I think the answer is:
B. Chemical Change.
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.
Explanation:
Given problem:
Find the molar mass of:
SO₃ and C₁₀H₈
Solution:
The molar mass of a compound is the mass in grams of one mole of the substance.
To solve this, we are going to add the individual atomic masses of the elements in the compound;
Atomic mass;
S = 32g/mol; O = 16g/mol; C = 12g/mol and H = 1g/mol
For SO₃;
= 32 + 3(16)
= 32 + 48
= 80g/mol
For C₁₀H₈
= 10(12) + 8(1)
= 120 + 8
= 128g/mol
1.)b
2.)true
3.)false
are the answer don't take me on my word
