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
1 / -0

A
16.761

B
26.383

C
13.260

D
23.484

Solution
Question 2
1 / -0

A
0.0072 mol L^-1 s^-1

B
0.0288 mol L^-1s^-1

C
0.0072 mol L^-1s^-1

D
0.0144 mol L^-1 s^-1

Solution
Question 3
1 / -0

The rate of a reaction can be increased in general by all the factors except by

A
using a catalyst

B
increasing temperature

C
increasing the activation energy

D
increasing the concentration of reactant

Solution

This option is correct because, it is the minimum amount of energy required by reactant molecule to get converted into product. Hence this option is correct.
Question 4
1 / -0

The concentration of reactant changes from 0.06 mol L^-1to 0.05 mol L^-1 in 25 minutes. Calculate the average rate of reaction.

A
4 x 10^-4mol L^-1min^-1

B
- 4 x 10^-4mol L^-1 min^-1

C
0.04 mol L^-1 min^-1

D
- 0.04 mol L^-1 min^-1

Solution

This option is correct because, rate of reaction is given by, Rate = - (change in concentration of reactants) / time. ( 0.06 - 0.05) / 25 = 4 x 10^-4 mol L^-1 min ^-1. Hence this option is correct.
Question 5
1 / -0

5A→ 2B

If there were 1.5 moles of 'A' 2 L solution in the beginning of the reaction, what is the concentration of 'B' when 'A' is 20% consumed?

A
0.75 M

B
0.06 M

C
0.15 M

D
0.03 M

Solution
Question 6
1 / -0

The rate constants were found to be K₁and K₂, when a reaction was carried out at 310 K and 290 K respectively.

The two rate constants are related as:

A
K₂ = 0.25 K₁

B
K₂ = 4 K₁

C
K₁ = 2 K₂

D
K₂ = 2 K₁

Solution

This option is correct because for every 10 degree rise in temprature, rate constant doubles.

Aince, T₁ = T₂ + 20

Hence, K₁ = 4 K₂

=> K₂ = 0.25 K₁

Hence, this option is correct.
Question 7
1 / -0

The following reaction takes place in one step:

\(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \leftrightharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\)

How will the rate of the above reaction change if the volume of the reaction vessel is diminished to one-third of its original volume?

A
\(27 \mathrm{r}_{1}\)

B
\(7 \mathrm{r}_{1}\)

C
\(97 \mathrm{r}_{1}\)

D
\(87 \mathrm{r}_{1}\)

Solution

The below reaction take place in one step, thus rate of reaction is:

\(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\)

Rate \(=\mathrm{k}[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\)

Suppose \(x\) moles of \(\mathrm{NO}\) and \(\mathrm{Y}\) moles of \(\mathrm{O}_{2}\) are taken in the vessle of volume \(\mathrm{V}\) litre, then:

\(r_{1}=k\left[\frac{x}{V}\right]^{2}\left[\frac{y}{V}\right]\)

If the volume of vessel is reduced to \(\mathrm{V} / 3\), then for the same moles of \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\).

\( r_{2} =k\left[\frac{x}{V / 3}\right]^{2}\left[\frac{y}{V / 3}\right]^{2}\)

\( r_{2} =27 k\left[\frac{y}{V}\right]^{2}\left[\frac{y}{V}\right] \)

\(\frac{r_{2}}{r_{1}}=27\) or \(r_{2}=27 r_{1}\)

The rate of the reaction becomes 27 times the initial rate.
Question 8
1 / -0

Which of the following covalent bonds has the highest ionic character?

A
H-Cl

B
H-NH₂

C
H-OH

D
H-SH

Solution

Among the given options, the difference in electronegativity between the H and O is the highest. So, this bond will have the highest ionic character.
Question 9
1 / -0

Which of the following statements correctly represents one of the characteristic of the lanthanides?

A
Most of the their ions are coloured.

B
They have greater tendency towards complex formation.

C
Their compounds are more basic.

D
They are generally non radioactive

Solution

This is correct as except promethium, lanthanides are non radioactive.
Question 10
1 / -0

Which of the following statements is correct about potassium permanganate?

A
It acts as strong reducing agent.

B
It is not suitable to be used in volumetric analysis.

C
It is used as a disinfectant

D
In it, four oxygen atoms are arranged in square planar arrangment.

Solution

Potassium permanganate is also widely used as a disinfectant and germicide.

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

16.761

26.383

13.260

23.484

Solution

Question 2

0.0072 mol L-1 s-1

0.0288 mol L-1 s-1

0.0072 mol L-1 s-1

0.0144 mol L-1 s-1

Solution

Question 3

using a catalyst

increasing temperature

increasing the activation energy

increasing the concentration of reactant

Solution

Question 4

4 x 10-4mol L-1min-1

- 4 x 10-4 mol L-1 min-1

0.04 mol L-1 min-1

- 0.04 mol L-1 min-1

Solution

Question 5

0.75 M

0.06 M

0.15 M

0.03 M

Solution

Question 6

K2 = 0.25 K1

K2 = 4 K1

K1 = 2 K2

K2 = 2 K1

Solution

Question 7

\(27 \mathrm{r}_{1}\)

\(7 \mathrm{r}_{1}\)

\(97 \mathrm{r}_{1}\)

\(87 \mathrm{r}_{1}\)

Solution

Question 8

H-Cl

H-NH2

H-OH

H-SH

Solution

Question 9

Most of the their ions are coloured.

They have greater tendency towards complex formation.

Their compounds are more basic.

They are generally non radioactive

Solution

Question 10

It acts as strong reducing agent.

It is not suitable to be used in volumetric analysis.

It is used as a disinfectant

In it, four oxygen atoms are arranged in square planar arrangment.

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.

0.0072 mol L^-1 s^-1

0.0288 mol L^-1s^-1

0.0072 mol L^-1s^-1

0.0144 mol L^-1 s^-1

4 x 10^-4mol L^-1min^-1

- 4 x 10^-4mol L^-1 min^-1

0.04 mol L^-1 min^-1

- 0.04 mol L^-1 min^-1

K₂ = 0.25 K₁

K₂ = 4 K₁

K₁ = 2 K₂

K₂ = 2 K₁

H-NH₂