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.
Oil molecules like to stick to other oil molecules more than they like to stick to water molecules. This is because mineral oil has nonpolar molecules and water has polar ones.
<u>Answer:</u> The concentration of the solution is 0.25 M
<u>Explanation:</u>
Let the volume of solution of 2.5 M NaCl be 10 mL
We are given:
Dilution ratio = 1 : 10
So, the solution prepared will have a volume of =
To calculate the molarity of the diluted solution, we use the equation:
where,
are the molarity and volume of the concentrated NaCl solution
are the molarity and volume of diluted NaCl solution
We are given:
Putting values in above equation, we get:
Hence, the concentration of the solution is 0.25 M
Answer:
renewable resources are abundant???