reacts with ethanoic acid in the presence of concentrated Sulphuric acid to form a sweet smelling compound. The type of reaction is known as. The reaction between an alcohol (ethanol) and a carboxylic acid (ethanoic acid) to form an ester (sweet-smelling compound) and water, catalyzed by concentrated sulfuric acid, is called esterification. C. Esterification. 20. Carbon dioxide of volume 400 cm³ diffuse through a porous medium and 140 cm³ of nitrogen gas diffuse through the same medium for 3.5 seconds. Find how long it took carbon dioxide gas to diffuse. This problem uses Graham's Law of Diffusion: Rate_1Rate_2 = sqrt(Molar Mass)_2Molar Mass_1. Let gas 1 be CO_2 and gas 2 be N_2. Molar mass of $CO_2 = 12 + (2 × 16) = 44 ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._