## CHEMISTRY SOLUTIONS

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USE THIS OBJ FOR A1.. FEW CORRECTION MADE

CHEMISTRY OBJ
11-20: BAEDDBDBAE
31-40: EBEEBEBCEE
51-60: DABBDEAECA

COMPLETED

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(1ai)
(i) It is used in the production of sulfuric acid
(ii) It is used in the vulcanization process of rubber
(iii) It is used in the formulation of pesticides and fungicides to control pests

(1aii)
(i) Hydrogen sulfide has a characteristic foul odor of rotten eggs
(ii) It is soluble in water

(1aiii)
Soaps are usually made from natural fats or oils (such as animal fats or vegetable oils) and an alkali (such as sodium hydroxide) WHILE Detergents, on the other hand, are synthetic compounds that are derived from petroleum products.

(1aiv)
Detergents are widely used as cleaning agents for various surfaces, fabrics, and dishes.

(1bi)

I. ²³₁₁X
The atomic number (Z) of X is 11, and the mass number (A) is 23.
Number of neutrons (N) = Mass number (A) – Atomic number (Z)
N = 23 – 11
N = 12

So, there are 12 neutrons in the atom ²³₁₁X.

II. ³⁹₁₉Y
The atomic number (Z) of Y is 19, and the mass number (A) is 39.
Number of neutrons (N) = Mass number (A) – Atomic number (Z)
N = 39 – 19
N = 20

So, there are 20 neutrons in the atom ³⁹₁₉Y.

(1bii)
Molar mass:
Na = 22.99 g/mol
O₂ = 2 * 16.00 g/mol = 32.00 g/mol

Now, let’s calculate the mass of oxygen needed:

First, calculate the number of moles of sodium (Na) in 9.2g:
Number of moles = Mass / Molar mass
Number of moles of Na = 9.2g / 22.99 g/mol ≈ 0.4002 mol

Since the mole ratio of Na to O₂ is 4:1, the number of moles of O₂ needed is:
Number of moles of O₂ = 0.4002 mol / 4 ≈ 0.1001 mol

Now, calculate the mass of oxygen needed:
Mass of O₂ = Number of moles of O₂ * Molar mass of O₂
Mass of O₂ = 0.1001 mol * 32.00 g/mol ≈ 3.204 g

Therefore, approximately 3.204 grams of oxygen are needed to burn 9.2 grams of sodium.

(1biii)
CaCO₃(s) + 2 HCl(aq) —> CaCl₂(aq) + CO₂(g) + H₂O(l)

From the balanced equation, 1 mole of calcium carbonate (CaCO₃) reacts with 2 moles of HCl to produce 1 mole of calcium chloride (CaCl₂).

Molar masses:
CaCO₃ = Ca(40.08) + C(12.01) + 3O(16.00) = 100.09 g/mol
CaCl₂ = Ca(40.08) + 2Cl(35.45) = 110.98 g/mol

Now, let’s calculate the number of moles of CaCO₃ in 50g:

Number of moles of CaCO₃ = Mass / Molar mass
Number of moles of CaCO₃ = 50g / 100.09 g/mol ≈ 0.4998 mol

Since the mole ratio of CaCO₃ to CaCl₂ is 1:1, the number of moles of CaCl₂ that can be obtained is also approximately 0.4998 mol.

Thus, about 0.4998 moles of calcium chloride can be obtained from 50g of limestone in the presence of excess hydrogen chloride.

(1ci)
(i) Sol: A sol is a colloidal solution in which solid particles are dispersed in a liquid medium. The particle size of the dispersed phase in a sol ranges from about 1 nanometer to 1 micrometer.

(ii) Aerosol: An aerosol is a colloidal solution in which liquid or solid particles are dispersed in a gas medium. The dispersed phase in an aerosol can be either solid or liquid, and the continuous phase is a gas.

(1cii)
The law of definite proportions, also known as the law of constant composition, states that a chemical compound always contains the same elements combined in fixed and definite proportions by mass. In other words, the ratio of the masses of the constituent elements in a compound is always constant.

