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Instruction
JAMB Chemistry 1994
JAMB (UTME)
Chemistry
1994
Q8401.In an electrolytic set-up to protect iron from corrosion, the iron is
Amade the cathode
Bmade the anode
Cused with a metal of lower electropositive potential
Dinitially coated with tin
Correct Answer: A
In cathodic protection, iron is made the cathode, and a more reactive metal (like zinc or magnesium) is made the anode (sacrificial anode).
JAMB (UTME)
Chemistry
1994
Q8402.Which of the following is NOT true of metals?
AThey are good conductors of electricity
BThey ionize by electron loss
CTheir oxides are acidic
DThey have high melting points
Correct Answer: C
The oxides of metals are generally basic, not acidic. Some metal oxides are amphoteric.
JAMB (UTME)
Chemistry
1994
Q8403.Which of the following is the correct order of decreasing activity of the metals Fe, Ca, Al and Na?
AFe > Ca > Al > Na
BNa > Ca > Al > Fe
CAl > Fe > Na > Ca
DCa > Na > Fe > Al
Correct Answer: B
The reactivity series order is Na > Ca > Al > Fe.
JAMB (UTME)
Chemistry
1994
Q8404.The IUPAC name of the compound above is
A2,2-dimethylbut-1-yne
B2,2-dimethylbut-1-ene
C3,3-dimethylbut-1-ene
D3,3-dimethylbut-1-yne
Correct Answer: B
The structure is CH₂=C(CH₃)-CH₂-CH₃. The longest chain containing the double bond has 4 carbons (butene). The double bond is at position 1, with two methyl groups at position 2. The correct name is 2,2-dimethylbut-1-ene.
JAMB (UTME)
Chemistry
1994
Q8405.When sodium is added to ethanol, the products are
Asodium hydroxide and water
Bsodium hydroxide and hydrogen
Csodium ethoxide and water
Dsodium ethoxide and hydrogen
Correct Answer: D
2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂. Sodium ethoxide and hydrogen gas are produced.
JAMB (UTME)
Chemistry
1994
Q8406.The general formula of alkanones is
ARCHO
BR₂CO
CRCOOH
DRCOOR
Correct Answer: B
Alkanones (ketones) have the general formula R₂CO, where R can be an alkyl group.
JAMB (UTME)
Chemistry
1994
Q8407.Which of the following is a primary amine?
A(CH₃)₃N
BCH₃NH₂
CCH₃CONH₂
D(CH₃)₂NH
Correct Answer: B
A primary amine has one alkyl group attached to the nitrogen atom. CH₃NH₂ (methylamine) is a primary amine.
JAMB (UTME)
Chemistry
1994
Q8408.Which of the following is NOT a monomer?
AC₆H₅CH=CH₂
BCH₂=CH₂
CCH₂=CHCl
DC₆H₆
Correct Answer: D
Benzene (C₆H₆) is not a monomer. The others (styrene, ethene, vinyl chloride) are monomers.
JAMB (UTME)
Chemistry
1994
Q8409.CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (in acid). The purpose of H⁺ in the reaction above is to
Aincrease the yield of products
Bmaintain the solution at a constant pH
Cincrease the rate of the hydrolysis
Ddecrease the rate of the reverse reaction
Correct Answer: C
H⁺ acts as a catalyst to increase the rate of hydrolysis of the ester. It does not affect the equilibrium position (yield).
JAMB (UTME)
Chemistry
1994
Q8410.An organic compound contains 60% carbon, 13.3% hydrogen and 26.7% oxygen. Calculate the empirical formula (C = 12, H = 1, O = 16)
AC₅H₁₂O
BC₃H₈O
CC₆H₁₃O₂
DC₄H₉O
Correct Answer: B
Moles ratio: C = 60/12 = 5, H = 13.3/1 = 13.3, O = 26.7/16 = 1.67. Divide by smallest (1.67): C = 3, H = 8, O = 1. Empirical formula = C₃H₈O.
JAMB (UTME)
Chemistry
1994
Q8411.When Fehling's solution is added to two isomeric carbonyl compounds X and Y with the molecular formula C₅H₁₀O, compound X gives a red precipitate while Y does not react. It can be inferred that X is
ACH₃COCH(CH₃)CH₂CH₃
BCH₃CH₂CH₂CH₂CHO
CCH₃CH₂COCH₂CH₃
DCH₃CH(CH₃)CH₂CHO
Correct Answer: B
Fehling's solution gives a red precipitate with aldehydes (alkanals), not ketones (alkanones). X is an aldehyde. CH₃CH₂CH₂CH₂CHO (pentanal) is an aldehyde.
JAMB (UTME)
Chemistry
1994
Q8412.The compound above contains
Asp³ hybridized carbon atoms only
Bsp³ hybridized carbon atoms only
Csp³ and sp hybridized carbon atoms
Dsp³ and sp² hybridized carbon atoms
Correct Answer: D
The compound is methylbenzene (toluene). The methyl carbon is sp³ hybridized, while the ring carbons are sp² hybridized.
JAMB (UTME)
Chemistry
1994
Q8413.The compound above is the product of the oxidation of
A2-methylbutan-2-ol
B2-methylbutan-1-ol
C2,3-dimethylpropan-1-ol
DPentan-2-ol
Correct Answer: D
The compound is a ketone (pentan-2-one). Ketones are produced by the oxidation of secondary alcohols. Pentan-2-ol is a secondary alcohol.
Instruction
JAMB Chemistry 1995
JAMB (UTME)
Chemistry
1995
Q8414.The addition of water to calcium oxide leads to
Aa physical change
Ba chemical change
Cthe formation of a mixture
Dan endothermic change
Correct Answer: B
CaO + H₂O → Ca(OH)₂. This is a chemical change because a new substance (calcium hydroxide) is formed.
JAMB (UTME)
Chemistry
1995
Q8415.A mixture of iron and sulphur can be separated by dissolving the mixture in
Asteam
Bdilute hydrochloric acid
Cdilute sodium hydroxide
Dbenzene
Correct Answer: D
Iron is soluble in dilute HCl, but sulphur is not. (Note: The question asks for separation by dissolving. Using a magnet is the simplest physical method. However, if dissolving is required, CS₂ can dissolve sulphur, leaving iron. Benzene is a non-polar solvent and does not dissolve either. The provided answer D is a typo).
JAMB (UTME)
Chemistry
1995
Q8416.8.0 g of an element X reacted with an excess of copper (II) tetraoxosulphate (VI) solution to deposit 21.3 g of copper. The correct equation for the reaction is
AX(s) + CuSO₄(aq) → Cu(s) + XSO₄(aq)
BX(s) + 2CuSO₄(aq) → 2Cu(s) + X(SO₄)₂(aq)
C2X(s) + CuSO₄(aq) → Cu(s) + X₂SO₄(aq)
D2X(s) + 3CuSO₄(aq) → 3Cu(s) + X₂(SO₄)₃(aq)
Correct Answer: A
Moles of Cu deposited = 21.3/64 = 0.333 mol. Moles of X = 8/M. For a 1:1 ratio (equation A), M = 8/0.333 = 24 (magnesium). This fits.
JAMB (UTME)
Chemistry
1995
Q8417.C₃H₈(g) + 5O₂(g) → 4H₂O(g) + 3CO₂(g). From the equation above, the volume of oxygen at s.t.p. required to burn 50 cm³ of propane is
A250 cm³
B150 cm³
C100 cm³
D50 cm³
Correct Answer: A
From the equation, 1 volume of propane requires 5 volumes of oxygen. 50 cm³ of propane requires 5 × 50 = 250 cm³ of oxygen.
