A gas at a temperature of 300 K occupies a 5 L flexible container at a pressure of 150 kPa. What volume will it occupy at STP?

6.73 L

This one requires the combined gas law.

P1V1/T1 = P2V2/T2

P1 = 150 kPa = 1.48 atm

V1 = 5 L

T1 = 300 K

P2 = 1 atm

V2 = This is what we’re solving for.

T2 = 273 K

P2 and T2 are based on the conditions of STP. I also changed P1 to atm to keep the units consistent.

Solve the equation for V2:

V2 = P1V1T2 / T1P2

V2 = (1.48 atm)(5 L)(273 K) / (300 K)(1 atm) = 6.73 L

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.