JEE Advanced Chemistry Test 17

Result

JEE Advanced Chemistry Test 17

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Weekly Quiz Competition

Question 1
4 / -1

An excited electron of H-atoms emits of photon of wavelength λ and returns in the ground state, the principal quantum number of excited state is given by:

A

B

C

D

Solution
Question 2
4 / -1

The correct order of acidic strength of the compounds, 1 – 3 is

A
1 > 2 > 3

B
2 > 3 >1

C
3 > 1 > 2

D
2 > 1 > 3
Question 3
4 / -1

The mechanism involved in the following conversion is

A
E₂ – elimination

B
E₁ – elimination

C
Syn – elimination

D
E₁cb – elimination
Question 4
4 / -1

The correct order of bond angle of the following compound is

A
NH₃ > NH₂F > NHF₂ > NF₃

B
NF₃ > NH₂F > NHF₂ > NH₃

C
NH₂F > NHF₂ > NH₃ > NF₃

D
NH₃ > NHF₂ > NH₂F > NF₃

Solution
Question 5
4 / -1

Directions For Questions

The rate law for one of the mechanism of the pyrolysis of CH₃CHO at 520^oC and 0.2 bar is

...view full instructions

The overall activation energy Ea in terms of rate law is:

A
Ea (2) + Ea (1)− 2 Ea (4)

B

C

D

Solution

From Arrhenius equation, K = Ae^Ea/RT

From rate equation given in question

Hence (C) is correct answer.
Question 6
4 / -1

The correct IUPAC name of compound given below is

A
5-ethoxy-3-methyl-5-oxopentanoic acid

B
1-ethoxy-3-methyl-5-oxopentanoic acid

C
ethyl-3-methyl-5-oxopentanoic acid

D
5-ethoxy-3-methyl-5-oxopentanon-1-ol
Question 7
4 / -1

Directions For Questions

The stopping potential (V₀) of photosensitive surface varies with the frequency of incident radiation as shown in the following figure.

...view full instructions

The work function of photosensitive surface.

A
16.5 eV

B
12.4 eV

C
10.8 eV

D
18.4 eV

Solution
Question 8
4 / -1

5.6 litre of an unknown gas at NTP requires 12.5 calorie to raise its temperature by 10^oC at constant volume.

A
1.67

B
1.20

C
1.40

D
1.33

Solution
Question 9
4 / -1

The value of (PV_m)_P→0 of a real gas is independent of the nature of gas because

A
As P → 0 , there is no molecular attractions and volume occupied by one molecule is negligible in compression to the total volume of gas.

B
As P → 0 , there is molecular attractions and volume occupied by one molecule is negligible in compression to the total volume of gas.

C
As P → 0 , there is no molecular attractions and volume occupied by one molecule can not be negligible in compression to the total volume of gas.

D
As P → 0 , gaseous molecule are free to move all the possible directions and volume occupied by one molecule is not negligible in compression to the total volume of gas

Solution

At very low pressure the molar volume of gas is very high therefore; V_m >>>b , and molecular attraction are also negligible. Thus gas behaves as an ideal gas.

Hence (A) is correct answer.
Question 10
4 / -1

Which of the following oxide cannot be reduced by carbon reduction?

A
Fe₃O₄

B
MnO₂

C
Cr₂O₃

D
Option B and option C are correct

Solution

MnO₂ and Cr₂O₃ are reduced by aluminum.