Chemistry
Paper 1 | Objectives | 50 Questions
WASSCE/WAEC MAY/JUNE
Year: 2017
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 50 Questions
WASSCE/WAEC MAY/JUNE
Year: 2017
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
|---|---|---|
| 1. |
Which of the following species determines the chemical properties of an atom? A. Electron B. Neutron C. Nucleus D. Proton
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Detailed SolutionAn element's chemical properties depend on its valence electrons.The chemical properties of elements depend on an element's electron configuration. |
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| 2. |
The following glasswares are used to measure the volume of liquids EXCEPT A. graduated beaker B. pipette C. test tube D. burette
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Detailed SolutionTest tubes are convenient containers for heating small amounts of liquids or solids with a Bunsen burner or alcohol burner. |
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| 3. |
Pauli exclusion principle is related to A. quantity of electrons in the valence shell B. filling the orbitals with lower energy first C. the filling of degenerated orbitals D. quantum numbers of electrons
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Detailed SolutionPauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers. |
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| 4. |
The species with no electron in the 3d-orbital is A. \(Zn^{2+}\) B. \(Sc^{3+}\) C. Cu D. \(Mn\)
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Detailed SolutionScandium has electronic configuration \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{1}4s^{2}\) hence, \(Sc^{3+}\) has configuration \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\). Hence, no electron in the 3d-orbital. |
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| 5. |
The seperation of Nitrogen from oxygen using fractional distillation of air is possible because A. nitrogen is less dense than oxygen B. oxygen is more reactive than nitrogen C. of the difference in their boiling points D. they belong to the same period
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Detailed SolutionThe various gases vaporize as their boiling points are reached, then they are pumped away and separately re-liquefied for convenient storage. |
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| 6. |
The purity of a solid sample can best be determined by its A. boiling point B. melting point C. conductivity D. solubility
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Detailed SolutionThe purity of a solid can be determined by measuring the melting point. A pure solid will have a single, sharp melting point. |
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| 7. |
The mass of one mole of \((NH_{4})_{2}CO_{3}\) is [O=16.0, N=14.0, C=12.0, H=1.0] A. 66.0g B. 76.0g C. 80.0g D. 96.0g
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Detailed SolutionThis is same as asking for the molar mass of the compound= \((2\times 14.0)+ (8\times 1.0) + 12.0+(16.0\times 3)\) = 28.0+8.0+12.0+48.0 = 96.0g |
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| 8. |
Atomic orbital is A. the circular path through which electrons revolve round the nucleus B. a region around the nucleus where electrons are most likely to be found C. the path around the nucleus through which electrons move D. the path around the nucleus through which protons move
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Detailed SolutionThe orbital is a path or region around the nucleus. The electrons move in the orbitals and not through it. |
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| 9. |
The property of elements which increases down a group of the periodic table is A. electronegativity B. electron affinity C. ionic radius D. ionization energy
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Detailed SolutionThe ionic radius decreases across a period but increases along a group. |
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| 10. |
The reason for the decrease in the atomic size of elements across a period is that A. nuclear charge increases while the outermost electrons are drawn closer to the nucleus B. nuclear charge decreases while the outermost electrons are drawn closer to the nucleus C. valence electrons increase across the period while the valence shell remains constant D. nuclear charge decreases while the distance of the valence shell from the nucleus is increasing
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Detailed SolutionThis is caused by the increase in the number of protons and electrons across a period. |
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