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.

2,996.

A reaction takes places spontaneously if?

A.

∆G = O

B.

∆S < O and ∆H > O

C.

∆H < T ∆S

D.

∆G > O

Correct answer is C

No explanation has been provided for this answer.

2,997.

The oxidation states of chlorine in HOCL, HCLO3 and HCLO4 are respectively?

A.

-1, +5, and +7

B.

-1, -5 and 7

C.

+1, +3 and +4

D.

+1, +5 and +7

Correct answer is D

No explanation has been provided for this answer.

2,998.

Which of the reactions occurs at the anode during the electrolysis of a very dilute aqueous solution of sodium chloride?

A.

OH- - e → OH

B.

CIsup>- - e → CI

C.

OHsup>- + CIsup>-HCI + O2

D.

Na+ - esup>- Hg (Na)/(Hg) amalgam

Correct answer is A

No explanation has been provided for this answer.

2,999.

During the electrolysis of a salt of a metal M, a current of 0.5 A flows from for 32 minutes 10 seconds and deposits 0.325 g of M, What is the charge of the metal ion?

(M = 65, If = 96,500 C per mole of electron)

A.

1

B.

2

C.

3

D.

4

Correct answer is B

Ampere = (Coulomb) x (Sec) x 1930 secs = 965 coulombs (unit of current)
But quantity of electricity = current x time = (Coulomb)/(Sec) x time
If 0.5 A for 32 mins 10 secs; Quantity of electricity = (0.5 coulomb)/(Sec x 1930) secs
= 965 coulombs
Since 965 coulombs give 0.325 g M
i.e1 coulombs = (0.325)/(965) g
∴96500 coulombs = (0.30)/(965) x 96500 =32.5 g
∴(65 g)/(32.5) = 2

inference: it takes 2 Faraday of electricity to deposit 1 mole of M. Therefore, the change on M ion = 2.

2,000.

4.0 g of sodium hydroxide in 250 cm3 of solution contains?

A.

0.40 moles per dm3

B.

0.10 moles per dm3

C.

0.04 moles per dm3

D.

0.02 moles per dm3

Correct answer is A

Unit of concentration = (moles)/(dm3)
If 250 cm3 solution NaOH contains 4.0 g NaOH
i.e -1cm3 solution NaOH (4.0)/(250) g
∴ 1000 cm3 (1 dm3) will contain (40 x 1000) / (250 x 1) = 16 g/dm3
16 g/dm3 = 16g/dm3 + molar mass NaOH
because no of moles = (mass of substance given) / (molar mass)
∴ (16 g) / (40 g) dm3 = (16 g)/(dm3 x (1 mole)/ (40g) = 0.4 (moles)/(dm3)