The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.
To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.
From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.
1) From pV = nRT, n = pV / RT
Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K
n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol * 308.15K] =0.00472 mol
mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042
=> MM = mass/n = 0.042 / 0.00472 = 8.90 g/mol
Now from a periodic table or a table you get that the molar mass of He is 4g/mol
So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.
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Answer:
34.28 L ( 1.5*22.4 L)
Explanation:
Calculation of the moles of aluminum as:-
Mass = 55 g
Molar mass of aluminum = 26.981539 g/mol
The formula for the calculation of moles is shown below:
Thus,
According to the reaction:-
4 moles of aluminum react with 3 moles of oxygen gas
1 mole of aluminum react with 
moles of oxygen gas
2.0384 moles of aluminum react with 
moles of oxygen gas
Moles of oxygen gas = 1.5288 moles
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K
⇒V = 34.28 L ( 1.5*22.4 L)
Answer:
44.8 litres is the answer to the question