JAMB (UTME)
Chemistry
1995
Q8418.30 cm³ of hydrogen was collected over water at 27°C and 780 mm Hg. If the vapour pressure of water at the temperature of the experiment was 10 mm Hg, calculate the volume of the gas at 760 mm Hg and 7°C.
A40.0 cm³
B35.7 cm³
C28.4 cm³
D25.2 cm³
Correct Answer: C
P₁ = 780 - 10 = 770 mm Hg, V₁ = 30 cm³, T₁ = 300 K. P₂ = 760 mm Hg, T₂ = 280 K. V₂ = (P₁V₁T₂)/(P₂T₁) = (770 × 30 × 280)/(760 × 300) = 28.4 cm³.
JAMB (UTME)
Chemistry
1995
Q8419.A given amount of gas occupies 10.0 dm³ at 4 atm. and 273°C. The number of moles of the gas present is
A0.089 mol
B1.90 mol
C3.80 mol
D5.70 mol
Correct Answer: A
Using ideal gas law: PV = nRT. P = 4 atm, V = 10.0 dm³, T = 273 + 273 = 546 K, R = 0.0821 dm³ atm K⁻¹ mol⁻¹. n = (4 × 10.0)/(0.0821 × 546) = 40/44.83 = 0.89 mol.
JAMB (UTME)
Chemistry
1995
Q8420.If sulphur (IV) oxide and methane are released simultaneously at the opposite ends of a narrow tube, the rates of diffusion R(SO₂) and R(CH₄) will be in the ratio
A4:1
B2:1
C1:2
D1:4
Correct Answer: C
Molar mass of SO₂ = 64, CH₄ = 16. Rate_SO₂/Rate_CH₄ = √(M_CH₄/M_SO₂) = √(16/64) = √(1/4) = 1/2. So the ratio is 1:2.
JAMB (UTME)
Chemistry
1995
Q8421.A solid begins to melt when
Aconstituent particles acquire a greater kinetic energy
Benergy of vibration of particles of the solid is less than the intermolecular forces
Cconstituent particles acquire energy above the average kinetic energy
Denergy of vibration of particles of the solid equals the intermolecular forces
Correct Answer: D
Melting occurs when the energy of vibration of the particles overcomes the intermolecular forces holding them in the solid lattice.
JAMB (UTME)
Chemistry
1995
Q8422.The diagram above represents an atom that can combine with chlorine to form
Aan ionic bond
Ba covalent bond
Ca dative bond
Da metallic bond
Correct Answer: A
The atom has 2 valence electrons (electronic configuration 2,8,2), so it is magnesium. It loses 2 electrons to form an ionic bond with chlorine.
JAMB (UTME)
Chemistry
1995
Q8423.Which of the following electron configurations indicates an atom with the highest ionization energy?
A2,8,7
B2,8,8,1
C2,8,8,2
D2,8,8,7
Correct Answer: A
Configuration 2,8,7 is chlorine, which has the highest ionization energy among the options because it is a non-metal and has a strong tendency to gain an electron.
JAMB (UTME)
Chemistry
1995
Q8424.The lines observed in the simple hydrogen spectrum are due to emission of
Aelectron from the atom
Benergy by proton transition
Cenergy by electron transition
Dneutrons from the atom
Correct Answer: C
The lines in the hydrogen spectrum are due to energy emitted when an electron transitions from a higher energy level to a lower energy level.
JAMB (UTME)
Chemistry
1995
Q8425.If an element X of atomic number Z and mass number Y is irradiated by an intense concentration of neutrons, the relevant nuclear equation is
Aᵞ_Z X + ¹₀n → ᵞ⁻¹_(Z+1)X
Bᵞ_Z X + ¹₀n → ᵞ⁺¹_Z X
Cᵞ_Z X + ¹₀n → ᵞ_(Z+1)X
Dᵞ_Z X + ¹₀n → ᵞ⁺¹_(Z-1)X
Correct Answer: B
When a nucleus absorbs a neutron, the mass number increases by 1 while the atomic number remains the same. The correct equation is ᵞ_Z X + ¹₀n → ᵞ⁺¹_Z X.
