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JAMB Chemistry 2010 Objective Past Questions

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.

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Chemistry 2010 Objective — Question 1

1. Which Chemistry Question Paper Type is given to you? A. Type A B. Type B C. Type C D. Type D

  • A. Type A
  • B. Type B
  • C. Type C
  • D. Type DCorrect

Explanation

This is a Type D Chemistry Question Paper, as indicated on the paper's header.

Chemistry 2010 Objective — Question 2

2. Which of the following is an example of a mixture? A. Sand B. Washing soda C. Common salt D. Calcium

  • A. SandCorrect
  • B. Washing soda
  • C. Common salt
  • D. Calcium

Explanation

Sand is a naturally occurring mixture of mineral particles, unlike washing soda and common salt which are pure compounds.

Chemistry 2010 Objective — Question 3

3. Calculate the percentage by mass of nitrogen in calcium trioxonitrate (V). A. 17.1% B. 27.6% C. 8.5% D. 13.1% [Ca=40, N=14, O=16]

  • A. 17.1%Correct
  • B. 27.6%
  • C. 8.5%
  • D. 13.1%

Explanation

Ca(NO3)2 has molar mass = 40 + 2(14+48) = 164. %N = (2x14/164) x 100 = 17.1%.

Chemistry 2010 Objective — Question 4

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.

Chemistry 2010 Objective — Question 5

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.

Chemistry 2010 Objective — Question 6

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.

Chemistry 2010 Objective — Question 7

7. How many unpaired electron(s) are there in the nitrogen sub-levels? A. 1 B. 0 C. 3 D. 2

  • A. 1
  • B. 0
  • C. 3Correct
  • D. 2

Explanation

Nitrogen has the electron configuration 1s2 2s2 2p3, and Hund's rule shows the three 2p electrons are unpaired, each occupying a separate orbital.

Chemistry 2010 Objective — Question 8

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.

Chemistry 2010 Objective — Question 9

9. The maximum number of electrons in the L shell of an atom is A. 18 B. 32 C. 2 D. 8

  • A. 18
  • B. 32
  • C. 2
  • D. 8Correct

Explanation

The maximum number of electrons in a shell is 2n^2; for the L shell (n=2), this gives 2x2^2 = 8.

Chemistry 2010 Objective — Question 10

10. Elements in the same period in the periodic table have the same A. chemical properties B. physical properties C. number of shells D. atomic number

  • A. chemical properties
  • B. physical properties
  • C. number of shellsCorrect
  • D. atomic number

Explanation

Elements in the same period have the same number of electron shells, while their chemical/physical properties vary progressively across the period.

Chemistry 2010 Objective — Question 11

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.

Chemistry 2010 Objective — Question 12

12. A noble gas with a high power of fog penetration used in aerodrome beacons is A. helium B. neon C. krypton D. argon

  • A. helium
  • B. neonCorrect
  • C. krypton
  • D. argon

Explanation

Neon has a high power of fog penetration and is used in aerodrome beacons due to its characteristic bright glow.

Chemistry 2010 Objective — Question 13

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.

Chemistry 2010 Objective — Question 14

14. Substances employed as drying agents are usually A. efflorescent B. acidic C. amphoteric D. hygroscopic

  • A. efflorescent
  • B. acidic
  • C. amphoteric
  • D. hygroscopicCorrect

Explanation

Hygroscopic substances are those that absorb moisture, making them suitable as drying agents.

Chemistry 2010 Objective — Question 15

15. Calculate the solubility in mol dm-3 of 40g of CuSO4 dissolved in 100g of water at 120C. A. 0.40 B. 0.25 C. 4.00 D. 2.50 [Cu=64, S=32, O=16]

  • A. 0.40
  • B. 0.25
  • C. 4.00
  • D. 2.50Correct

Explanation

Molar mass of CuSO4 = 160g/mol. Moles = 40/160 = 0.25mol. Volume of H2O = 100cm3 = 0.1dm3. Solubility = 0.25/0.1 = 2.5 mol dm-3.

Chemistry 2010 Objective — Question 16

16. Coffee stains can best be removed by A. a solution of borax in water B. ammonia solution C. kerosene D. turpentine

  • A. a solution of borax in waterCorrect
  • B. ammonia solution
  • C. kerosene
  • D. turpentine

Explanation

A solution of borax in water is the best agent for removing coffee stains.

Chemistry 2010 Objective — Question 17

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.

Chemistry 2010 Objective — Question 18

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.

Chemistry 2010 Objective — Question 19

19. What volume of NaOH will neutralize 25cm3 of 0.1 mol dm-3 H2SO4? A. 0.4cm3 B. 0.1cm3 C. 5.0 D. 2.5cm3

  • A. 0.4cm3
  • B. 0.1cm3
  • C. 5.0
  • D. 2.5cm3Correct

Explanation

Using CaVa/na = CbVb/nb: for a 1:1 ratio scenario as given in the original working, Va = 2.5cm3.

Chemistry 2010 Objective — Question 20

20. The colour of methyl orange in alkaline medium is A. orange B. red C. yellow D. pink

  • A. orange
  • B. red
  • C. yellowCorrect
  • D. pink

Explanation

Methylorange is red in acidic medium, orange in neutral medium, and yellow in alkaline medium.

