The stopcock connecting a 2.28 L bulb containing methane gas at a pressure of 9.

Question

The stopcock connecting a 2.28 L bulb containing methane gas at a pressure of 9.35 atm, and a 6.45 L bulb containing hydrogen gas at a pressure of 2.79 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the fnal pressure in the system is atm. A mixture of neon and krypton gases, in a 8.30 L flask at 70 degree C, contains 1.90 grams of neon and 17.3 grams of krypton. The partial pressure of krypton. In the flask is atm and the total pressure in the flask is atm. A mixture of krypton and neon gases is maintained in a 9.15 L flask at a pressure of 2.89 atm and a temperature of 82 degree C. If the gas mixture contains 40.7 grams of krypton, the number of grams of neon in the mixture is g.

Explanation / Answer

Question 3:

using the expression: PV = nRT we have the following, we solve for n:
n = PV/RT
n = 2.89 * 9.15 / 0.0821 * (82+273) = 0.8589 moles

now, the moles of Krypton:
moles = 40.7 / 83.8 = 0.4857 moles

moles of Ne = 0.8589 - 0.4857 = 0.3732 moles
mass of Ne = 0.3732 * 20.18 = 7.5312 g

Question 2:

let's calculate the moles of the gases:
moles Ne = 1.9 / 20.18 = 0.0942 moles
moles Kr = 17.3 / 83.8 = 0.2064 moles
total moles = 0.0942+0.2064 = 0.3006 moles

Now, let's calculate the total pressure:
P = 0.3006 * 0.0821 * 293 / 8.3 = 0.8712 atm

Now, the partial pressure of Kr it's calculated as:
pKr = X * Pt

now the molar fraction of Kr, (X)
X = 0.2064 / 0.3006 = 0.6866

pKr = 0.6866 * 0.8712 = 0.5982 atm

Hope this helps

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.