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Crank
3 years ago
5

Container A holds 722 mL 722 mL of an ideal gas at 2.40 atm. 2.40 atm. Container B holds 169 mL 169 mL of a different ideal gas

at 4.60 atm. 4.60 atm. If the gases are allowed to mix together, what is the resulting pressure?
Chemistry
1 answer:
iren [92.7K]3 years ago
6 0

Answer:

The resulting pressure is 2.81 atm

Explanation:

According to Dalton's Law of Partial Pressure, each of the gases (A and B) will exert their pressure independently. If we use Boyle's Law to calculate the pressure of each of the gases separately we have:

Pressure of gas A:

p1V1 = p2V2

p1 = 2.4 atm

V1 = 722 mL

V2 = 722 + 169 = 891 mL

p2 =?

Clearing p2:

p2 = (p1V1)/V2 = (2.4*722)/891 = 1.94 atm

Pressure of gas B:

p1 = 4.6 atm

V1 = 169 mL

V2 = 169+722 = 891 mL

p2=?

Clearing p:

p2 = (4.6*169)/891 = 0.87 atm

Dalton's expression for total partial pressures is equal to:

ptotal = pA + pB = 1.94+0.87 = 2.81 atm

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Answer:

Laws governing gas behavior.

Explanation:

Boyle's law:

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According to this law:

"The volume of a fixed amount of gas at constant temperature is inversely proportional to its pressure".

P  \alpha   V.

Charle's law:

It relates the volume and absolute temperature of an ideal gas at a constant pressure.

According to this law:

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According to this law:

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R is called universal gas constant.

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Answer:

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Explanation:

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In this scenario, we can infer and logically deduce that a statement from the laboratory report which best represents a conclusion for this investigation is that the group of students who spend less time studying after school would get lower math grade.

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