X = (619kPa)(20.0L)/(8.314)(298K) X = 5.00 mol CO 5.00 mol CO (28.01 g/1mol) = 140.05 g

4
12.0 grams of CH 4 has a pressure of 2.5 atm and a temperature of 30 C. Calculate the volume of the gas.

description

12.0 grams of CH 4 has a pressure of 2.5 atm and a temperature of 30 C. Calculate the volume of the gas. 12.0 g CH 4 (1mole / 16.05 g) = .747 mole CH 4 (2.5atm)(x) = (.747 mole CH 4 )(.0821 L atm /mol K)(303K) X = 7.4 L CH 4. - PowerPoint PPT Presentation

Transcript of X = (619kPa)(20.0L)/(8.314)(298K) X = 5.00 mol CO 5.00 mol CO (28.01 g/1mol) = 140.05 g

Page 1: X = (619kPa)(20.0L)/(8.314)(298K) X = 5.00 mol CO 5.00 mol CO (28.01 g/1mol) = 140.05 g

12.0 grams of CH4 has a pressure of 2.5 atm and a temperature of 30 C. Calculate the volume of

the gas.

Page 2: X = (619kPa)(20.0L)/(8.314)(298K) X = 5.00 mol CO 5.00 mol CO (28.01 g/1mol) = 140.05 g

12.0 g CH4 (1mole / 16.05 g) = .747 mole CH4

(2.5atm)(x) = (.747 mole CH4)(.0821 L atm /mol K)(303K)

X = 7.4 L CH4

Page 3: X = (619kPa)(20.0L)/(8.314)(298K) X = 5.00 mol CO 5.00 mol CO (28.01 g/1mol) = 140.05 g

A sample of carbon monoxide has a volume of 20.0 L at 25 C

and 619 kPa. Calculate the mass of the carbon monoxide gas.

Page 4: X = (619kPa)(20.0L)/(8.314)(298K) X = 5.00 mol CO 5.00 mol CO (28.01 g/1mol) = 140.05 g

X = (619kPa)(20.0L)/(8.314)(298K)

X = 5.00 mol CO

5.00 mol CO (28.01 g/1mol) = 140.05 g


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