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This material is 100% free and it contains the 2021 WAEC Chemistry OBJ questions and answers sample and the possible WAEC 2021 Chemistry theory questions and answers that may likely appear on the exam day.
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WAEC 2021 Chemistry Questions and Answers for Theory and OBJ.
Exam Scheme, Format, and instructions:
This exam paper follows the following format and exam scheme: Paper 1 (Objective) and paper 2 (theory).
Paper one contains 50 OBJ questions (to answers all questions) while paper 2 is essay/theory questions (candidates are to answer the number of questions as indicated on the exam paper)
- Go to your exam hall early (1 hour before exam)
- Follow the WAEC timetable daily to avoid missing a paper.
- Pray before and after each paper.
- Follow the exam instructions on the question paper.
- Shade your answers properly using an HB pencil.
- Fill in your details properly before you start answering your questions.
- Attempt all questions before submitting them.
Sample WAEC Chemistry OBJ questions and Answers 2021 for paper 1.
Below are possible WAEC Chemistry Objective questions 2021 that may appear in this year’s exam.
There are not the actual chemistry questions but, sample questions for practices which might possibly be repeated in this year’s exam.
1. Which of the statements below is correct about Isotopes of the same element?
- Have the Same number of Protons, neutrons, and electrons.
- Same Protons and neutrons but a different number of electrons.
- Same number Protons and electrons but a different number of neutrons.
- The number of Neutrons and electrons is the same but the number of protons differs.
2. The chemical bond that is formed by the transfer of electrons is known as?
- Covalent Bond
- Dative Bond
- Ionic bond
- Metallic bond
3. Two electrons can fill the same orbital if only?
- They have different Angular momentum quantum numbers.
- Magnetic quantum numbers differ
- Principal quantum numbers are different.
- Spin quantum numbers not the same
4. Which of the substances listed below is not a hydrocarbon?
5. The complete ionization of A substance into hydroxonium ions indicates:
- Strong acid.
- Strong base.
- Weak acid.
- Weak base.
6. One of the solutions below that can resist changes in pH when a small quantity of a base or acid is added is?
- Buffer solution
- Neutral solution
- Saturated solution
- Supersaturated solution
7. Write a balanced chemical equation for a 2.47g dry pure copper (II) oxide which is completely reduced copper using a laboratory gas. If the mass of the residue left is found to be 1.97g.
- Cu + O → CuO.
- 2Cu+O2 → 2CuO
- Cu → O
- 2Cu →O2
The correct answer is B. 2Cu+O2 → 2CuO.
See how this question is solved in the number one (No.1) question in the theory section.
Use this question to answer question 2 & 3
8. Calculate the mass of anhydrous sodium-trioxocarbonate (IV) present in a 300cm3 of 0.1M; (Na =23, C=12, O=16).
The correct answer is A. 3.18g. See solving in No.2 ques. in the theory section.
9. The number of Na2CO3 particles present in the solution
- 6.02 X 1023
- 0.1 X 6.02 1023
- 1.81 X 1022
- 1.81 X 1023
- 6.02 X 1022
The correct answer is C. 1.81 X 1022.
See how to solve this question in the number No.2 question in the theory section.
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WAEC Chemistry Theory Questions and answers 2021 samples
These are possible WAEC Chemistry theory questions and answers 2021 that may appear in this year’s exam.
There are not actual questions and answers but, sample questions for practices.
2.47g of dry pure copper (II) oxide was completely reduced to a copper using laboratory gas. The mass of the residue left was found to be 1.97g. Write a chemical equation for this reaction.
Molar mass of copper atoms =63.5
No. of moles of copper atoms =1.97
=63.5 = 0.03 mole
Molar mass of oxygen atoms = 16
No. of moles Oxygen atom = (2.47 – 1.97) = 0.5
Mole ratio of Cu to O = 0.03 : 0.03
= 1 : 1
Therefore, the equation is Cu + O → CuO.
But Oxygen is diatomic, so we have 2Cu+O2 → 2CuO
2. Calculate the;
- mass of anhydrous sodiumtrioxocarbonate (IV) present in a 300cm3 of 0.1M;
- Number of Na2CO3 particles present in the solution (Na =23, C=12, O=16)
a. Molar concentration of Na2CO3 =0.1M
Molar mass of Na2CO3 = 106gmol-1
Mass concentration = Molar concentration X Molar mass
=0.1 X 106
i.e 1000cm3 of 0.1M solution contain 10.6g of Na2CO3
Therfore,300cm3 of 0.1M solution will contain
= 300 X 10.6/1000
=3.18g of Na2CO3
b. Number of Na2CO3 particles
=Molar concentration X 6.02 X 1023
=0.1 X 6.02 1023
=6.02 X 1022
Now, 100cm3 of 0.1M solution contains 6.02X1022 Na2CO3 particles
Therefore, 300cm3 of 0.1M solution will contain 300X6.02 X1022/1000
= 1.81 X 1022 particular of Na2CO3.
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