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OverLord2011
2 months ago
13

A 226.4-l cylinder contains 65.5% he(g) and 34.5% kr(g) by mass at 27.0°c and 1.40 atm total pressure. what is the mass of he in

this container?
Chemistry
2 answers:
KiRa [2.9K]2 months ago
8 0
The helium (He) mass in the container is calculated to be 0.2 grams

Solution:

First, we need to determine the gas moles in the container using the ideal gas equation:

given that: P = 1.40 atm (pressure of the gas)

V = 226.4 L (volume) R = 0.0821 Latm/moleK (gas constant)

T = PV=nRT\\\\n=\frac{RT}{PV}

n =? (moles of gas)

Substituting these values into the equation provides the total moles of gas present.

Next, we will calculate the moles of helium.

Then the mass of helium in the container can be ascertained.

27^oC=273+27=300K

Thus, the mass of helium (He) in this container is 0.2 grams

Tems11 [2.7K]2 months ago
5 0
Initially, we calculate the moles of gas using the ideal gas law:

PV = nRT
n = PV / RT
n = (1.4 * 226.4) / (0.082 *(27 + 273.15))
n = 12.88

Next, we apply the given percentages to estimate the moles of helium:
Moles of helium = 0.655 * 12.88
Moles of helium = 8.44

We then use the formula:
Mass = moles * molar mass

Mass of helium = 8.44 * 4

Mass of helium = 33.76 grams.
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