When an electron jumps from n 5 to n 1 in a hydrogen atom the number of spectral lines obtained is?

The charge on a proton is

A: -1.602×10-19C

B: 1.602×10-19C

C: 0

D: None of the above

Solution:

Explanation:

Protons are the positively charged particles that make up a hydrogen atom's nucleus. A proton has a mass of 1.6726219×10-27 kilograms.

A proton's mass is the same as that of a hydrogen atom. One electron and one proton make up a hydrogen atom. Because the mass of an electron is deemed unimportant, the mass of a hydrogen atom can be said to be equal to the mass of a proton. A proton has a mass of 1840 times that of an electron.

A proton's charge is the same as and opposite that of an electron. As a result, it has a single positive charge having the value 1.602×10-19C.

Final answer:

Option B is the correct answer, The charge on a proton is 1.602×10-19C.

The hybridisation of Ni inNi(CO)4 is

A:sp2

B:dsp2

C:sp3

D:sp3p

Solution:

Explanation:-

• The outer electronic configuration of Ni is 3d84s2 .
• Now ,Ni is in zero oxidation state. The 2 electrons from 4s orbitals moves to the 3d orbital and we get completely filled 3d orbital and vacant 4s and 4 p orbital. 

3d104s04d0

• This way we get sp3  hybridization and shape is tetrahedral.
• It does not have any unpaired electrons so it is diamagnetic in nature.

Final Answer:-

The hybridization of Ni in Ni(CO)4   complex  is  (C ) sp3

Express the following in the scientific notation

(i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

Solution:

Explanation:-

• In scientific notation the number is represented as 

• The scientific notations will be:

Final Answer:-

• The Scientific notations are

The cell in which the following reaction occurs

2Fe3+  (aq) + 2I- (aq) → 2Fe3+ (aq) + I2 (s) has Ecell  = 0.236 V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Solution:

Explanation:

Eº= 0.236 V

T = 293  k

We know that G =- nFEº

F = 96500

n = 2 (no. of electron)

ΔG = -2965000.236 = -45.548kJ

From nernst equation

Ecell = Eº- 0.591/n logk

Ecell = 0 (at equilibrium) 

Logk =  nEº/0.591= 2×0.236/0.591

Logk = 7.98

K = 107.98= 9.616 ×107

Final Answer:

Hence standard Gibbs energy = -45.548kJ & equilibrium constant of the cell reaction K = 9.616 107.

The SI unit of weight is

A: kilogram

B: gram

C: newton

D: milligram

Solution:

Explanation:

substance's mass can be defined as a measurement of its physical 3D structure and quantity.

A substance's mass is usually expressed in kilograms or grams. The kilograme is a metric unit of mass used by the International System of Units. 'kg' is another way of writing it.

Final Answer :

Thus, option A is the correct answer. A(Kilogram)

Page 2

The charge on a proton is

A: -1.602×10-19C

B: 1.602×10-19C

C: 0

D: None of the above

Solution:

Explanation:

Protons are the positively charged particles that make up a hydrogen atom's nucleus. A proton has a mass of 1.6726219×10-27 kilograms.

A proton's mass is the same as that of a hydrogen atom. One electron and one proton make up a hydrogen atom. Because the mass of an electron is deemed unimportant, the mass of a hydrogen atom can be said to be equal to the mass of a proton. A proton has a mass of 1840 times that of an electron.

A proton's charge is the same as and opposite that of an electron. As a result, it has a single positive charge having the value 1.602×10-19C.

Final answer:

Option B is the correct answer, The charge on a proton is 1.602×10-19C.

The hybridisation of Ni inNi(CO)4 is

A:sp2

B:dsp2

C:sp3

D:sp3p

Solution:

Explanation:-

• The outer electronic configuration of Ni is 3d84s2 .
• Now ,Ni is in zero oxidation state. The 2 electrons from 4s orbitals moves to the 3d orbital and we get completely filled 3d orbital and vacant 4s and 4 p orbital. 

3d104s04d0

• This way we get sp3  hybridization and shape is tetrahedral.
• It does not have any unpaired electrons so it is diamagnetic in nature.

