Answer:could u expand on the question
Explanation:
Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen 
if it occupies 31.6 mL at 
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.
Temperature = 
As molar mass of 
is 32 g/mol. Hence, the number of moles of are calculated as follows.
Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.
Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen if it occupies 31.6 mL at .
.50 M KCl because 5% is the same as .05, which makes the .50M more concentrated.
Answer: 83.11 torr
Explanation:
According to Dalton's Law of partial pressure, the total pressure of a mixture of gases is the sum of the pressure of each individual gas.
i.e Ptotal = P1 + P2 + P3 + .......
In this case,
Ptotal = 384 torr
P1 = 289 torr
P2 = 11.89 torr
P3 = ? (let the partial pressure of the remaining gas be Z)
Ptotal = P1 + P2 + Z
384 torr = 289 torr + 11.89 torr + Z
384 torr = 300.89 torr + Z
Z = 384 torr - 300.89 torr
Z = 83.11 torr
Thus, the partial pressure of the remaining gas is 83.11 torr.
Answer:
The answer is supposed to be "Electron cloud" or "Electon".
