Waec Chemistry Practical Answers 2023 for Thursday 18th May 2023
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VERIFIED ANSWERS BELOW:
TEST; C + distilled water + shake
OBSERVATIONS; It dissolves to give a colourless or clear solution
INFERENCE; Soluble salt suspected
TEST; Solution C + litmus paper
OBSERVATIONS; It has no effect on both blue and red litmus paper
INFERENCE; Neutral solution
TEST; solution C + Fehling solution A and B + heat
OBSERVATIONS; It gives a black red colour
INFERENCE; Reading agent confirmed
TEST; D + heat
OBSERVATIONS; It turns yellow on heating and white on cooling
INFERENCE; ZnO is present
TEST; D + dilute Hcl + heat
OBSERVATIONS; It dissolves completely to give a clear or colourless solution
INFERENCE; Solution chlorides of Zn³+, Al³+ is present
TEST; Solution D + NaOH in drops and then in excess
OBSERVATIONS; It gives white gelatinous precipitate. The precipitate is soluble in excess.
INFERENCE; Al³+, Zn²+ present.
Al³+, Zn²+ present .
TEST; Solution D + NH3 in drops and then in excess
OBSERVATIONS; It gives white gelatinous precipitate. The precipitate is soluble in excess NH3
INFERENCE; Al³+, Zn²+ is present.
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Waec Chemistry Practical Answers 2023
Thursday, 18th May, 2023
- Chemistry 3 (Practical) (Alternative A) – 09:30am – 11:30am (1st Set)
- Chemistry 3 (Practical) (Alternative A) –12:00pm – 2:00pm (2nd Set)
Attention all WAEC candidates! Are you prepared for your upcoming Chemistry Practical exam on May 18th, 2023? Don’t leave anything to chance – make sure you have the advantage you need to ace this test with our WAEC Chemistry Practical Answers and Questions for 2023!
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WAEC Chemistry Practical Specimen 2023
Welcome to our guide on the WAEC Chemistry Practical Specimen for the 2023/2024 exams. If you’re a secondary school student preparing for the Chemistry practical exam, you’re in the right place! In this article, we’ll provide you with easy-to-understand information and useful tips to help you do well in this important part of your Chemistry examination. Let’s get started and explore the specimen together.
- Great care should be taken to ensure that the information given in the items 2 and 3 below does not reach the candidate either directly or indirectly before the examination.
- In addition to the fittings and reagents normally found in a chemistry laboratory, the following apparatus and materials will be required by each candidate:
- (a) One burette of 50 cm3 capacity;
- (b) One pipette, either 20 cm3 or 25 cm3. (All candidates at one centre must use pipettes of the same volume. These should be clean and free from grease)
- (c) The usual apparatus for titration;
- (d) The usual apparatus and reagents for qualitative work including the following with all reagents appropriately labelled;
- (i) Dilute sodium hydroxide solution;
- (ii) Dilute hydrochloric acid;
- (iii) Dilute trioxonitrate(V) acid;
- (iv) Silver trioxonitrate(V) solution;
- (v) Acidified potassium dichromate solution;
- (vi) Aqueous ammonia;
- (vii) Lime water;
- (viii) Red and blue litmus paper
- (ix) Dilute tetraoxosulphate (VI) acid;
- (x) Fehlings solution A & B.
- (e) Spatula;
- (f) Filtration apparatus;
- (g) One beaker
- (h) One boiling tube;
- (i) Four test tubes;
- (j) Methyl orange as indicator;
- (k) Glass rod;
- (l) Wash bottle containing distilled/deionized water;
- (m) Burning splint;
- (n) Watch glass;
- (o) Bunsen burner/source of heat;
- (p) Droppers;
- (q) Mathematical table/calculator.
- Each candidate should be supplied with the following, where ‘n’ is the candidiate’s serial number
- (a) 150 cm3 of a solution of HCl in a corked flask or bottle labelled ‘An’. These should all be the same containing 8.5 cm3 of concentrated HCl per dm3 of solution.
- (b) 150 cm3 of Na2CO3.10 H2O in a corked flask or bottle labelled ‘Bn’. These should all be the same containing 5.0 g of the hydrated salt per dm3 of solution.
- (c) One spatulaful of glucose in a specimen bottle labelled ‘Cn’. This must be the same for all candidates.
- (d) One spatulaful of zinc oxide powder in a specimen bottle labelled ‘Dn’. This must be same for all candidate.
Past Chemistry Practical Answers
Titration; | rough | 1 | 2 | 3 |
Final; | 20.50 | 14.00 |14.00 | 14.00 |
Initial; | 0.00 | 0.00 | 0.00 | 0.00 |
Volume of A used; | 20.50 | 14.00 |14.00 | 14.00 |
Volume of pipette used
Average volume of A used = 14.00+14.00+14.00/3
2HNO³ + X²CO³ —> 2XNO³ + H²O + CO²
Conc – of B in molldm³ = ?
Molar mass of B =?
Conc(molldm³)= Conc(gldm³)/molar mass
0.042/1 * 4.20/m.m
Value of x in x²(i)³=?
