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:
The standard enthalpy of combustion of solid urea ((CO(NH2)2) is -632 kJ mol-1 at 298 K and its standard molar entropy is 104.60 J K-1 mol-1
Explanation:
Answer:
hola soy jess, tu respuesta esta aqui
¿cuantos moles de CO2 se requiere para reaccionar 2 moles de Ba(OH)2
2 mol Ba(OH)₂ × \frac{1molCO_{2} }{1molBa (OH)_{2}}
1molBa(OH)
2
1molCO
2
= 2 moles CO₂
Explanation:
espero que pueda ayudarte
hermana/hermano
lo que
hahahaha
B the atomic number is 20 and then you subtract atomic mass to find the neutrons
The wording of your question doesn't quite make sense, but a mole of an element has the same mass in grams as a single atom of that element has in amu. The mole is defined as 6.02 x10^22 things, whether they be atoms or molecules or even moles! 6.02x10^22 atoms of carbon has a mass of 12.01 g, and a single atom of carbon has a mass of 12.01 amu. Hope this helps!
