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Wewaii [24]
3 years ago
11

Hydrogen gas H2 is the least dense of all gases. A sample of H2 gas is found to occupy a volume of 1.23 L at 755 mmHg and 0 ◦C.

What volume , in liters, will this same gas occupy at .97 atm and a temperature of 50 ◦C.(10 points)
Please include equation used, variable you solved for and answer with correct units
Chemistry
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

V₂ = 1.48 L

Explanation:

Given data:

Initial volume = 1.23 L

Initial pressure = 755 mmHg (755/760 =0.99 atm)

Initial temperature = 0 °C (0 +273 = 273 K)

Final temperature = 50°C (50+273 = 323 K)

Final volume =  ?

Final pressure = 0.97 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 0.99 atm × 1.23 L × 323 K / 273 K × 0.97 atm

V₂ = 393.32 atm .L. K / 264.81 atm.K

V₂ = 1.48 L

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The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.

Given that the total pressure of the mixture of gases containing oxygen and water vapor = 749 Torr

Vapor pressure of pure water at 26.4^{0}C=25.81mmHg

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According to Dalton's law of partial pressures,

Total pressure = Partial pressure of Oxygen gas + Partial pressure of water

  749 Torr = Partial pressure of Oxygen gas + 25.81 Torr

Partial pressure of Oxygen gas = 749 Torr - 25.81 Torr =  723.19 Torr

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6 0
4 years ago
One gram is equal to ____ nanograms?
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Answer:

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notka56 [123]

Answer:

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mRT

​  

 

where m is the mass and M is the molar mass.

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​  

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3 years ago
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Answer:

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is the correct answer

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