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 

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

  • The scientific notations will be:

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

Final Answer:-

  • The Scientific notations are

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

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 

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

  • The scientific notations will be:

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

Final Answer:-

  • The Scientific notations are

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

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 

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

  • The scientific notations will be:

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

Final Answer:-

  • The Scientific notations are

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

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

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

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

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

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

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

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

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

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