(1ciii)

I. Sodium trioxonitrate (V) is also known as sodium nitrate, with the chemical formula NaNO₃.

The atomic masses are as follows:
Na (Sodium) = 22.99 g/mol
N (Nitrogen) = 14.01 g/mol
O (Oxygen) = 16.00 g/mol

Relative molecular mass of NaNO₃ = (1 * Na) + (1 * N) + (3 * O)
Relative molecular mass of NaNO₃ = (1 * 22.99 g/mol) + (1 * 14.01 g/mol) + (3 * 16.00 g/mol)
Relative molecular mass of NaNO₃ = 22.99 g/mol + 14.01 g/mol + 48.00 g/mol
Relative molecular mass of NaNO₃ = 85.00 g/mol

Therefore, the relative molecular mass of sodium nitrate (NaNO₃) is 85.00 g/mol.

II. Copper (II) trioxosulphate (VI) pentahydrate is also known as copper (II) sulfate pentahydrate, with the chemical formula CuSO₄ · 5H₂O.

The atomic masses are as follows:
Cu (Copper) = 63.55 g/mol
S (Sulfur) = 32.06 g/mol
O (Oxygen) = 16.00 g/mol
H (Hydrogen) = 1.01 g/mol

Relative molecular mass of CuSO₄ · 5H₂O = (1 * Cu) + (1 * S) + (4 * O) + (10 * H) + (5 * O)

Relative molecular mass of CuSO₄ · 5H₂O = (1 * 63.55 g/mol) + (1 * 32.06 g/mol) + (4 * 16.00 g/mol) + (10 * 1.01 g/mol) + (5 * 16.00 g/mol)

Relative molecular mass of CuSO₄ · 5H₂O = 63.55 g/mol + 32.06 g/mol + 64.00 g/mol + 10.10 g/mol + 80.00 g/mol

Relative molecular mass of CuSO₄ · 5H₂O = 249.71 g/mol

Therefore, the relative molecular mass of copper (II) sulfate pentahydrate (CuSO₄ · 5H₂O) is 249.71 g/mol.

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(2ai)
Mass of silver deposited (in grams) = (Current in Amperes × Time in seconds × Atomic mass of silver) / (1 Faraday)

Given:
Current = 4.6 A
Time = 90 minutes = 90 × 60 seconds = 5400 seconds
Atomic mass of silver (Ag) = 108g/mol

Substituting the values to calculate the mass of silver deposited:

Mass of silver deposited = (4.6 A × 5400 s × 108 g/mol) / 96,500 C

Mass of silver deposited ≈ (2,682,720 g·s/mol) / 96,500 C

Mass of silver deposited ≈ 27.8g

(2aii)
(i) Electrode Material
(ii) Electrolyte Concentration

(2aiii)
(i) The oxidizing agent is MnO₄⁻(aq) because it gains electrons and undergoes reduction, changing from MnO₄⁻ to Mn²⁺.
(ii) The reducing agent is Fe²⁺(aq) because it loses electrons and undergoes oxidation, changing from Fe²⁺ to Fe³⁺.

(2aiv)
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ —-> Mn²⁺ + 4H₂O(l)

(2bi)

(2bii)
(i) Gases have no fixed shape or volume.
(ii) Gases have low density compared to solids and liquids.
(iii) Gases have high kinetic energy and are in constant motion.

(2biii)
Faraday’s second law of electrolysis states that the mass of a substance deposited (or liberated) during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte.

(2biv)
(i) Charcoal
(ii) Coal

(2bv)
Na (Sodium) > Ca (Calcium) > Mg (Magnesium) > Al (Aluminum)

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(3ai)
(i) Butan-2-ol – Secondary alkanol
(ii) 2-methylpropanol – Tertiary alkanol
(iii) 2-methylpropan-2-ol – Tertiary alkanol

(3aii)
(i) Fermentation: In this method, ethanol is produced by the fermentation of sugars or carbohydrates using microorganisms such as yeast or bacteria to convert sugars into ethanol and carbon dioxide.

(ii) Ethylene hydration: In this method, ethanol is produced by the hydration of ethylene gas. Ethylene is reacted with water in the presence of a catalyst usually a strong acid like phosphoric acid or sulfuric acid to produce ethanol.

(3aiii)
Let the relative molecular mass of gas Z be M.
(Rate of diffusion of hydrogen)/(Rate of diffusion of gas Z) = √(molar mass of gas Z)/√(molar mass of hydrogen)
6/1 = (√M)/(√2)
36 = M/2
M = 2×36
M = 72

(3bi)
1s², 2s², 2p⁴

(3bii)
(i) It is a colorless and odorless gas.
(ii) It is slightly soluble in water.
(iii) It supports combustion allowing materials to burn more readily.

