WAEC Chemistry Answers 2023 – Wednesday 24th May 2023

WAEC Chemistry Answers 2023: Are you a 2023 Waec candidate searching for 100% verified questions and answers for both objective and essay that will guarantee you A or B in the WAEC Chemistry 2023 examination? look no further, we have all it takes to make you score A in this Waec examination.

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VERIFIED CHEMISTRY ANSWERS BELOW:

CHEMISTRY-OBJ
1-10: CCCDCCBDAC
11-20: BCACBBCCBA
21-30:ACBABDCCCB

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NUMBER 1

1A
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-.

1C.

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.

1d.

-methane (CH4)
-propane (C3H8).

1e.

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.

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NUMBER 4

a)

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) → 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.

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Waec Chemistry Answers 2023

Wednesday, 24th May, 2023

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

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Past Chemistry Answers

1)
Tabulate
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
=25.00cm³

(1a)
Average volume of A used = 14.00+14.00+14.00/3
=14.00cm³ (Va)

2HNO³ + X²CO³ —> 2XNO³ + H²O + CO²

(1bi)
Conc – of B in molldm³ = ?
Given;
Ca= 0.15molldm³
Vb= 25.00cm³
Va= 14.00cm³
Cb= ??
Na= 2
Nb= 1

Using;
CaVa/CbVb=Na/Nb
Cb= CaVaNb/VbNa
Cb= 0.15*14.00*1/25.00*2
Cb= 2.1/50
Cb= 0.042molldm³.

(1bii)
Molar mass of B =?
Using;
Conc(molldm³)= Conc(gldm³)/molar mass
0.042/1 * 4.20/m.m
m.m=4.20/0.042
m.m=100glmol

(1biii)
Value of x in x²(i)³=?
(2x)+12 + (16*3)= 100
2x+12+48=100
2x+60=100
2x=100-60=40
2x/2=40/2
x=20g/mol

(1ci)
(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

(1cii)
Pink
==============================

(2)
Tabulate

(2ai)
Under Observation;
An effervescence occurs in which a colorless and odourless gas which burns blue litmus red is evolved.

Under Inference;
CO² gas from CO3^²is suspected

(2aii)
Under Observation;
The gas turns lime water milky.

Under Inference;
CO² gas is confirmed.

(2b)
Under Observation;
Y dissolves completely in water

Under Inference;
Soluble salt

(2bi)
Under Observation;
White gelatinous PPT is formed.

Under Inference;
Al³+, Zn²+, Pb²+, present

(2bii)
Under Observation;
The white gelatinous PPT dissolve in excess NaOH

Under Inference;
Al³+, Zn²+, Pb²+, present

(2biii)
Under Observation;
White gelatinous PPT formed.

Under Inference;
Al³+, Zn²+, Pb²+, present

(2biv)
Under Observation;
The white gelatinous PPT remain insoluble in excess NH³

Under Inference;
Al³+, Pb²+, present.
==============================

(3ai)
It increases the boiling point of water

(3aii)
It requires heat to occur.

(3bi)
This is because it is deliquescent.

(3bii)
I. Calcium oxide
II. Anhydrous copper(ii) tetraoxosulphate(vi)

(3ci)
Hydrogen Chloride

(3cii)
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)

+++++++++++++++++++++++++++++++++
TAKE NOTE

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
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CHEMISTRY 

==> Chemistry+Obj!
1ABACCABBCD
11DCDBCCDCDB
21BDCADAABCA
31CBDCCDCBBD
41CACCCDADCA

CHEMISTRY
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(1ai)
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.

(1aii)
AlCl₃

(1b)
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.

(1c)
Reagent – Silver trioxonitrate(v)
Condition – The reaction takes place at high temperatures and pressure.

(1d)
This is the proportion at which isotopes of an element are to each other in its composition.

(1e)
(i) It has a stable configuration
(ii) This is because they have a covalent bond.

(1f)
They can be differentiated using barium chloride. When barium chloride is added to concentrated H₂SO₄ a white precipitate is formed with concentrated HNO₃

(1g)
[TABULATE]

-ELECTROCHEMICAL CELL-
(i) There are two electrolytes
(ii) Salt bridge is present.

-ELECTROLYTIC CELL-
(i) There is only one electrolyte
(ii) There is no salt bridge present.

(1h)
The lower the ionisation energy, the higher the reactivity of metals. Since it decreases down the group, the reactivity increases down the group I.

(1i)
This can be defined as a formula that shows the actual number of atoms in a compound.

(1j)
(i) NH₃
(ii) This is because its interaction is perfectly inelastic.

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(2ai)
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.

(2aii)
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.

(2b)
I = 0.12A, t = 500seconds
m = 0.015g, F =96500C
M = 48.0, Charge = ?
Using;
m = MIt/CF
C = MIt/mF
C = (48×0.12×500)/(0.015×96500)
C = 2

(2ci)
Al₂O₃

(2cii)
(i) Electricity supply
(ii) Nearness to source of material

(2di)
This can be defined as a property of metal which are weakly attracted to an applied magnetic field

(2dii)
(i) I → ₂₄Cr²⁺ → 2
₂₄Cr⁶⁺ → 0

(ii) ₂₄Cr²⁺

(iii) It has unpaired electrons.
●●●●●●●●●●●●●●●●●●●
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(3ai)
This can be defined as the number of atoms or molecules in one mole of a substance which is equal to 6.02 ×10²³

(3aii)
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

(3aiii)
Since the molar mass = 46g/mol
The formula is NO₂
:. 14+(16×2)
14+32 = 46g/mol
:. NO₂

(3bi)
(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.

(3bii)
The reaction is different because CCl₄ is an organic substance while SiCl₄ is not.

(3ci)
Extraction of copper

(3cii)
(i) CuSO₄
(ii) CuCl₂

(3ciii)
Cu²⁺ + 2e⁻ —> Cu

(3civ)
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

(3d)
https://i.imgur.com/wnfOVYP.jpg.

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
●●●●●●●●●●●●●●●●●●●
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(5a)
(i) It melts to clear mobile liquid
(ii) A brownish gas is evolved.

(5bi)
(i) Carbon(ii)oxide
(ii) Lead(ii)oxide

(5bii)
Methane(CH₄)

(5biii)
(i) It leads to erosion
(ii) It causes pollution

(5biv)
This is because it has more surface area than it occupies leading to a faster rate of reaction.

(5bv)
Ammoniacal liquor

(5ci)
When water is added to white anhydrous CuSO₄ it turns blue.

(5cii)
(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).

(5ciii)
Ca(HCO₃)₂ → CaCO₃ + H₂O + CO₂

(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⁻

(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

(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)

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Conclusion

Waec Chemistry Answers 2023 for 24th May 2023 (Objective and Essay)

f you want to succeed in your Chemistry exam, you need to prepare thoroughly and have access to accurate and reliable answers and questions. Our WAEC Chemistry Answers and Questions for 2023 offer just that. With our expertly crafted material and our unbeatable price of just N600, you can give yourself the best possible chance of success in your upcoming exam.

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