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.
They are called reactants.
That statement is true because to make a diluted solution, you have to add more solvent than the solute in a concentrated solution. This is extra information, but to make a diluted solution more concentrated, you have to add more solute since a concentrated solution has an excess amount of solute as stated before.
Answer:
Molar mass of MgCl2 is 95 g/mol
Mg = 24 g/mol and Cl = 35.5 ×2 = 71 g/mol
moles = mass given/ molar mass
= 2.7/95 = 0.028 mol
volume = 250/1000 = 0.25 dm3 (ml is the same as dm3)
molarity of MgCl2 = moles/volume
= 0.028/0.25
= 0.112 mol/dm3
Theyre the big bunched up group in the middle of the periodic table
