owLv2 | Online teaching chemistry question I che x C cvg.cengagenow.com/ilrn/tak
ID: 552944 • Letter: O
Question
owLv2 | Online teaching chemistry question I che x C cvg.cengagenow.com/ilrn/takeAssignment/takeCovalent Activity do?locator-assignment-take&takeAssignmentSessionLocator-assignment-take; : Apps : kober portal D 2004 Saturn L300 H: web chemucsbedu -Baltimore Classificat Acute Viral Hepatitis Hepatitis A Treatme Jaundice: MedlinePl Other bookmarks Time Remaining; 1:29:23 UNIT TEST Use the References to access important values if needed for this question. 1 pt 1 pr 1 pt 1 pt 1 pt 1 pt 1 pr 1 pt 1 pt 1 pt 1 pt 1 pr 1 pt 1 pt 1 pt 1 pt Question 3 Question 4 Question 5 Question 6 Question 7 A mixture of oxygen and methane gases, in a 6.24 L flask at 1S °C, contains 7.09 grams of oxygen and 3.49 grams of methane. The partial pressure of methane in the flask is atm and the total pressure in the flask is atm Submit Answer Try Another Version 1 item attempt remaining Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Question 16 Question 17 Progress: 0/25 items Due Nov 2 at Previous Next 11:00 PM '.il &54 PM 11/2/2017Explanation / Answer
Solution
(7.09 g Oxygen) / (15.999 g O2/mol) = 0.4431 mol Oxygen
(3.49 g Methane) / (16.04 g Ch4/mol) = 0.2176 mol Ch4
Now using formula-
P = nRT / V
putting the values in the formula
= (0.4431 mol + 0.2176 mol) x (0.08205746 Latm/Kmol) x (18 + 273)K / (6.24 L)
= 2.53 atm total
(2.53 atm) x (0.2176 mol Ch4) / (0.4431 mol + 0.2176 mol) = 0.833 atm Ch4
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