Year :
2012
Title :
Chemistry
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
11 - 20 of 44 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
The complex compound formed by the combination of two elements with a large electronegativity difference is likely to be A. polar covalent B. giant molecular C. covalent D. ionic Detailed SolutionIonic bonds are also formed when there is a large electronegativity difference between two atoms. |
|
| 12. |
The complex compound formed when aluminum dissolves in aqueous sodium hydroxide is A. Na3Al(OH)4. B. NaAl(OH)4 C. NaAl(OH)3 D. Na2Al(OH)3 Detailed SolutionWhen aluminium is treated with dilute sodium hydroxide it forms a salt called sodium tetrahydroxoaluminate and hydrogen gas. |
|
| 13. |
The vapour pressure of a liquid depends on l. temperature. ll. rate of condensation. III. cohesive forces holding the particles together A. l only B. l and ll only C. l and lll only D. ll and lll only Detailed SolutionThree common factors that influence the vapor pressure of a liquid are the surface area, intermolecular forces and the temperature. |
|
| 14. |
MgO does not readily dissolve in water because A. of its high melting point B. it is a covalent compound C. it forms a hydroxide when dissolved in water D. its lattice energy is higher than its hydration energy |
D |
| 15. |
Consider the following reaction equation: CaCO\(_{3(s)}\) + 2HCI\(_{(aq)}\) → CaCl\(_{2(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\). What mass of CaCI\(_2\) would be obtained when 25.0g of CaCO\(_{3(s)}\) is reacted with excess HCI\(_{(aq)}\)? [CaCO\(_3 = 100; CaCI\(_2\) = 111] A. 4.00g B. 4.44 g C. 18.9 g D. 27.8 g Detailed SolutionCaCO\(_3\) + 2HCI → CaCI\(_2\) + H\(_2\)O + CO\(_2\)100g CaCO\(_3\) will produce 111g CaCI\(_2\) 25g CaCO\(_3\) will produce x , x = \(\frac{25 x 111}{100}\) = 27.75 \(\approxeq\) 27.8g |
|
| 16. |
The number of sulphur atoms in 3.20g of SO\(_{2(g)}\) is [O = 16.0;S = 32.0; Avogadro constant = 6.02 x 10\(^{23}\)] A. 3.01 x 1022 B. 6.02 x 1022 C. 6.02 x1023 D. 1.20 x 1024 Detailed Solution32g of sulphur = 64g / 6.02 x 10\(^{23}\) atoms of SO\(_2\)xg of sulphur = 3.2g of SO\(_2\) : 1.6g of sulphur is present in 3.2g / 3.01 x 10\(^{23}\) atoms of SO\(_2\) |
|
| 17. |
Consider the reaction represented by the following equation: xCH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)O. The values of x,y and z respectively, are A. 2,3 and 4 B. 2, 4 and 3 C. 1,2 and 3 D. 1, 3 and 5 Detailed SolutionxCH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)OBalancing the equation, we have C = x (on the left) and 2 (on the right) ⇒ x = 2 2CH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)O On the left, H = 8 and on the right = 2z ⇒ 2z = 8 ; z = 4 2CH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + 4H\(_2\)O On the right, O = 4 + 4 = 8. On the left, O = 2 + 2y ⇒ 2 + 2y = 8 2y = 6 ; y = 3 x, y, z = 2, 3, 4. |
|
| 18. |
The formula of mercury (l) dioxonitrate (III) is A. HgNO3 B. Hg2NO2 C. Hg2(NO2)2 D. Hg(NO3)2 |
C |
| 19. |
A sample of a gas may be identified as chlorine if it turns A. damp blue litmus paper red B. lime water miky C. lead ethanoate paper black D. starch iodide paper blue-black Detailed SolutionA test for chlorine is to place damp starch-iodide paper into a test tube of chlorine. A positive result for chlorine is indicated by the paper turning from white to blue-black. |
|
| 20. |
A metal M forms two types of chlorine, MCl\(_2\) and MCl\(_3\) which of the following laws best explains the relationship between the chlorides? Law of A. conservation of mass B. definite proportion C. multiple proportion D. reciprocal proportion Detailed SolutionLaw of multiple proportions, statement that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers. |