(3biii)
(i) The parent chain should contain the double bond and it should be the longest possible chain.
(ii) The suffix “-ene” is added to the name of the parent chain to indicate the presence of a double bond.

(3biv)
C₂H₄ + O₂ —> 2CO₂ + 2H₂O

(3ci)
An endothermic reaction is a chemical reaction that absorbs heat from its surroundings resulting in a decrease in the temperature of the surroundings.

(3cii)
Zn(s) + H₂SO₄(aq) —> ZnSO₄(aq) + H₂(g)

(3ciii)
Displacement reaction or a redox reaction.

(3iv)
(i) Hydrogen is used as a fuel for various applications including powering fuel cells and as a clean-burning fuel for vehicles.
(ii) Hydrogen is used in the production of ammonia for fertilizers and the production of methanol and other chemicals.
(iii) Hydrogen is used in the petroleum industry for the hydrogenation of vegetable oils and fats.

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(4ai)
A super saturated solution is a solution that contains more solute than it can normally dissolve at a given temperature.

(4aii)
Molar mass of lead (II) trioxonitrate (IV) (Pb(NO₃)₂)
= [207 + 2 (14 + 3×16)]g/mol
= 331 g/mol
Number of moles of Pb(NO₃)₂
= 15g/331g/mol
= 0.045 mol
Volume of solution = 25 cm³
= 25/1000 dm³
= 0.025 dm³
Solubility of Pb(NO₃)₂ in moldm-³
= 0.045 mol/0.025 dm³
= 1.8 moldm-³

(4aiii)
(i) Hexacyanoferrate(II) ion
(ii) Tetrachloroaurate(III) ion
(iii) Tetraamminecopper(II) ion

(4bi)
(i) Air does not have a fixed composition and can vary in its proportions of different gases.
(ii) Air can be separated into its various components by methods such as fractional distillation or chromatography.
(iii) Air can be polluted or contaminated with other substances such as pollutants or particles.

(4bii)
(i) Nitrogen (N₂)
(ii) Oxygen (O₂)

(4biii)
Given: 0.1 mole of H₂SO₄
H₂SO₄(aq) —> 2H⁺(aq) + SO₄²⁻(aq)
1 mole of H₂SO₄ produces 2 moles of H⁺ ions.
Moles of H⁺ ions = (0.1 mole H₂SO₄) x (2 moles H⁺ ions)/ (1 mole H₂SO₄)
= 0.2 moles H⁺ ions
Number of H⁺ ions = (0.2 moles H⁺ ions) x (6.0 x 10²³ H⁺ ions)/ (1 mole H⁺ ions)
= 1.2 x 10²³ H⁺ ions

(4biv)
(i) Ionic bonding
(ii) Covalent bonding

(4bv)
(i) BRASS:
Constituent: Copper and zinc.
Use: Brass is used in the production of musical instruments decorative items and plumbing fixtures.

(ii) BRONZE:
Constituent: Copper and tin.
Use: Bronze is used in the production of statues coins and various machinery.

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(5ai)
A base is a type of chemical substance that can react with an acid to form a salt and water. Bases are often characterized by their ability to accept protons (H⁺ ions) or donate pairs of electrons. When a base reacts with an acid, it undergoes a chemical reaction known as neutralization, producing a salt and water.

(5aii)
(i) Calcium oxide (CaO)
(ii) Sodium oxide (Na₂O)

(5aiii)
(i) Iodine is used in medicine as an antiseptic, particularly in disinfecting wounds.
(ii) It is an essential element in the production of thyroid hormones, which are crucial for regulating the body’s metabolism.
(iii) Iodine is used in the production of various organic compounds and in certain industrial applications, such as in the manufacture of dyes and pharmaceuticals.

(5aiv)
Aliphatic hydrocarbons consist of straight chains, branched chains, or non-aromatic rings of carbon atoms, while aromatic hydrocarbons contain at least one aromatic ring, which is a stable and planar (flat) ring of carbon atoms with alternating single and double bonds.

(5av)
Assuming the atomic mass of chlorine (Cl) is 35.5 (approximately, since it varies with isotopes).

