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laiz [17]
3 years ago
8

nert xenon actually forms many compounds, especially with highly electronegative fluorine. The ΔH o f values for xenon difluorid

e, tetrafluoride, and hexafluoride are −105, −284, and −402 kJ/mol, respectively. Find the average bond energy of the XebondF bonds in each fluoride.
Physics
1 answer:
loris [4]3 years ago
3 0

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

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Answer:

10^80 i think, scientists say so.

Explanation:

i pretty much just need points and i'm pretty sure you can look this up

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3 years ago
How does the resistance change as you add bulbs to a series circuit?
Ronch [10]
The resistance is increased when more and more bulbs are added to the circuit.
7 0
3 years ago
Read 2 more answers
What is the speed of sound in air at 50°F (in ft/s)?
gizmo_the_mogwai [7]

Answer:

Speed of air = 1106.38 ft/s

Explanation:

Speed of sound in air with temperature

v_{air}=331.3\sqrt{1+\frac{T}{273.15}} \\

Here speed is in m/s and T is in celcius scale.

T = 50°F

T=(50-32)\times \frac{5}{9}=10^0C \\

Substituting

v_{air}=331.3\sqrt{1+\frac{10}{273.15}}=337.31m/s \\

Now we need to convert m/s in to ft/s.

1 m = 3.28 ft

Substituting

v_{air}=337.31\times 3.28=1106.38ft/s \\

Speed of air = 1106.38 ft/s

6 0
2 years ago
When do phase transitions occur in molecules?
jekas [21]
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5 0
3 years ago
Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?
Debora [2.8K]

493 \; \text{W}\cdot \text{m}^{-2}.

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

\dfrac{P}{A} = \sigma \cdot T^{4}

where,

  • P the total power emitted,
  • A the surface area of the body,
  • \sigma the Stefan-Boltzmann Constant, and
  • T the temperature of the body in degrees Kelvins.

\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}.

T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}.

\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}.

Keep as many significant figures in T as possible. The error will be large when T is raised to the power of four. Also, the real value will be much smaller than 493\; \text{W}\cdot \text{m}^{-2} since the emittance of a human body is much smaller than assumed.

5 0
3 years ago
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