Im Having a really Hard time with this problem :(
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A sample containing 5.60g O2 gas has a volume of 31.0L . Pressure and temperature remain constant.
1 .) Oxygen is released until the volume is 8.00L . How many moles of O2 are removed?
PV = nRT; so if P, R and T are constant, that means V = n*k and n = grams/molar mass
You had 5.60/32 = 1.75 mols O2 to start. You end up with 8.00 which means you lost 31.0-8.00 = 23 L. How many mols is that? 1.75 x 23/31 = ?
To solve this problem, you need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
In this case, the pressure and temperature remain constant, so we can rewrite the equation as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
Let's plug in the given values:
P1 = P2 (since pressure remains constant)
V1 = 31.0 L
V2 = 8.00 L
Now we can solve for the unknown, which is the number of moles (n) of oxygen gas that is removed.
P1V1 = P2V2
nRT1 / V1 = nRT2 / V2
Since n and R are the same on both sides of the equation, we can cancel them out:
T1 / V1 = T2 / V2
Now we can solve for n (the number of moles of oxygen gas removed):
n = (T1 / V1) * V2
Given that pressure and temperature remain constant, you can assume that T1 / V1 = T2 / V2.
Therefore, the number of moles of oxygen gas removed can be calculated as:
n = V2 / V1
Let's plug in the values:
n = 8.00 L / 31.0 L
Now you can calculate this division to find the number of moles of O2 removed.