B. The partial pressure of N2 is 101 kPa
<h3>Further explanation</h3>
Given
volume = 22.4 L
1.0 mol of nitrogen and 2.0 mol of hydrogen at 0°C
Required
Total pressure and partial pressure
Solution
Ideal gas law :
PV = nRT
n total = 3 mol
T = O °C + 273 = 273 K
P = nRT/V
P = 3 x 0.08205 x 273 / 22.4
P total = 3 atm = 303,975 kPa
P Nitrogen = 1/3 x 303.975 = 101.325 kPa
P Hydrogen = 2/3 x 303.975 = 202.65 kPa
Wind ,ice,and water eroded it over years
Answer:
The answer to your question is P2 = 0.78 atm
Explanation:
Data
Temperature 1 = T1 = 263°K Temperature 2 = T2 = 298°K
Volume 1 = V1 = 24 L Volume 2 = V2 = 35 L
Pressure 1 = P1 = 1 Pressure 2 = P2 = ?
Process
1.- To solve this problem use the Combined gas law
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
-Substitution
P2 = (1)(24)(298) / (263)(35)
-Simplification
P2 = 7152 / 9205
-Result
P2 = 0.777
or P2 = 0.78 atm
Potential energy and height; best guess;)
Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called <u>halohydrin formation</u>. This is a <u>markovnikov reaction</u> with <u>anti configuration</u>. Therefore the halogen in this case "Br" and the "OH" must have <u>different configurations</u>. Additionally, in this molecule both carbons have the <u>same substitution</u>, so the "OH" can go in any carbon.
Finally, in the product we will have <u>chiral carbons</u>, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
I hope it helps!