All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Chemistry 2010 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
4. The droplets of water observed around a bottle of milk taken out of the refrigerator is due to the fact that A. the saturated vapour pressure of the milk is equal to the atmospheric pressure B. water vapour in the air around the bottle loses some of its energy to the bottle C. water vapour in the air around the bottle gains some energy from the bottle D. the milk is colder than the air
A. the saturated vapour pressure of the milk is equal to the atmospheric pressure
B. water vapour in the air around the bottle loses some of its energy to the bottleCorrect
C. water vapour in the air around the bottle gains some energy from the bottle
D. the milk is colder than the air
Explanation
Water vapour in the air condenses on the cold bottle because it loses energy (heat) to the cold surface, cooling below its dew point.
5. The volume of a given gas is Vcm3 at P mmHg. What is the new volume of the gas if the pressure is reduced to half at constant temperature? A. V/2 cm3 B. V cm3 C. 4Vcm3 D. 2Vcm3
A. V/2 cm3
B. V cm3
C. 4Vcm3
D. 2Vcm3Correct
Explanation
By Boyle's Law, P1V1 = P2V2. Halving the pressure at constant temperature doubles the volume: 2Vcm3.
6. Moving from left to right across a period, the general rise in the first ionization energy can be attributed to the A. decrease in screening effect B. increase in screening effect C. decrease in nuclear charge D. increase in nuclear charge
A. decrease in screening effect
B. increase in screening effect
C. decrease in nuclear charge
D. increase in nuclear chargeCorrect
Explanation
Across a period, the number of protons (nuclear charge) increases while shielding stays roughly constant, so the increase in nuclear charge causes ionization energy to rise.
8. The stability of the noble gases is due to the fact that they A. belong to group zero B. are volatile in nature C. have no electron in their outermost shells D. have duplet or octet electron configurations
A. belong to group zero
B. are volatile in nature
C. have no electron in their outermost shells
D. have duplet or octet electron configurationsCorrect
Explanation
Noble gases are stable because they have duplet (Helium) or octet electron configurations in their outermost shells.
11. 2/1 D + 3/1 T -> 4/2 He + 1/0 n + energy. The nuclear reaction above illustrates A. nuclear fusion B. nuclear fission C. alpha decay D. artificial transmutation
A. nuclear fusionCorrect
B. nuclear fission
C. alpha decay
D. artificial transmutation
Explanation
Two light nuclei (deuterium and tritium) combining to form a heavier nucleus (helium) with release of energy is nuclear fusion.
13. Permanent hardness of water can be removed by A. adding caustic soda B. boiling C. filtration D. adding slaked lime
A. adding caustic sodaCorrect
B. boiling
C. filtration
D. adding slaked lime
Explanation
Permanent hardness of water (caused by calcium/magnesium sulphates and chlorides) can be removed by adding washing soda; boiling only removes temporary hardness.
17. Carbon (II) oxide is considered dangerous if inhaled mainly because it A. competes with carbon (IV) oxide in the blood B. can cause lung cancer C. can cause injury to the nervous system D. competes with oxygen in the blood
A. competes with carbon (IV) oxide in the blood
B. can cause lung cancer
C. can cause injury to the nervous system
D. competes with oxygen in the bloodCorrect
Explanation
Carbon (II) oxide (CO) is dangerous because it competes with oxygen for attachment to haemoglobin, resulting in insufficient availability of oxygen in the blood.
18. Hydrochloric acid is used for removing rust. Boric acid is commonly employed as A. an eyewash B. a rust remover C. a bleaching agent D. a drying agent
A. an eyewashCorrect
B. a rust remover
C. a bleaching agent
D. a drying agent
Explanation
Hydrochloric acid is used for removing rust; boric acid is commonly employed as an eyewash.
22. 6AgNO3(aq) + PH3(g) + 3H2O(l) -> 6Ag(s) + H3PO3(aq) + 6HNO3(aq). In the reaction above, the reducing agent is A. PH3(g) B. AgNO3(aq) C. HNO3(aq) D. H2O(l)
A. PH3(g)Correct
B. AgNO3(aq)
C. HNO3(aq)
D. H2O(l)
Explanation
The oxidation number of silver decreased from +1 (in AgNO3) to 0 in Ag; this is reduction, so AgNO3 is the oxidizing agent. The oxidation number of Phospho in PH3 increased to +4 (from PH3 to H3PO3); this is oxidation, so PH3 is the reducing agent.
