Chemistry Test 111

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Chemistry Test 111

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Weekly Quiz Competition

Question 1
4 / -1

In hexagonal close packing of sphere in three dimensions.

A
In one unit cell there are 12 octahedral voids and all are completely inside the hexagonal unit cell

B
In one unit cell there are six octahedral voids and all are completely inside the hexagonal unit cell.

C
In one unit cell there are six octahedral void and of which three are completely inside the unit cell and other three are from contributions of octahedral voids which are partially inside the hexagonal unit cell

D
In one unit cell 6 tetrahedral voids are completely inside the hexagonal unit cell.

Solution

hcp = AB AB AB ...............pattern repeat

For calculating voids between two layers A and B.

The dots are representing tetrahedral voids.

no. of tetrahdral voids = 3 + 1 +1/3 × 6 = 6

so in one complete unit cell total tetrahedral voids = 12, out of which 8 are completely inside but rest are shared by other unit cells.

Octahedral voids = 3 = represented by (x).

So total octahedral voids = 6 = All are completely inside.

[4R = √3 a]
Question 2
4 / -1

Bleaching action of H₂O₂ is due to its :

A
Oxidising nature

B
Reducing nature

C
Acidic nature

D
Thermal instability
Question 3
4 / -1

Which of the following is Bacteriostatic antibiotic.

A
Penicillin

B
Chloramphenicol

C
Aminoglycosides

D
Ofloxacin
Question 4
4 / -1

Which of the following kinds of catalysis can be explained by the adsorption theory?

A
Heterogeneous catalysis

B
Enzyme catalysis

C
Homogeneous catalysis

D
Acid base catalysis

Solution

Adsorption of gas on solid surface.
Question 5
4 / -1

The product obtained in the reaction of diborane with excess of ammonia at low temperature is :

A
B₂H₆. NH₃

B
B₂H₆. 2NH₃

C
(BN)_x

D
Borazine
Question 6
4 / -1

A
enantiomers

B
diastereomers

C
mesomers

D
all of these

Solution

The given two compounds are stereoisomers and they are not mirror image of each other hence they are diastereomers.
Question 7
4 / -1

A proton accelerated from rest through a potential difference of ‘V’ volts has a wavelength λ associated with it. An alpha particle in order to have the same wavelength must be accelerated from rest through a potential difference of

A
V volt

B
4V volt

C
2V volt

D
V/8 volt

Solution
Question 8
4 / -1

Calamine is:-

A
BaCO₃

B
ZnCO₃

C
ZnS

D
ZnO
Question 9
4 / -1

Which of the following have maximum dipole moment.

A
(1)

B
(2)

C
(3)

D
(4)

Solution
Question 10
4 / -1

On reacting with KMnO₄, FeSO₄ converts into Fe⁺³ in acidic medium. If 250 ml of FeSO₄ reacts with 32 ml of 5 M solution of KMnO₄. The molarity of FeSO₄ solution is :

A
3.2 M

B
1 M

C
4/3 M

D
2/3 M

Solution

32 × 5 × 5 = 250 × M × 1 M = 3.2

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Chemistry Test 111

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Question 1

In one unit cell there are 12 octahedral voids and all are completely inside the hexagonal unit cell

In one unit cell there are six octahedral voids and all are completely inside the hexagonal unit cell.

In one unit cell there are six octahedral void and of which three are completely inside the unit cell and other three are from contributions of octahedral voids which are partially inside the hexagonal unit cell

In one unit cell 6 tetrahedral voids are completely inside the hexagonal unit cell.

Solution

Question 2

Oxidising nature

Reducing nature

Acidic nature

Thermal instability

Question 3

Penicillin

Chloramphenicol

Aminoglycosides

Ofloxacin

Question 4

Heterogeneous catalysis

Enzyme catalysis

Homogeneous catalysis

Acid base catalysis

Solution

Question 5

B2H6. NH3

B2H6. 2NH3

(BN)x

Borazine

Question 6

enantiomers

diastereomers

mesomers

all of these

Solution

Question 7

V volt

4V volt

2V volt

V/8 volt

Solution

Question 8

BaCO3

ZnCO3

ZnS

ZnO

Question 9

(1)

(2)

(3)

(4)

Solution

Question 10

3.2 M

1 M

4/3 M

2/3 M

Solution

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B₂H₆. NH₃

B₂H₆. 2NH₃

(BN)_x

BaCO₃

ZnCO₃