Calculate the volume of the H2 gas obtained when the water vapor is removed from 289 mL of H2 gas collected by displacing water from a flask at 15.0°C and 0.988 atm.
Added by Sarah L.
Step 1
0°C and 0.988 atm. PV = nRT Where: P = pressure (0.988 atm) V = volume (289 mL = 0.289 L) n = number of moles R = ideal gas constant (0.0821 L.atm/mol.K) T = temperature in Kelvin (15.0°C = 15.0 + 273.15 = 288.15 K) Substitute the values into the Show more…
Show all steps
Your feedback will help us improve your experience
Ronald Prasad and 91 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
How many moles of H2 are contained in a balloon that contains hydrogen and water vapor (wet gas), has a volume of 28.3 cm3 at 25.0oC and a total pressure of 1.001 atm?
Ronald P.
When H2 gas was released by the reaction of HCl with Zn, the volume of H2 collected was 75.4 mL at 23 °C and 748 mmHg. What is the volume of the H2 at 0 °C and 1.00 atm pressure (STP)?
Madhur L.
When H2 gas was released by the reaction of HCl with Zn , the volume of H2 collected was 69.7 mL at 34 ∘C and 742 mmHg .What is the volume of the H2 at 0 ∘C and 1.00 atm pressure (STP)?
Chareen G.
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD