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

1. For the foundation shown below: Qapp = 60 kips (Load obtained from structural engineer) 1.5 ft G.W.T. 3 ft Poorly Graded Sand (SP): Ym 115 pcf (above G.W.T.) Ysat 125 pcf (below G.W.T.) c' = 0, ' = 35° K Square footing, 4' x 4' Foundation Dimension Information: 1-ft x 1-ft square concrete column. 1-ft thick "foot" flanges. Yconc=150 pcf *Assume weight of reinforcing steel included in unit weight of concrete. *Assume compacted backfill weighs the same as in-situ soil. Assume this foundation is being designed for a warehouse that had a thorough preliminary soil exploration. Using the general bearing capacity equation: a. Calculate the gross applied bearing pressure, the gross ultimate bearing pressure, and determine if the foundation system is safe using a gross bearing capacity ASD approach. Please include the weight of the foundation, the weight of the backfill soil, and the effect of the uplift pressure caused by the presence of the water table in your bearing capacity…

٢٥ ٠٥:٤٠١٠ 2025 ChatGPT VivaCut Onet Puzzle مسلم X Excel JPG I❤> PDF Copilot Chat Bot PDF2IMG iLovePDF NokoPrint O.O StudyX ☑ W CapCut Candy Crush DeepSeek Word ☐ Saga 啡 AcadAl ل TikTok

Refer to the figure below. Given: L = 7 m, y = 16.7 kN/m², and ø' = 30°. L L3 ση Sand γ $' D T LA L σε σε IN P Sand 1. Calculate the theoretical depth of penetration, D. (Enter your answer to three significant figures.) D= m 2. Calculate the maximum moment. (Enter your answer to three significant figures.) Mmax kN-m/m

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