Chemistry 2010 Objective — Question 21

21. Which of the following salts is slightly soluble in water? A. Na2CO3 B. PbCl2 C. AgCl D. CaSO4

  • A. Na2CO3
  • B. PbCl2Correct
  • C. AgCl
  • D. CaSO4

Explanation

PbCl2 is insoluble in cold water but slightly soluble in hot water, while AgCl is insoluble in water and CaSO4 is only sparingly soluble.

Chemistry 2010 Objective — Question 22

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.

Chemistry 2010 Objective — Question 23

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).

Chemistry 2010 Objective — Question 24

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.

Chemistry 2010 Objective — Question 25

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.

Chemistry 2010 Objective — Question 26

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.

Chemistry 2010 Objective — Question 27

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

Diagram for question 27
  • 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.

Chemistry 2010 Objective — Question 28

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

Diagram for question 28
  • 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.

Chemistry 2010 Objective — Question 29

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

Diagram for question 29
  • 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.

Chemistry 2010 Objective — Question 30

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.

Chemistry 2010 Objective — Question 31

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)).

Chemistry 2010 Objective — Question 32

32. In the diagram above, the gas produced is A. N2O B. N2O4 C. NO D. NO2

Diagram for question 32
  • A. N2O
  • B. N2O4
  • C. NOCorrect
  • D. NO2

Explanation

With dilute HNO3 and copper turnings, the gas produced directly is NO, which turns brown (forming NO2) on contact with air.

Chemistry 2010 Objective — Question 33

33. Which of the following is used as a rocket fuel? A. H2SO4 B. HCl C. HNO3 D. CH3COOH

  • A. H2SO4
  • B. HCl
  • C. HNO3Correct
  • D. CH3COOH

Explanation

Trioxonitrate (V) acid (HNO3) is very useful; it is also used as a rocket fuel and in the production of dyes, explosives, and fertilizers.

Chemistry 2010 Objective — Question 34

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

Diagram for question 34
  • 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.

Chemistry 2010 Objective — Question 35

35. A constituent common to bronze and solder is A. copper B. tin C. lead D. silver

  • A. copper
  • B. tinCorrect
  • C. lead
  • D. silver

Explanation

Bronze = Copper + Tin; Solder = Lead + Tin. The common element in both is tin.

Chemistry 2010 Objective — Question 36

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.

Chemistry 2010 Objective — Question 37

37. A compound gives an orange-red colour to a non-luminous flame. This compound is likely to contain A. Fe3+ B. Fe2+ C. Na+ D. Ca2+

  • A. Fe3+
  • B. Fe2+
  • C. Na+
  • D. Ca2+Correct

Explanation

Sodium compounds give a golden yellow colour to a non-luminous flame; calcium compounds give an orange-red colour to a non-luminous flame.

Chemistry 2010 Objective — Question 38

38. Stainless steel is used for making A. coins and medals B. moving parts of clocks C. magnets D. tools

  • A. coins and medals
  • B. moving parts of clocks
  • C. magnets
  • D. toolsCorrect

Explanation

Stainless steel is used for making tools and cutlery.

Chemistry 2010 Objective — Question 39

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.

Chemistry 2010 Objective — Question 40

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.

Chemistry 2010 Objective — Question 41

41. Which of the following is used as fuel in miners' lamp? A. Ethene B. Ethane C. Ethanal D. Ethyne

  • A. Ethene
  • B. Ethane
  • C. Ethanal
  • D. EthyneCorrect

Explanation

The gas used as fuel in miners' lamp is ethyne (C2H2).

Chemistry 2010 Objective — Question 42

42. Which of the following organic compounds is very soluble in water? A. C2H4 B. CH3COOC2H5 C. CH3COOH D. C2H2

  • A. C2H4
  • B. CH3COOC2H5
  • C. CH3COOHCorrect
  • D. C2H2

Explanation

Ethanoic acid (CH3COOH) is very soluble in water because it is a polar compound.

Chemistry 2010 Objective — Question 43

43. Benzene reacts with hydrogen in the presence of nickel catalyst at 180C to give A. cyclopentane B. cyclohexane C. xylene D. toluene

  • A. cyclopentane
  • B. cyclohexaneCorrect
  • C. xylene
  • D. toluene

Explanation

Benzene reacts with hydrogen in the presence of nickel catalyst at 180C to give cyclohexane: C6H6 + 3H2 -> C6H12.

Chemistry 2010 Objective — Question 44

44. Which of the following is used to hasten the ripening of fruits? A. Ethyne B. Ethane C. Ethene D. Ethanol

  • A. Ethyne
  • B. Ethane
  • C. EtheneCorrect
  • D. Ethanol

Explanation

Ethene is the gas used to hasten the ripening of fruits.

Chemistry 2010 Objective — Question 45

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.

Chemistry 2010 Objective — Question 46

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.

Chemistry 2010 Objective — Question 47

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.

Chemistry 2010 Objective — Question 48

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.

Chemistry 2010 Objective — Question 49

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.

Chemistry 2010 Objective — Question 50

50. The compound that is used as an anaesthetic for surgical operations is A. CH2Cl2 B. CH3Cl C. CCl4 D. CHCl3

  • A. CH2Cl2
  • B. CH3Cl
  • C. CCl4
  • D. CHCl3Correct

Explanation

Trichloromethane (CHCl3) is suitable as an anaesthetic for surgical operations.

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