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:
This would support Dalton's postulates that proposed the atoms are indivisible because no small particles are involved.
Explanation:
Experiment using the gas discharge tube by J.J Thomson led to the discovery of cathode rays which are now known as electrons.
Primarily, Thomson's experiment led to the discovery of cathode rays, electrons, as subatomic particles.
If the size of the atoms observed at the cathode is the same as that of the rays,we can conclude that the particles of the rays are the simplest form of matter we can have. This would suggest that the atom is indeed the smallest indivisible particle of a matter according to Dalton.
Answer: The oxidation state of selenium in SeO3 is +6
Explanation:
SeO3 is the chemical formula for selenium trioxide.
- The oxidation state of SeO3 = 0 (since it is stable and with no charge)
- the oxidation number of oxygen (O) IN SeO3 is -2
- the oxidation state of selenium in SeO3 = Z (let unknown value be Z)
Hence, SeO3 = 0
Z + (-2 x 3) = 0
Z + (-6) = 0
Z - 6 = 0
Z = 0 + 6
Z = +6
Thus, the oxidation state of selenium in SeO3 is +6
273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit
