2025 NECO GCE Chemistry Answers

2025 NECO GCE CHEMISTRY

NECO CHEMISTRY
01-10: DAEEBADCEE
11-20: EDDCDDECAD
21-30: EDCBABACEA
31-40: CAEDBECACC
41-50: DCABDCEEDD
51-60: AECECAAABC

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INSTRUCTIONS:. ANSWER FOUR (4), QUESTIONS IN ALL

NECO GCE 2025 CHEMISTRY ANSWERS (ESSAY)

INSTRUCTION: ANSWER FOUR(4) QUESTIONS ONLY

(1ai)
Activation energy is the minimum energy required to activate atoms or molecules for a chemical reaction to occur.

(1aii)
The balanced chemical equation for the reaction is:
Zn(s) + H₂SO₄(aq) -> ZnSO₄(aq) + H₂(g)

(1aiii)
Convert temperature to kelvin = 100 + 273.15= 373.15K
Entropy change = 318.2 J K⁻¹ mol⁻¹
ΔH = T × ΔS
∆H = 373  × 318.2 
= 118,736.33J mol⁻¹

Convert to kJ mol⁻¹;
∆H = 118.74 kJ mol⁻¹

(1bi)
(i) Magnetic separation: use a magnet to remove the iron filings.
(ii) Sublimation: gently heat the mixture; iodine sublimes leaving iron behind.

(1bii)
A physical change alters only physical properties like shape without forming new substances, while a chemical change forms new substances with altered composition.

(1biii)
I. Physical changes: melting of ice; dissolving sugar in water.
II. Chemical changes: rusting of iron; combustion of methane.

(1ci)
The periodic law states that the properties of elements are periodic functions of their atomic numbers.

(1cii)
Molar mass of Mg(NO₃)₂ = 24 + 2(14) + 6(16) = 148gmol⁻¹
Mass of nitrogen = 2 × 14 = 28g

% N = (28/148) × 100 = 18.92%

(1ciii)
(i) Calcium chloride (CaCl₂)
(ii) Calcium oxide (quicklime, CaO)

(1civ)
(i) Phenolphthalein
(ii) Methyl orange
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(2ai)
Nitrogen is chemically inert due to its strong triple covalent bond in N₂ molecules.

(2aii)
(i) Used in the manufacture of ammonia for fertilizers via the Haber process.
(ii) Employed as an inert atmosphere to prevent oxidation in food preservation and welding.

(2aiii)
Butan-1-ol

(2aiv)
(i) Amino group (-NH₂).
(ii) Carboxyl group (-COOH).

(2bi)
A standard solution is a solution of accurately known concentration used in titrations for quantitative analysis.

(2bii)
(i) Rinse burette with the solution to be used before filling.
(ii) Read the meniscus at eye level against a white background to avoid parallax error.
(iii) Ensure no air bubbles are trapped in the burette tip.

(2biii)
2Kl(aq) + Cl₂(g) —> 2KCl(aq) + l₂(aq)

(2ci)
Faraday’s Second Law of electrolysis states that when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements discharged are inversely proportional to the charges on the ions of the elements.

(2cii)
Charge passed(Q) = I × t = 5 × (2 × 60) = 5 × 120 = 600C

Moles of electrons = Q/F = 600/96500 = 6.22 × 10⁻³ mol

At the anode: 2 H₂O –> O₂ + 4H⁺ + 4e⁻, so 4e⁻ produce 1 mol O₂.

Moles of O₂ = (6.22 × 10⁻³)/4 = 1.55 × 10⁻³ mol

Volume at STP = moles × 22.4 dm³ mol⁻¹ = 1.55 × 10⁻³ × 22.4
Volume at STP = 0.035dm³ (35cm³).

(2di)
I. In a water molecule: Polar covalent bond
II. Between water molecules: Hydrogen bond

(2dii)
I. Hydrogen from water gas: Water–gas shift process.
II. Ammonia: Haber process.
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(3ai)
(i) Calcium chloride (CaCl₂)
(ii) Magnesium sulfate (MgSO₄)

(3aii)
(i) Detergents work effectively in both hard and soft water, while soap forms scum in hard water.
(ii) Detergents are generally more soluble and have better lathering properties in cold water compared to soap.

(3aiii)
When the stable form of sulfur is heated above approximately 370K, it transitions into the monoclinic form of sulfur, a different crystalline structure.

(3aiv)
(i) Manufacture of sulphuric acid.
(ii) Vulcanization of rubber.

(3bi)
I. Ethanoic acid, Citric acid
II. Citrus fruits, Vinegar
III. Reaction of an acid with a metal, Reaction of an acid with a base.

(3bii)
I. KOH: Used in the manufacture of soft soap.
II. NH₄OH: Used in removing grease and stains in household cleaning.
III. Al: Used in making cooking utensils and overhead cables.

