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galina1969 [7]
2 years ago
14

A balloon at sea level on earth (1 atm pressure, 19°C) takes up 14.5 L of space. The balloon travels to Mars where atmospheric p

ressure is 4.55 torr and the temperature is -55°C What is the volume of the balloon on Mars?
Chemistry
1 answer:
Firlakuza [10]2 years ago
3 0

Answer:

1807.24L

Explanation:

Using combined gas law equation:

P1V1/T1 = P2V2/T2

Where;

P1 = pressure on Earth

P2 = Pressure on Mars

V1 = volume on Earth

V2 = volume on Mars

T1 = temperature on Earth

T2 = temperature on Mars

According to the information provided of the balloon in this question;

P1 = 1 atm

P2 = 4.55 torr = 4.55/760 = 0.00599atm

V1 = 14.5L

V2 = ?

T1 = 19°C = 19 + 273 = 292K

T2 = -55°C = -55 + 273 = 218K

Using P1V1/T1 = P2V2/T2

1 × 14.5/292 = 0.00599 × V2/218

14.5/292 = 0.00599V2/218

Cross multiply

14.5 × 218 = 292 × 0.00599V2

3161 = 1.74908V2

V2 = 3161 ÷ 1.74908

V2 = 1807.24L

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A mixture of noble gases [helium (MW 4), argon (MW 40), krypton (MW 83.8), and xenon (MW 131.3)] is at a total pressure of 150 k
kodGreya [7K]

Answer:

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

Explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

  • Helium: 0,25 moles ×\frac{4 g}{1 mol} = 1 g of Helium
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Total grams: 1g+10g+20,85g+30,825g= 62,675 g

Mass fraction of helium: \frac{1 gHelium}{62,675 g} × 100 = <em>1,6%</em>

<em />

<em>The mass fraction of Helium is 1,6%</em>

<em />

<em>b)</em><em>  </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:

\frac{4+40+83,8+131,3}{4} = 64,775 g/mol

c) The molar concentration is possible to know ussing ideal gas law, thus:

\frac{P}{R.T} = M

Where:

P is pressure: 150 kPa

R is gas constant: 8,3145\frac{L.kPa}{K.mol}

T is temperature: 500 K

And M is molar concentration. Replacing:

M = 3,6×10⁻² M

d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:

3,6×10⁻² \frac{mol}{L} × \frac{64,775 g}{mol} × \frac{1L}{1000 mL} =

2,3×10⁻³ g/mL

I hope it helps!

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