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 .
V₁ = initial Volume of the balloon after it is blown up = 365 L
V₂ = new Volume of the balloon after it is taken outside = ?
T₁ = initial temperature of the balloon = 283 K
T₂ = new temperature of the balloon = 300 K
using the equation
V₁/V₂ = T₁/T₂
365/V₂ = 283/300
V₂ = 387 L
Answer:
38.75 L
Explanation:
From the question,
Applying Boyles Law,
PV = P'V'....................... Equation 1
Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' = Final Volume of Argon gas.
make V the subject of the equation
V = P'V'/P.................... Equation 2
Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.
Substitute these values into equation 2
V = (456×2.94)/34.6
V = 38.75 L
If you are referring to to the molecule, CO, it contains two atoms , C1+O1
Answer:
?
Explanation:
can you describe the question a little more please?
