The only products will be carbon dioxide and water.
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Answer: Wind, in climatology, the movement of air relative to the surface of the Earth. Winds play a significant role in determining and controlling climate and weather.
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:
a. NH₃ : base
CH₃COOH (acetic acid) : acid
NH₄⁺ : conjugate acid
CH₃COO⁻ : conjugate base
b. HClO₄ (perchloric acid) : acid
NH₃ : base
ClO₄⁻ : conjugate base
NH₄⁺ : conjugate acid
Hope this helps.
Answer:
so the answer that you get isn't wrong? i dont know
