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Atmosphere Problem: Tightly capped 5 liter container of oxygenated water

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Assume that you have a tightly capped 5 liter container of oxygenated water. The temperature of the water is 25 degrees celcius, the oxygen partial pressure being 2.4 atmospheres . This water is degassed(outgassed) such that 96% of the oxygen is removed. How many moles of oxygen gas are removed from the sample? How many molecules? How many grams? How many drams? And, getting biblical, how many cubic cubits (STP)?

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Solution Preview

V = 5 lit = 5*10^(-3) m^3
T = 25+273 K = 298 K
P = 2.4 Atm.= 2.4*1.013*10^5 N/m^2
R = 8.31 J/mole/k
because,
PV = nRT
=>2.4*1.013*10^5*5*10^(-3) = n*8.31*298
=> n ...

Solution Summary

A tightly capped container carrying oxygenated water is analyzed. How many moles of oxygen gas were removed from the sample is found. With good explanations and computations, the problem is solved.

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