Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

4. A helium gas tank has a pressure of 81.6 atm at 25°C. The volume of the tank

ID: 992616 • Letter: 4

Question

4. A helium gas tank has a pressure of 81.6 atm at 25°C. The volume of the tank is 29.5L. How many moles of helium are in the tank?

5. If all of the helium was released into a large plastic balloon at a constant temperature and the pressure of the gas drops to 6.00 atm, what would be the volume of the balloon after it expands?

6. A car tire at has a pressure of 2.5 atm at 30°C. What would the tire pressure be at 20°C, assuming the volume of the tire and the amount of air within stays constant?

7. What is the density of oxygen (O2) gas at 25°C and 1 atm? Questions 8 & 9 focus on the use of baking soda in breads.

8. The release of CO2 from the decomposition of baking soda causes breads to rise. 2NaHCO3 (s) Na2CO3(s) + H2O (l) + CO2(g) If the recipe uses 5.00g of baking soda, what number of moles (n = ?) of CO2 will be formed?

9. If the oven is at 200°C and the pressure is 1 atm, what is the volume of CO2 released?

10. A lab studying air pollution made a mix of 2.00g nitrous oxide (N2O), 3.00g sulfur dioxide (SO2), and 1.00g carbon monoxide (CO) in a sealed container. If the mixture has a pressure of 0.22 atm, what is the partial pressure of sulfur dioxide?

Explanation / Answer

4)

P = 81.6 atm

T = 25 + 273 = 298 K

V = 29.5 L

P V = n R T

81.6 x 29.5 = n x 0.0821 x 298

n = 98.4

moles = 98.4

5)

P1 = 81.6 atm

P2 = 6.00 atm

V1 = 29.5 L

V2 = ?

P 1 V1 = P2 V2

81.6 x 29.5 = 6.00 x V2

V2 = 401.2 L

volume = 401.2 L

6)

P1 = 2.5 atm

T1 = 30 + 273 = 303 K

T2 = 20 +273 = 293 K

P2 = ?

P1 / T1 = P2 / T2

2.5 / 303 = P2 / 293

P2 = 2.42 atm

tire pressure = 2.42 atm

7)

P V = n R T

P = d RT/M

d = P M / R T

d = 1 x 32 / 0.0821 x 298

d = 1.31 g/ L

density of O2 = 1.31 g/L

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote