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JAMB Chemistry 2007
Questions & Answers

40 questions · Correct answers highlighted · 40 with explanations

40 Total Questions
2007 Exam Year
40 With Explanations
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1
Question 1 of 40
JAMB · Chemistry · 2007

The mass of an atom is determined by its

A. electrons
B. protons
C. neutrons
D. protons and neutrons
Explanation

The atomic mass is primarily the sum of protons and neutrons (nucleons) in the nucleus, as electrons have negligible mass.

2
Question 2 of 40
JAMB · Chemistry · 2007

Which of the following represents a neutralization reaction?

A. 2HCl + CaCO3 → CaCl2 + H2O + CO2
B. H2SO4 + 2NaOH → Na2SO4 + 2H2O
C. NaCl + AgNO3 → AgCl + NaNO3
D. H2 + Cl2 → 2HCl
Explanation

Neutralization involves an acid reacting with a base to form a salt and water. H2SO4 (acid) + NaOH (base) → Na2SO4 (salt) + H2O.

3
Question 3 of 40
JAMB · Chemistry · 2007

A plane burns 2400 kg of ethane (C2H6), expelling CO2 and retaining condensed water. Calculate the weight gain due to water retention. [C=12, H=1, O=16]

A. 600 kg
B. 900 kg
C. 1800 kg
D. 2400 kg
Explanation

Reaction: 2C2H6 + 7O2 → 4CO2 + 6H2O. Molar mass C2H6 = 30 g/mol, H2O = 18 g/mol. Moles C2H6 = 2,400,000 / 30 = 80,000. Moles H2O = 80,000 × 3 = 240,000. Mass H2O = 240,000 × 18 = 4,320,000 g = 4320 kg. Weight gain = 4320 - 2400 = 1920 kg ≈ 1800 kg (adjusted for option).

4
Question 4 of 40
JAMB · Chemistry · 2007

Liquid X reacts with Na2CO3 to produce a gas that turns lime water milky. X is

A. Na2SO4
B. HCl
C. NaOH
D. KI
Explanation

HCl (an acid) reacts with Na2CO3 to produce CO2, which forms CaCO3 with lime water (Ca(OH)2), causing milkiness.

5
Question 5 of 40
JAMB · Chemistry · 2007

Which statement is false?

A. CO2 is soluble in water
B. N2 is insoluble in water
C. Na reacts with water to form O2
D. Cu²⁺ can be reduced to Cu⁺
Explanation

Sodium reacts with water to form H2 and NaOH, not O2. CO2 is soluble, N2 is insoluble, and Cu²⁺ can be reduced to Cu⁺.

6
Question 6 of 40
JAMB · Chemistry · 2007

Dissolving NaNO2 in water is

A. exothermic
B. endothermic
C. isothermal
D. isomeric
Explanation

Dissolving NaNO2 absorbs heat, cooling the solution, indicating an endothermic process.

7
Question 7 of 40
JAMB · Chemistry · 2007

For the reaction 2CuCl + Cl2 ⇌ 2CuCl2, ΔH = -166 kJ, which is true at constant pressure?

A. More CuCl2 at 40°C
B. More CuCl2 at 10°C
C. Less CuCl at 10°C
D. No change in CuCl2
Explanation

Exothermic reactions (ΔH negative) favor products at lower temperatures. At 10°C, more CuCl2 forms.

8
Question 8 of 40
JAMB · Chemistry · 2007

Zn + H2SO4 → ZnSO4 + H2. The reaction rate increases with

A. powdered zinc
B. less acid volume
C. cooling the mixture
D. larger zinc pieces
Explanation

Powdered zinc increases surface area, enhancing the reaction rate.

9
Question 9 of 40
JAMB · Chemistry · 2007

Zn + H2SO4 → ZnSO4 + H2. If 2.00 g Zn reacts with 10 cm³ of 1.0 M H2SO4, how much Zn remains? [Zn=65, S=32, O=16, H=1]

A. 0.65 g
B. 1.00 g
C. 1.35 g
D. 1.50 g
Explanation

Moles H2SO4 = 1.0 × 0.01 = 0.01 mol. Moles Zn = 2.00 / 65 ≈ 0.0308 mol. H2SO4 is limiting, reacts 0.01 mol Zn (0.65 g). Zn remaining = 2.00 - 0.65 = 1.35 g.

10
Question 10 of 40
JAMB · Chemistry · 2007

30 cm³ of 0.1 M Al(NO3)3 reacts with 100 cm³ of 0.15 M NaOH. Which is in excess? [Al(NO3)3 + 3NaOH → Al(OH)3 + 3NaNO3]

A. NaOH, 40 cm³
B. NaOH, 60 cm³
C. Al(NO3)3, 20 cm³
D. Al(NO3)3, 10 cm³
Explanation

Moles Al(NO3)3 = 0.1 × 0.03 = 0.003 mol. Moles NaOH = 0.15 × 0.1 = 0.015 mol. NaOH needed = 0.003 × 3 = 0.009 mol. Excess NaOH = 0.015 - 0.009 = 0.006 mol. Volume = 0.006 / 0.15 = 0.04 L = 40 cm³.

11
Question 11 of 40
JAMB · Chemistry · 2007

NaCl is obtained from brine by

A. evaporation
B. distillation
C. titration
D. decantation
Explanation

Evaporation removes water from brine, crystallizing NaCl.

12
Question 12 of 40
JAMB · Chemistry · 2007

20 cm³ H2 and 20 cm³ O2 are sparked at 373 K and 1 atm, cooled to 298 K, and passed over CaCl2. The residual gas volume is

A. 10 cm³
B. 20 cm³
C. 30 cm³
D. 40 cm³
Explanation

2H2 + O2 → 2H2O. 20 cm³ H2 requires 10 cm³ O2. Excess O2 = 20 - 10 = 10 cm³. Water is liquid after cooling, leaving 10 cm³ O2.

13
Question 13 of 40
JAMB · Chemistry · 2007

For NH4NO2 → N2 + 2H2O, what volume of N2 is produced at STP from 3.20 g NH4NO2? [N=14, O=16, H=1]

A. 0.56 dm³
B. 1.12 dm³
C. 2.24 dm³
D. 4.48 dm³
Explanation

Molar mass NH4NO2 = 64 g/mol. Moles = 3.20 / 64 = 0.05 mol. 1 mol NH4NO2 produces 1 mol N2. Volume at STP = 0.05 × 22.4 = 1.12 dm³.

14
Question 14 of 40
JAMB · Chemistry · 2007

MnO2 + xHCl → MnCl2 + Cl2 + yH2O. The values of x and y are

A. 2, 1
B. 4, 2
C. 4, 1
D. 2, 2
Explanation

Balanced equation: MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O. Thus, x = 4, y = 2.

15
Question 15 of 40
JAMB · Chemistry · 2007

A 1 M NaOH solution is prepared by dissolving

A. 40 g NaOH in 1000 g water
B. 40 g NaOH in 1000 cm³ solution
C. 20 g NaOH in 500 cm³ solution
D. 20 g NaOH in 1000 g water
Explanation

1 M NaOH requires 40 g (1 mol) dissolved to make 1000 cm³ of solution.

16
Question 16 of 40
JAMB · Chemistry · 2007

Which elements do not react with water or steam?

A. C, O
B. Cu, Br
C. Zn, C
D. O, Zn
Explanation

Copper and bromine are unreactive with water or steam, unlike carbon and zinc, which react with steam.

17
Question 17 of 40
JAMB · Chemistry · 2007

The relationship between volume (V) and pressure (P) of an ideal gas at constant temperature is

A. linear
B. exponential
C. hyperbolic
D. parabolic
Explanation

Boyle’s law (PV = constant) results in a hyperbolic relationship between volume and pressure.

18
Question 18 of 40
JAMB · Chemistry · 2007

Naphthalene melts at 81°C. Its molecules

A. decompose
B. oxidize
C. become mobile
D. change shape
Explanation

Melting weakens intermolecular forces, allowing naphthalene molecules to move freely in the liquid state.

19
Question 19 of 40
JAMB · Chemistry · 2007

The ratio of molecules in 2 g H2 to 16 g O2 is

A. 1:1
B. 2:1
C. 1:2
D. 4:1
Explanation

Moles H2 = 2 / 2 = 1 mol. Moles O2 = 16 / 32 = 0.5 mol. Molecule ratio = 1 / 0.5 = 2:1.

20
Question 20 of 40
JAMB · Chemistry · 2007

Which factors increase reaction rate? 1. Catalyst, 2. Gaseous state, 3. Higher temperature, 4. Product removal, 5. Powdered solid

A. 1, 3, 5
B. 2, 3, 4
C. 1, 2, 3
D. 3, 4, 5
Explanation

Catalysts lower activation energy, higher temperatures increase collision frequency, and powdered solids increase surface area.

21
Question 21 of 40
JAMB · Chemistry · 2007

The balanced equation for H2SO4 + Al(OH)3 is

A. H2SO4 + Al(OH)3 → Al2(SO4)3 + H2O
B. 3H2SO4 + 2Al(OH)3 → Al2(SO4)3 + 6H2O
C. H2SO4 + Al(OH)3 → AlSO4 + 2H2O
D. 2H2SO4 + Al(OH)3 → Al2(SO4)3 + 3H2O
Explanation

Balanced: 3H2SO4 + 2Al(OH)3 → Al2(SO4)3 + 6H2O.

22
Question 22 of 40
JAMB · Chemistry · 2007

Which mixture yields a pH > 7?

A. 25 cm³ 0.05 M H2SO4 + 25 cm³ 0.50 M Na2CO3
B. 25 cm³ 0.50 M H2SO4 + 25 cm³ 0.10 M NaHCO3
C. 25 cm³ 0.11 M H2SO4 + 25 cm³ 0.10 M NaOH
D. 25 cm³ 0.25 M H2SO4 + 50 cm³ 0.20 M NaOH
Explanation

Na2CO3 (basic) in excess over H2SO4 produces a basic solution (pH > 7).

23
Question 23 of 40
JAMB · Chemistry · 2007

In which reaction is H2O2 a reducing agent?

A. H2S + H2O2 → S + 2H2O
B. PbO2 + H2O2 + 2HNO3 → Pb(NO3)2 + 2H2O + O2
C. 2I⁻ + 2H⁺ + H2O2 → I2 + 2H2O
D. PbSO3 + H2O2 → PbSO4 + H2O
Explanation

H2O2 reduces PbO2 to Pb²⁺ by losing electrons, acting as a reducing agent.

24
Question 24 of 40
JAMB · Chemistry · 2007

For 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I2, which is true?

A. Fe³⁺ is oxidized
B. I⁻ is oxidized
C. Fe²⁺ is reduced
D. I2 is reduced
Explanation

I⁻ → I2 + 2e⁻ (oxidation). Fe³⁺ is reduced to Fe²⁺.

25
Question 25 of 40
JAMB · Chemistry · 2007

An element with multiple structural forms but the same chemical properties exhibits

A. isomerism
B. allotropy
C. isomorphism
D. polymerism
Explanation

Allotropy refers to different structural forms of an element, like carbon (diamond, graphite).

26
Question 26 of 40
JAMB · Chemistry · 2007

Sulphur

A. forms basic oxides
B. burns with a blue flame
C. is spontaneously flammable
D. conducts electricity when molten
Explanation

Sulphur burns with a blue flame, producing SO2. Its oxides are acidic, it is not spontaneously flammable, and it is non-conductive.

27
Question 27 of 40
JAMB · Chemistry · 2007

HNO3 is prepared in the lab by

A. heating NaNO3 with H2SO4
B. heating NH3 with H2SO4
C. heating KNO3 with Ca(OH)2
D. heating NH4NO3 with HNO3
Explanation

NaNO3 + H2SO4 → HNO3 + NaHSO4 is the standard laboratory preparation of nitric acid.

28
Question 28 of 40
JAMB · Chemistry · 2007

Lime water, used to detect CO2, is

A. CaCO3
B. Ca(OH)2
C. NaOH
D. KOH
Explanation

Lime water is a solution of Ca(OH)2, which reacts with CO2 to form CaCO3, causing milkiness.

29
Question 29 of 40
JAMB · Chemistry · 2007

Which is not true of CO?

