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Irina-Kira [14]
3 years ago
15

A gaseous fluorinated methane compound has a mass of 8.00 g at 2.81 atm and 300. K in a 1.00 L container.

Chemistry
1 answer:
trasher [3.6K]3 years ago
8 0

Answer:

Look at my answer to the last one, and see if you can solve this one on your own!

After you try it, I can let you know if you got it right :)

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5. What is metallic bonding?<br> 6. Why are metals shiny?
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Answer:

Metallic bonding is found in metals and their alloys. When the atoms give up their valence electrons, they form ions. These ions are held together by the electron cloud surrounding them. Metals are shiny because they have a lot of free (i.e. delocalized) electrons that form a cloud of highly mobile negatively charged electrons on and beneath the smooth metal surface in the ideal case. ... In the absence of any external EM field, the charges in the plasma are uniformly distributed within the metal.

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Alex saw a rusty old car in an abandoned lot. Because he had recently read about chemical reactions, he began to wonder how the
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The answere is A .......


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Fluids that have a lower viscosity than water
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Syrup, molasses, and honey have a lower viscosity than water

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Which of the following can you determine from mass spectrometry data
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different isotopes of na element

7 0
4 years ago
The value of Kp for the reaction NO(g) 1 1 2 O2(g) 4 NO2(g) is 1.5 3 106 at 25°C. At equilibrium, what is the ratio of PNO2 to P
expeople1 [14]

Answer : The ratio of p_{NO_2} to p_{NO} is, 6.87\times 10^5

Solution :  Given,

K_p=1.5\times 10^6

p_{O_2} = 0.21 atm

The given equilibrium reaction is,

NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)

The expression of K_p will be,

K_p=\frac{(p_{NO_2})}{(p_{NO})\times (p_{O_2})^{\frac{1}{2}}}

Now put all the given values in this expression, we get:

1.5\times 10^6=\frac{(p_{NO_2})}{(p_{NO})\times (0.21)^{\frac{1}{2}}}

\frac{(p_{NO_2})}{(p_{NO})}=(1.5\times 10^6)\times (0.21)^{\frac{1}{2}}

\frac{(p_{NO_2})}{(p_{NO})}=6.87\times 10^5

Therefore, the ratio of p_{NO_2} to p_{NO} is, 6.87\times 10^5

5 0
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