Paper 1 | Objectives | 48 Questions
WASSCE/WAEC MAY/JUNE
Year: 1997
Level: SHS
Time:
Type: Question Paper
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# | Question | Ans |
---|---|---|
1. |
The phenomenon observed when dust particles collide randomly in a beam of sunlight is known as A. Tyndal effect B. diffusion C. osmosis D. brownian movement E. dialysis |
A |
2. |
The mass number of an element is the sum of its A. electrons, neutrons and protons B. electrons and protons C. protons and neutrons D. orbital electrons E. valence electrons |
C |
3. |
How many unpaired electrons are there in an atom of an element with the following electronic configuration? 1s22s22p6 A. o B. 1 C. 2 D. 6 E. 8 |
A |
4. |
Radon is used as a tracer in medical research because it A. is monatomic B. is radioactive C. is a noble gas D. has a complete valence shell E. has a low boiling point |
B |
5. |
What is the most probable group of an element which is a soft, silvery white solid and reacts violently with water? A. Group 0 B. Group 1 C. Group 4 D. Group 6 E. Group 7 |
B |
6. |
In linear molecules, the bond angle is A. 90o B. 140o C. 109o D. 120o E. 180o |
E |
7. |
A sodium atom and a sodium ion have the same A. number of neutrons B. number of electrons C. electric charge D. electronic structure E. chemical properties |
A |
8. |
If the atomic number of an element X is 11 and that of nitrogen is 7, the most likely formula of the nitride of X is A. X3N B. XN3 C. X3N2 D. N2X E. NX2 |
A |
9. |
An increase in the pressure of a gas results in a decrease in its A. mass B. vapour density C. volume D. temperature E. concentration |
C |
10. |
A given volume of methane diffuses in 20 seconds. How long will it take the same volume of sulphur (IV) oxide to diffuse under the same conditions? [CH4 = 16; SO2 = 64] A. 5 seconds B. 20 seconds C. 40 seconds D. 60 seconds E. 80 seconds
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Detailed SolutionApplying the Graham's law defination; i.e,\(\frac{R(SO_2)}{R(CH_4)} = \sqrt{\frac{M(CH_4)}{M(SO_2)}} = \frac{t(CH_4)}{t(SO_2)}\) Where t(SO\(_2\)) = ?, M(SO\(_2\)) = 64 t(Ch\(_4\)) = 20 sec, M(Ch\(_4\)) = 16 therefore, \(\frac{t(SO_2)}{20}\) = \(\sqrt{\frac{64}{16}}\) \(\frac{t(SO_2)}{20}\) = \(\frac{8}{4}\) t(SO\(_2\)) = 8 x \(\frac{20}{4}\) = 40 sec. |
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