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The following question based on general concepts of chemistry topic of general science mcq

Questions : An organic compound containing C, H and O gave on analysis C – 40% and H – 6.66%. Its empirical formula would be

(a) CH2O

(b) C3H6O

(c) CHO

(d) CH4O

The correct answers to the above question in:

Answer: (a)

Element%At.Wt.Rel.NumberRatio
C4012$40/12$= 3.33${3.33}/{3.33}=1$
H6.661${6.66}/1$= 6.66${6.66}/{3.33}=2$
O53.3416${53.34}/16$= 3.33${3.33}/{3.33}=1$
(% of O in organic compound
= 100 – (40 + 6.66) = 53.34% )
Empirical formula of organic compound = $CH_2$O.

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Question : 1

Which one is a decomposition reaction ?

a) 2H2O $⟶ ↖ {Electrolysis}$ 2H2 + O2

b) 2HgO $⟶ ↖ {heat}$ 2Hg + O2

c) CaCO3 $⟶ ↖ {heat}$ CaO + CO2

d) All the above reactions are decomposition reaction

Answer: (d)

In all these reactions the reactants decompose to form simpler products.

Question : 2

$Na_2S_2O_3$ represent the compound

a) sodium thiosulphate

b) sodium sulphate

c) sodium sulphite

d) None of these

Answer: (a)

Question : 3

The chromate and dicharomate ions are respectively

a) CrO-4 and CrO-5

b) CrO2-4 and Cr2O2-7

c) Cr2O2-7and CrO4-

d) CrO2-4 and Cr2O2-5

Answer: (b)

Question : 4

Valency of an atom or radicals is

a) its combining capacity

b) ionisation energy

c) electron affinity of atom

d) size of atom

Answer: (a)

Question : 5

Which of the following reaction is endothermic?

a) CH4 + 2O2 ⟶ CO2 + 2H2O

b) C + O2 ⟶ CO2

c) CaCO3 ⟶ CaO + CO2

d) CaO + H2 O ⟶ Ca(OH)2

Answer: (c)

Heat is required to decompose calcium carbonate. Thus this reaction proceeds with absorption of heat therefore it is endothermic reaction.

Question : 6

Which one is a combination reaction ?

a) Ca(OH)2 + Na2CO3 ⟶ 2NaOH + CaCO3

b) Formation of a mixture of carbon monoxide and hydrogen. When steam is passed over red hot iron.

c) Reaction of water with sodium metal to form sodium hydroxide and hydrogen.

d) Preparation of stannic chloride (Tin (iv) chloride) by passing chlorine gas into molten tin (Sn).

Answer: (d)

Sn + 2Cl2 SnCl4.

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