The value of B for an ion pair.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Construction general requirements can be categorized in three-way different ways, business requirements, solution requirements and stakeholder requirements, can some items appear in more than one category?

Send me the solution to the following question based on the source and I do not want the solution from artificial intelligence

Answer 2

Question

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

Answer 6

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

Answer 7

Chapter 2, Problem 2.25P

To determine

The value of B for an ion pair.

Answer 8

Expert Solution & Answer

Answer to Problem 2.25P

The value of constant B is 8.3497×10−134 J⋅m12.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The value of m is 12.

The energy required to separate one pair of K+ and Cl− ions is 5.0 eV.

The equilibrium interionic separation is 0.266 nm.

Formula used:

The formula to calculate the inter-ionic potential is given by,

Vpion(r)=−( Ze)24πε0r+Brm ......... (I)

Here, Vpion(r) is the inter-ionic potential, Ze is the electron charge of K+ ion, ε0 is the permittivity of vacuum, r is inter atomic separation and B, and m are constants.

Calculation:

The inter-ionic potential is calculated as,

Substitute 5 eV for Vpion(r), 1.602×10−19 C for Ze, 8.85×10−12 F/m for ε0, 0.266×10−9 m for r and 12 for m in equation (I).

5 eV=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 125 eV( 1.60218× 10 −19 J 1 eV)=− ( 1.602× 10 −19 C )24π( 8.85× 10 −12 F/m )( 0.266× 10 −9 m)+B ( 0.266× 10 −9 m ) 12−8.01×10−19 J=2.307× 10 −280.266× 10 −9 m+B ( 0.266× 10 −9 m ) 12B=(−6.654× 10 −20)(1.2548× 10 −115 m 12)

Solve further as,

B=8.3497×10−134 J⋅m12

Conclusion:

The value of constant B is 8.3497×10−134 J⋅m12.

Answer 12

Ch. 2 - Prob. 1CQ Ch. 2 - Prob. 2CQ Ch. 2 - Prob. 3CQ Ch. 2 - Prob. 4CQ Ch. 2 - Prob. 5CQ Ch. 2 - Prob. 6CQ Ch. 2 - Prob. 7CQ Ch. 2 - Prob. 8CQ Ch. 2 - Prob. 9CQ Ch. 2 - Prob. 10CQ

The value of B for an ion pair.

Videos

The value of B for an ion pair.

Answer to Problem 2.25P

Explanation of Solution

Want to see more full solutions like this?

Chapter 2 Solutions