(2x)+12 + (16*3)= 100
(i)I ensure that the readings are taken at the lower meniscus
(ii)I ensure that my saliva did not enter the pipette while taking the base
An effervescence occurs in which a colorless and odourless gas which burns blue litmus red is evolved.
CO² gas from CO3^²is suspected
The gas turns lime water milky.
CO² gas is confirmed.
Y dissolves completely in water
White gelatinous PPT is formed.
Al³+, Zn²+, Pb²+, present
The white gelatinous PPT dissolve in excess NaOH
Al³+, Zn²+, Pb²+, present
White gelatinous PPT formed.
Al³+, Zn²+, Pb²+, present
The white gelatinous PPT remain insoluble in excess NH³
Al³+, Pb²+, present.
It increases the boiling point of water
It requires heat to occur.
This is because it is deliquescent.
I. Calcium oxide
II. Anhydrous copper(ii) tetraoxosulphate(vi)
I. To separate a mixture of immiscible liquids
II. To collect gases.
Teacher’s Report filling
1: A=0.15, B= 4.20
2 (i=14.00, ii=14.00, iii=14.00)
For question 1
Use your school’s endpoint and use it wherever you see our endpoint (13.20cm³) …and calculate with it.
Use the answer you got in 1b(i) to solve in 1b(ii) i.e we got 0.0396 in 1bi after calculating with our average titre (endpoint) value 0.0396, as u can see, we then used the 0.0396 in 1bii which we got from our calculation in 1bi.
NB: If you use our endpoint and your school teacher submits a different one with a wide margin to what u used, you will either fail or lose some mark in this particular question. Your school teacher should have in one way given you the average titre value(endpoint) or you might have gotten it in the lab from your past experiment.
If your school wants to use mine, then the chemistry teacher must submit endpoint similar to ours without much difference…eg 13.10, 13.20, 13.30, etc
According to Lewis, an acid can be defined as a substance that can accept a pair of non-bonding electrons i.e an electron pair acceptor.
Salting out is a process through which soap is precipitated as solid from the suspension by adding common salt to the suspension. It is a purification process.
Reagent – Silver trioxonitrate(v)
Condition – The reaction takes place at high temperatures and pressure.
This is the proportion at which isotopes of an element are to each other in its composition.
(i) It has a stable configuration
(ii) This is because they have a covalent bond.
They can be differentiated using barium chloride. When barium chloride is added to concentrated H₂SO₄ a white precipitate is formed with concentrated HNO₃
(i) There are two electrolytes
(ii) Salt bridge is present.
(i) There is only one electrolyte
(ii) There is no salt bridge present.
The lower the ionisation energy, the higher the reactivity of metals. Since it decreases down the group, the reactivity increases down the group I.
This can be defined as a formula that shows the actual number of atoms in a compound.
(ii) This is because its interaction is perfectly inelastic.
Ionization energy can be defined as the energy required to remove a loosely bound electron from the outermost shell of a gaseous covalently bonded atom.
This is because B has more shells than Be which makes the valence electrons to be far from the nucleus making it require less energy due to lesser nuclear attraction.
I = 0.12A, t = 500seconds
m = 0.015g, F =96500C
M = 48.0, Charge = ?
m = MIt/CF
C = MIt/mF
C = (48×0.12×500)/(0.015×96500)
C = 2
(i) Electricity supply
(ii) Nearness to source of material
This can be defined as a property of metal which are weakly attracted to an applied magnetic field
(i) I → ₂₄Cr²⁺ → 2
₂₄Cr⁶⁺ → 0
(iii) It has unpaired electrons.
This can be defined as the number of atoms or molecules in one mole of a substance which is equal to 6.02 ×10²³
Mass/m.m = no. of molecules/Avogadro’s number
(2.30/m.m) × (3.01×10²²/6.02×10²³)
m.m = (2.30×6.02×10²³)/3.01×10²²
m.m = 46g/mol
Since the molar mass = 46g/mol
The formula is NO₂
14+32 = 46g/mol
(i) There is no reaction but rather forms a layer underneath the water
(ii) It reacts violently with water forming white silicon dioxide and hydrogen chloride gas.
The reaction is different because CCl₄ is an organic substance while SiCl₄ is not.
Extraction of copper
Cu²⁺ + 2e⁻ —> Cu
Using; m = ZIt
Z = m/It
Z = 3.2/(50x(3×60)+13)
Z = 3.2/(50×193)
Z = 3.2/9650
Z = 0.000332g/Asec
Oxygen can be prepared by heating KClO₃ in the presence of manganese(iv)oxide which acts as a catalyst to produce KCl and oxygen gas. The reaction takes place at a lower temperature and much faster rate.
KClO₃ (MnO₂)heat—–> KCl + O₂
(i) It melts to clear mobile liquid
(ii) A brownish gas is evolved.
(i) It leads to erosion
(ii) It causes pollution
This is because it has more surface area than it occupies leading to a faster rate of reaction.
When water is added to white anhydrous CuSO₄ it turns blue.
(i) It removes the hardness by precipitating the trioxocarbonate(iv)
(ii) It removes the hardness by removing the calcium or magnesium ion and precipitating trioxocarbonate(iv).
Ca(HCO₃)₂ → CaCO₃ + H₂O + CO₂
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Waec Chemistry Practical Answers 2023 for 18th May 2023
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