Neco 2023 Chemistry Essay & Objective Answers

 

CHEMISTRY QUESTIONS PAPER

 

 

USE THIS CHEMISTRY OBJ
01-10: DEADADECAD
11-20: BAEDDBDBAE
21-30: CCDCAADECD
31-40: BBEEBABCEE
41-50: BCCECEDADE
51-60: DABBDEAECA

(1ai) (CHOOSE ANY BEST 3)
(i) Sulphur is used to produce sulphur(IV)oxide for manufacturing tetraoxosulphate(VI)acid
(ii) Sulphur is used in the vulcanization of rubber.
(iii) Sulphur and some of its products are used as fungicides and insecticides for spraying crops.
(iv) Sulphur is used to manufacture the bleaching agent used in the pulp and paper industry.

(1aii) (CHOOSE ANY BEST 2)
(i) Hydrogen Sulphide is a colorless gas with a repulsive smell like that of a rotten egg.
(ii) It is very poisonous.
(iii) It is about 1.18 times denser than air.
(iv) It burns with a pale blue flame.

(1aiii)Soaps are made from natural ingredients, typically fats or oils, and an alkali (such as sodium hydroxide or potassium hydroxide) in a process called saponification. WHILE Detergents are synthetic (man-made) cleaning agents composed of various chemicals, including surfactants.

OR

Soaps are effective at removing dirt, grease, and oil, but they may not perform as well in hard water conditions. WHILE Detergents are generally more effective than soaps in all water conditions, including hard water, due to their ability to work with the mineral ions present.

(1aiv) (CHOOSE ANY BEST 1)
(i) Detergents are used in various personal care products, such as shampoos, body washes, and hand soaps.
(ii) Detergents are specifically designed for cleaning dishes, cutlery, glasses, and cookware.
(iii) Detergents are used for general household cleaning tasks.

(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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(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) DIAGRAM

(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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(3a)(i) Classifying the alkanols:
– Butan-2-ol: secondary alkanol
– 2-methylpropanol: secondary alkanol
– 2-methylpropan-2-ol: tertiary alkanol

3a(ii) Two methods to prepare ethanol commercially:
– Fermentation: Ethanol can be produced by the fermentation of sugars using yeast. This is commonly used to produce alcoholic beverages and biofuels.
– Hydration of ethene: Ethanol can be produced by the hydration of ethene (ethylene) in the presence of a catalyst, such as phosphoric acid.

3a(iii) To calculate the relative molecular mass of Z, we need to compare the rates of effusion or diffusion of Z and hydrogen. The rate of effusion/diffusion is inversely proportional to the square root of the molar mass. Since hydrogen diffuses 6 times as fast, the molar mass of Z is 6 times larger than the molar mass of hydrogen. Therefore, the relative molecular mass of Z is 6 * 2 = 12.

(3bi) The electronic configuration of oxygen using s, p, d, f notation is 1s2 2s2 2p4.

(3bii) Three physical properties of oxygen:
– Oxygen is a colorless and odorless gas.
– It is slightly soluble in water.
– Oxygen supports combustion and is necessary for respiration.

(3biii) Two rules for naming alkenes:
1)Alkenes must have the ending “-ene” in their name.
2)The longest continuous carbon chain containing the double bond is used as the base name, and the position of the double bond is indicated by the lowest possible number.

(3biv) Equation for the oxidation reaction of ethene:
Ethene + Oxygen → Carbon Dioxide + Water
C2H4 + O2 → CO2 + H2O

(3ci) Endothermic reaction: A reaction that absorbs heat (energy) from the surroundings, resulting in a decrease in temperature.

(3cii) Equation for the laboratory preparation of hydrogen using dilute tetraoxosulphate(VI) acid and Zinc:
Zn + H2SO4 → ZnSO4 + H2

(3ciii) The type of reaction involved in (cii) is a displacement reaction, where zinc displaces hydrogen from the acid to form hydrogen gas.

(3civ) Three uses of hydrogen:
(i) Hydrogen is used as a fuel for combustion engines and fuel cells.
(ii) It is used in the production of ammonia for fertilizer and other chemicals.
(iii) Hydrogen is used in the hydrogenation of oils and fats in the food industry.
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(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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(5ai) A base is a substance that can accept protons or donate pairs of electrons. Bases typically have a pH greater than 7 and can neutralize acids.

(5aii) Examples of basic oxides include
i) calcium oxide (CaO) and
ii) sodium oxide (Na2O).

(5aiii) Two uses of iodine are:
(i)Iodine is used as an antiseptic to disinfect wounds and prevent infections.
(ii) Iodine is also used in the production of X-ray contrast agents, which helps in visualizing organs and tissues during medical imaging.

(5aiv) One difference between an aliphatic and aromatic hydrocarbon is the structure. Aliphatic hydrocarbons are linear or branched chains of carbon and hydrogen atoms, while aromatic hydrocarbons contain a ring structure with alternating double bonds.

(5av)To calculate the vapour density of XCl3, we need to know the molar mass of X. Without additional information on the element X, we cannot calculate the vapour density.

(5bi)Three factors that affect the rate of a chemical reaction are:
(i)Concentration: Increasing the concentration of reactants generally increases the rate of reaction.
(ii)Temperature: Higher temperatures usually increase the rate of reaction as particles have more energy and move faster.
(iii) Catalysts: Catalysts are substances that can speed up a reaction by providing an alternative reaction pathway with a lower activation energy.

(5bii)The law of conservation of energy states that energy cannot be created or destroyed in an isolated system; it can only be transferred or transformed from one form to another.

(5biii) Two examples of chemical change are:
(i)Combustion, where a substance reacts with oxygen and releases heat and light.
(ii)Rusting of iron, where iron reacts with oxygen in the presence of water and forms iron oxide.

(5c) DIAGRAM

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6ai). Pick any two
The same general formula.
Same functional group.
Similar chemical properties

(aii)
*trans–1,4–dibromo but–2–ene

Br
\
CH2 H
\ /
C–C
/ \
H CH2
\
Br
For trans Isomer

*Cis–1,4–dibromo but–2–ene
H H
\ /
C–C
/ \
CH2 CH2
/ \
Br Br
For Cis Isomer.

(aiii) pick any 3
-Wrought Iron.
– White Iron.
– Gray Iron.
-Ductile Iron.

(bi) molar mass= 64+32+16×4= 160g/mol

Conc. In mol/dm³ = 0.25 per 500cm³ = 0.5mol per 1000cm³ =0.5mol/dm³

Recall ,
Molar mass = (conc. in g/dm³)/(conc. in mol/dm³)

=> conc. in g/dm³=molar mass × conc. in mol/dm³ = 160×0.5=80g/dm³

=> 80g needed to make 1dm³
=> 80g needed to make 1000cm³

=> amount needed to make 500cm³ = ½(80g)=40g
(bii)
-Crude oil
-Natural gas
(biii) PICK ANY TWO
-high product selectivity
– high yields of gasoline
– high octane number
– high aromatic yield

(biv)All elements in a row have the same number of electron shells

(bv)The family name of group 2 elements is alkaline earth metals

(ci)S+2H2SO4=> 3SO2+2H2O
(cii)2H2S+SO2 ⟶2H20+3S
(ciii) oxidizing agent is S02
Reducing agent is H2S
(civ)
i) 10 electrons
ii) 18 electrons