The relative molecular mass of XCl₃ is:
Relative molecular mass of XCl₃ = Atomic mass of X + (3 * Atomic mass of Cl)
Relative molecular mass of XCl₃ = 10.8 + (3 * 35.5)
Relative molecular mass of XCl₃ = 10.8 + 106.5
Relative molecular mass of XCl₃ = 117.3

The vapor density is the ratio of the relative molecular mass of a compound to 2 (as the vapor density of hydrogen gas, H₂, is 2). Therefore, the vapor density of XCl₃ is:
Vapor density of XCl₃ = Relative molecular mass of XCl₃ / 2
Vapor density of XCl₃ = 117.3 / 2
Vapor density of XCl₃ = 58.65

(5bi)
(i) Reactant concentration
(ii) Presence of a catalyst
(iii) Nature of reactants and products

(5bii)
The law of conservation of energy states that energy cannot be created or destroyed in an isolated system. The total energy of a closed system remains constant over time. Energy can only change forms or be transferred from one part of the system to another, but the total amount of energy within the system remains constant.

(5biii)
(i) Combustion of gasoline
(ii) Rusting of iron

(5c)

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(6ai)
(CHOOSE ANY 2)
(i) Same functional group: All the members of a homologous series have the same functional group. For example, the alkanes have the functional group -CH₂-.

(ii) Gradation in physical properties: The members of a homologous series exhibit a regular gradation in physical properties such as boiling point, melting point, and density. This is due to the gradual increase in the size of the carbon chain.

(iii) Similar chemical properties: The members of a homologous series have similar chemical properties due to the presence of the same functional group. For example, the alkanes are all relatively unreactive and undergo similar types of reactions.

(iv) Same general formula: The members of a homologous series have the same general formula. For example, the general formula for the alkanes is CnH₂n+2.

(v) Structural isomers: The members of a homologous series can exhibit structural isomerism. This means that compounds with the same molecular formula can have different structures. For example, butane and isobutane are structural isomers of each other.

(6aii)

(6aiii)
(i) Wrought Iron: Wrought iron is a type of iron that contains a very low carbon content (less than 0.1%). It is tough, ductile, and malleable, making it suitable for use in forging, welding, and various decorative purposes.

(ii) Cast Iron: Cast iron contains a higher carbon content (typically between 2% to 4%). It is brittle and cannot be forged or shaped when heated. However, it has excellent casting properties, and its ability to hold intricate shapes makes it useful for applications like engine blocks, pipes, and cookware.

(iii) Steel: Steel is an alloy of iron that contains carbon (usually less than 2%) and other elements. The carbon content in steel is intermediate between wrought iron and cast iron. Steel is known for its strength, durability, and versatility, and it is widely used in construction, manufacturing, and many other industries.

(6bi)
Mass (g) = molarity (mol/dm³) × volume (dm³) × molar mass (g/mol)

Molar mass of CuSO₄ = (1 × molar mass of Cu) + (1 × molar mass of S) + (4 × molar mass of O)
Molar mass of CuSO₄ = (1 × 64) + (1 × 32) + (4 × 16) = 64 + 32 + 64 = 160 g/mol

Substituting the values into the formula:

Mass (g) = 0.25 mol/dm³ × 500 cm³ × 160 g/mol

Converting 500 cm³ to dm³:

1 dm³ = 1000 cm³

500 cm³ = 500/1000 dm³ = 0.5 dm³

Mass (g) = 0.25 mol/dm³ × 0.5 dm³ × 160 g/mol
Mass (g) = 20 g

(6bii)
(i) Petroleum
(ii) Natural gas

(6biii)
(i) Catalytic cracking produces more branched and cyclic hydrocarbons, which have higher octane ratings and burn more cleanly than straight-chain hydrocarbons produced by thermal cracking.
(ii) Catalytic cracking requires lower temperatures and pressures than thermal cracking, which reduces energy consumption and equipment costs.

(6biv)
Elements in the same period have the same number of electron shells.

(6bv)
Alkaline earth metals.

(6ci)
2H₂SO₄(aq) + S(s) —> 2H₂O(l) +3SO₂(g)

(6cii)
3H₂S(g) + SO₂(g) —> 3S(s) + 3H₂O(l)

(6ciii)
I. Oxidizing agent: Sulfur dioxide (SO₂) is the oxidizing agent because it undergoes reduction, gaining electrons to form sulfur (S).

II. Reducing agent: Hydrogen sulfide (H₂S) is the reducing agent because it undergoes oxidation, losing electrons to form sulfur (S).

(6civ)
I. ₁₂²⁴Mg²⁺
Number of electrons = Atomic number – Charge
Number of electrons = 12 – 2 = 10 electrons

II. ₁₆³²S²⁻
Number of electrons = Atomic number + Charge
Number of electrons = 16 + 2 = 18 electrons

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