JAMB Past Questions and Answers - Page 1166

5,826.

What type of reaction is involved in the formation of alkanols from alkenes?

A.

Elimination reaction

B.

Redox reaction

C.

Substitution reaction

D.

Addition reaction

Correct answer is D

Alkanols can be formed from alkenes through an addition reaction. In this reaction, the carbon-carbon double bond of the alkene is broken, and a hydroxyl group (-OH) is added to one of the carbon atoms, resulting in the formation of an alcohol.

5,827.

What is the common name for ethanoic acid?

A.

Acetic acid

B.

Butyric acid

C.

Propionic acid

D.

Formic acid

Correct answer is A

Ethanoic acid is commonly known as acetic acid. It is the main component of vinegar and is used in various industrial processes.

5,828.

Which separation technique is used to separate different pigments in a mixture based on their affinity for a stationary phase and a mobile phase?

A.

Chromatography

B.

Filtration

C.

Decantation

D.

Distillation

Correct answer is A

Chromatography is a separation technique used to separate and analyze components of a mixture based on their differential partitioning between a stationary phase (e.g., paper or a column) and a mobile phase (e.g., solvent). Different components move at different rates, leading to separation.

5,829.

What is the empirical formula of a compound containing 40.00% carbon, 6.67% hydrogen, and 53.33% oxygen by mass?

A.

C\(_4\)H\(_8\)O\(_4\)

B.

CH\(_2\)O

C.

C\(_2\)H\(_4\)O\(_2\)

D.

C\(_3\)H\(_6\)O\(_3\)

Correct answer is B

To determine the empirical formula, we first need to convert the percentages to moles by assuming a convenient mass for the sample. Let's assume we have 100 grams of the compound.

Mass of carbon = 40.00 grams
Mass of hydrogen = 6.67 grams
Mass of oxygen = 53.33 grams

Now, we calculate the moles of each element:
Moles of carbon = 40.00 g / molar mass of carbon = 40.00 g / 12.01 g/mol ≈ 3.33 mol
Moles of hydrogen = 6.67 g / molar mass of hydrogen = 6.67 g / 1.01 g/mol ≈ 6.60 mol
Moles of oxygen = 53.33 g / molar mass of oxygen = 53.33 g / 16.00 g/mol ≈ 3.33 mol

Now, we find the smallest whole-number ratio of the moles:
C ≈ 3.33 / 3.33 ≈ 1
H ≈ 6.60 / 3.33 ≈ 2
O ≈ 3.33 / 3.33 ≈ 1

Thus, the empirical formula is CH2O.

5,830.

What is the main source of carbon monoxide (CO) in urban areas?

A.

Volcanic eruptions

B.

Vehicle emissions

C.

Forest fires

D.

Industrial processes

Correct answer is B

The main source of carbon monoxide (CO) in urban areas is vehicle emissions, especially from incomplete combustion of gasoline and diesel in engines.