Final Answer:-

The hybridization of Ni in Ni(CO)4   complex  is  (C ) sp3

Express the following in the scientific notation

(i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

Solution:

Explanation:-

• In scientific notation the number is represented as 

• The scientific notations will be:

Final Answer:-

• The Scientific notations are

The cell in which the following reaction occurs

2Fe3+  (aq) + 2I- (aq) → 2Fe3+ (aq) + I2 (s) has Ecell  = 0.236 V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Solution:

Explanation:

Eº= 0.236 V

T = 293  k

We know that G =- nFEº

F = 96500

n = 2 (no. of electron)

ΔG = -2965000.236 = -45.548kJ

From nernst equation

Ecell = Eº- 0.591/n logk

Ecell = 0 (at equilibrium) 

Logk =  nEº/0.591= 2×0.236/0.591

Logk = 7.98

K = 107.98= 9.616 ×107

Final Answer:

Hence standard Gibbs energy = -45.548kJ & equilibrium constant of the cell reaction K = 9.616 107.

The SI unit of weight is

A: kilogram

B: gram

C: newton

D: milligram

Solution:

Explanation:

substance's mass can be defined as a measurement of its physical 3D structure and quantity.

A substance's mass is usually expressed in kilograms or grams. The kilograme is a metric unit of mass used by the International System of Units. 'kg' is another way of writing it.

Final Answer :

Thus, option A is the correct answer. A(Kilogram)

Page 3

The charge on a proton is

A: -1.602×10-19C

B: 1.602×10-19C

C: 0

D: None of the above

Solution:

Explanation:

Protons are the positively charged particles that make up a hydrogen atom's nucleus. A proton has a mass of 1.6726219×10-27 kilograms.

A proton's mass is the same as that of a hydrogen atom. One electron and one proton make up a hydrogen atom. Because the mass of an electron is deemed unimportant, the mass of a hydrogen atom can be said to be equal to the mass of a proton. A proton has a mass of 1840 times that of an electron.

A proton's charge is the same as and opposite that of an electron. As a result, it has a single positive charge having the value 1.602×10-19C.

Final answer:

Option B is the correct answer, The charge on a proton is 1.602×10-19C.

The hybridisation of Ni inNi(CO)4 is

A:sp2

B:dsp2

C:sp3

D:sp3p

Solution:

Explanation:-

• The outer electronic configuration of Ni is 3d84s2 .
• Now ,Ni is in zero oxidation state. The 2 electrons from 4s orbitals moves to the 3d orbital and we get completely filled 3d orbital and vacant 4s and 4 p orbital. 

3d104s04d0

• This way we get sp3  hybridization and shape is tetrahedral.
• It does not have any unpaired electrons so it is diamagnetic in nature.

Final Answer:-

The hybridization of Ni in Ni(CO)4   complex  is  (C ) sp3

Express the following in the scientific notation

(i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

Solution:

Explanation:-

• In scientific notation the number is represented as 

• The scientific notations will be:

Final Answer:-

• The Scientific notations are

The cell in which the following reaction occurs

2Fe3+  (aq) + 2I- (aq) → 2Fe3+ (aq) + I2 (s) has Ecell  = 0.236 V at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Solution:

Explanation:

Eº= 0.236 V

T = 293  k

We know that G =- nFEº

F = 96500

n = 2 (no. of electron)

ΔG = -2965000.236 = -45.548kJ

From nernst equation

Ecell = Eº- 0.591/n logk

Ecell = 0 (at equilibrium) 

Logk =  nEº/0.591= 2×0.236/0.591

Logk = 7.98

K = 107.98= 9.616 ×107

Final Answer:

Hence standard Gibbs energy = -45.548kJ & equilibrium constant of the cell reaction K = 9.616 107.

The SI unit of weight is

A: kilogram

B: gram

C: newton

D: milligram

Solution:

Explanation:

substance's mass can be defined as a measurement of its physical 3D structure and quantity.

A substance's mass is usually expressed in kilograms or grams. The kilograme is a metric unit of mass used by the International System of Units. 'kg' is another way of writing it.

Final Answer :

Thus, option A is the correct answer. A(Kilogram)

Page 4

Follow Us:

Physics Wallah is India's top online ed-tech platform that provides affordable and comprehensive learning experience to students of classes 6 to 12 and those preparing for JEE and NEET exams. We also provide extensive NCERT solutions, sample papers, NEET, JEE Mains, BITSAT previous year papers, which makes us a one-stop solution for all resources.

Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.

We Stand Out because

We successfully provide students with intensive courses by India's top faculties and personal mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn’t dream of a good career in engineering and medical field earlier.

Our Key Focus Areas

Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.

What Makes Us Different

Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others.

The vision of PW Tuitions is to prepare students for a better future both academically and skill based at the most affordable prices.