What is the volume of 19 grams of flourine gas at stp?
The molecular mass of Fluorine (note the spelling please, Fluorine is not Flourine, F2, is 36. So 19 grams is
moles=19/36
volume at stp=22.4liters*19/36
To calculate the volume of a gas at STP (Standard Temperature and Pressure), you can use the ideal gas law equation: PV = nRT.
Where:
P = pressure (usually given as 1 atm at STP)
V = volume
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (usually given as 273 K at STP)
First, we need to find the number of moles of fluorine gas. We can use the molar mass of fluorine (F2 = 2 x 19.00 g/mol) to convert grams to moles:
n = mass / molar mass
n = 19 g / (2 x 19.00 g/mol)
n = 0.5 mol
Now, we can solve for the volume:
PV = nRT
V = nRT / P
V = (0.5 mol)(0.0821 L·atm/(mol·K))(273 K) / 1 atm
V ≈ 11.24 L
Therefore, the volume of 19 grams of fluorine gas at STP is approximately 11.24 liters.
To calculate the volume of a gas at STP (Standard Temperature and Pressure), you need to know the moles of the gas. Here's how you can calculate it step by step:
1. Convert the given mass of fluorine gas to moles using the molar mass of fluorine. The molar mass of fluorine (F) is 19 grams per mole. So, divide the mass (19 grams) by the molar mass (19 g/mol):
Moles = Mass / Molar mass
= 19 g / 19 g/mol
= 1 mole
2. At STP, 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume of 1 mole of fluorine gas at STP is 22.4 liters.
3. Since you have 1 mole of fluorine gas, the volume will be the same, which is 22.4 liters.
So, the volume of 19 grams of fluorine gas at STP is 22.4 liters.