WAEC GCE 2018/2019 Chemistry Essay and Objective Questions and Answers

WAEC GCE 2018/2019 Chemistry Essay and Objective Questions and Answers January/February series expo Questions and Answers runz runs posted for free now available for free on WhatsAppWAEC GCE EXPO

WAEC GCE 2018/2019 Chemistry Essay and Objective Questions and Answers

1a)
Gas > liquid > Solid
(1aii)
Solid > liquid > gas
(1bi)
PH>7
(1bii)
PH>7
(1ci)
(I) Magnesium from 0 to +2
(II) Hydrogen changes from 1 to 0
(1cii)
I —> Magnesium is oxidized
II —> Hyfrogen is reduced
(1ciii)
HCL
(1di)
(i) Boiling occurs throughout the entire volume while evaporation occurs at the surface
(ii)It involves the formation of bubbles while evaporation does not form bubbles
(1dii)
It decreases the boiling point as the boiling point of water is proportional to the magnitude of atmosphere pressure.
(1e)
It dictates the rate of reaction and also optimize the amount of product formed.
(1f)
C3H7CooH —> Butanoic acid
(CH3)3COH —> Butan-1-ol
(1g)
Functional groups are responsible for their different chemical behavior or properties so that organic compounds with smaller functional group would behave similarly.
(1h)
Solids in liquid:
(i) It depends on the nature of temperature changes or exothermic or endothermic as affect by the temperature of the liquid.
(ii) The surface area of solid is lesser, they dissolves less as compound to gases in the same volume.
(iii) Pressure does not affect the solubility of the solids
Gas in liquid:
(i) It depends on the average kinetic energy as affect by temperature of the liquid.
(ii) The surface area of gases are higher so they dissolve more in a given volume of water.
(iii) The solubility of gases is a direct consequences of pressure changes due to the partial pressure of the gas molecules.
(1i)
3F2 + 3H2O -> 6HF + O3
(1j)
Basicity of an acid is the number of replaceable hydrogen ions in the acid.
============
(2i)
C = 0.48 H = 0.8 Cl = 1.42
H
C = 1 H = 1 Cl = 35.5
C = 0.48/12
H = 0.08/1
Cl = 1.42/35.5
0.04/0.04 , 0.08/0.04 , 0.04/0.04
Therefore CH2CL
The electrical formular = CH2CL
(2aii) The molar mass of the compound is 99
Hence (CH2CL)n = 99
(12 + 2C1) + 35.5)n = 99
12 + 2 + 35.5
(495)n = 99
Therefore n = 2
Hence the molar formula = (CH2CL)2
C2H4CL2
(2b)
(I) it is soluble in water and in other polar solvents.
(II) Nacl(s) when dissolve in water ionized to form Na+ and Cl-
(III) the aqueous solution of Nacl(s) can be use as n electrolyte
(2ci) DRAW THE DIAGRAM
(2cii)
(i)Eq1- actuation energy for in catalyzed reaction
(ii)Eq2 – actuation energy for catalyzed reaction
2di)
Paraffin wax
2dii)
To obtain the fraction the crude oil must undergo fraction distillation. In this process the crude oil is separated into different portion and the paraffin’s wax district in the temperature range of 35°c to 500°c
2diii)
i)ethene
ii)propane
2div)
They can be obtained from the cracking of crude oil. Cracking which is the breaking down of the crude oil into smaller components through the action of heat
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