Free Live 2023 NECO Chemistry OBJ/ Essay Questions And Answers

2023 Get Free NECO Chemistry (CHEM) OBJ & Essay Questions and Answers fo June/July | NECO Chemistry (Objectives and Theory) Questions and Answers for June/July EXPO Room, July 13, 2023.

June/July 2023 NECO NO COST CHEMISTRY OBJECTIVE & ESSAY (CHEM) QUESTION ROOM 

Thursday, July 13, 2023 Chemistry (Expository & Objective)

10:00am – 1:00pm

NECO Chemistry OBJ/ Essay Questions And Answers 2023 

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2023 NECO CHEMISTRY OBJ

1-10: DEADADECAD

11-20: BAEDDBDBAE

21-30: CCDCABDDCD

31-40: EBEECEBCEE

41-50: BCCECDDADD

51-60: DABBDEAECA


COMPLETED!!!

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2023 NECO CHEMISTRY ESSAY


NUMBER 1


(1ai)

(i) Manufacturing sulfuric acid

(ii) Vulcanization of rubber

(iii) Formulation of Pesticides and fungicides


(1aii)

(i) It is a colorless gas that has a distinct smell of rotten eggs

(ii) Hydrogen sulphide is soluble in water to some extent


(1aiii)

Soaps are made from natural products while detergents are made from synthetic products.


(1aiv)

Detergents is for household cleaning and laundry purposes


(1bi)

Number of neutrons = Mass number (A) - Atomic number (Z)

I. ²³₁₁X

A = 23 (mass number)

Z = 11 (atomic number)


Number of neutrons = 23 - 11 = 12

II. ³⁹₁₉Y

A = 39 (mass number)

Z = 19 (atomic number)

Number of neutrons = 39 - 19 = 20


(1bii)

Molar mass:

Na = 22.99 g/mol

O₂ = 2 * 16.00 g/mol = 32.00 g/mol

Now, let's calculate the mass of oxygen needed:


First, calculate the number of moles of sodium (Na) in 9.2g:

Number of moles = Mass / Molar mass

Number of moles of Na = 9.2g / 22.99 g/mol ≈ 0.4002 mol

Since the mole ratio of Na to O₂ is 4:1, the number of moles of O₂ needed is:

Number of moles of O₂ = 0.4002 mol / 4 ≈ 0.1001 mol

Now, calculate the mass of oxygen needed:

Mass of O₂ = Number of moles of O₂ * Molar mass of O₂

Mass of O₂ = 0.1001 mol * 32.00 g/mol ≈ 3.204 g

Therefore, approximately 3.204 grams of oxygen are needed to burn 9.2 grams of sodium.


(1biii)

CaCO₃(s) + 2 HCl(aq) ---> CaCl₂(aq) + CO₂(g) + H₂O(l)

From the balanced equation, 1 mole of calcium carbonate (CaCO₃) reacts with 2 moles of HCl to produce 1 mole of calcium chloride (CaCl₂).

Molar masses:

CaCO₃ = Ca(40.08) + C(12.01) + 3O(16.00) = 100.09 g/mol

CaCl₂ = Ca(40.08) + 2Cl(35.45) = 110.98 g/mol

Now, let's calculate the number of moles of CaCO₃ in 50g:

Number of moles of CaCO₃ = Mass / Molar mass

Number of moles of CaCO₃ = 50g / 100.09 g/mol ≈ 0.4998 mol

Since the mole ratio of CaCO₃ to CaCl₂ is 1:1, the number of moles of CaCl₂ that can be obtained is also approximately 0.4998 mol.

Thus, about 0.4998 moles of calcium chloride can be obtained from 50g of limestone in the presence of excess hydrogen chloride.


(1ci)

(i) Sol: A sol is a colloidal solution in which solid particles are dispersed in a liquid medium.

(ii) Aerosol: An aerosol is a colloidal solution in which liquid or solid particles are dispersed in a gas medium.


(1cii)

The law of definite proportions, also known as the law of constant composition, states that a given chemical compound always contains its constituent elements in fixed and definite proportions by mass. This means that the ratio of the masses of the elements in a compound is constant, regardless of the compound's origin or method of preparation.


(1ciii)

(I) Sodium trioxonitrate (V) is also known as sodium nitrate, with the chemical formula NaNO₃.

The atomic masses are as follows:

Na (Sodium) = 22.99 g/mol

N (Nitrogen) = 14.01 g/mol

O (Oxygen) = 16.00 g/mol

Relative molecular mass of NaNO₃ = (1 * Na) + (1 * N) + (3 * O)

Relative molecular mass of NaNO₃ = (1 * 22.99 g/mol) + (1 * 14.01 g/mol) + (3 * 16.00 g/mol)

Relative molecular mass of NaNO₃ = 22.99 g/mol + 14.01 g/mol + 48.00 g/mol

Relative molecular mass of NaNO₃ = 85.00 g/mol

Therefore, the relative molecular mass of sodium nitrate (NaNO₃) is 85.00 g/mol.


(II) Copper (II) trioxosulphate (VI) pentahydrate is also known as copper (II) sulfate pentahydrate, with the chemical formula CuSO₄ · 5H₂O.

