JAMB Chemistry Past Questions & Answers - Page 57

281.

Which of the compounds is composed of Al, Si, O and H?

A.

Epson salt

B.

Limestone

C.

Clay

D.

Urea

Correct answer is C

Between the five main minerals found in clay, kaolinite is the most common. Kaolinite holds the chemical composition Al\(_{2}\) Si\(_{2}\)O\(_{5}\)(OH)\(_{4}\) and is an aluminum silicate material with a low “shrink-swell” capacity. It is a soft, white mineral but is often colored orange or red by iron oxide found in the soil.

282.

H\(_2\) S\(_{(g)}\) + Cl\(_{2(g)}\) → 2HCl\(_{(g)}\) + S\(_{(g)}\) In the reaction above, the substance that is reduced is

A.

H\(_2\)S

B.

S

C.

HCl

D.

Cl\(_2\)

Correct answer is D

H \(_2\) S + Cl \(_2\) → 2HCl + S

  Please note: substance oxidized or reduced are always the reactants. So, one can easily cross out the two products.

  Reduction is

  1. Addition of hydrogen

  2. Removal of Oxygen

  3. Addition of Electron

  Oxidation is

  1. Removal of hydrogen

  2. Addition of hydrogen

  3. Removal of Electron

  From the question we can see that H was added to Cl in the product, thereby making hydrogen to be removed from S in the reactant hence S is oxidized as a result of the removal of hydrogen and Cl \(_2\) has been reduced due to addition of hydrogen.

Cl \(_2\) is reduced

283.

The elements in the periodic table are listed in order of increasing

A.

relative atomic number

B.

atomic mass

C.

relative isotopic mass

D.

nuclear charge

Correct answer is A

The periodic number table is arranged in order of atoms increasing atomic number

284.

The alkanoic acid found in human sweat is

A.

CH\(_3\)-COOH

B.

CH\(_3\)—CH\(_2\)—COOH

C.

H-COOH

D.

CH\(_3\)

Correct answer is B

Bacteria of the genus Propionibacterium produce propionic acid (Propanoic Acid) as the end-product of their anaerobic metabolism. This class of bacteria is commonly found in the stomach of ruminants and the sweat glands of humans, and their activity is partially responsible for the odour of both Swiss cheese and sweat.

285.

The situation obtained when a perfect gas expands into a vacuum is

A.

ΔH is positive and TΔ S is zero

B.

ΔH is positive and T Δ S is negative

C.

ΔH is negative and TΔ S is zero

D.

ΔH is zero and T Δ S is positive

Correct answer is D

The definition of enthalpy change is

  ΔH = ΔU + Pex(ΔV)

  The enthalpy change is zero because both terms on the right are zero in free expansion of an ideal gas.

  There is no external pressure in a free expansion, so P (ex) = 0. And the internal energy change, ΔU of free expansion in a closed system is zero since no inter molecular forces have to be overcome. The temperature remains constant and no heat is absorbed or released.