Waec Chemistry Questions and Answers 2021/2022

CHEMISTRY OBJ
1-10: CDBAAAACAD
11-20: CBBCDCCADB
21-30: CACCABACBC
31-40: DBCBDCBABD
41-50: CBBCCABDDA

 

(1a)
A molecular formula consists of the chemical symbols for the constituent followed by numeric subscripts showing the number of atoms of each element present while structural formula consists of symbols for the atoms connected by short lines that represent chemical bonds.

(1b)
(i) Screening effect
(ii) Size of atom
(iii) Nuclear change

(1c)
(i) The reaction must proceed both forward and backward i.e reversible
(ii) DG = O

(1d)
(i) B
(ii) It has the least ionization energy among the three

(1e)
Graham’s law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of their densities provided other conditions remained constant

(1f)
(i) Na2SO4. Mg(NO3)2
(ii) CaCO3

(1g)
(i) Protein – Addition polymer
(ii) Perspex – Condensation polymer
(iii) Nylon – Condensation polymer

(1h)
Atomic radius can be defined as the half distance between the nucleus of two covalently bonded atoms.

(1i)
Ethanol has a higher boiling point than propane because it has stronger force of attraction than propane which has a weak van der waal’s force.

(1j)
(i) It aids the digestion of food
(ii) It aids the action of enzymes
(iii) It controls the availablity of nutrients

==================================

(2ai)
(i) They have the same method of preparation
(ii) They have the same chemical properties

(2aii)
(i) Ethane has a single bond while Ethene has a double bond
(ii) They belong to different homogeneous series
(iii) They have different molecular formula and structural formula

(2bi)
HCl

(2bii)
– It is colourless
– It has an irritating odour

(2biii)
2NaCl(aq) + H2SO4(aq) —> Na2SO4(aq) + 2HCl(g)

(2ci)
Hydrocarbons are compounds having hydrogen and carbon atoms only. Examples are alkanes, alkenes, alkynes etc

(2cii)
(i) Coal
(ii) Crude oil

(2ciii)
C H
83% 100.83
83/12 17/1
6.92/6.92 17/6.92
1 2

Empirical formula = CH2

(2d)

===================================

(3ai)
H2SO4 (aq) + 2KOH(aq) —> K2SO4(aq) + 2H2O(l)

(3aii)
Given, Va = 25.0cm³, Vb = 24.0cm³, Cb = 0.15mol/dm³, Ca =?, Na = 1 , Nb = 2
Using,
CaVa/CbVb = Na/Nb
Ca = CbVbNa/VaNb
Ca = 0.15 × 24.0 × 1/25.0×2
Ca = 3.6/50 = 0.072mol/dm³

(3bi)
(i) 2Mg(s) + CO2(g) —> 2MgO(aq) + C(g)
(ii) This is because magnesium can reduce carbon (iv) oxide to black carbon and producing magnesium oxide

(3bii)
Mg(NO3)2
M of Mg(NO3)2 = 24+(2×14) + (6×16)
= 24 + 28 + 96
= 148g/mol
% of N = 28/148 × 100/1
N = 18.9%

(3ci)
(i) Reaction with KMnO4
(ii) Reaction with bromine water

(3cii)
It turns them colourless

(3d)
Boil the vegetable oil in sodium hydroxide solution, it breaks down releasing the organic acid and the alkanol. The process is known as saponification. The organic ester is immediately neutralized by the sodium hydroxide solution to form the sodium salt of the organic acid which is soap
=================================

(4ai)
(i) Graphite – Hexagonal
(ii) Diamond – Octahedral

(4aii)
Diamond us hard because of the strong force of attraction that hold them together and it can not conduct electricity because all the available electrons are used for bonding while Graphite is soft because it’s particles are in layers and can conduct electricity because not all the electrons are used for bonding

(4bi)
(i) It helps to remove dissolved gases and oxidizes dissolved metals
(ii) It helps to remove objects such as rags, paper, plastics, and clogging of downstream equipment
(iii) It reduces particle concentration in water and minimizes the need for coagulation and flocculation

(4bii)
(i) Calcium tetraoxosulphate (vi)
(ii) Calcium hydrogen trioxocarbonate

(4biii)
(i) Filtration
(ii) Boiling

(4biv)
It has a sweet taste and it builds the shells of lower organisms

(4ci)
The ore (SnO2) is wasted out of the ground with water, powdered, washed and roasted to drive off oxides sulphur, arseric and antimony. The roasted one is now reduced by heating with coke in a reverberatory furnace and molten tin is tapped off from the bottom of the furnace. The tin obtained is impure and it is remelted on a slopping furnace in which the tin melts and run its moulds. The tin obtained is 99.9% pure

(4cii)
SnO2(g) + 2C(s) —> Sn(l) + 2CO(g)

(4ciii)
(i) Sn(g) + O2 —> SnO2(g)
(ii) Sn(g) + 2Cl(g) —> SnCl4(g)

 

 

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