Answer:
41.17g
Explanation:
We are given the following parameters for Flourine gas(F2).
Volume = 5.00L
Pressure = 4.00× 10³mmHG
Temperature =23°c
The formula we would be applying is Ideal gas law
PV = nRT
Step 1
We find the number of moles of Flourine gas present.
T = 23°C
Converting to Kelvin
= °C + 273k
= 23°C + 273k
= 296k
V = Volume = 5.00L
R = 0.08206L.atm/mol.K
P = Pressure (in atm)
In the question, the pressure is given as 4.00 × 10³mmHg
Converting to atm(atmosphere)
1 mmHg = 0.00131579atm
4.00 × 10³ =
Cross Multiply
4.00 × 10³ × 0.00131579atm
= 5.263159 atm
The formula for number of moles =
n = PV/RT
n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K
n = 1.0834112811moles
Step 2
We calculate the mass of Flourine gas
The molar mass of Flourine gas =
F2 = 19 × 2
= 38 g/mol
Mass of Flourine gas = Molar mass of Flourine gas × No of moles
Mass = 38g/mol × 1.0834112811moles
41.169628682grams
Approximately = 41.17 grams.
Answer:
Percent composition by element
Element Symbol # of Atoms
Hydrogen H 5
Carbon C 3
Nitrogen N 3
Oxygen O 9
The correct answer is c. Please give me brainlest let me know if it’s correct or not thanks bye
The standard ambient temperature and pressure are
Temperature =298 K
Pressure = 1atm
The density of gas is 1.5328 g/L
density = mass of gas per unit volume
the ideal gas equation is
PV = nRT
P = pressure = 1 atm
V = volume
n = moles
R= gas constant = 0.0821 Latm/mol K
T = 298 K
moles = mass / molar mass
so we can write
n/V = density / molar mass
Putting values



Thus molar mass of gas is 37.50g/mol
Answer:
When a mixture of methane and chlorine is exposed to ultraviolet light - typically sunlight - a substitution reaction occurs and the organic product is chloromethane. CH 4 + Cl 2 → CH 3 Cl + HCl However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms.
Explanation:
hope it help i try best