Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
Answer:
Increasing Surface Area
Explanation:
A greater surface area (meaning more, smaller particles) allows for more opportunity for particles to collide. On the other hand, decreasing temperature and removing a catalyst would only decrease the number of collisions, and the clumping option doesn't make much sense. Hope this helps!
Answer:
Explanation:
The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
Data:
p₁ =5.7 atm; T₁ = 100.0 °C
p₂ = ?; T₂ = 20.0 °C
Calculations:
1. Convert the temperatures to kelvins
T₁ = (100.0 + 273.15) K = 373.15
T₂ = (20.0 + 273.15) K = 293.15
2. Calculate the new pressure
Answer: Option (a) is the correct answer.
Explanation:
A miscible solution is defined as the one in which two or more than two components are soluble with each other.
So, when hexane is added to heptane then both of them will mix with each other and hence they are miscible. This mixing will lead to formation of more number of ions into the solution as a result, more will be the disorder present in the solution.
As entropy is the degree of randomness present.
Hence, we can conclude that a solution of hexane and heptane will lead to an increase in entropy.