| 11. |
The complex compound formed by the combination of two elements with a large electronegativity difference is likely to be A. polar covalent B. giant molecular C. covalent D. ionic Detailed SolutionIonic bonds are also formed when there is a large electronegativity difference between two atoms. |
|
| 12. |
The complex compound formed when aluminum dissolves in aqueous sodium hydroxide is A. Na3Al(OH)4. B. NaAl(OH)4 C. NaAl(OH)3 D. Na2Al(OH)3 Detailed SolutionWhen aluminium is treated with dilute sodium hydroxide it forms a salt called sodium tetrahydroxoaluminate and hydrogen gas. |
|
| 13. |
The vapour pressure of a liquid depends on l. temperature. ll. rate of condensation. III. cohesive forces holding the particles together A. l only B. l and ll only C. l and lll only D. ll and lll only Detailed SolutionThree common factors that influence the vapor pressure of a liquid are the surface area, intermolecular forces and the temperature. |
|
| 14. |
MgO does not readily dissolve in water because A. of its high melting point B. it is a covalent compound C. it forms a hydroxide when dissolved in water D. its lattice energy is higher than its hydration energy |
D |
| 15. |
Consider the following reaction equation: CaCO\(_{3(s)}\) + 2HCI\(_{(aq)}\) → CaCl\(_{2(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\). What mass of CaCI\(_2\) would be obtained when 25.0g of CaCO\(_{3(s)}\) is reacted with excess HCI\(_{(aq)}\)? [CaCO\(_3 = 100; CaCI\(_2\) = 111] A. 4.00g B. 4.44 g C. 18.9 g D. 27.8 g Detailed SolutionCaCO\(_3\) + 2HCI → CaCI\(_2\) + H\(_2\)O + CO\(_2\)100g CaCO\(_3\) will produce 111g CaCI\(_2\) 25g CaCO\(_3\) will produce x , x = \(\frac{25 x 111}{100}\) = 27.75 \(\approxeq\) 27.8g |
| 16. |
The number of sulphur atoms in 3.20g of SO\(_{2(g)}\) is [O = 16.0;S = 32.0; Avogadro constant = 6.02 x 10\(^{23}\)] A. 3.01 x 1022 B. 6.02 x 1022 C. 6.02 x1023 D. 1.20 x 1024 Detailed Solution32g of sulphur = 64g / 6.02 x 10\(^{23}\) atoms of SO\(_2\)xg of sulphur = 3.2g of SO\(_2\) : 1.6g of sulphur is present in 3.2g / 3.01 x 10\(^{23}\) atoms of SO\(_2\) |
|
| 17. |
Consider the reaction represented by the following equation: xCH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)O. The values of x,y and z respectively, are A. 2,3 and 4 B. 2, 4 and 3 C. 1,2 and 3 D. 1, 3 and 5 Detailed SolutionxCH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)OBalancing the equation, we have C = x (on the left) and 2 (on the right) ⇒ x = 2 2CH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)O On the left, H = 8 and on the right = 2z ⇒ 2z = 8 ; z = 4 2CH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + 4H\(_2\)O On the right, O = 4 + 4 = 8. On the left, O = 2 + 2y ⇒ 2 + 2y = 8 2y = 6 ; y = 3 x, y, z = 2, 3, 4. |
|
| 18. |
The formula of mercury (l) dioxonitrate (III) is A. HgNO3 B. Hg2NO2 C. Hg2(NO2)2 D. Hg(NO3)2 |
C |
| 19. |
A sample of a gas may be identified as chlorine if it turns A. damp blue litmus paper red B. lime water miky C. lead ethanoate paper black D. starch iodide paper blue-black Detailed SolutionA test for chlorine is to place damp starch-iodide paper into a test tube of chlorine. A positive result for chlorine is indicated by the paper turning from white to blue-black. |
|
| 20. |
A metal M forms two types of chlorine, MCl\(_2\) and MCl\(_3\) which of the following laws best explains the relationship between the chlorides? Law of A. conservation of mass B. definite proportion C. multiple proportion D. reciprocal proportion Detailed SolutionLaw of multiple proportions, statement that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers. |