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anzhelika [568]
3 years ago
10

a sample of hydrogen gas (h2) is mixed with water vapor (h2o (g)). the make sure has a total pressure of 811 torr, and the water

vapor has a partial pressure of 12 torr. how many moles of hydrogen gas are present in a 10.0 l mixture at 298k?
Chemistry
1 answer:
Alborosie3 years ago
6 0

Answer:

n=0.430molH_2

Explanation:

Hello!

In this case, considering the partial Dalton's law of partial pressures, we can notice that the total pressure equals the pressure of steam and the pressure of hydrogen, which can be determined as shown below:

p_T=p_H+p_w\\\\p_H=811torr-12torr=799torr*\frac{1atm}{760torr}\\\\p_H=1.05atm

Thus, by using the ideal gas law, we can compute the moles of hydrogen as shown below:

PV=nRT\\\\n= \frac{PV}{RT}=\frac{1.05atm*10.0L}{0.082\frac{atm*L}{mol*K}*298K}\\\\n=0.430molH_2

Best regards!

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Answer:

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Explanation:

3 0
2 years ago
A 3.42 gram sample of an unknown gas is found to occupy a volume of 1.90 L at a pressure of 547 mm Hg and a temperature of 33 °C
liq [111]
Use the Ideal Gas Law to find the moles of gas first.

Be sure to convert T from Celsius to Kelvin by adding 273.

Also I prefer to deal with pressure in atm rather than mmHg, so divide the pressure by 760 to get it in atm.

PV = nRT —> n = PV/RT
P = 547 mmHg = 547/760 atm = 0.720 atm
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T = 33°C = 33 + 273 K = 306 K
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n = (0.720 atm)(1.90 L) / (0.08206 L atm / mol K)(306 K) = 0.0545 mol of gas

Now divide grams by mol to get the molecular weight.

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Answer:

1.69.

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So, the solution should contain 3 significant figures.

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Hope This Helps You!

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