JAMB (UTME)
Chemistry
1995
Q8426.The property used in obtaining oxygen and nitrogen industrially from air is the
Aboiling point
Bdensity
Crate of diffusion
Dsolubility
Correct Answer: A
Oxygen and nitrogen are obtained from liquid air by fractional distillation, which relies on the difference in their boiling points.
JAMB (UTME)
Chemistry
1995
Q8427.Excess phosphorus was burnt in a gas jar and the residual gas passed successively over concentrated KOH solution and concentrated H₂SO₄ before being collected in a flask. The gases collected are
Acarbon (IV) oxide, nitrogen and the rare gases
Bnitrogen (IV) oxide and the rare gases
Cnitrogen and the rare gases
Dcarbon (IV) oxide, nitrogen (IV) oxide and the rare gases
Correct Answer: C
Phosphorus burns in air to remove oxygen. The remaining gas (mostly nitrogen) is passed over KOH to remove CO₂, and then over H₂SO₄ to dry it. The gas collected is nitrogen and the rare (noble) gases.
JAMB (UTME)
Chemistry
1995
Q8428.Potassium tetraoxomanganate (VII) is often added to impure water to
Areduce organic impurities
Breduce inorganic impurities
Cdestroy bacteria and algae
Dremove permanent hardness
Correct Answer: C
Potassium tetraoxomanganate(VII) (KMnO₄) is an oxidizing agent used in water treatment to destroy bacteria and algae and to oxidize organic impurities.
JAMB (UTME)
Chemistry
1995
Q8429.The soil around a battery manufacturing factory is likely to contain a high concentration of
ACa²⁺ salts
BPb²⁺ salts
CMg²⁺ salts
DAl³⁺ salts
Correct Answer: B
Lead-acid batteries contain lead. The soil around a battery factory is likely to be contaminated with lead (Pb²⁺) salts.
JAMB (UTME)
Chemistry
1995
Q8430.90.0 g of MgCl₂ was placed in 50.0 cm³ of water to give a saturated solution at 298 K. If the solubility of the salt is 8.0 mol dm⁻³ at the same temperature, what is the mass of the salt left undissolved at the given temperature?
A52.0 g
B58.5 g
C85.5 g
D88.5 g
Correct Answer: A
Molar mass of MgCl₂ = 24 + 71 = 95 g/mol. Maximum moles that can dissolve in 50 cm³ (0.05 dm³) = 8.0 × 0.05 = 0.4 mol. Mass that can dissolve = 0.4 × 95 = 38.0 g. Mass left undissolved = 90.0 - 38.0 = 52.0 g.
JAMB (UTME)
Chemistry
1995
Q8431.Soap leather is an example of a colloid in which a
Aliquid is dispersed in gas
Bsolid is dispersed in liquid
Cgas is dispersed in liquid
Dliquid is dispersed in liquid
Correct Answer: D
Soap lather is a foam, which is a colloid of a gas dispersed in a liquid. (Note: The provided answer D is a typo; the correct answer is C).
JAMB (UTME)
Chemistry
1995
Q8432.The pH of a solution obtained by mixing 100 cm³ of a 0.1 M HCl solution with 100 cm³ of a 0.2 M solution of NaOH is
A1.3
B7.0
C9.7
D12.7
Correct Answer: D
Moles of HCl = 0.1 × 0.1 = 0.01 mol. Moles of NaOH = 0.2 × 0.1 = 0.02 mol. NaOH is in excess by 0.01 mol. Total volume = 200 cm³ = 0.2 dm³. [OH⁻] = 0.01/0.2 = 0.05 M. pOH = -log(0.05) = 1.3. pH = 14 - 1.3 = 12.7.
JAMB (UTME)
Chemistry
1995
Q8433.In the conduction of aqueous potassium tetraoxosulphate (VI) solution, the current carriers are the
Aions
Belectrons
Chydrated ions
Dhydrated electrons
Correct Answer: C
In aqueous solutions, ions are hydrated (surrounded by water molecules). These hydrated ions carry the current.
JAMB (UTME)
Chemistry
1995
Q8434.What volume of 0.1 mol dm⁻³ solution of tetraoxosulphate (VI) acid would be needed to dissolve 2.86 g of sodium trioxocarbonate (IV) decahydrate crystals?
Q8435.1.2 F of electricity are passed through electrolytic cells containing Na⁺, Cu²⁺ and Al³⁺ in series. How many moles of each metal would be formed at the cathode of each cell?
A0.6 mole of Na, 1.2 moles of Cu and 1.2 moles of Al
B1.2 moles of Na, 0.6 mole of Cu and 0.4 mole of Al
C1.3 moles of Na, 2.4 moles of Cu and 2.4 moles of Al
D1.2 moles of Na, 2.4 moles of Cu and 3.6 moles of Al
Q8436.What mass of gold is deposited during the electrolysis of gold (III) tetraoxosulphate (VI) when a current of 15 A is passed for 193 seconds?
A1.97 g
B3.94 g
C5.91 g
D19.70 g
Correct Answer: A
Q = It = 15 × 193 = 2895 C. Au³⁺ + 3e⁻ → Au. 3 × 96500 C deposits 197 g Au. Mass = (2895 × 197)/(3 × 96500) = 1.97 g.
JAMB (UTME)
Chemistry
1995
Q8437.Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s). From the reaction above it can be inferred that
AFe is the oxidizing agent
BFe is reduced
CCu²⁺ loses electrons
DCu²⁺ is the oxidizing agent
Correct Answer: D
Cu²⁺ gains electrons (is reduced) and causes Fe to be oxidized. Therefore, Cu²⁺ is the oxidizing agent.
JAMB (UTME)
Chemistry
1995
Q8438.2FeCl₂(s) + Cl₂(g) → 2FeCl₃(g). The reducing agent in the reaction above is
AFeCl₂
BCl₂
CFeCl₃
DFe
Correct Answer: A
Fe²⁺ in FeCl₂ is oxidized to Fe³⁺ in FeCl₃, so FeCl₂ is the reducing agent. Cl₂ is the oxidizing agent.
JAMB (UTME)
Chemistry
1995
Q8439.The reaction that is accompanied by a decrease in entropy when carried out at constant temperature is
AN₂O₄ → 2NO₂
BN₂ + 3H₂ → 2NH₃
CCaCO₃ → CaO + CO₂
D2N₂H₄ → 3N₂ + 4H₂O
Correct Answer: B
Entropy decreases when the number of gas molecules decreases. In N₂ + 3H₂ → 2NH₃, 4 moles of gas become 2 moles, so entropy decreases.
JAMB (UTME)
Chemistry
1995
Q8440.32 g of anhydrous copper (II) tetraoxosulphate (VI) dissolved in 1 dm³ of water generated 13.0 kJ of heat. The heat of solution is
A26.0 kJ mol⁻¹
B65.0 kJ mol⁻¹
C130.0 kJ mol⁻¹
D260.0 kJ mol⁻¹
Correct Answer: B
Molar mass of CuSO₄ = 63.5 + 32 + 64 = 159.5 g/mol. Moles = 32/159.5 = 0.2 mol. Heat released = 13.0 kJ. Heat of solution per mole = 13.0/0.2 = 65.0 kJ mol⁻¹.
JAMB (UTME)
Chemistry
1995
Q8441.Mg²⁺(aq) + 2e⁻ → Mg(s); E° = -2.37 V
ACu(s)
BCd(s)
CZn(s)
DMg(s)
Correct Answer: D
The strongest reducing agent is the metal with the most negative standard electrode potential, which is Mg (-2.37 V).
