2024 - JAMB Chemistry Past Questions and Answers - page 4

The terms "strong" and "weak" acids refer to the extent to which an acid ionizes in a solution:

• Strong acids: Ionize almost completely in water, dissociating nearly 100% into H⁺ and the corresponding anion.
• Weak acids: Ionize only partially in water, typically less than 10%.

To solve this problem:

1. Understand half-life: The half-life is the time it takes for half the substance to decay.
2. Calculate number of half-lives: 200 days ÷ 50 days/half-life = 4 half-lives.
3. Calculate remaining amount: After each half-life, the amount halves:
   After 1 half-life: 5 ÷ 2 = 2.5 g
   After 2 half-lives: 2.5 ÷ 2 = 1.25 g
   After 3 half-lives: 1.25 ÷ 2 = 0.625 g
   After 4 half-lives: 0.625 ÷ 2 = 0.3125 g ≈ 0.31 g

Butene (C₄H₈) has different forms. Considering 2-butene (CH₃-CH=CH-CH₃):

• 2-butene exhibits geometrical (cis-trans) isomerism because the double-bonded carbons have two different groups attached.
• Two isomers exist: cis-2-butene and trans-2-butene.

Therefore, there are 2 geometrical isomers.

Group VIII elements, or noble gases, are inert because they have completely filled valence shells. This makes them stable and unreactive, with no tendency to gain, lose, or share electrons.

A solubility curve graphically shows the relationship between solubility (amount that can dissolve in a solvent) and temperature.

The liquid hydrocarbon from coal tar used in the pharmaceutical industry is:

• Benzene: Used as a solvent and in drug synthesis.
• Xylene and Toluene: Obtained from coal tar but primarily for other industrial uses.
• Hexane: Not from coal tar distillation.

The Biuret test detects proteins. Copper(II) ions react with peptide bonds in an alkaline solution, producing a color change from blue to violet if proteins are present.

Fractional distillation separates components by boiling point. The fraction distilling at 200-250°C is diesel.

Nitrogen from air contains other components:

• Oxygen: Present in traces.
• Carbon (IV) oxide: Present in air.
• Inert gases: Present (exception option depends on the question context).
• Water vapor: Present in air.

The exception is the component not present; based on the options, the inert gases may be treated as the exception in context.

Determine the solution type:

1. Molar mass of KNO₃: 39.1 + 14 + 3×16 = 101.1 g/mol
2. Solubility in g/dm³: 3.10 × 101.1 = 313.41 g/dm³
3. Compare given concentration: 303 g/dm³ < 313.41 g/dm³
4. Since the concentration is close to maximum, the solution is likely supersaturated, with potential precipitation of excess solute.