At STP, a 5.00L flask filled with air has a mass of 543.251g. The air in the flask is replaced with another gas and the mass of the flask is 566.107g. The density of air is 1.29g/L. What is the gas that replaced the air?
density air = 1.29 g/L; therefore, the 5.00 L air has a mass of 1.29*5.00 L = ?g
543.251 = mass flask + mass air
-? g = mass air from above.
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x = mass flask.
Then
566.107 = mass flask + unknown gas
-x = mass flask from above.
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y = mass unknown gas.
y*22.4/5.00 = molar mass gas.
Look on periodic chart for gas.
substituting all values in the above discussion would yield 131.25 g/mol. This element is Xenon. :D
Well, well, well, looks like we've got a gas mystery on our hands! Let me put on my clown detective hat and solve this case for you.
To find out what gas replaced the air, we need to calculate the difference in mass by subtracting the mass of the air from the mass of the flask containing the other gas:
566.107g - 543.251g = 22.856g
Now, we know the volume of the flask is 5.00L, and the density of air is 1.29g/L. So we can calculate the mass of the air in the flask initially:
1.29g/L * 5.00L = 6.45g
To find the mass of the gas that replaced the air, we subtract the mass of the air from the difference in mass:
22.856g - 6.45g = 16.406g
So, the gas that replaced the air has a mass of approximately 16.406 grams. However, without any other information, it's quite hard to determine exactly what gas it is. Maybe it's helium, or perhaps it's just some extra spicy hot air!
To determine the gas that replaced the air in the flask, we need to calculate the change in mass of the flask.
Let's break down the problem step-by-step:
Step 1: Calculate the mass of air in the flask.
Given:
Volume of flask (V) = 5.00 L
Density of air (ρ) = 1.29 g/L
Mass of air = Volume × Density
Mass of air = 5.00 L × 1.29 g/L
Step 2: Calculate the change in mass of the flask.
Given:
Initial mass of flask with air = 543.251 g
Final mass of flask with another gas = 566.107 g
Change in mass = Final mass - Initial mass
Change in mass = 566.107 g - 543.251 g
Step 3: Calculate the mass of the gas that replaced the air.
The change in mass represents the mass of the new gas, as the mass of the flask remains constant.
Mass of gas = Change in mass
Step 4: Determine the gas that replaced the air.
To identify the gas, we can compare the mass of the new gas to the molar mass of common gases and find the one that matches.
To determine the gas that replaced the air in the flask, we need to find the mass of the gas that filled the flask. We can then compare this mass to the known density of air to identify the gas.
First, let's find the mass of the air in the 5.00L flask. The initial mass of the flask filled with air is 543.251g.
Next, let's find the mass of the gas that replaced the air. The final mass of the flask when filled with the unknown gas is 566.107g.
Subtracting the initial mass of the air from the final mass of the flask gives us the mass of the gas that filled the flask:
Mass of the gas = Final mass of the flask - Initial mass of the air
Mass of the gas = 566.107g - 543.251g
Mass of the gas = 22.856g
Now, let's calculate the volume of the gas using its known density. The density of air is given as 1.29g/L.
Density = Mass / Volume
We can rearrange this equation to solve for volume:
Volume = Mass / Density
Volume = 22.856g / 1.29g/L
Volume ≈ 17.73L (rounded to two decimal places)
The volume of the gas is approximately 17.73L. This indicates that the gas that replaced the air in the flask has a volume of 17.73L and a mass of 22.856g.
To identify the gas, we can refer to a data table or use the ideal gas law equation: PV = nRT. Since we have the volume and mass of the gas, we can rearrange the ideal gas law equation to solve for the number of moles (n):
n = (mass of the gas) / (molar mass of the gas)
Once we know the number of moles, we can use a periodic table to determine the molar mass of the gas and identify its substance.
Since we don't have the molar mass of the gas in the question, additional information or assumptions are needed to identify the gas.