JAMB (UTME)
Chemistry
1995
Q8442.In the diagram above, the activation energy for the backward reaction is
A+5 kJ
B+15 kJ
C+25 kJ
D+30 kJ
Correct Answer: C
From the diagram, the transition state is at 40 kJ and the products are at 15 kJ. Activation energy for the backward reaction = 40 - 15 = 25 kJ.
JAMB (UTME)
Chemistry
1995
Q8443.Z(g) → Z(g). In the equation above, the rate of formation of Z is found to be independent of the concentration of Y and to quadruple when the concentration of X is doubled. The rate equation for the reaction is
AR = k[X][Y]
BR = k[X]²[Y]
CR = k[X]²[Y]²
DR = k[X]²[Y]⁰
Correct Answer: D
The rate is independent of [Y] (zero order). The rate quadruples when [X] doubles, so the reaction is second order with respect to X. Rate = k[X]²[Y]⁰.
JAMB (UTME)
Chemistry
1995
Q8444.2Cl₂(g) + 2H₂O(g) ⇌ 4HCl(g) + O₂(g); ΔH = +115 kJ mol⁻¹. In the above equilibrium reaction, a decrease in temperature will
Afavour the reverse reaction
Bfavour the forward reaction
Chave no effect on the equilibrium state
Ddouble the rate of the reverse reaction
Correct Answer: A
The forward reaction is endothermic (ΔH positive). Decreasing the temperature favours the exothermic (reverse) reaction.
In all three reactions, ammonia reduces the other substance (CuO, Cl₂, O₂) and is itself oxidized to nitrogen. This demonstrates the reducing properties of ammonia.
JAMB (UTME)
Chemistry
1995
Q8446.A gas that turns a filter paper previously soaked in lead ethanoate solution black is
Ahydrogen chloride
Bhydrogen sulphide
Csulphur (IV) oxide
Dsulphur (VI) oxide
Correct Answer: B
Lead ethanoate paper turns black in the presence of hydrogen sulphide (H₂S) gas due to the formation of black lead(II) sulphide (PbS).
JAMB (UTME)
Chemistry
1995
Q8447.A solution containing chloride gives a white precipitate with silver trioxonitrate (V) solution. The precipitate will be insoluble in dilute
AHNO₃ but soluble in ammonia solution
BHNO₃ and in ammonia solution
CHCl but soluble in ammonia solution
DHCl and in ammonia solution
Correct Answer: A
AgCl is a white precipitate that is insoluble in dilute HNO₃ but soluble in ammonia solution.
JAMB (UTME)
Chemistry
1995
Q8448.In the experiment above, X could be a solution of
ASodium trioxonitrate (V) and ammonium chloride
BSodium trioxonitrate (III) and ammonium chloride
CLead (II) trioxonitrate (V) and copper turnings
DPotassium trioxonitrate (V) and copper turnings
Correct Answer: B
The setup is for the laboratory preparation of nitrogen gas. It uses sodium trioxonitrate(III) (NaNO₂) and ammonium chloride (NH₄Cl).
JAMB (UTME)
Chemistry
1995
Q8449.The oxide that remains unchanged when heated in hydrogen is
ACuO
BFe₂O₃
CPbO₂
DZnO
Correct Answer: D
ZnO is not reduced by hydrogen under normal heating conditions because zinc is more reactive than hydrogen. CuO, Fe₂O₃, and PbO₂ are reduced.
JAMB (UTME)
Chemistry
1995
Q8450.Which of the following is observed when a solution of iron (III) chloride is mixed with a solution of sodium hydroxide?
AA white precipitate is formed
BA reddish-brown precipitate is formed
CA green precipitate is formed
DNo visible reaction occurs
Correct Answer: B
FeCl₃ reacts with NaOH to form a reddish-brown precipitate of iron(III) hydroxide, Fe(OH)₃.