The WAEC Chemistry 2022 Practical Answers -14th June 2022: have been outlined here. The West African Examination Council (WAEC) WAEC Chemistry 2022 Practical Answers -14th June 2022 Examination will hold on Tuesday 14th June 2022
The WAEC Chemistry 2022 Practical Answers -14th June 2022 will commence from 08:30 hrs. – 10:30 hrs. 10:30 hrs. -11:30 hrs. Below, we will be posting WAEC Chemistry 2022 Practical Answers -14th June 2022 for candidates that will participate in the examination.
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Answers Loading…………………. ====================== Meaning of Titre value °°°°°°°°°°°°°°°°°°°°°°°°°°°°°°° IN titration, the titer value is the ratio of actual to nominal concentration of a titrant, e.g a titer of 0.5 would require 1/0.5 = 2 times more titrant than nominal. This is to compensate for possible degradation of the titrant solution. (3a) Sulphur Dioxide Solution Turns Acidified Potassium Chromate Solution green Where As carbon dioxide Shows no Change (3bi) (I) H2 ; downward displacement of water (II) NH3; downward displacement of air (III) HCL; upward displacement of air (3bii) (I) Hydrogen gas(H2) is collected by the downward displacement of water because It is insoluble in water and It form an explosive mixture with air. (II) Ammonia gas(NH3) is collected by downward displacement of air because it is lighter than air (III) HCl gas is collected by upward displacement of air because it is 1.28 times heavier than air. (3c) (i) Distillation (ii) Filtration followed by evaporation to dryness (3d) This is because KCl react with NaHCO₃ to form two salts ============================= (2a) [TABULATE] =TEST= C + Distilled water =OBSERVATION= It dissolves completely to give a light green solution. =INFERENCE= Soluble salt. (2ai) =TEST= Solution C + NaOH in drops and in excess + Heat gently =OBSERVATION= A dirty green precipitate is formed which remains insoluble in excess. Effervescence occurs in which a colourless gas with a pungent smell which turns red litmus blue is given off. =INFERENCE= Fe²⁺ is present. NH₃ gas form. NH₄⁺ is present. (2aii) =TEST= Solution + BaCl₂ + dilute HCL in excess =OBSERVATION= A white precipitate is formed. The white precipitate remains insoluble and gives a white dense. =INFERENCE= SO₄²⁻, CO₃²⁻, SO₃²⁻, is present. SO₄²⁻ confirmed. (2b) Cations → Fe²⁺ and NH₄⁺ Anions → SO₄²⁻ ======================
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Sample WAEC Chemistry Practical Answers 2022 is Out – See Chemistry Titration Specimen Here.
The Waec chemistry practical 2022 paper for SSCE will now be written on Tuesday, 14th June 2022 from 08:30 hrs. to 10:30 hrs. for the first set and from 11:00 hrs. to 13:00 hrs. for the second set. The paper is Waec Chemistry practical alternative A.
We made this post to show apparatus, Waec chemistry practical specimen, graphs, and calculations that are used in the Chemistry Practical Examination questions and syllabus.
Here, we will be providing you with the Waec Chemistry practical 2022 questions and materials that will be used for Examination preparation. Read answers below.
Sample waec chemistry practical specimen 2019
WAEC Chemistry Practical 2022 Apparatus.
The following apparatus and materials will be required by each candidate for waec chemistry practical 2022;
(a) one burette of 50cm³ capacity.
(b) one pipette, either 20cm³ or 25cm³.
(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.
(f) filtration apparatus.
(g) one beaker
(h) one boiling tube
(i) four test tubes.
(j) Methyl orange as an indicator.
(k) mathematical table/calculator
(l) wash bottle containing distilled/deionized water.
(m) a burning splint
(n) watch glass
(o) bunsen burner/source of heat
Each candidate should be supplied with the following where n is the candidate’s serial number.
(a) 150cm₃ of iodine solution in a corked flask or bottle labeled ‘An’. These should all be the same containing 25.4g of I₂ in 6.0g KI per dm₃ of solution.
(b) 150cm₃ of sodium thiosulphate pentahydrate solution in a corked flask or bottle labeled ‘Bn’. These should all be the same containing 24.82g of Na₂S₂O₃ per dm₃ of solution.
(c) One spatula of ammonium ferrous sulphate, (NH₄)₂Fe(SO₄)₂.6H₂O in a specimen bottle labeled ‘Cn’. This must be the same for all candidates.
WAEC Chemistry Practical Questions & Answers.
1. A is 0.100 mol dm-3 solution of an acid. B is a solution NaOH containing 2.8 g per 500 cm3.
(a) Put A into the burette and titrate it against 25.0 cm3 portions B using methyl orange as an indicator. Repeat the titration to obtain consistent titres. Tabulate your readings and calculate the average volume of A used.
(b) From your results and the information provided above, calculate the:
(i) number of moles of acid in the average titre;
(ii) number of moles of NaOH in the volume of B pipetted;
(iii) mole ratio of acid to base in the reaction
[H = 1.00, O = 16.0, K = 39.0]
ANS: (a) titration answer soon.
(b) (i) number of moles of acid = 0.100 x VA
= X mole(s) [3sig. Fig to score]
1000cm3 contains 0.100 mole(s)
VA will contain 0.100 x VA
= X moles [3 Sig. Fig. to score]
(ii) Number of moles of NaOH in B
500cm3 of B contains 2.8g of NaOH
1000cm3 of B will contain 2.8 x 1000 = 5.6 NaOH
Molar mass of NaOH = 39 + 16 + 1 or 56 gmol-1
Conc of B = 5.6 = 0.100 mol dm-3
(iii) Mole ratio of acid to base = X: Y to nearest whole number ratio.
2. C is a mixture of two salts. Carry out the following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusion drawn from the result of each test.
(a) Put all of C into a boiling tube and add about 5cm3 of distilled water. Stir thoroughly and filter. Keep both the residue and the filtrate.
(b) To about 2 cm3 of the filtrate, add few drops of K2Cr2O7/H+. Boil the mixture and then allow to cool as (D).
(c) (i) Put the residue in a test tube and add dilute NaOH(aq). Shake the mixture and divide the solution into two portions.
(ii) To the first portion from (c)(i), add NaOH(aq) in drops and then in excess.
(d) What organic class does C belong to?
a) C+ burning splint
C burns when ignited with a burning splint.
C is flammable.
b) (i) C + Water
C forms a solution with water.
C is a mixture with water.
(ii) C + K2Cr2O7/H+ve
No visible reaction.
The K2Cr2O7/H+ve changes colour to green.
C is a reducing agent.
c) Residue from b (D) + C
D dissolves to form a reddish-brown solution.
A yellowish precipitate is formed and an antiseptic smell is given out.
D is a solution in D
Ethanol, ethanal or a secondary alkanol present.
C belongs to Alkanols
3. (a) Explain briefly the observations in each of the following processes:
(i) when carbon(IV) oxide is bubbled through lime water, it turns milky but the milkiness disappears when the gas is bubbled for a long time;
(ii) a precipitate of calcium hydroxide is insoluble in excess sodium hydroxide solution whereas that of lead (II) hydroxide is soluble.
(b) (i) What is a primary standard solution?
(ii) Calculate the mass of sodium trioxocarbonate(V) required to prepare 250 cm3 of 0.15 mol dm-3 solution.
[Na = 23.0; O = 16.0; C = 12.0]
(c) Name one gas that can be collected by:
(i) upward displacement of air;
(ii) downward displacement of air.
ANS: (a) (i): Insoluble CaCO3 formation is responsible for the milkiness produce when CO2 is bubbled through lime water while the disappearance of milkiness is due to the formation of soluble Ca(HCO3)2.
Lime water turns milky with CO2 because CaCO3/ CaCO3(s) is formed. Milkiness disappear when excess CO2 reacts with CaCO3 in water medium forming the soluble Ca(HCO3)2/ Ca(HCO3)2 (aq).
(ii) Calcium hydroxide is not amphoteric. Does not react with an alkali NaOH whereas lead (II) hydroxide is amphoteric so reacts with excess NaOH.
(b) (i) Primary standard solution is one whose concentration is known and can be used to standardize another solution.
(ii) M (Na2CO3) = 106 gmol-1
m(Na2CO3) = C x M x V
= 0. 15 x 106 x 0. 25
= 3. 98 g
(c) (i) Carbon(IV) oxide, sulphur (IV) oxide, hydrogen chloride, oxygen, nitrogen (IV) oxide, chlorine, hydrogen sulphide.
(ii) Ammonia, oxygen, hydrogen, methane.
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