WAEC 2023 Chemistry Questions and Answers (Essay and Objective)

WAEC 2023 Chemistry Questions and Answers (Essay and Objective)

Wednesday, 24th May 2023

  • Chemistry 2 (Essay) – 9:30 am – 11:30 am
  • Chemistry 1 (Objective) – 11:30 am – 12:30 pm

Click Here To Joint Our WhatsApp Group πŸ‘‡




(Q1) πŸ‘‡

A transmission element is a component that transfers data or energy from one point to another.

1bi. Element D-288 forms a doubly charged cation.
1bii. The soluble trioxocarbonate (IV) is CO32-.


The general decrease in the first ionization energies of the period in the periodic table is due to the increase in atomic radius and the decrease in effective nuclear charge.


-methane (CH4)
-propane (C3H8).


Alkanols are stronger bases than water because they have a higher tendency to donate a proton (H+) to an acid. This is because the alkyl group in alkanols is electron-donating, which increases the electron density on the oxygen atom in the hydroxyl group (-OH).

1f. The major raw materials used in the Solvay process are
-salt (NaCl)
-limestone (CaCO3)
-ammonia (NH3).

1g. Geometric isomerism is a type of stereoisomerism that arises when two or more compounds have the same molecular formula and connectivity, but differ in the spatial arrangement of their atoms due to restricted rotation around a double bond or ring.

1h. Water gas is a better fuel than producer gas because it has a higher calorific value and a higher percentage of hydrogen gas.

1hi. Heat of combustion is the amount of heat energy released when one mole of a substance undergoes complete combustion with oxygen under standard conditions.

1ji. Faraday’s second law of electrolysis states that the amount of substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. It can be expressed mathematically as:
m = (Q * M) / (n * F)

1jii. To calculate the amount of silver deposited, we can use Faraday’s law of electrolysis, which states that:
m = (Q * M) / (n * F)
Plugging in the values, we get:
m = (10920 C * 107.87 g/mol) / (1 * 96500 C/mol)
m = 12.17 g
Therefore, the amount of silver deposited when 10920 coulombs of electricity is passed through a solution of a silver salt is 12.17 grams.

(Q2) πŸ‘‡

(Q4) πŸ‘‡


Step 1) Prepare a solution of calcium chloride: Dissolve calcium chloride (CaCl2) in water to create a calcium chloride solution. This step involves measuring the appropriate amount of calcium chloride and adding it to a container of distilled water while stirring until the compound is completely dissolved.

Step 2) Add a source of carbonate ions: Introduce a source of carbonate ions to the calcium chloride solution. This can be achieved by adding sodium carbonate (Na2CO3) or sodium bicarbonate (NaHCO3) to the solution. The carbonate ions from these compounds will react with the calcium ions in the solution, forming calcium carbonate.

Step 3) Precipitation of calcium carbonate: A white precipitate of calcium carbonate will form as a result of the reaction between the calcium ions and carbonate ions. The reaction can be written as follows:

Ca2+ (aq) + CO3^2- (aq) β†’ CaCO3 (s)

The solid calcium carbonate will precipitate out of the solution.

Step 4) Filtration and washing: Use filtration to separate the solid calcium carbonate from the liquid solution. Set up a filter paper in a funnel and pour the mixture through it. The calcium carbonate will be retained on the filter paper while the liquid, containing any remaining calcium chloride and byproducts, passes through.

Step 5) Drying: After filtration, transfer the wet calcium carbonate onto a watch glass or a suitable container. Allow the solid to air dry or use gentle heat to remove any remaining moisture. The resulting dry solid is calcium trioxocarbonate (IV), commonly known as calcium carbonate.

(Q5) πŸ‘‡


Reaction between Iron (Fe) and Dilute H2SO4:

Iron reacts with dilute sulfuric acid to form iron sulfate (FeSO4) and hydrogen gas (H2). The balanced chemical equation for this reaction is:

3Fe(s) + 4H2SO4(aq) -> 3FeSO4(aq) + 2H2(g) + 2H2O(l)

In this reaction, iron displaces hydrogen from the acid, resulting in the production of iron sulfate and hydrogen gas. The iron sulfate is soluble in water and remains in solution, while the hydrogen gas is released as a gas.

Reaction between Aluminum (Al) and Dilute H2SO4:

Aluminum reacts differently with dilute sulfuric acid due to the protective oxide layer that forms on its surface. The oxide layer prevents further reaction by acting as a barrier between the aluminum metal and the acid. However, the reaction can occur if the protective layer is disrupted or removed

2Al(s) + 3H2SO4(aq) -> Al2(SO4)3(aq) + 3H2(g)

In this reaction, aluminum sulfate (Al2(SO4)3) and hydrogen gas are produced. The aluminum sulfate formed is soluble in water and remains in solution, while the hydrogen gas is liberated as a gas.

Please be patient and keep checking this page to wait for the Answers as soon as the time is right, the answer will be given as soon as possible

Please share it to your friends and invite them to visit our website πŸ‘‰ πŸ‘ˆ for more subjects

Leave a Reply

Your email address will not be published. Required fields are marked *

Back to top button