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3 years ago

Which of these best describes the particle motion taking place as CO2 gas is exposed to freezing temperatures?

Chemistry

2 answers:

nirvana33 [79]3 years ago

6 0

The options in the question are as follows

a. The motion of particles becomes random.

b. The motion of the particles is unchanged.

c. The particles move with more force.

d. The particles decrease in speed.

Answer:

The particles decrease in speed.

Explanation:

In a gas at room temperature, the particles of the gas move at a very high speed and in a random manner. A gas has no definite volume and it is also assumed that there is minimal interaction between gas particles. Recall that temperature is a measure of the average kinetic energy of the particles of a system. As the temperature is decreased and heat is withdrawn as the CO2 approaches freezing temperature, the speed and kinetic energy of the particles decreases and significant interaction between particles is observed. Hence the CO2 particles slow down.

mote1985 [20]3 years ago

5 0

CO2 is carbon dioxide which is most famous for being in gas form so i would figure if it was exposed to freezing temperatures it would turn into a liquid then maybe a solid<span />

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Answer:

Democritus

Explanation:

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3 years ago

if 30.0L of oxygen are cooled from 200 degrees celsius to 1 degree celsius at constant pressure, what is the new volume of oxyge

Sergio039 [100]

Answer: 18.65L

Explanation:

Given that,

Original volume of oxygen (V1) = 30.0L

Original temperature of oxygen (T1) = 200°C

[Convert temperature in Celsius to Kelvin by adding 273.

So, (200°C + 273 = 473K)]

New volume of oxygen V2 = ?

New temperature of oxygen T2 = 1°C

(1°C + 273 = 274K)

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

30.0L/473K = V2/294K

To get the value of V2, cross multiply

30.0L x 294K = 473K x V2

8820L•K = 473K•V2

Divide both sides by 473K

8820L•K / 473K = 473K•V2/473K

18.65L = V2

Thus, the new volume of oxygen is 18.65 liters.

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3 years ago

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malfutka [58]

C. NaOH ammmonia is also an base but not as strong as NaOH

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2 years ago

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Answer:

- False.

- True.

Explanation:

Hello, for each statement we state:

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The exponent of "2" powers the concentration to the second power, not doubles the rate law, thus, if C is 3, for k=1, r will be -9. On the other hand if the rate is like:

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The rate will be -3, that is why the rate is not doubled when the "2" in concentration is present.

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FALSE because for nonelemental chemical reactions, the exponents do not match with each species' stoichiometric coefficients in the rate law.

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TRUE, since the Arrhenius equation allows us to prove the effect of the activation energy and the temperature:

$k=Aexp(-\frac{Ea}{RT})$

- Differential rate laws allow us to compare concentration and time.

TRUE as they are given like:

$\frac{1}{\nu _A} \frac{dC_A}{dt} =\frac{1}{\nu _B} \frac{dC_B}{dt} =...$

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