False, they are all different because they help you know different things.
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.
We will use Arrehenius equation
lnK = lnA -( Ea / RT)
R = gas constant = 8.314 J / mol K
T = temperature = 25 C = 298 K
A = frequency factor
ln A = ln (1.5×10 ^11) = 25.73
Ea = activation energy = 56.9 kj/mol = 56900 J / mol
lnK = 25.73 - (56900 / 8.314 X 298) = 2.76
Taking antilog
K = 15.8
1. electrostatic interactions
<span>3. van de waals interactions </span>
<span>4. hydrogen bonding </span>
