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Chemistry questions and answers

Chemistry Questions and Answers

Learn more about the properties, composition, and structure of substances (elements and compounds) with these Chemistry questions and answers. This Test can be used by students preparing for Chemistry in JAMB, WAEC, NECO or Post UTME.

416.

Chromatography is used to separate components of mixture which differ in their rates of ?

A.

diffusion

B.

migration

C.

reaction

D.

sedimentation

Correct answer is B

Chromatography is a method of separating mixtures by using a moving solvent on filter paper. The rate at which each solute moves up the paper depends on how strongly it is absorbed by the paper, and how soluble it is in the solvent.

417.

A few drop of conc HCL are added to about 10cm3 of a solution of PH 3.4. The PH of the resulting mixture is?

A.

less than 3.4

B.

greater than 3.4

C.

unaltered

D.

the same as that of the pure water

Correct answer is A

The addition of conc HCL makes the solution more acidic and the solution falls below the original value

418.

Which of the following gases will rekindle a brightly splint?

A.

NO2

B.

NO

C.

N2O

D.

CL2

Correct answer is C

Nitrogen(I)Oxide rekindles a brightly glowing splinter but extinguishes a feebly glowing one which is not hot enough to decompose the gas to liberate oxygen

419.

The volume occupied by 1.58g of a gas at S.T.P is 500cm3. What is the relative molecular mass of the gas?  [ G.M.V at S.T.P = 22.4dm3 ]

A.

28

B.

32

C.

44

D.

71

Correct answer is D

500cm3 of the gas weighed 1.58

: 22400cm3 of the gas weight

 

[22400 X 1.58g] ÷ [ 500 ]

= 71

420.

The number of molecules of Carbon(iv)Oxide produced when 10.0g of CaCO3 is treated with 0.2dm3 of 1 Mole of HCL in the equation

CaCO3 + 2HCL ⇒  CaCl2  + H2 O + CO2  , is?

A.

1.00 X 1023

B.

6.02 X 1023

C.

6.02 X 1022

D.

6.02 X 1024

Correct answer is C

1 mole of CaCO3 = 100g or  6.02 X 1023

liberated 1 mole of CO2 or 44g or 6.02 X 1023

 

:100g =  6.02 X 1023

10g of CaCO3 = x

cross multiply

[10g X 6.02 X 1023] / 100g = x

⇒ x = 6.02 X 1022

 

Since 

 6.02 X 1023 of CaCO3 liberated 6.02 X 1023 of CO2

 

⇒ 6.02 X 1022 of CaCO3 will liberate 6.02 X 1022 of CO2