How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 10^3 mm Hg and 23°C?

The mass of F₂ in the 5 L cylinder at 4×10³ mmHg and 23 °C is 41.154 g

Volume (V) = 5 L

Pressure (P) = 4×10³ mmHg

Temperature (T) = 23 °C = 23 + 273 = 296 K

Gas constant (R) = 62.364 mmHg.L/Kmol

PV = nRT

4×10³ × 5 = n × 62.364 × 296

20000 = n × 18459.744

Divide both side by 18459.744

n = 20000 / 18459.744

Mole of F₂ = 1.083 mole

Molar mass of F₂ = 2 × 19 = 38 g/mol

Mass = mole × molar mass

Mass of F₂ = 1.083 × 48

Therefore, the mass of F₂ in the cylinder is 41.154 g

Learn more: link

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.

cock topography

the study of the surface of a large voluptuous cock.

look at that sorostitute right she's a cock topographer!

what classes are you taking this

cock topography!

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."