WAEC GCE 2020 – CHEMISTRY QUESTIONS AND ANSWERS ( Objective & Essay)

 

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Monday 3rd Dec. 2020

Paper II & I: Objective & Essay – Chemistry
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CHEMISTRY ANSWERS

 

OBJ

1ACBBDADCAD
11BCACCAABBB
21DBDCACBCAA
31CDACBBDBDD
41ACCBCDCDCC
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[Pls Trace Your Objective Any Were You See The Option That The Correct Answers.]
Chemistry-Obj!
1 Involves the loss and gain of electrons
2 Polymerisation
3 Global warming
4 3
5 Zinc ions
6 +1.56V
7 Ethene
8 NH3
9 Propanol
10 28
11 Hydrolysis
12 d-orbital
13 Lowering the activation energy
14 Closeness between reactant particles
15 Remaining the same with time
16 Reaction vessel Fels cool during the reaction
17 Faster
18 Solvent extraction
19 Saturated Solution
20 2.75mol/dm³
21 Partially dissociates in aqueous solution
22 138g
23 HI
24 2.00cm³
25 hydrogen chloride
26 strong electrovalent bond between ions
27 is not ductile
28 Electrons
29 C2H4
30 Aluminium
31 0.010mol/dm³
32 PbCO3
33 Linear
34 HCL and HOCL
35 +1
36 Ionic bond
37 have relatively low ionization energy
38 sour to taste
39 I,III and IV only
40 chrometography
41 2.00 dm
42 mole of solvent in 1dm³ of solution
43 does not contain neutron
44 1s²2s²2p⁶
45 Ammonium chloride
46 IV
47 I,II and IV only
48 Quantum numbers of Electrons
49 -273⁰C
50 Mass number

Chemistry+Obj


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The End
(4)

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(5)

(5rest)

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(2ai)
Percentage C5H12 of mass m = 7.2g
Volume of O2 = 20.0dm³
(i) from the general combustion equation
CxHy(g) + (x+y/4)O2 –> XCO2 + y/2H2O
C5H12(l) + 802(g) –> 5CO2(g) + 6H2O(l)

(2aii)
1 mole of C5H12(72g) = 5 moles of CO2
At stop 7.2gC5H12 = x volume of CO2
X = 7.2g×5×22.4dm³/72g
X = 5×2.224 = 11.2dm³ of CO2

(2aiii)
Volume of oxygen left after the reaction from the equation of reaction
1 mole of C5H12(72g) = 8(22.4)dm³
7.2g = x
X = 7.2×8×22.4/72 = 17.92dm³
Volume of O2 left after the reaction
= 20.0dm³ – 17.92dm³
= 2.08dm³

(2b)
When molecules collide with one another they possess kinetic energy. As most energetic molecules (those with greater kinetic energy) try to escape. Their escape may be facilitated by heat or by passing a wave of air over the container or by increasing the surface area of the container. As they try to do this, some molecules will loose energy on collision and fall back to the container; as such the average kinetic of the molecules of the liquid in the container reduces which results to cooling effect.

(2ci)
Avogadro’s Law states that the total number of atoms/molecules of a gas (i.e. the amount of gaseous substance) is directly proportional to the volume occupied by the gas at constant temperature and pressure.

(2cii)
N2(g) + 3H2—> 2NH3
Where 1 mole = 30cm³ of gas

At constant temperature, the volume of a fixed mass of gas. When the volume of a cylinder or a container is increased, the gases have more space to travel and collide hence the pressure is reduced but as the volume is decreased or compressed the gases have less space to travel therefore more pressure is built up.

(2a)

(2b)

(2rest)

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(3)

(3ai)
¹³R, ⁸Q
¹³R=1s²,2s²,2p⁶,3s²,3p¹
⁸Q=1s²,2s²,2p⁴

(3aii)
¹³R=2,8,3
Valency of ¹³R is 3
⁸Q= 2,6
Valency of ⁸Q=2

(3di)
2H² SO4(aq)+4NaOH(aq)—>2Na² SO4(aq)+4H²O(s)

(3dii)
Sodium teraoxosulphate (iv) salt and water

(3diii)
The resulting solution NaSO4 is basic and will have no effect on litmus paper

(3div)
When heated to dryness it can be used as a dehydrating agent