Help: WhatsApp 08131323935 | Call Naijaclass 08131323935

ENTER NECO PIN


Neco 2026 Chemistry Questions And Answers – Objectives & Essay

Are you searching for the correct Neco 2026 Chemistry Questions And Answers? Do you need genuine and verified NECO  Questions and Answers 2026 for the June/July examination? Welcome to Naijaclass.com, your trusted platform for accurate NECO updates and examination assistance.

Neco 2026 Chemistry Questions And AnswersTuesday, 14th July 2026
Chemistry 3 & 2 (Objective & Essay) 10:00am – 1:00pm

 


NECO CHEMISTRY-OBJ
01-10: CBADDBACEE
11-20: DDCDBCDAEB
21-30: BEEDADCEED
31-40: BBBCBBCDCE
41-50: DEEAAAEACC
51-60: ACCDEBCCDD
==================

NUMBER 1
csLuv2y
avvcd7L

NUMBER 2
EvyDQP6
cfeDoAX

NUMBER 3
yCieaQm
J6lJcmW

NUMBER 4
AEYkBcy
5G7f9zg

NUMBER 5
CVFukPs
Xtgwdrj

NUMBER 6
ttIuPsV

(1ai) I. Dissolution of salt in water — Physical
II. Burning of paper — Chemical
III. Magnetization of iron — Physical
IV. Rusting of iron — Chemical

(1aii) Liquid state

(1bi) Relative molecular mass:
I. Ca(OH)₂ = 40 + 2(16+1) = 74

II. Pb(NO₃)₂ = 207 + 2(14+48) = 331

(1bii)
Isotopes: atoms of the same element with the same atomic (proton) number but different mass numbers (different neutron numbers).

(1biii)
I. Ionic pair: Na & S (11 & 16), since it’s a metal + nonmetal

II. Compound name: Sodium sulfide (Na₂S)

(1ci)
Charles’ Law state At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute (Kelvin) temperature.

(1cii)
P₁=7.50×10⁴ Nm⁻², V₁=15 cm³, T₁=323 K;

s.t.p.: P₂=1.0×10⁵ Nm⁻², T₂=273 K
V₂ = P₁V₁T₂ / (T₁P₂)

= (7.50×10⁴×15×273)/(323×1.0×10⁵)
≈ 9.51 cm³

(1di)
(i)Brownian motion.
(ii)Diffusion of gases.

(1dii)
Ionization energy is the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state

(1diii)
TABULATE;
UNDER; Property
(i)Atomic radius
(ii)Electron affinity
(iii)Electronegativity

UNDER; Down a group
(i)Increases
(ii)Decreases
(iii)Decreases

UNDER; Across a period
(i)Decreases
(ii)Increases
(iii)Increases
=========================================

(2ai)
(i) Haematite (Fe₂O₃)
(ii) Magnetite (Fe₃O₄).

(2aii)
(i)Environmental pollution: Air, water, and soil contamination from industrial waste.
(ii)Health hazards: Toxic exposure leading to respiratory issues, skin irritation, or chronic diseases.
(iii)Global warming: Greenhouse gases like carbon dioxide trap heat and alter global climates.

(2bi)
Li > Na > K.

(2bii)
(i)Aufbau principle; electrons fill lowest-energy orbitals first.
(ii)Pauli exclusion principle; an orbital holds a maximum of 2 electrons, with opposite spins.
(iii)Hund’s rule; electrons singly occupy orbitals of equal energy before pairing up.

(2ci)
(i)Oxidizing agent; a substance that gains electrons (is itself reduced) and oxidizes another species.
(ii) Reducing agent; a substance that loses electrons (is itself oxidized) and reduces another species.

(2cii)
I. KOH → K⁺ + OH⁻

II. Pb(NO₃)₂ → Pb²⁺ + 2NO₃⁻

III. Na₂CO₃ → 2Na⁺ + CO₃²⁻

(2di)
I. Anode; the electrode where oxidation occurs (positive electrode in electrolysis).

II. Cathode; the electrode where reduction occurs (negative electrode in electrolysis).

(2dii)
Q = It = 1.0×900 = 900 C
mol e⁻ = 900/96500 = 0.009326
2e⁻ + 2H⁺ → H₂, so mol

H₂ = 0.009326/2 = 0.004663

V(H₂) = 0.004663 × 22.4 dm³ ≈ 0.104 dm³
(104 cm³)
==================================================

(3a)
DIAGRAM

(3bi)
Exothermic reactions release heat to the surroundings (ΔH negative). WHILE
Endothermic reactions absorb heat from the surroundings (ΔH positive).

(3bii)
Le Chatelier’s principle state if a system at equilibrium is subjected to a change (in concentration, pressure, or temperature), the system shifts in the direction that opposes/minimizes that change, establishing a new equilibrium.

(3biii)
(i)Concentration.
(ii)Pressure.
(iii)Temperature.

(3ci)
I. Saturated solution: A solution that contains the maximum amount of solute that can be dissolved in a given volume of solvent at a specific temperature in the presence of undissolved solute particles.

II. Molar solution: A solution containing one mole of solute dissolved per one decimeter cubed (dm³) of solution (i.e., a solution of concentration 1 mol/dm³).

