If a 22.4 L volume of a sample of gas has a density of 0.900 grams/L at 1.00 atm and 0.00°C. Given
1 answer:
Answer:
Formula Weight of gas sample = 20.1 g/mole => Neon (Ne)
Explanation:
Use Ideal Gas Law formula to determine formula weight and compare to formula weights of answer choices.
PV = nRT = (mass/fwt)RT => fwt = (mass/Volume)RT = Density x R x T
Density = 0.900 grams/L
R = 0.08206 L·atm/mole·K
T = 0.00°C = 273Kfwt = (0.900g/L)(0.08206L·atm/mole·K )(273K)
= 20.1 g/mol => Neon (Ne)
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Solution here,
Volume(V)=67.4 L
Pressure(P)=1 atm
Temperature(T)=(0+273)K=273K
Universal gas constant(R)=0.0821 L.atm.mol^-1K^-1
No. of moles(n)=?
Now,
PV=nRT
or, 1×67.4=n×0.0821×273
or, 67.4=22.4n
or, n=67.4/22.4
or, n=3
therefore, required no. of mole is 3.
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Answer:
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Answer:
False
Explanation:
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Answer:it all above
Explanation:
it all above because all the answer are truth so it all above