The atomic masses are as follows:

Cu (Copper) = 63.55 g/mol

S (Sulfur) = 32.06 g/mol

O (Oxygen) = 16.00 g/mol

H (Hydrogen) = 1.01 g/mol

Relative molecular mass of CuSO₄ · 5H₂O = (1 * Cu) + (1 * S) + (4 * O) + (10 * H) + (5 * O)

Relative molecular mass of CuSO₄ · 5H₂O = (1 * 63.55 g/mol) + (1 * 32.06 g/mol) + (4 * 16.00 g/mol) + (10 * 1.01 g/mol) + (5 * 16.00 g/mol)

Relative molecular mass of CuSO₄ · 5H₂O = 63.55 g/mol + 32.06 g/mol + 64.00 g/mol + 10.10 g/mol + 80.00 g/mol

Relative molecular mass of CuSO₄ · 5H₂O = 249.71 g/mol

Therefore, the relative molecular mass of copper (II) sulfate pentahydrate (CuSO₄ · 5H₂O) is 249.71 g/mol.


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NUMBER 2


(2ai)

Mass of silver deposited (in grams) = (Current in Amperes × Time in seconds × Atomic mass of silver) / (1 Faraday)


Given:

Current = 4.6 A

Time = 90 minutes = 90 × 60 seconds = 5400 seconds

Atomic mass of silver (Ag) = 108g/mol

1 Faraday = 96,500C

Substituting the values to calculate the mass of silver deposited:

Mass of silver deposited = (4.6 A × 5400 s × 108 g/mol) / 96,500 C

Mass of silver deposited ≈ (2,682,720 g·s/mol) / 96,500 C

Mass of silver deposited ≈ 27.8g


(2aii)

(i) Electrode surface area

(ii) Electrolyte temperature


(2aiii)

(i) The oxidizing agent is MnO₄⁻(aq)

(ii) The reducing agent is Fe²⁺(aq)


(2aiv)

MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ----> Mn²⁺ + 4H₂O(l)


(2bi)


(2bii)

(i) Gases have no fixed shape or volume.

(ii) Gases have low density compared to solids and liquids.

(iii) Gases have high kinetic energy and are in constant motion.


(2biii)

Faraday's second law of electrolysis states that the mass of a substance deposited (or liberated) during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte.


(2biv)

(i) Charcoal

(ii) Coal


(2bv)

Na (Sodium) > Ca (Calcium) > Mg (Magnesium) > Al (Aluminum)


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NUMBER 3


(3ai)

(i) Butan-2-ol - Secondary alkanol

(ii) 2-methylpropanol - Primary alkanol

(iii) 2-methylpropan-2-ol - Tertiary alkanol


(3aii)

(i) Fermentation

(ii) Ethylene hydration


(3aiii)

Let the relative molecular mass of gas Z be M.

(Rate of diffusion of hydrogen)/(Rate of diffusion of gas Z) = √(molar mass of gas Z)/√(molar mass of hydrogen)

6/1 = (√M)/(√2)

36 = M/2

M = 2x36

M = 72


(3bi)

1s², 2s², 2p⁴


(3bii)

(i) It is a colorless

(ii) It is soluble in water.

(iii) It is tasteless


(3biii)

(i) Identify the longest chain.

(ii) Name the substituents alphabetically


(3biv)

C₂H₄ + O₂ ---> 2CO₂ + 2H₂O


(3ci)

Endothermic reaction can be defined as a form of heat reaction in which heat is absorbed from the surrounding into the reacting system.


(3cii)

Zn(s) + H₂SO₄(aq) ---> ZnSO₄(aq) + H₂(g)


(3ciii)

Redox reaction.


(3iv)

(i) For refining petrol

(ii) For food processing

(iii) For producing fertilizer


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NUMBER 4


(4ai)

A super saturated solution is a solution that contains more than the maximum amount of solute that is capable of being dissolved at a given temperature.


(4aii)

15/345 = Solubility *25/1000

Solubility =1000*15/25*345=15000/8625

Solubility = 1.79mol/dm³


(4aiii)

(i) H₃0⁺

(ii) NH₄⁺

(iii) [CN]⁻₆


(4bi)

(i) It has no chemical formula

(ii) It can be separated physically

(iii) Freezing air slowly yields different liquids at different temperatures


(4bii)

(i) Noble gases

(ii) Carbon (iv) oxide


(4biii)

H₂SO₄ ----> 2H+ + SO₄²⁻

1 mole of H₂SO₄ = 2 mole of H⁺

0.1 mole of H₂SO₄ = 0.2 mole of H⁺

Mole = no. of H⁺/Avogadro's constant

No. of H⁺ = Mole * Avogadro's constant

= 0.2 * 6.0*10²³

= 1.2*10²³ ions


(4biv)

(i) Dative bonding

(ii) Hydrogen bonding


(4bv)

(i) BRASS:

Constituent: Copper and zinc.

Use: Brass is used in the production of musical instruments decorative items and plumbing fixtures.


(ii) BRONZE:

Constituent: Copper and tin.

Use: Bronze is used in the production of statues coins and various machinery.


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NUMBER 5


(5ai)

A base is a substance which when disolve produce hydroxyl ion (OH⁻) as the only negative ion


(5aii)

(i) K₂O

(ii) MgO


(5aiii)

(i) it is used in printing inks and dyes

(ii) it is used in making photographic chemicals


(5aiv)

Aliphatic does not have good odour while an aromatic hydrocarbon has


(5av)

M.m of XCl₃=10-8+(35-5*3)

=10.8+106.5

=117.3

Vapour density =117.3/2=58.65


(5bi)

(i) Temperature

(ii) concentration

(iii) surface area


(5bii)

The law states that energy can neither be created nor destroyed in and isolated system.


(5biii)

(i) burning of wood

(ii) neutralization reaction


5C

Free 2023 NECO Chemistry OBJ/ Essay Questions And Answers


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NUMBER 6

Free 2023 NECO Chemistry OBJ/ Essay Questions And Answers


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