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JAMB Chemistry Past Questions & Answers - Page 138

686.

In the above experiment X is?

A.

pure nitrogen

B.

a mixture of nitrogen and oxygen

C.

a mixture of nitrogen and carbondioxide

D.

a mixture of oxygen and inert gases

E.

a mixture of nitrogen and inert gases

Correct answer is E

Alkaline solutions of pyrogallol absorb oxygen efficiently and are used in determining the oxygen content of gas mixtures.

 calcium chloride absorbs sufficient moisture/water from the air.

So,Nitrogen and Inert gases will be left 

687.

Which of the structural formal above is ethanoic acid is

A.

A

B.

B

C.

C

D.

D

E.

E

Correct answer is E

Ethanoic acid is CH3COOH

688.

The IUPAC name for

A.

2-methylbut-3-ene

B.

2-methylbut-4-ene

C.

3-methylbut-2-ene

D.

3-methylbut-1-ene

E.

3-methylpent-1-ene

Correct answer is D

No explanation has been provided for this answer.

689.

30 cm3 of 0.1 M Al (NO3)3 SOLUTION IS RECTED WITH 100cm3 of 0.15M of NaOH solution. Which reactant is in excess, and by now much ?

A.

NaOH solution by 70 cm3

B.

NaOH solution, by 60 cm3

C.

NaOH solution by 40 cm3

D.

Al(NO3)3 solution by 20 cm3

E.

Al(NO3)3 solution by 10 cm3

Correct answer is C

No explanation has been provided for this answer.

690.

Zn+H2SO4ZnSO4+H2

in the above reaction, how much zinc will be left undissolved if 2.00g of zinc is treated with 10cm^3 of 1.0 M of H2SO4?
[Zn =65, s = 32, O = 16, H = 1]

A.

1.35g

B.

1.00g

C.

0.70g

D.

0.65g

E.

0.00g

Correct answer is A

Zn+H2SO4ZnSO4+H2

1 mole of Zn = 1 mole H2SO4

no of moles of H2SO4 = C X V = 1 X 10/1000 = 0.01moles.

0.01mole H2SO4 = 0.01 mole Zn

mass of Zn = 0.01 x 65(given) = 0.65g

therefore, the mass of undissolved Zn = 2.00(given) - 0.65 = 1.35g.