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CHM 101 CBT
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CHM 101 CBT
1-35 QUESTION
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Question 1 of 28
1. Question
The number of electrons in the K. L. M and N — shells of calcium are respectively
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Hint
Explanation
K-shells contain maximum of 2-electron, L shells can only contain maximum of 8 electrons, M-shells contain
maximum of 18 electrons, N- contains 32 electrons. Calcium with atomic number 20, shows that electrons are added
to a shell until a stable duplet (for the K-shell) and an octet for L-shells. The increase in the number of electron result
in the electrons occupying the shell with next highest energy level so calcium has 2, 8, 8, 2 (A). -
Question 2 of 28
2. Question
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Hint
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Question 3 of 28
3. Question
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Hint
MnO2- = -2
X+(-2×2) =-2
X + -4 = -2
X = +2
MnO4 = -1
X+(-2×4)=-1
X = -1+8
X=+7 (B) -
Question 4 of 28
4. Question
The empirical formula of a hydrocarbon that contains 93.3% carbon is C 12, H = 13
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Hint
Explanation.
C H % composition Mole ratio 92.3% 7.7% 92.3/12 7.7/1 Mole ratio of element 7.67 7.7 Smallest mole ratio 7.67 7.67 = 1 1 Empirical formula = CH (A)
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Question 5 of 28
5. Question
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Hint
Saturated hydrocarbons have single bond around its carbon, so they exhibit SP3 hybridization (A)
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Question 6 of 28
6. Question
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Question 7 of 28
7. Question
A metallic element X forms an on X2 what will be the formula of tetraoxosulphate of x
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Question 8 of 28
8. Question
When air is passed through alkaline pyragallol. the component of air absorbed by the pyragallol is
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Question 9 of 28
9. Question
When water drops are added to calcium carbide in a container and the gas produced is passed through jet and lighted, the resultant flame is called an
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Hint
CaC2 + 2H2O → C2H2 + Ca(OH)2
Ethyne (ethylene) produced when passed through oxygen give oxyacetylene flame, used for welding (D) -
Question 10 of 28
10. Question
If the relative formula mass of a hydroxide M(OH)3 is 78gmo1-1 what is the relative atomic mass, M?
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Hint
M(OH)3 = 78gmol-1
M+(16+1)3 = 78
M+(17×3) = 78
M=78—51 = 27(D) -
Question 11 of 28
11. Question
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Hint
E.g. X → X + e (loss of electron) (A)
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Question 12 of 28
12. Question
CH3COOH + CH3OH —> CH3COOCH3 + H2O
The reaction represented by the equation above isget materials/pq Here
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Hint
The reaction is acid catalysed and reversible produce methylethanoate (ester) in the process called esterification (B)
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Question 13 of 28
13. Question
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Question 14 of 28
14. Question
At atom of an element represent by X with A 17 and Z = 8. How many neutron does X contains?
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Hint
AZX when A = 17 mass number
X = 8 atomic number
Atomic No. = no. of proton + no. of neutron = 17–8 = 9 (D) -
Question 15 of 28
15. Question
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Question 16 of 28
16. Question
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Question 17 of 28
17. Question
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Question 18 of 28
18. Question
Two gas cylinder contain ethylene and acetylene respectively one test, which can be used to distinguish between them is by
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Hint
Ethylene (C2H4) and acetylene (C2H2). All alkynes with a terminal triple bond undergo substitution reaction with CuCl
to give reddish brown colouration of Cu2C2 (B) -
Question 19 of 28
19. Question
Brass and bronze are both metallic alloys, which of the following constituents is common to be alloys
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Question 20 of 28
20. Question
An organic compound has the empirical formula CH,O. What is its molecular formula if its vapour density is 90? (H = 1, C =12.0 = 16)
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Question 21 of 28
21. Question
A technique that can be used to show that chlorophyll pigment in a mixture of chemical compound and not a single coloured
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Question 22 of 28
22. Question
Which of the following statements is not true as we move from left to right along the periodic table?
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Question 23 of 28
23. Question
What mass of a divalent metal. M (atomic mass 40) would react with excess hydrochloric acid to liberate 22.4cm3 of dry hydrogen gas measured as s.t.p.?
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Hint
M + 2HCI → MCI2 + H2
1 mole: 2 mole 1mole : 1mole
No. of mole = 0.224.m4
22.4dm
= 0.01 mole of hydrogen gas
Since 1 mole of H2 is liberated by 1mole of M 0.0lmole of H2 will liberate 1mole of M Mass of M = 40×0.01 = 0.4g (C) -
Question 24 of 28
24. Question
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Question 25 of 28
25. Question
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Hint
An element with atomic number 12 is metallic divalent and a reducing agent, ionize by losing 2 electrons to combine with
non-metals to form electrovalent bond (B) -
Question 26 of 28
26. Question
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Hint
Oxygen is 21% of oxygen
% oxygen = Vol of oxygen x 100
Total Vol of air
21%= 10cm3 x 100
Total volume of air
Total vol. of air = (10cm3 x 100%) /21%=47.61
=50(C)
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Question 27 of 28
27. Question
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Question 28 of 28
28. Question
The energy change accompanying the addition of an electron to a gaseous atom is called
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