Chemistry Test 133

Result

Chemistry Test 133

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Weekly Quiz Competition

Question 1
4 / -1

Consider the following equilibrium

N₂O₄ (g) ⇔ 2NO₂(g)

Then select the correct graph: (graph represent relation between concertation of NO₂ and N₂O₄ at equilibrium)

A

B

C

D

Solution
Question 2
4 / -1

If two different non-axial d-orbitals having XZ nodal plane form π bond by overlapping each other, then internuclear axis will be;

A
X

B
Y

C
Z

D
No π-bond will form

Solution

Two different non-axial d-orbitals will lie in planes perpendicular to each other, such d-orbital will not formed.
Question 3
4 / -1

The given osazone can be formed by

A
D-glucose

B
D-fructose

C
D- Idose

D
Both a and b

Solution

Formation of osazone involves only C₁ and C₂ carbon atoms. The rest of the carbon atoms remains uneffected.
Glucose and fructose differ only in the configurations of first and second carbon atoms remaining positions are similar. So they form same osazone.
Question 4
4 / -1

The cyclotrimetaphosphoric acid is

A
(HPO₃)₃ and contains 9 σ – bonds

B
H₃P₃O₆ and contains 12 σ – bonds

C
(HPO₃)₃ and contains 15 σ – bonds

D
H₃P₃O₉ and contains 18 σ – bonds

Solution
Question 5
4 / -1

In a face centered cubic lattice of edge length 'a' number of next to next nearest neighbor to corner atom and distance of corner atom to next to next nearest atom is respectively-

A

B
6,a

C

D

Solution
Question 6
4 / -1

CrO₄^2(-) (yellow) changes to Cr₂O₇^2(-) (orange) in pH= x and Cr₂O₇^2(-) (orange) changes to CrO₄^2(-) (yellow) in ph = y. Then x and y can be:

A
6, 8

B
6, 5

C
8, 6

D
7, 7

Solution
Question 7
4 / -1

Identify position most favorable for aromatic substitution (EAS)

A
A

B
B

C
C

D
A and C

Solution

is activating group so it would activate ortho and para position for E.A.S but only para position is empty so EAS will take place there.
Question 8
4 / -1

Polymer of prop – 2 – ene nitrile is-

A
saran

B
orlon

C
Dacron

D
Teflon

Solution
Question 9
4 / -1

Molar mass of a substance, 1g of which when dissolved in 100g of water gave a solution whose boiling point is 100.1^oC at a pressure of 1 atm (k_b of water = 0.52 k kg mol^–1) is-

A
50

B
52

C
60

D
62

Solution

The given values
W_B = 1g k_b = 0.5 2 k kg mol^–1
W_A = 100
Tb = 100. 1–100 = 0. 1
Using equation
Question 10
4 / -1

Arrange the following in increasing order of their P_ka values

A
y < x < Z

B
x < y < Z

C
y < Z < x

D
x < Z < y

Solution

Because acidic character depends on stability of conjugate ion.