Question

A mixture of 6.00 g O2(g) and 9.00 g CH4(g) is placed in a 15.0-L vessel at 0°C. What is the partial pressure of each gas, and what is the total pressure in the vessel? A) PO2 = 0.281 atm; PCH4 = 0.841 atm; PT = 1.122 atm B) PO2 = 0.281 atm; PCH4 = 0.841 atm; PT = 1.12 atm C) PO2 = 0.281 atm; PCH4 = 0.80 atm; PT = 1.122 atm D) PO2 = 0.290 atm; PCH4 = 0.841 atm; PT = 1.122 atm

Name: a mixture of 600 g o2g and 900 g ch4g is placed in a 150 l vessel at 0c what is the partial pressure of each gas and what is the total pressure in the vessel a po2 0281 atm pch4 0841 atm pt 04247
Uploaded: 2022-04-28T12:24:42-08:00
Duration: 3 min 56 s
Channel: Sri K
Description: a mixture of 600 g o2g and 900 g ch4g is placed in a 150 l vessel at 0c what is the partial pressure of each gas and what is the total pressure in the vessel a po2 0281 atm pch4 0841 atm pt 04247

          A mixture of 6.00 g O2(g) and 9.00 g CH4(g) is placed in a 15.0-L vessel at 0°C. What is the partial pressure of each gas, and what is the total pressure in the vessel? 
A) PO2 = 0.281 atm; PCH4 = 0.841 atm; PT = 1.122 atm 
B) PO2 = 0.281 atm; PCH4 = 0.841 atm; PT = 1.12 atm 
C) PO2 = 0.281 atm; PCH4 = 0.80 atm; PT = 1.122 atm 
D) PO2 = 0.290 atm; PCH4 = 0.841 atm; PT = 1.122 atm

Added by Tara T.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Sri K

04/28/2022

Step 1

Moles of O2 = 6.00 g / 32.00 g/mol = 0.1875 mol Moles of CH4 = 9.00 g / 16.04 g/mol = 0.5617 mol Show more…

Show all steps

Thanks for your feedback!

A mixture of 6.00 g O2(g) and 9.00 g CH4(g) is placed in a 15.0-L vessel at 0°C. What is the partial pressure of each gas, and what is the total pressure in the vessel? A) PO2 = 0.281 atm; PCH4 = 0.841 atm; PT = 1.122 atm B) PO2 = 0.281 atm; PCH4 = 0.841 atm; PT = 1.12 atm C) PO2 = 0.281 atm; PCH4 = 0.80 atm; PT = 1.122 atm D) PO2 = 0.290 atm; PCH4 = 0.841 atm; PT = 1.122 atm