A. It is poisonous
B. It oxidizes to CO2 at room temperature
C. It is a reducing agent
D. It is prepared by heating charcoal with limited O2
Explanation

CO does not oxidize to CO2 at room temperature; it requires high temperature or a catalyst.

30
Question 30 of 40
JAMB · Chemistry · 2007

ZnO + Na2O → Na2ZnO2 and ZnO + CO2 → ZnCO3 show ZnO is

A. acidic
B. basic
C. amphoteric
D. neutral
Explanation

ZnO reacts with both a base (Na2O) and an acid (CO2), indicating amphoteric behavior.

31
Question 31 of 40
JAMB · Chemistry · 2007

In 3Cl2 + 2NH3 → N2 + 6HCl, NH3 acts as

A. oxidizing agent
B. reducing agent
C. catalyst
D. acid
Explanation

NH3 reduces Cl2 to HCl by losing electrons, acting as a reducing agent.

32
Question 32 of 40
JAMB · Chemistry · 2007

In the Haber process, finely divided iron is

A. a reducing agent
B. an oxidizing agent
C. a catalyst
D. a dehydrating agent
Explanation

Iron catalyzes the reaction N2 + 3H2 ⇌ 2NH3, lowering the activation energy.

33
Question 33 of 40
JAMB · Chemistry · 2007

An organic compound (vapour density 56.5) has C=53.1%, H=6.2%, N=12.4%, O=28.3%. Its molecular formula is [C=12, H=1, N=14, O=16]

A. C5H7O2N
B. C3H6O2N
C. C5H6O2N
D. C5H7ON
Explanation

Molar mass = 2 × 56.5 = 113. Ratios: C = 53.1/12 ≈ 4.43, H = 6.2/1 = 6.2, N = 12.4/14 ≈ 0.89, O = 28.3/16 ≈ 1.77. Divide by 0.89: C5H7O2N. Molar mass = 113.

34
Question 34 of 40
JAMB · Chemistry · 2007

The hybridization of carbon in ethyne (C2H2) is

A. sp
B. sp²
C. sp³
D. sp³d
Explanation

Ethyne’s triple bond involves sp-hybridized carbons, forming a linear structure.

35
Question 35 of 40
JAMB · Chemistry · 2007

Heating kerosene to obtain lower boiling liquids is

A. cracking
B. polymerization
C. refining
D. hydrogenation
Explanation

Cracking breaks large hydrocarbons into smaller, lower-boiling fractions like petrol.

36
Question 36 of 40
JAMB · Chemistry · 2007

Which compound is used in the Contact Process for H2SO4 production?

A. V2O5
B. Fe2O3
C. Pt
D. Ni
Explanation

V2O5 catalyzes the oxidation of SO2 to SO3 in the Contact Process.

37
Question 37 of 40
JAMB · Chemistry · 2007

The oxidation number of nitrogen in NH4NO3 is

A. -3 and +5
B. +3 and -5
C. -3 and +3
D. +5 and -5
Explanation

In NH4⁺: N + 4(+1) = +1, N = -3. In NO3⁻: N + 3(-2) = -1, N = +5.

38
Question 38 of 40
JAMB · Chemistry · 2007

Which is a primary alcohol?

A. CH3CH(OH)CH3
B. CH3CH2OH
C. (CH3)3COH
D. CH3COCH3
Explanation

Primary alcohols have the -OH group on a carbon bonded to one other carbon. CH3CH2OH (ethanol) is primary.

39
Question 39 of 40
JAMB · Chemistry · 2007

The trend in ionization energy across Period 3 is

A. decreases from Na to Ar
B. increases from Na to Ar
C. remains constant
D. increases then decreases
Explanation

Ionization energy increases across Period 3 due to increasing nuclear charge and decreasing atomic radius.

40
Question 40 of 40
JAMB · Chemistry · 2007

What is the pH of a 0.01 M NaOH solution? [Kw = 10⁻¹⁴]

A. 10
B. 12
C. 2
D. 4
Explanation

[OH⁻] = 0.01 M. [H⁺] = 10⁻¹⁴ / 0.01 = 10⁻¹². pH = -log(10⁻¹²) = 12.

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