(3cii)
mol Na₂CO₃ = 0.2 × 0.25 = 0.05 mol;
M(Na₂CO₃)
= 2(23)+12+3(16) = 106
Mass = 0.05 × 106
= 5.3 g

(3di)
I. Paper chromatography.

II. (3) Three components (indicated by the three distinct separated spots emerging from the starting line).

III. Chromatography paper (or Filter paper).

IV. Label IV (the baseline or origin line where the ink spot was applied).

(3dii)
I. Filtration
II. Evaporation to dryness (or Crystallization).
=========================================

(4ai) A buffer solution resists changes in pH when small amounts of acid or base are added to it.

(4aii) (i)Maintaining pH in biological systems (e.g., human blood).
(ii)Calibrating pH meters in industrial and analytical chemistry labs.

(4bi)
Haber-Bosch Process.

(4bii)
(i)Colorless, odorless, and tasteless gas.
(ii)Insoluble (or very slightly soluble) in water.

(4biii)
(i)diamond
(ii)graphite.

(4biv)
(i)Diamond: giant covalent structure each carbon tetrahedrally bonded to 4 others in a rigid 3‑D lattice.
(ii)Graphite: layered structure each carbon bonded to 3 others in hexagonal sheets, held together by weak van der Waals forces, with delocalised electrons between layers.

(4bv)
Used in gas masks to adsorb poisonous gases (or used as a decolorizing agent in sugar refining)

(4ci)
I. Alkenes: CₙH₂ₙ
II. Alkynes: CₙH₂ₙ₋₂

(4cii)
(i)n-butane (CH₃CH₂CH₂CH₃)
(ii) Isobutane (2-methylpropane)

(4ciii)
(i)Butane
(ii) 2-methylpropane

(4civ)
Composition: 60% C, 13.3% H, 26.7% O, molar mass 60

(propanol composition):
Moles: C = 60/12 = 5, H = 13.3/1 = 13.3,

O = 26.7/16 ≈ 1.67

Ratio ÷1.67: C=3, H=8, O=1 →

C₃H₈O
Empirical mass
= 36+8+16 = 60
= molar mass →

Molecular formula: C₃H₈O
======================================

(5ai) Octane rating. Is a measure of a fuel’s ability to resist engine knocking or pre-ignition during combustion compared to a mixture of isooctane and heptane.

(5aii)
TABULATE
UNDER; CRACKING
(i)Breaks long-chain alkanes into smaller, more useful hydrocarbons.
(ii)Primarily used to increase the yield of gasoline (petrol).

UNDER; REFORMING
(i)Rearranges linear hydrocarbons into branched or cyclic hydrocarbons.
(ii)Primarily used to improve the quality (octane rating) of gasoline.

(5bi)
DIAGRAM

(5bii)
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

(5biii)
(i)Fire extinguishers: It does not support combustion and is denser than air, smothering flames.
(ii)Carbonated drinks: It dissolves under pressure to give beverages their fizzy taste.

(5ci)
Vulcanization: heating rubber with sulphur to form cross-links between polymer chains, improving elasticity and strength.

(5cii)
I. Natural rubber:
Example: latex (from Hevea brasiliensis)

II. Synthetic rubber: Example; Neoprene (or SBR)

(5ciii)
Dissolved at 100°C in 80 g water: 39.8 × 80/100
= 31.84 g

Dissolved at 15°C in 80 g water: 35.9 × 80/100
= 28.72 g

Mass precipitated = 31.84 − 28.72 = 3.12 g
=============================================

(6ai) Peroxide: is a compound containing an O–O single bond, with oxygen in the −1 oxidation state.

(6aii) I. Example: Hydrogen peroxide (H₂O₂)
II. Example: HCl (hydrochloric acid)
III. (i)Sour taste
(ii)Soluble in water to form aqueous solutions.

(6aiii) Strong electrolyte: a substance that ionizes/dissociates completely in solution (or molten state), giving a high concentration of ions and conducting electricity strongly.

(6aiv) Hydrochloric acid (HCl).

(6bi) pH is a measure of hydrogen ion concentration in a solution: pH = −log₁₀[H⁺].

(6bii) pH: A=2, B=5, C=7, D=8, E=12
I. Most acidic: A
II. Most basic: E
III. Neutral: C

(6biii) Deliquescent substances absorb water vapour from the air and dissolve into a solution. WHILE Efflorescent substances lose their water of crystallisation to the air and crumble to a powder.

(6biv) (i)Deliquescent
Example: NaOH;
(ii)Efflorescent
Example: Na₂CO₃·10H₂O.

(6ci) (i)It is a clear, colourless, odourless, and tasteless liquid at room temperature.
(ii)It boils at 100°C and freezes at 0°C under standard atmospheric pressure.

(6cii) Mass of X = 28.92 − 26.42 = 2.50 g; mass of H₂
(same vessel, same T,P) = 0.31 g
Equal volumes at same T & P → equal moles, so:
RMM(X) = (mass X / mass H₂) × RMM(H₂)
= (2.50/0.31) × 2 ≈ 16.1
(Close to CH₄, RMM 16.)

=======================================
r2r24pX
Da42j5r
OCzqLPz
=======================================

Leave a Comment

Your email address will not be published. Required fields are marked *