2018 - JAMB Chemistry Past Questions and Answers - page 7

61

The reaction between an organic acid and an alcohol in the presence of an acid catalyst is known as;

A
saponification
B
dehydration
C
esterification
D
hydrolysis
correct option: c

Esterification is a chemical reaction that forms at least one ester (= a type of compound produced by reaction between acids and alcohols). Esters are produced when acids are heated with alcohols in a process called esterification. An ester can be made by an esterification reaction of a carboxylic acid and an alcohol.

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62

The constituent common to duralumin and alnico is

A
Co
B
Mn
C
Al
D
Mg
correct option: c

Constituents of duralumin: Al, Cu, Mg, Mn.

Constituents of Alnico: Al, Ni and Co

The alloy of duralumin was discovered by Alfred Wilon consisting of 94% Al, 4% Cu, 1% Mg and 1% Mn

Alnico is composed primarily of Al, Ni and Co.

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63

The solubility of the solids that dissolves in a given solvent with the liberation of heat will

A
increase with an increase in temperature
B
decrease with an increase in temperature
C
decrease with a decrease in temperature
D
not be affected by changes in temperature
correct option: a

For many solids dissolved in liquid water, the solubility increases with temperature. The increase in kinetic energy that comes with higher temperature allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions.

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64

What mass of Cu would be produced by the cathodic reduction of Cu\(^{2+}\)  when 1.60A of current passes through a solution of CuSO\(_4\) for 1 hour. (F=96500Cmol\(^{-1}\) , Cu=64)

A
7.64g
B
3.82g
C
1.91g
D
0.96g
correct option: c

M = \(\frac{MmIT}{96500n}\)

  Where

  M=mass

  Mm = Molar mass = 64g/mol

  I = current = 1.6A

  T= Time  =1h r=3600s

  N= No of Charge = +2

  M = \(\frac{64x \times 1.6 \times 3600}{96500 \times 2}\)

  M=1.91g

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65

Calculate the pH of 0.05 moldm\(^{-3}\) H\(_2\)SO\(_4\)

A
1.30
B
1.12
C
1.00
D
0.30
correct option: c

pH = -Log [H+]

  H2SO4 → 2H+ + SO42-

  2H+ → 2 X0.05 = 0.1

  pH= -Log [0.1] = 1

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66

The choice of method for extracting a metal from its ores depends on the

A
strength of the core
B
position of the metal in the electrochemical series
C
source of the core
D
position of the metal in the periodic table
correct option: b

The method used to extract a metal from its ore depends upon the stability of its compound in the ore, which in turn depends upon the reactivity of the metal:

  Oxides of very reactive metals, such as aluminum, form stable oxides and other compounds. A lot of energy is needed to reduce them to extract the metal.

  Oxides of lesser reactive metals, such as iron, form less stable oxides and other compounds. Relatively little energy is needed to reduce them to extract the metal.

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67

At what temperature is the solubility of potassium trioxonitrate(V ) equal to that of sodium trioxonitrate (V)?

A
0\(^o\)c
B
30\(^o\)c
C
40\(^o\)c
D
45\(^o\)c
correct option: d

Potassium Trioxonitrate (V), KNO\(_3\), and Sodium Trioxonitrate (V), NaNO\(_3\)... the point of intersection of this two compound on the solubility graph is traced downward to the temperature axis and the result gives 45 \(^o\)C

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68

Which of the following produces relatively few ions in solution?

A
NaOH
B
Ca(OH)\(_2\)
C
KOH
D
AI(OH)\(_3\)
correct option: d

In other words, Aluminium Hydroxide is not soluble in water because it is polymeric and it also have high lattice energy and is therefore it is hard to break bonds in this compound. So, water cannot dissolve such types of compounds.

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69

The figure above shows the electrolysis of molten sodium chloride. Z is the

A
anode where the Cl\(^-\) ions are oxidized
B
cathode where the Cl\(^-\) ions are reduced
C
anode where the Na\(^+\) ions are reduced
D
cathode where the Na\(^+\) ions are oxidized
correct option: a

During the electrolysis of brine, the chlorine ions are discharged at the anode and are oxidized. While at the cathode Hydrogen ion is discharged.

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70

The knowledge of half-life can be used to

A
Create an element
B
Detect an element
C
Split an element
D
Irradiate an element
correct option: c

Half-life is the time taken for the radioactivity of a specified isotope to fall to half its original value."iodine-131 has a half-life of 8.1 days"

It can also be defined as the time required for any specified property (e.g. the concentration of a substance in the body) to decrease by half.

A useful application of half-lives is radioactive dating. ... It takes a certain amount of time for half the atoms in a sample to decay. It then takes the same amount of time for half the remaining radioactive atoms to decay, and the same amount of time for half of those remaining radioactive atoms to decay, and so on.

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