23. The IUPAC name of the compound LiAlH4 is A. tetrahydrido lithium aluminate (III) B. lithium aluminium hydride C. lithium tetrahydridoaluminate (III) D. aluminium tetrahydrido lithium
A. tetrahydrido lithium aluminate (III)
B. lithium aluminium hydride
C. lithium tetrahydridoaluminate (III)Correct
D. aluminium tetrahydrido lithium
Explanation
The IUPAC name of the compound LiAlH4 is lithium tetrahydridoaluminate (III). Water of NaBH4 is sodium tetrahydridoborate (II).
24. Iron can be protected from corrosion by coating the surface with A. copper B. zinc C. gold D. silver
A. copper
B. zincCorrect
C. gold
D. silver
Explanation
Zinc is higher than iron in the reactivity series; therefore it is used to galvanize (protect) iron. Copper, gold and silver are inappropriate as they are lower than iron in the series.
25. What quantity of aluminium is deposited when a current of 10A is passed through a solution of an aluminium salt for 1930s? A. 5.4g B. 14.2g C. 0.2g D. 1.8g [Al=27, F=96500 C mol-1]
A. 5.4g
B. 14.2g
C. 0.2g
D. 1.8gCorrect
Explanation
Al3+ + 3e- -> Al. 1 mole of Al (27g) requires 3F (289500C). Charge passed = 10A x 1930s = 19300C. Mass deposited = 19300/289500 x 27g = 1.8g.
26. In which of the following is entropy change positive? A. Addition of concentrated acid to water B. Dissolution of sodium metal in water C. Thermal dissociation of ammonium chloride D. Reaction between an acid and a base
A. Addition of concentrated acid to water
B. Dissolution of sodium metal in water
C. Thermal dissociation of ammonium chlorideCorrect
D. Reaction between an acid and a base
Explanation
The thermal dissociation of ammonium chloride (a solid decomposing into gases) shows an increase in disorder, giving a positive entropy change.
27. If a reaction is exothermic and there is a great disorder, it means that A. there will be a large increase in free energy B. there will be a large decrease in free energy C. the reaction is static D. the reaction is in a state of equilibrium
A. there will be a large increase in free energy
B. there will be a large decrease in free energyCorrect
C. the reaction is static
D. the reaction is in a state of equilibrium
Explanation
An exothermic reaction has a negative delta H. The free energy change delta G = delta H - T delta S. If delta H is negative and there is a great disorder (delta S is large), delta G will unequivocally be negative, i.e. there will be a large decrease in free energy.
28. In the preparation of oxygen by heating KClO3 in the presence of MnO2, only moderate heat is needed because the catalyst acts by A. increasing the rate of the reaction B. lowering the energy barrier of the reaction C. lowering the pressure of the reaction D. increasing the surface area of the reactant
A. increasing the rate of the reaction
B. lowering the energy barrier of the reactionCorrect
C. lowering the pressure of the reaction
D. increasing the surface area of the reactant
Explanation
All positive catalysts act by lowering the activation energy required for the respective reactions to occur, which allows the reaction to proceed with only moderate heat.
29. The graph above demonstrates the effect of A. pressure on the rate of reaction B. concentration on the rate of reaction C. surface area on the rate of reaction D. catalyst
A. pressure on the rate of reaction
B. concentration on the rate of reaction
C. surface area on the rate of reactionCorrect
D. catalyst
Explanation
The graph shows that the reaction will be faster when powdered marble is used than when marble chips are used, because the powdered marble offers a larger surface area for the reaction to occur.
30. 2H2(g) + O2(g) <-> 2H2O(g) delta H = -ve. What happens to the equilibrium constant of the reaction above if the temperature is increased? A. It decreases B. It increases C. It is unaffected D. It becomes zero
A. It decreasesCorrect
B. It increases
C. It is unaffected
D. It becomes zero
Explanation
Since the reaction is exothermic, an increase in temperature will favour the backward (endothermic) reaction. Thus, the value of K decreases.
31. To a solution of an unknown compound, a little dilute tetraoxosulphate (VI) acid was added with some freshly prepared iron (II) tetraoxosulphate (VI) solution. The brown ring observed after the addition of a stream of concentrated tetraoxosulphate (VI) acid confirmed the presence of A. SO4 2- B. NO3- C. CO3 2- D. Cl-
A. SO4 2-
B. NO3-Correct
C. CO3 2-
D. Cl-
Explanation
This illustrates the brown ring test, which is a test that confirms the presence of NO3- (trioxonitrate(V)).
