Chemistry, 2004, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2004

Title :

Chemistry

Exam :

JAMB Exam

Paper 1 | Objectives

1 - 10 of 50 Questions

#	Question	Ans
1.	When a solid substance disappears completely as a gas on heating, the substance is said to have undergone A. evaporation B. distillation C. crystallization D. sublimation	D
2.	A chemical reaction is always associated with A. an increase in the composition of one of the substance B. a change in the volume of the reactants C. a change in the nature of the reactants D. the formation of new substance	D
3.	If a solution contains 4.9g of tetraoxosulphate (VI) acid, calculate the amount of copper (II) oxide that will react with it [Cu = 64, O = 16, S = 32, H = 1] A. 0.8g B. 4.0g C. 40.0g D. 80.0g Show Content Detailed Solution Equation of the reaction H₂SO_4(g) + CuO_(s) → Cu_{4(aq) + H₂O Relative molecular mass of H₂SO₄ = 1 + 2 + 32 + 16 + 4 = 98 Relative molecular mass of CuO = 64 + 16 =80 From the above equation, 98_s of H₂SO₄ reacts with 80_s of CuO. ∴ 4.9g of H₂SO₄ will react with = (4.9 * 80)/98 of CuO = 392/98 = 4.0g}	B
4.	What volume of gas is evolved at S.t.p if 2g of calcium trioxocarbonate (IV) is added to a solution of hydrochloric acid? [Ca = 40, C = 12, O = 16, Cl = 35.5, H = 1 Molar volume of agas at s.t.p = 22.4 dm³ A. 112 cm² B. 224 cm² C. 448 cm² D. 2240 cm² Show Content Detailed Solution CaCO_{3(s) + 2HCl_(aq) → CaCl₂ + H₂O_(l) + CO₂ The relative molecular mass of CaCO₃ = 40 + 12 + 16 * 3 = 52 +48 = 100 ; Relative molecular mass of CO₂ = 12 + 10 * 2 = 44 From the above equation, 100g of CaCO₃reacts with 44g of CO₂ at s.t.p. i.e 22.4 dm³ 2g of CaCO₃ will react with (222.4)/100 of CO₂ = 0.448 dm³ 0.448 dm³ = 0.448 1000 = 448cm³}	C
5.	According to charles' law, the volume of a gas becomes zero at A. 0 ^oC B. -100 ^oC C. -273 ^oC D. -373 ^oC	C
6.	It is difficult to achieve an orderly arrangement of the molecules of a gas because they A. have no definite shape B. have little force of attraction between them C. can collide with one another in the container D. are too small in size	C
7.	A given volume of methane diffuses in 20s. How long will it take the same volume of sulphur (IV) oxide to diffuse under the same conditions? [C = 12, H = 1, S = 32, O = 16 A. 5s B. 20s C. 40s D. 60s Show Content Detailed Solution Relative molecular mass of CH₄ = 12 * 1 + 4 = 16 Relative molecular mass of SO₂ = 32 + 10 2 = 64 tSO₂/tCH₄ = √MSO₂/MCH₄ t/20 = √64/16 = √4 t/20 = 2 ∴ t = 20 2 = 40sec	C
8.	An electron can be added to a halogen atom to form a halide ion with A. 2 valence electrons B. 3 valence electrons C. 7 valence electrons D. 8 valence electrons	D
9.	The shape of s-orbitals is A. spherical B. elliptical C. spiral D. circular	A
10.	²²⁶₈₈Ra → ^x₈₆Rn + α - particle What is the value of x in the nuclear reaction above? A. 220 B. 222 C. 226 D. 227 Show Content Detailed Solution ²²⁶₈₈Ra → ²²²₈₆Rn + α - particle	B

1.	When a solid substance disappears completely as a gas on heating, the substance is said to have undergone A. evaporation B. distillation C. crystallization D. sublimation	D
2.	A chemical reaction is always associated with A. an increase in the composition of one of the substance B. a change in the volume of the reactants C. a change in the nature of the reactants D. the formation of new substance	D
3.	If a solution contains 4.9g of tetraoxosulphate (VI) acid, calculate the amount of copper (II) oxide that will react with it [Cu = 64, O = 16, S = 32, H = 1] A. 0.8g B. 4.0g C. 40.0g D. 80.0g Show Content Detailed Solution Equation of the reaction H₂SO_4(g) + CuO_(s) → Cu_{4(aq) + H₂O Relative molecular mass of H₂SO₄ = 1 + 2 + 32 + 16 + 4 = 98 Relative molecular mass of CuO = 64 + 16 =80 From the above equation, 98_s of H₂SO₄ reacts with 80_s of CuO. ∴ 4.9g of H₂SO₄ will react with = (4.9 * 80)/98 of CuO = 392/98 = 4.0g}	B
4.	What volume of gas is evolved at S.t.p if 2g of calcium trioxocarbonate (IV) is added to a solution of hydrochloric acid? [Ca = 40, C = 12, O = 16, Cl = 35.5, H = 1 Molar volume of agas at s.t.p = 22.4 dm³ A. 112 cm² B. 224 cm² C. 448 cm² D. 2240 cm² Show Content Detailed Solution CaCO_{3(s) + 2HCl_(aq) → CaCl₂ + H₂O_(l) + CO₂ The relative molecular mass of CaCO₃ = 40 + 12 + 16 * 3 = 52 +48 = 100 ; Relative molecular mass of CO₂ = 12 + 10 * 2 = 44 From the above equation, 100g of CaCO₃reacts with 44g of CO₂ at s.t.p. i.e 22.4 dm³ 2g of CaCO₃ will react with (222.4)/100 of CO₂ = 0.448 dm³ 0.448 dm³ = 0.448 1000 = 448cm³}	C
5.	According to charles' law, the volume of a gas becomes zero at A. 0 ^oC B. -100 ^oC C. -273 ^oC D. -373 ^oC	C

6.	It is difficult to achieve an orderly arrangement of the molecules of a gas because they A. have no definite shape B. have little force of attraction between them C. can collide with one another in the container D. are too small in size	C
7.	A given volume of methane diffuses in 20s. How long will it take the same volume of sulphur (IV) oxide to diffuse under the same conditions? [C = 12, H = 1, S = 32, O = 16 A. 5s B. 20s C. 40s D. 60s Show Content Detailed Solution Relative molecular mass of CH₄ = 12 * 1 + 4 = 16 Relative molecular mass of SO₂ = 32 + 10 2 = 64 tSO₂/tCH₄ = √MSO₂/MCH₄ t/20 = √64/16 = √4 t/20 = 2 ∴ t = 20 2 = 40sec	C
8.	An electron can be added to a halogen atom to form a halide ion with A. 2 valence electrons B. 3 valence electrons C. 7 valence electrons D. 8 valence electrons	D
9.	The shape of s-orbitals is A. spherical B. elliptical C. spiral D. circular	A
10.	²²⁶₈₈Ra → ^x₈₆Rn + α - particle What is the value of x in the nuclear reaction above? A. 220 B. 222 C. 226 D. 227 Show Content Detailed Solution ²²⁶₈₈Ra → ²²²₈₆Rn + α - particle	B