JAMB Chemistry Past Questions & Answers - Page 349

1,741.

The atom of an element X is represented as \(_z^yX\). The basic chemical properties of X depends on the value of

A.

Y

B.

Z

C.

Y - Z

D.

Z - Y

Correct answer is B

The chemical properties of elements depend on their atomic numbers. The atomic number is represented by Z.

1,742.

The experiment that showed that atoms have tiny positively charged nucleus was first carried out by

A.

Moseley

B.

Rutherford

C.

Millikan

D.

Dalton

Correct answer is B

The experiment that showed that atoms have tiny positively charged nuclei was first carried out by Rutherford. Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively charged nucleus.

1,743.

Diffusion is slowest in solid particles than in particles of liquid and gases because

A.

solid particles have more kinetic energy than the particles of liquid and gases

B.

solid particles have less kinetic energy than the particles of liqiud and gases

C.

solid particles have less restriction in their movement than liquid and gas particles

D.

the particles in solids are far apart and the cohesive forces between them are negligible

Correct answer is B

Diffusion is faster in gases owing to their higher kinetic energy i.e increase in kinetic energy of molecules leads to increase in rate of diffusion.

1,744.

300 cm3 of gas has a pressure of 800 mmHg. If the pressure is reduced to 650 mmHg, find its volume

A.

243.75 cm3

B.

369.23 cm3

C.

738.46 cm3

D.

1733.36 cm3

Correct answer is B

P1V1= P2V2
800 * 300 = 650 * V2
V2 = (800 * 300)/650 = 369.23 cm3

1,745.

16.8 g of sodium hydrogen trioxocarbonate (IV) is completely decomposed by heat. Calculate the volume of carbon(IV) oxide given off at s.t.p
[Na 23, C = 12, O = 16, H = 1, Molar volume of a gas at s.t.p = 22.4 dm3]

A.

22.40 dm3

B.

11.20 dm3

C.

2.24 dm3

D.

1.12 dm3

Correct answer is C

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2
Relative molecular mass of NaHCO3 = 23 +1 + 12 + 16 * 3 = 84gmol-1
2 * 84g of NaHCO3 will be liberated 22.4dm3 or CO2 at s.t.p
16.8g of NaHCO3 will liberate (16.8 * 22.4)/(2 * 84) = 2.24 dm3