Answer:
1.7 bar
Explanation:
We can use the <em>Ideal Gas Law</em> to calculate the individual gas pressure.
pV = nRT Divide both sides by V
p = (nRT)/V
Data: n = 1.7 × 10⁶ mol
R = 0.083 14 bar·L·K⁻¹mol⁻¹
T = 22 °C
V = 2.5 × 10⁷ L
Calculations:
(a) <em>Change the temperature to kelvins
</em>
T = (22 + 273.15) K
= 295.15 K
(b) Calculate the pressure
p = (1.7 × 10⁶ × 0.083 14 × 295.15)/(2.5× 10⁷)
= 1.7 bar
I would believe it to be 55g. A -> B YIELDS AB. So, 10g + 45g = 55g.
Answer:
0.047 %
Explanation:
Step 1: Given data
- Partial pressure of ozone (pO₃): 0.33 torr
- Total pressure of air (P): 695 torr
Step 2: Calculate the %v/v of ozone in the air
Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.
<em>%v/v = pO₃/P × 100%</em>
%v/v = 0.33 torr/695 torr × 100%
%v/v = 0.047 %
About 255.3 grams is the answer I believe. I just had a school work packet and had that question on it, and that is the answer that I had put on it I believe.