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Chemistry
UTME / JAMB 2024
Acids, Bases, and Salts
Solution
A major effect of oil pollution in coastal water is
A
destruction of aquatic life
B
nitrification of the water body
C
detoxification of water
D
disinfection of the water body
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
Aquatic animals need oxygen to survive. Oil spills can have a severe impact on aquatic life, causing a drastic reduction in the number of aquatic animals. The oil spills tend to hinder dissolved oxygen transfer in the water.Even small amounts of hydrocarbons can prevent oxygen from transferring in the water column.The oil layer directly can also cause the dissolved oxygen level in the water to become low due to the obstruction of the diffusion process from the air.
Nitrification in water bodies is caused by nitrifying bacteria converting ammonia and other nitrogen compounds into nitrite and nitrate. This process is affected by several environmental factors, including temperature, oxygen, pH, and light.
Detoxification of water is the process of removing harmful components from wastewater. This process is necessary to treat water and wastewater that contain toxic compounds that can harm human health and the environment.
Water disinfection occurs when physical or chemical processes disrupt the cell walls or membranes of microorganisms. This prevents the microorganisms from multiplying and causes them to die.
Source: Myschool Past Questions Archive (Record #70383)
Chemistry
UTME / JAMB 2024
Acids, Bases, and Salts
Solution
The substance that reacts with sodium to form alkali and changes white anhydrous copper(II) tetraoxosulphate (VI) to blue is
A
base
B
acid
C
water
D
solvent
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
When water reacts with alkali metals such as sodium, alkali are formed. The reason they are called alkali metals. Also, the colour change from white anhydrous copper(II) tetraoxosulphate(VI) to blue hydrated copper(II) tetraoxosulphate(VI) is a way of testing for the presence of water. This colour change is a simple and effective way to detect the presence of water in a substance or environment.
Source: Myschool Past Questions Archive (Record #70384)
Chemistry
UTME / JAMB 2024
Atomic Structure and Periodic Table
Solution
\(^{14}_7\)N + X → \(^{14}_6\)C + \(^1_1\)P
In the reaction above, X is
A
a proton
B
an alpha particle
C
a beta particle
D
a neutron
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
A nuclear reaction that occurs when an atom is bombarded with a neutron is called a neutron bombardment reaction. When a nucleus is hit by a neutron, it can split into two or more smaller nuclei, releasing energy in a process called nuclear fission.Neutron has a mass number of 1 and no atomic number.
Source: Myschool Past Questions Archive (Record #70385)
Chemistry
UTME / JAMB 2024
Matter & Separation Techniques
Solution
An example of a substance that does not change directly from solid to gas when heated is
A
NH\(_4\)Cl\((_s\))
B
I\(_2\)\((_s\))
C
CaCO\(_3\) \((_s\))
D
S\(_8\) \((_s\))
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
NH\(_4\)Cl, I\(_2\), S\(_8\) all undergo sublimation. i.e changes from solid to gaseous substances without first changing to liquid, when heated but CaCO\(_3\) does not.
Source: Myschool Past Questions Archive (Record #70386)
Chemistry
UTME / JAMB 2024
Atomic Structure and Periodic Table
Solution
When a specie undergoes oxidation, its
A
oxidation number increases
B
oxidation number decreases
C
oxidation number remains constant
D
acts as oxidizing agent
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
Oxidation is a chemical reaction that occurs when the oxidation number of an atom increases. This means that the atom loses electrons and has a greater positive charge or a lesser negative charge. The oxidation number of an atom is a number that represents the degree of oxidation of the atom in a chemical compound.
Source: Myschool Past Questions Archive (Record #70387)
Chemistry
UTME / JAMB 2024
Acids, Bases, and Salts
Solution
Hydrochloric acid is regarded as a strong acid because it
A
ionizes partially
B
ionizes completely
C
is a mineral acid
D
reacts with a base
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
An acid is considered strong if it dissociates completely into ions when mixed with water. When hydrochloric acid (HCl) dissolves in water, it dissociates into H\(^+\) and Cl\(^-\) . This is different from weak acids, like acetic acid, which only partially dissociate in water.
Source: Myschool Past Questions Archive (Record #70388)
Chemistry
UTME / JAMB 2024
Stoichiometry & Mole Concept
Solution
The IUPAC nomenclature of the complex K\(_4\)Fe(CN)\(_6\) is
A
potassium hexacyanoferrate(II)
B
potassium hexacyanoferrate(III)
C
potassium hexacyanoferrate(IV)
D
potassium hexacyanoferrate(VI)
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
The naming starts with the first metal which is potassium, followed by the hexacyano (6 moles of Cyanide) and then ferrate (II).
Source: Myschool Past Questions Archive (Record #70389)
Chemistry
UTME / JAMB 2024
Acids, Bases, and Salts
Solution
An example of an amphoteric oxide is
A
P\(_4\)O\(_{10}\)
B
Al\(_2\)O\(_3\)
C
CuO
D
K\(_2\)O
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Amphoteric oxides are metal oxides that can behave as both acidic and basic oxides. They react with acids and bases to produce water and salts. When dissolved in water, amphoteric oxides form alkaline solutions that contain hydroxide ions. E.g ZnO, Al\(_2\)O\(_3\),PbO etc
The chemical reactions
Al\(_2\)O\(_3\) + 6HCl → 2AlCl\(_3\) + 3H\(_2\)O and Al\(_2\)O\(_3\) + 2NaOH → 2NaAlO\(_2\) + H\(_2\)O show that Al\(_2\)O\(_3\) is amphoteric.
Source: Myschool Past Questions Archive (Record #70390)
Chemistry
UTME / JAMB 2024
Stoichiometry & Mole Concept
Solution
The term that is not associated with petroleum industry is ?
A
Cracking
B
Saponification
C
Polymerization
D
Fermentation
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
Cracking, saponification and polymerization are all terminologies associated with the petroleum industry but fermentation is associated with the brewery industry.
Cracking is a chemical process that breaks down heavy hydrocarbon molecules into lighter, more useful ones.
Saponification is a chemical reaction that converts fats and oils into soap. It's not directly involved in petroleum, but it can be used to analyze petroleum products.
Polymerization is a process in the petroleum industry that converts light olefin gases into higher molecular weight hydrocarbons.
Fermentation is the process in which a substance breaks down into a simpler substance. Microorganisms like yeast and bacteria usually play a role in the fermentation process, creating beer, wine, bread,yogurt and other foods.
Source: Myschool Past Questions Archive (Record #70391)
Chemistry
UTME / JAMB 2024
Matter & Separation Techniques
Solution
Kerosene is used as solvent for
A
Paints
B
Sulphur
C
Gums
D
Fats
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
Kerosene can thin synthetic and synthetic oil paints to make them spread more easily and shine more.
Source: Myschool Past Questions Archive (Record #70392)
Chemistry
UTME / JAMB 2024
Kinetics, Energetics and Equilibrium
Solution
A typical chemical reaction will be spontaneous if
A
enthalpy change is negative
B
free energy change is greater than one
C
entropy change is zero
D
enthalpy and free energy change are equal
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
A reaction which is exothermic (\DeltaH negative) and results in an increase in the entropy of the system (\DeltaS positive) will always be spontaneous. Also, if \DeltaG is negative, the reaction will be spontaneous because the reaction can take place on it's own without any external source.
Source: Myschool Past Questions Archive (Record #70393)
Chemistry
UTME / JAMB 2024
Stoichiometry & Mole Concept
Solution
The indicator used in a titration between strong acid and weak base is
A
methyl orange
B
phenolphthalein
C
bromothymol blue
D
methyl red
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
The indicator used in a titration between strong acid and weak base is methylorange while phenolphthalein is used in a titration between strong base and weak acid.
Source: Myschool Past Questions Archive (Record #70394)
Chemistry
UTME / JAMB 2024
Acids, Bases, and Salts
Solution
Nitrogen obtained from air is not absolutely pure because it contains the following except
A
oxygen
B
carbon (IV)oxide
C
inert gases
D
water vapour
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Nitrogen obtained from air is generally considered "impure" compared to chemically synthesized nitrogen because it contains trace amounts of atmospheric constituents that are difficult to separate entirely.
Carbon (IV) oxide (CO₂) is the correct answer because it is typically removed early in the industrial process of separating nitrogen from air (such as passing air through caustic soda/sodium hydroxide). Therefore, the final nitrogen obtained is considered to no longer contain CO₂.
Source: Myschool Past Questions Archive (Record #70395)
Chemistry
UTME / JAMB 2024
Acids, Bases, and Salts
Solution
Solubility curve is a plot of solubility against
A
temperature
B
pressure
C
volume
D
concentrate
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
A solubility curve is a graph that plots the relationship between the solubility of a substance and temperature:
Temperature: Plotted on the X-axis
Solubility: Plotted on the Y-axis, measured in grams of solute per 100 grams of water
Source: Myschool Past Questions Archive (Record #70396)
Chemistry
UTME / JAMB 2024
Stoichiometry & Mole Concept
Solution
If 11.0g of a gas occupies 5.6 dm\(^3\) at s.t.p., calculate its vapour density (1 mole of a gas occupies 22.4 dm\(^3\)).
A
22
B
44
C
66
D
88
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
If 11.0g occupies 5.6 dm\(^3\)
x will occupy 22.4 dm\(^3\)
x = \(\frac{11\times 22.4}{5.6}\)
x = 44g
But Molecular mass = 2 x Vapour density
44 = 2 x Vapour density
Vapour density = \(\frac{44}{2}\) = 22
Source: Myschool Past Questions Archive (Record #70397)