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SOVA2 [1]
4 years ago
13

which gas will have the most collisions between its particles, methane at 340K, carbon dioxide at 80 C, argon at 265K or helium

at 20 C?
Chemistry
2 answers:
enyata [817]4 years ago
8 0

Answer: Option (b) is the correct answer.

Explanation:

As it is known that more is the kinetic energy present between the molecules of a gas, more will be the number of collisions take place.

Also, relation between kinetic energy and temperature is as follows.

                     K.E = \frac{3}{2}kT

This means that more is the temperature of gas more will be the kinetic energy present withing its molecules or vice-versa.

As methane is at 340 K.

Carbon dioxide is at 80 ^{o}C = (80 +273) K = 353 K

Argon is at 265 K and helium is at 20^{o}C = (20 + 273)K = 293 K

Therefore, temperature of carbon dioxide gas is the maximum.

Thus, we can conclude that the carbon dioxide gas at 80^{o}C will have the most collisions between its particles.

blsea [12.9K]4 years ago
6 0

Hey there!:

Option B,CO2 at 353K

*Since most collisions means that highest kinetic energy.

*As we know that kinetic energy of gas molecules is directly proportional to absolute temperature

*More the temperature, More the kinetic energy.

* Hence B  is the correct answer as it has highest temperature.

Hope this helps!

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3 years ago
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide.
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Answer:

[H_2]_{eq}=0.183M

[I_2]_{eq}=0.183M

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Explanation:

Hello.

In this case, for this equilibrium problem, we first realize that at the beginning there is just HI, it means that the reaction should be rewritten as follows:

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Whereas the law of mass action (equilibrium expression) is:

Kc=\frac{[H_2][I_2]}{[HI]^2}

That in terms of initial concentrations and reaction extent or change x turns out:

Kc=\frac{x*x}{([HI]_0-2x)^2}\\\\54.3=\frac{x^2}{(0.391M-2x)^2}

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x_1=0.183M\\\\x_2=0.210M

Whereas the correct answer is 0.183 M since the other value yield a negative concentration of HI at equilibrium (0.391-2*0.210=-0.029M).This, the equilibrium concentrations are:

[H_2]_{eq}=0.183M

[I_2]_{eq}=0.183M

[HI]_{eq}=0.391M-2*0.183M=0.025M

Regards.

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