34. In the diagram above, the purpose of the platinized asbestos is to A. solidify the gas B. dry the gas C. absorb impurities D. catalyze the reaction
A. solidify the gas
B. dry the gas
C. absorb impurities
D. catalyze the reactionCorrect
Explanation
The purpose of the platinized asbestos is to catalyze the reaction, as used in the Contact Process for sulphuric acid production.
36. When iron is exposed to moist air, it gradually rusts. This is due to the formation of A. an hydrous iron (II) oxide B. hydrated iron (II) oxide C. hydrated iron (III) oxide D. anhydrous iron (III) oxide
A. an hydrous iron (II) oxide
B. hydrated iron (II) oxide
C. hydrated iron (III) oxideCorrect
D. anhydrous iron (III) oxide
Explanation
When iron is exposed to water and oxygen, it gradually rusts; rusting of iron is the formation of hydrated iron (III) oxide.
39. The residual solids from the fractional distillation of petroleum are used as A. fuel for driving tractors B. fuel for jet engines C. coatings for pipes D. raw materials for the cracking process
A. fuel for driving tractors
B. fuel for jet engines
C. coatings for pipesCorrect
D. raw materials for the cracking process
Explanation
Residues from the fractional distillation of petroleum are used as coating for pipes (bitumen), for making water-proof roofs.
40. CH3(CH2)3CHC2H5, with C3H7 branch. The IUPAC nomenclature of the compound above is A. 5-propylheptane B. 3-propylheptane C. 4-ethyloctane D. 5-ethyloctane
A. 5-propylheptane
B. 3-propylheptane
C. 4-ethyloctaneCorrect
D. 5-ethyloctane
Explanation
The longest continuous chain has 8 carbon atoms; the ethyl group ('-CH2CH3') is attached to the fourth carbon atom. Thus, the correct IUPAC name is 4-ethyloctane.
45. The final products of the reaction between methane and chlorine in the presence of ultraviolet light are hydrogen chloride and A. tetrachloromethane B. chloromethane C. trichloromethane D. dichloromethane
A. tetrachloromethaneCorrect
B. chloromethane
C. trichloromethane
D. dichloromethane
Explanation
The reaction between methane and chlorine ultimately gives tetrachloromethane; this occurs in steps: CH4 -> CH3Cl -> CH2Cl2 -> CHCl3 -> CCl4.
46. The correct order of increasing boiling points of the compounds C7H16, C4H10 and C3H7OH is A. C7H16 -> C3H7OH -> C4H10 B. C4H10 -> C3H7OH -> C7H16 C. C3H7OH -> C4H10 -> C7H16 D. C4H10 -> C7H16 -> C3H7OH
A. C7H16 -> C3H7OH -> C4H10
B. C4H10 -> C3H7OH -> C7H16
C. C3H7OH -> C4H10 -> C7H16
D. C4H10 -> C7H16 -> C3H7OHCorrect
Explanation
C3H7OH has the highest boiling point because it has a polar (OH) group, which allows hydrogen bonding. C4H10 -> C7H16 -> C3H7OH is the correct order of increasing boiling points.
47. One of the major uses of alkanes is A. in the textile industries B. in the production of plastics C. as domestic and industrial fuels D. in the hydrogenation of oils
A. in the textile industries
B. in the production of plastics
C. as domestic and industrial fuelsCorrect
D. in the hydrogenation of oils
Explanation
One of the major uses of alkanes is as domestic and industrial fuels.
48. The haloalkanes used in the dry-cleaning industries are A. trichloroethene and chloroethane B. chloroethane and trichloromethane C. trichloromethane and tetrachloromethane D. chloroethane and tetrachloromethane
A. trichloroethene and chloroethane
B. chloroethane and trichloromethane
C. trichloromethane and tetrachloromethaneCorrect
D. chloroethane and tetrachloromethane
Explanation
Trichloromethane (CHCl3) and tetrachloromethane (CCl4) are common haloalkanes used in the dry cleaning industries.
49. Two hydrocarbons X and Y were mixed with bromine water. X decolourised the solution and Y did not. Which class of compound does Y belong to? A. Alkenes B. Alkynes C. Benzene D. Alkanes
A. Alkenes
B. Alkynes
C. Benzene
D. AlkanesCorrect
Explanation
Alkanes do not decolourize bromine water, while alkenes and alkynes do, so the hydrocarbon Y that did not decolourise the water belongs to the alkanes.