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A person breathes in 6.0 L of pure oxygen at 298 K and
1,000 kPa to fill their lungs
How many moles of oxygen did they take in?
Use the ideal gas law: PV = nRT where R = 8.31 L – kPa /mol – K

A) 0.05 mole
B) 0.41 mole
C) 2.42 moles
D) 20.0 moles

Respuesta :

ardni313

Moles of Oxygen they took in : C. 2.42

Further explanation

The gas equation can be written  

[tex]\large{\boxed{\bold{PV=nRT}}}[/tex]

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

R = gas constant = 0.08205 l.atm / mol K  

T = temperature, Kelvin  

P=1000 kPa

V = 6 L

T = 298 K

moles of Oxygen :

[tex]\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1000\times 6}{8.31\times 298}\\\\n=2.42~moles[/tex]

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