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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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John recently suffered a blow to his head. Since then, he finds it difficult to comprehend what others say to him. He also finds
katovenus [111]

Answer: ​Wernicke's aphasia

Explanation:

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3 years ago
Explain, using the kinetic theory of matter, why liquids and solids are more denser than gases
AlekseyPX
Solid and liquids are much more denser than gas because their molecules are close to each other and with that the molecules of them can't move that freely unlike the gas molecules. Also, because of being near to each other the molecules of solid and liquids became heavy making them dense.
5 0
3 years ago
A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

N=n N_A

where

n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

The energy emitted by each atom (the energy of one photon) is

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

4 0
3 years ago
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