Chemistry

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New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

Due to inert pair effect, elements of group 14 exhibit oxidation states of +2 and +4. Thus, option (b) is correct.

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

Thermodynamically the most stable form of carbon is graphite, i.e., option (b) is correct.

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

In B2H6, B is sp3-hybridized. Therefore, option (c) is correct.

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Boric acid is polymeric due to the presence of H-bonds. Therefore, option (b) is correct.

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

Borax is a salt of a strong base (NaOH) and a weak acid (H3BO3), therefore, it is basic in nature, i.e., option (c) is correct.

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

Laboratory preparation of carbon monoxide:

Formic acid is dehydrated with concentrated sulphuric acid at 373 K.

HCOOH → H2O + CO↑

Commercial preparation of CO:

Steam is passed over hot coke.

C + H2O → CO + H2

Laboratory preparation of carbon dioxide:

Calcium carbonate reacts with dilute HCl to form carbon dioxide.

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

Industrial preparation of carbon dioxide:

Limestone is heated to produce carbon dioxide.

CaCO3 →→CaO + CO2

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

The balanced equations are given below:

(i) 2BF3 + 6LiH → B2H6 + 6LiF

(ii) B2H6 + 6H2O → 2H3BO3 (orthoboric acid)

(iii) 2NaH + B2H6 →2Na [BH4] (sodium borohydride)

(iv) H3BO3 →HBO2 (metaboric acid) + H2O
     4HBO2 →H2B4O7→ 2B2O3 (boron trioxide) + H2O

(v) Al+3NaOH→Al (OH)3 + 3Na
 3 B2H6 + 6NH3 → 2B3N3H6 + 12H2

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

The compounds X, Y and Z are borax, sodium metaborate + boric anhydride and boric acid respectively.

When borax is heated, it first swells and then forms a transparent glass like bead of sodium meta borate and boric anhydride.

Na2B4O7 à2NaBO2+B2O3+10H2O

(Borax)       (sodium metaborate) (Boric anhydride)

Aqueous solution of borax is alkaline due to formation of strong base NaOH.

Hence, it turns red litmus blue.

Na2B4O7 +7 H2O → 4H3BO3 + 2NaOH

Borax reacts with sulphuric acid to form boric acid and sodium sulphate.

Na2B4O7 + H2SO4 + 5 H2O → 4H3BO3 + Na2SO4

New answer posted

a year ago

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V
Vishal Baghel

Contributor-Level 10

(a) Inert pair effect: When the pair of electrons in the valence shell does not take part in bond formation, then this effect is called as inert pair effect.

(b)Allotropy: It is the property of the element by which an element can exist in two or more forms which have same chemical properties but different physical properties due to their structures.

(c)Catenation: The tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation.
For example, carbon forms chains with (C-C) single bonds and also with multiple bonds (C = C or C = C).

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

The compounds X, A, B, C and D are aluminium, aluminium hydroxide, sodium tetrahydrozoaluminate (III), aluminium chloride and alumina.

Aluminium reacts with NaOH to form white PPT of Al (OH)3→.

2Al + 3NaOH → Al (OH)3→ + 3Na+

Al (OH)3→ reacts with NaOH to form Na+ [Al (OH)4→]→.

Al (OH)3 + NaOH → Na+ (Al (OH)4)→
Al (OH)3→ reacts with HCL to form AlCl3→.

Al (OH)3 + 3HCl → AlCl3 + 3H2O

When Al2→O3→ is heated,  Al2→O3→ is obtained. 

2Al (OH)3 → Al2O3 + 3H2O

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