Chemistry Test

Result

Chemistry Test - 20

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

In 2HI ⇌ H₂ + I₂, the equilbrium is not affected by:

A
number of moles of reactants

B
pressure

C
volume

D
temperature

Solution

Since the numbers of moles of reactants and products are the same, there will not be any effect of pressure on the equilibrium.
Question 2
4 / -1

When a number 3.56923 *10³ is expressed in four significant figures, it becomes

A
3.5702 *10³

B
3.569 *10³

C
3.560 *10³

D
3.5692 *10³

Solution

When a number 3.56923 x 10³ is expressed in four significant figures, it becomes 3.569 x 10³.
Question 3
4 / -1

Insulin regulates the metabolism of

A
minerals

B
amino acids

C
glucose

D
vitamins

Solution

Insulin is a hormone produced by the pancreas which is central to regulating carbohydrate and fat metabolism in the body. Insulin causes cells in the liver, muscle, and fat tissue to take up glucose from the blood, storing it as glycogen in the liver and muscle.
Question 4
4 / -1

Consider the following skeleton equation:

__As₂O₃ + __SnCl₂+ __HCl → __SnCl₄ + __As + __H₂O

What are the stoichiometric coefficients of As₂O₃, SnCl₂, HCl, SnCl₄, As and H₂O, when the above equation is balanced?

A
1, 6, 3, 2, 3, 3

B
3, 2, 1, 6, 1, 2

C
1, 3, 6, 3, 2, 3

D
3, 2, 3, 1, 2, 6

Solution

The stoichiometric coefficients of As₂O₃, SnCl₂, HCl, SnCl₄, As and H₂O are 1, 3, 6, 3, 2 and 3, respectively.

As₂O₃ + 3SnCl₂ + 6HCl → 3SnCl₄ + 2As + 3H₂O
Question 5
4 / -1

The rate constant of a first order reaction is 3 * 10^-6 per second and initial concentration is 0.10 M. The initial rate of reaction is

A
3 * 10^-6 Ms^-1

B
3 * 10^-8 Ms^-1

C
3 * 10^-7 Ms^-1

D
3 * 10^-9 Ms^-1

Solution

Rate = K[A]¹

= 3 x 10^-6 x 0.1 = 3 x 10^-7 Ms^-1
Question 6
4 / -1

Directions: The following question has four choices out of which ONLY ONE is correct.

Dissolution of ammonium chloride in water is an endothermic change. At constant temperature, it is accompanied by

A
no change in enthalpy

B
decrease in entropy

C
no change in entropy

D
increase in entropy

Solution

Entropy in liquid state is more than that in solid state. So, dissolution of ammonium chloride in water is accompanied by increase in entropy.
Question 7
4 / -1

Coagulation of lyophobic colloids can be carried out by

A
boiling

B
persistent dialysis

C
electrophoresis

D
All of the above

Solution

The phenomenon of precipitation of a colloidal solution by the addition of excess of an electrolyte is called coagulation or flocculation.

The coagulation of lyophobic colloids can be carried out by the following methods:

(1) By electrophoresis: In electrophoresis, the colloidal particles move towards oppositely charged electrode. When these come in contact with the electrode for long, these are discharged and precipitated.

(2) By mixing two oppositely charged sols: When oppositely charged sols are mixed in almost equal proportions, their charges are neutralised. Both sols may be partially or completely precipitated as the mixing of ferric hydroxide (+ve sol) and arsenious sulphide (–ve sol) brings them in precipitated form. This type of coagulation is called mutual coagulation or meteral coagulation.

(3) By boiling: When a sol is boiled, the adsorbed layer is disturbed due to increased collisions with the molecules of dispersion medium. This reduces the charge on the particles and ultimately they settle down to form a precipitate.

(4) By persistent dialysis: On prolonged dialysis, the traces of electrolyte present in the sol are removed almost completely and the colloids become unstable.
Question 8
4 / -1

Which of the following pairs has the highest heat of neutralisation?

A
NH₄OH and CH₃COOH

B
NH₄OH and HCl

C
KOH and CH₃COOH

D
KOH and HCl

Solution

Since both acid and base are strong in this case. So heat of neutralization is highest in this case.
Question 9
4 / -1

Directions: The following question has four choices out of which ONLY ONE is correct.

The enthalpy of neutralization of HCN by NaOH is - 12kJ/mol. The enthalpy of ionization of HCN is

A
- 12 KJ/ mol

B
69 KJ/ mol

C
12 KJ/ mol

D
- 45 KJ/ mol

Solution

The enthalpy of ionization of HCN is reverse of the enthalpy of neutralization of HCN. So enthalpy of ionization of HCN is 12 kJ/mol.
Question 10
4 / -1

Hydrolysis of DDT is a first order reaction. Its half-life is 10 year. Time to hydrolyse 10 g DDT to half of its original amount is

A
100 year

B
50 year

C
5 year

D
10 year

Solution

Key Idea : The half-life of reaction is the time it takes to reduce to half of its original amount.
\(\because\) The half-life of reaction of hydrolysis of DDT is 10 yr.
\(\therefore \quad\) Time required to hydrolyse \(10 \mathrm{g}\) DDT to half \(=10\) yrs.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemistry Test - 20

Result

Chemistry Test - 20

Score

-

Rank

-

SHARING IS CARING

Question 1

number of moles of reactants

pressure

volume

temperature

Solution

Question 2

3.5702 *103

3.569 *103

3.560 *103

3.5692 *103

Solution

Question 3

minerals

amino acids

glucose

vitamins

Solution

Question 4

1, 6, 3, 2, 3, 3

3, 2, 1, 6, 1, 2

1, 3, 6, 3, 2, 3

3, 2, 3, 1, 2, 6

Solution

Question 5

3 * 10-6 Ms-1

3 * 10-8 Ms-1

3 * 10-7 Ms-1

3 * 10-9 Ms-1

Solution

Question 6

no change in enthalpy

decrease in entropy

no change in entropy

increase in entropy

Solution

Question 7

boiling

persistent dialysis

electrophoresis

All of the above

Solution

Question 8

NH4OH and CH3COOH

NH4OH and HCl

KOH and CH3COOH

KOH and HCl

Solution

Question 9

- 12 KJ/ mol

69 KJ/ mol

12 KJ/ mol

- 45 KJ/ mol

Solution

Question 10

100 year

50 year

5 year

10 year

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

3.5702 *10³

3.569 *10³

3.560 *10³

3.5692 *10³

3 * 10^-6 Ms^-1

3 * 10^-8 Ms^-1

3 * 10^-7 Ms^-1

3 * 10^-9 Ms^-1

NH₄OH and CH₃COOH

NH₄OH and HCl

KOH and CH₃COOH