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:It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume.
Explanation:
However, pressure is commonly measured in one of three units: kPa, atm, or mmHg. Therefore, can have three different values.
Answer:They are small because they don't need larger bodies to enable them attach to the females. They don't have fully formed gut because when they reach adulthood their digestive system stops functioning.
Explanation:they then find a female angler fish, attach by bitting into her flesh and fusing to her body. So whatever she eats they eat too.
Answer:
The law of definite proportions. I had the same question for chemistry and this is what they said was right so I got 100%.
Explanation:
