Particles in a gas are far apart compared to a solid or liquid, allowing it not to have a definitive shape or volume. This also means that gases can fill any container and be easily compressed.
Grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline is 45.5g
Octane is a hydrocarbon which burns in gasoline in presence of oxygen according to the given balanced chemical equation,
2C₈H₁₈ + 25O₂------> 16CO₂ + 18H₂0
Molar mass of octane = 114.23g/mol
Molar mass of Oxygen = 32g/mol
According to the stiochiometry of the balanced equation the mole ratio of Octane and Oxygen is 2:25
2 mole of octane needs 25 mole of oxygen
1 mole of octane needs 12.5 moleof oxygen
114.23g of octane needs 400g of oxygen
13g of octane needs 45.5g of oxygen
Mass of oxygen needed =45.5g
Hence, the Mass of oxygen needed is 45.5g for the combustion of octane in gasoline.
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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.
That they both will be the same average kinetic energy
