Chemistry 2007 Jamb Examination

Chemistry 2007 Jamb Examination
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Question 41

The IUPAC nomenclature of compound CH3CH2COOCH2CH3 is

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The given compound is an ester, RCOOR' type. Identify the alkyl group attached to oxygen and the acyl group:

CH3CH2COOCH2CH3 → acyl = CH3CH2CO- (propanoate), alkyl = CH2CH3 (ethyl)

Thus the IUPAC name is ethyl propanoate.

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Question 42

When temporary hard water is boiled for sometime in a kettle, the inner surface of the kettle becomes coated with a deposit of

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Temporary hard water contains soluble calcium hydrogencarbonate, Ca(HCO3)2.

Upon boiling: Ca(HCO3)2 → CaCO3(s) + CO2(g) + H2O(l)

The insoluble CaCO3 (calcium carbonate) precipitates and forms a scale on the kettle surface. In IUPAC naming, it is calcium hydrogentrioxocarbonate (IV).

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Question 43

A supersaturated solution is said to contain
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A supersaturated solution contains more solute than the equilibrium solubility at that temperature.

Unlike saturated solutions, supersaturated solutions are unstable, and addition of a seed crystal can trigger precipitation.

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Question 44

The presence of nitrogen in air is to slow down
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Nitrogen is chemically inert and dilutes oxygen in air (~21%).

This reduces the rate of oxidation reactions, thus slowing down combustion and corrosion processes. It does not prevent respiration but moderates the availability of O2 for combustion.

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Question 45

A good drying agent should be
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A drying agent absorbs moisture from gases or liquids.

Hygroscopic substances readily absorb water from the surroundings, making them effective drying agents. Examples include CaCl2 and P2O5.

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Question 46

Smog usually occurs over cities which
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Smog forms when smoke and fog combine, especially in cities with high combustion of fossil fuels. These pollutants accumulate in the air, causing haze and respiratory hazards.

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Question 47

What weight of NaCl is needed to make 2.0 dm³ of a 1.5M solution? [Na = 23, Cl = 35.5]

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Calculate moles of NaCl: n = C × V = 1.5 × 2.0 = 3.0 mol.

Calculate mass: m = n × M = 3.0 × (23 + 35.5) = 3 × 58.5 = 175.5 g.

Check: The closest correct option considering rounding is 175 g.

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Question 48

A. CH3 — CH2 — CH — CH2 — OH
                               |
                              CH3

B. CH3 — CH — CH2 — CH2 — OH
                   |
                 CH3

                CH3
                 |
C. CH  — C — CH2 — CH3
                 |
               OH

                 H
                  |
D. CH2 — C — CH3
                  |
                OH

Which of the compounds above is a tertiary alkanol?

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In order to identify the different types of alcohol

  • Identify the Carbon with the OH Group: Find the carbon atom that has the hydroxyl group (OH) attached to it.
  • Count the Carbon Bonds: Count how many carbon atoms are directly bonded to that carbon atom.
  • Classify the Alcohol:
    • If the carbon with the OH group is bonded to one other carbon, it's a primary alcohol.
    • If the carbon with the OH group is bonded to two other carbons, it's a secondary alcohol.
    • If the carbon with the OH group is bonded to three other carbons, it's a tertiary alcohol.
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Question 49

Use the diagram above to answer the question: The fuel gas labelled L is

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The diagram shows the producer gas setup, where fuel gas L is a mixture of CO and H₂, commonly referred to as water gas. 

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Question 50

The alternative raw material labelled M is

Use the diagram above to answer the question:
The alternative raw material labelled M is

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The raw material M in the producer gas diagram is solid carbon (coke), which reacts with steam and air to produce the fuel gas. Option A is correct.

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Chemistry 2007 Jamb Examination
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