(3ci)
A suspension is a heterogeneous mixture in which insoluble solid particles are dispersed in a liquid and can settle on standing.

(3cii)
(i) Used in car batteries and steam irons.
(ii) Used in the laboratory for preparing solutions and reagents.

(3ciii)
Volume = 200cm³ = 200/1000 = 0.200dm³.
Concentration = 0.10 mol·dm⁻³.

Moles of Na₂CO₃ = concentration × volume = 0.10 × 0.200 = 0.0200 mol.

Molar mass Na₂CO₃ = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106g·mol⁻¹.

Mass = moles × molar mass = 0.0200 × 106 = 2.12g.
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(4ai)
A flame is the visible, gaseous region of a combustion reaction where fuel vapour and oxidant react, producing heat, light and reaction products.

(4aii)
(i) Luminous flame: bright, yellow, incomplete combustion.
(ii) Non-luminous flame : hot, complete combustion.

(4aiii)
Nitrogen (about 78%) and Oxygen (about 21%), also small amounts of argon and other trace gases.

(4bi)
X: Ammonification.
Y: Nitrification.
Z: Denitrification.

(4bii)
(i) It is a colourless and odourless gas.
(ii) It is less dense than air

(4ci)
(i) Iodine
(ii) Dry ice

(4cii)
(i) Boiling point determination.
(ii) Chromatography.

(4ciii)
I. Mining: Froth flotation.
II. Petroleum: Fractional distillation.
III. Water purification: Filtration.

(4di)
Graham’s Law of Diffusion states that the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its molar mass, provided the temperature and pressure are constant

(4dii)
Using Graham’s Law of Diffusion:

(tCl₂/tQ) = √(MCl₂/MQ)

Given:
tQ = 100 seconds
tCl₂ = 90 seconds
MCl₂ = 71g/mol

Substitute the values:
(90/100) = √(71/MQ)
0.9 = √(71/MQ)
0.9² = (√(71/MQ))²
0.81 = (71/MQ)
MQ = 71/0.81
MQ = 87.65g/mol
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(5ai)
(i) They are good conductors of heat and electricity.
(ii) They are malleable and ductile.

(5aii)
(i) Thermal cracking.
(ii) Catalytic cracking.

(5aiii)
MgSO₄(aq) + 2NaOH(aq) –> Mg(OH)₂(s) + Na₂SO₄(aq)

(5aiv)
(i) It is a Colourless and odourless gas.
(ii) It is lightly soluble in water

(5bi)
Metals are elements that readily lose electrons to form positive ions and have characteristic properties such as conductivity, malleability, and metallic lustre.

(5bii)
I. Uses of oxygen:
-Used in welding and cutting metals.
-Used in hospitals for respiration of patients.

II. Uses of copper:
-Making electrical wires.
-Making alloys such as bronze and brass.

(5biii)
Given: NO₂ = 55

Vapour density = Molar mass/2
Vapour density = 55/2 = 27.5
Vapour density = 27.5

(5c)
I. Substances produced from cast iron;
-Steel
-Wrought iron

II. Ores of iron;
-Haematite (Fe₂O₃)
-Magnetite (Fe₃O₄)

III. Conditions necessary for rusting;
-Presence of oxygen
-Presence of water

IV. Methods of preventing rusting;
-Painting
-Galvanization
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(6ai)
(i) Photosynthesis by green plants
(ii) Absorption of CO₂ by the oceans (dissolution)

(6aii)
CO(g) + NaOH(aq) –> HCOONa(aq)

(6aiii)
It is a colourless, odourless gas that is only slightly soluble in water.

(6aiv)
(i) Used as fuel for cooking and heating
(ii) Used as adsorbent in water‑filtration and gas masks

(6bi)
I. Equation for the reaction:
  C₅H₁₂(l) + Cl₂(g) –> C₅H₁₁Cl(l) + HCl(g)

II. Products: Chloropentane (C₅H₁₁Cl) and hydrogen chloride (HCl)

III. Role of light: Provides the energy to split Cl₂ into chlorine radicals, initiating the chain reaction.

(6bii)
Polymerisation is the chemical process in which small molecules called monomers join together to form a large molecule (a polymer), either by addition polymerisation or condensation polymerisation.

(6biii)
(i) Coke
(ii) Coal gas
(iii) Coal tar

(6ci)
(i) Inorganic (metal) peroxides, e.g., sodium peroxide Na₂O₂
(ii) Organic peroxides, e.g., benzoyl peroxide

(6cii)
Electrolysis of water
2H₂O –> 2H₂ + O₂

(6ciii)
Given:
P = 2.0atm
V = 2.7dm³
T = −4°C = 269.15K
R = 0.082 atm·dm³K⁻¹mol⁻¹

PV = nRT
n = PV/RT
n = (2 × 2.7) / (0.082 × 269.15)
n = 5.4/22.07
n = 0.245mol
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