NECO 2022 CHEMISTRY ESSAY AND OBJ ANSWERS

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Chemistry-Essay-Answers
(1ai)
(i) KAl(SO₄)₂. 12H₂O
(ii) Na₂SO₄

(1aii)
Number of moles of sodium reascted = reacting mass/molar mass
= 3.6/23
= 0.1565moles
Number of moles of oxygen reacted = (1/4) x Number of sodium
= 1/4 x (3.6/23)
= 0.03913
Reacting mass of oxygen = Number of moles x molar mass
= 0.03913 x (16×2)
= 1.252grams

(1bi)
Air in gaseous form is first passed through caustic soda to remove CO₂. It is compressed and cooled until ut becomes a liquid at -200°C. It is then led to a fractionating column. On distillation, Nitrogen which has a lowet boiling point of -196°C is evolced first leaving behind Oxygen in liquid form. Further heating converts the liquid to a gas at -183°C.

(1bii)
(i) It is used for combustion
(ii) It is used for artificial respiration

(1biii)
This is because room temperature is warmer

(1biv)
(i) Methyl orange
(ii) Phenophthalein

(1ci)
An ion is an atom or molecule which has a net electric charge due to the loss or gain of one or more electrons.

(1cii)
HCl gas doesn’t contain ions

(1cii)
(i) Noble gases: Neon, Helium
(ii) Halogens: Iodine, Fluorine
(iii) Alkaline earth metals: Calcium, Magnesium

(1civ)
Halogens attain stable octet configuration by accepting an electron from donors Group I elements.

(1di)
A sol is a colloid where solid particles are dispersed in liquid medium.

(1dii)
Jelly

(1diii)
(i) Acid-base titrations are used to determine the percentage purity of a substance
(ii) Acid-base titrations are important to determine the number of water molecules in a hydrate.
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(2ai)
A molar solution of a compound contains one mole or the molar mass of the compound in one dm³ of the solution.

(2aii)
(i) Identification of gases
(ii) Identification of acidic and metallic radicals

(2aiii)
(i) Purification of bauxite
(ii) Electrolysis of alumina

(2bi)

(2bii)
This is because of the presence of copper which is a transition metal

(2bii)
(i) Methanamide is a liquid while the rest amides are crystalline solids
(ii) Melting points and boiling points of aides are much higher than expected.

(2ci)
(i) Heavy chemicals are produced in very large quantities in industries While fine chemicals are produced in small quantities for specific purposes
(ii) Heavy chemicals are in a crude state While fine chemicals are purified.

(2cii)
(i) Million’s reagent
(ii) Feeling’s solution

(2ciii)
Graham’s law of diffusion states that at a constant temperature and pressure the rate of diffusion of a gas is inversely proportional to the square root of its density.

(2civ)
RO₂/RCH₄ = √(MCH₄/MO₂)
5/(200/t) = √[(12+4)/(16×2)]
5t/200 = √16/32
t/40 = 1/√2
t = 40/√2
t = (40/√2) × (√2/√2)
t = 20√2 seconds
Time taken = 28.28 seconds.


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(3)


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(4ai)
A hydrocarbon is an organic compound composed only of two elements carbon and hydrogen.

(4aii)
(i) Alkenes
(ii) Alkynes

(4aiii)
They are unsaturated compounds

(4aiv)
(i) Decrease in pressure
(ii) Finely divided zinc
(iii) Removal of hydrogen gas

(4bi)
(i) Electrovalent compounds conduct electricity While covalent compounds do not.
(ii) Electrovalent compounds dissolve in polar solvents While covalent compounds dissolve in non-polar solvents.

(4bii)
(i) High melting and boiling points
(ii) Exist mainly as solids at room temperature

(4biii)
(i) Ammonia, NH₃
(ii) Hydrogen chloride gas, HCl

(4ci)
(i) Potassium hexacyanoferrate(II)
(ii) It is a complex salt

(4cii)
(i) Wrought iron
(ii) Cast iron

(4ciii)
2H₂ + O₂ →2H₂O
2 : 1 : 2
Reacting volumes;
40cm³ : 20cm³ : 40cm³
Residual volumes;
(50 – 40) (20 – 20)
= 10cm³ = 0cm³ = 40cm³
Total residual volume = 10 + 40
= 50cm³
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5ai)
(i)Chlorofluorocarbons
(ii) Hydrofluorocarbon

(5aii)
I – C₂H₂OH + H₂SO₄ —>C₂H₅HSO₄ + H₂O
II – C₂H₅HSO₄ —> C₂H₄ +H₂SO₄

(5aiii)
(i) They incompletely filled d-orbitals
(ii) They exhibit variable oxidation state

(5bi)
(i) It is used in making aeroplane parts due to its tensile strength
(ii) It is used in making window frames because it cannot be easily oxidised

(5bii)
A white precipitate is formed with evolution of brown gas

(5biii)
(i) Hydrogen bond
(ii) Covalent bond

(5ci)
(i) Oxidation
(ii) Electrolysis

(5cii)
(i)Temperature
(ii)Concentration

(5ciii)
H₂SO₄ + 2NaOH —> Na₂SO₄ +2H₂O
CaVa/CbVb = Na/NB
Va = CbVbna/Can
Va = 0.1×25×1/0.5×2
Va = 2.5cm³
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(6ai)
Rate of a chemical reaction is defined as the number of moles of reactant converted or product formed per unit time.

(6aii)
Mass of CaCO3 = 3g
Molar mass of CaCO3= 100g/mol
No. of moles = 3/100 = 0.03
Time = 2×60 = 120s
Rate of reaction = No. of moles/Time
= 0.03/120 = 2.5 × 10-⁴

(6aiii)
(i) They are more resistant to corrosion
(ii) They are more durable
(iii) They have lower melting point

(6aiv)
2Pb(NO₃)₂ —> 2PbO + 4NO₂ +O₂

(6bi)
I =Hard water=
(i)It has better taste than soft water
(ii) It can be supplied in lead pipes as it does not dissolve lead

II =Soft water=
(i) It does not deposit scale in boilers and fur in kettles
(ii) It does not waste soap

(6bii)
I – CO
II – SO₂

(6biii)
I – Hardness
II – Lustre appearance
III – Conductivity
IV – Ability to absorb

(6c)
4.3/2.12 = 142+18x/142
2.12(142+18x)/2.12 = 4.3*142/2.12
142+18x = 288
18x = 288-142
x = 146/18
x = 8
(5)

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(6a)

(6b)

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