A 6.9 gram sample of a gaseous substance
occupies 10 L at 63�C and 537 torr. What is
the density of the gas under these conditions?
Answer in units of g/L
@Ms.Sue ok boomer
Refer to your problem below in density of a gas.
To find the density of the gas, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, we need to convert the given temperature from Celsius to Kelvin by adding 273.15:
63°C + 273.15 = 336.15 K
Next, we need to convert the given pressure from torr to atm by dividing by 760:
537 torr ÷ 760 torr/atm = 0.70658 atm
Now, we can rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
n = (0.70658 atm)(10 L) / (0.0821 L·atm/mol·K)(336.15 K)
n ≈ 0.245 mol
Now, we can calculate the density by dividing the mass (given as 6.9 grams) by the volume:
Density = mass / volume
Density = 6.9 g / 10 L
Density ≈ 0.69 g/L
Therefore, the density of the gas under these conditions is approximately 0.69 g/L.
Your 13 questions that show no work or effort on your part make me wonder why on earth you're taking AP chemistry.
We call it homework dumping.