Please help me out...
The pressure of a container of fluorine gas with a volume of 1.5L is 678mmHg at 55.0*C.
a. If the volume of the fluorine gas increases to 5.0L, what would be the pressure of the gas in atm? The amount and tempature of gas are consistant.
b.How many grams of fluorine gas are in the container?
a. P1V1 = P2V2. Solve for P2 and convert to atm. atm = mmHg/760
b. Use PV = nRT and solve for n = number of moles of fluorine. Then n = grams/molar mass. Solve for grams.
whats the force of electrical attraction between a proton and an electron that are 9.5x 10-11
To answer these questions, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Let's solve these questions step by step:
a. To find the pressure of the gas in atm when the volume increases to 5.0L:
Step 1: Convert the given pressure from mmHg to atm.
1 atm = 760 mmHg
678 mmHg / 760 mmHg/atm ≈ 0.892 atm
Step 2: Calculate the new pressure using the ideal gas law.
P1V1 = P2V2
P1 = 0.892 atm
V1 = 1.5 L
V2 = 5.0 L (the new volume we want to find)
P2 = What we'll solve for
Using the equation:
P2 = (P1 * V1) / V2
P2 = (0.892 atm * 1.5 L) / 5.0 L
P2 ≈ 0.267 atm
Therefore, the pressure of the gas in atm would be approximately 0.267 atm when the volume increases to 5.0L.
b. To find the number of grams of fluorine gas in the container:
Step 1: Convert the given temperature from Celsius to Kelvin.
T(K) = T(°C) + 273.15
T(K) = 55.0°C + 273.15 = 328.15 K
Step 2: Calculate the number of moles using the ideal gas law.
PV = nRT
P = 0.892 atm (from part a)
V = 1.5 L (given volume)
n = What we'll solve for
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 328.15 K
Using the equation:
n = PV / RT
n = (0.892 atm * 1.5 L) / (0.0821 L·atm/(mol·K) * 328.15 K)
n ≈ 0.0654 mol
Step 3: Find the molar mass of fluorine (F₂).
Using the periodic table, the molar mass of fluorine (F₂) is approximately 38.0 g/mol.
Step 4: Calculate the number of grams.
Number of grams = number of moles * molar mass
Number of grams = 0.0654 mol * 38.0 g/mol
Number of grams ≈ 2.48 g
Therefore, there would be approximately 2.48 grams of fluorine gas in the container.