Answer:
For Xenon fluoride, the average bond energy is 132kj/mol
For tetraflouride,the average bond energy is 150.5kj/mol.
For hexaflouride, the average bond energy is 146.5 kj/mol
Explanation:
For xenon fluoride
105/2 = 52.5
For F-F
159/2 = 79.5
Average bond energy of Xe-F = 79.5 + 52.5 = 132kj/mole
For tetraflouride
284/4 = 71
For F-F
159/2 = 79.5
Average bond energy = 79.5 + 71 = 150.5kj/mol
For hexaflouride
402/6 = 67
F-F = 159/2 = 79.5
Average bond energy = 67 + 79.5 = 146.5kj/ mol
Answer:
Hello some part of your question is missing below is the missing part
2. What is the force on the charged particle if it is now located at the 0V potential difference line? (mN) (hint: The electric field can be obtained as above using the 0V and -10V equipotential lines.)
answer :
1) 0.8 mN
2) 0.8 mN
Explanation:
Given data:
1) Calculate the force on the charged particle
q = 80 μC , Va = 30v , Vb = 40v, ∝ = 1 m
E = ( Δv ) / ∝
= ( Vb - Va ) / ∝
F = qE
= 80 μC * ( 40 - 30 ) / 1 m
= 800 μC
F = 0.8 mN
<u>2) Calculate the force on the charged particle when it is located at 0V</u>
Va = -10V , Vb = 0V, q = 80 μC, ∝ = 1 m
F = qE
where E = ( 0 - ( -10 ) / 1
F = 80 μC * ( 0 - ( -10 ) / 1
= 800 μC = 0.8 mN
A greater weight so the object has greater weight
Subtract one from the to find the magnitude. Whether it's positive or negative will determine the direction.