Answer:
0.2 is conpound Co2 STP.
Explanation:
on combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical formula of compound.?
∆H ° rxn =-2855.56 kJ
<h3>Further explanation</h3>
Given
ΔHf CO₂ = -393.5 kJ/mol
ΔHf H₂O = -241.82 kJ/mol
ΔHf C₂H₆ = - 84.68 kJ/mol
Reaction
2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g)
Required
ΔHrxn=
Solution
<em>∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants) </em>
∆H ° rxn = (4.-393.5+6.-241.82)-(2.-84.68)
∆H ° rxn = (-1574-1450.92)-(-169.36)
∆H ° rxn =-3024.92+169.36
∆H ° rxn =-2855.56 kJ
Cellular respiration occurs in animals.. what is it asking you
Answer:
Composition of the mixture:
%
%
Composition of the vapor mixture:
%
%
Explanation:
If the ideal solution model is assumed, and the vapor phase is modeled as an ideal gas, the vapor pressure of a binary mixture with 
and 
molar fractions can be calculated as:
Where 
and 
are the vapor pressures of the pure compounds. A substance boils when its vapor pressure is equal to the pressure under it is; so it boils when 
. When the pressure is 0.60 atm, the vapor pressure has to be the same if the mixture is boiling, so:
With the same assumptions, the vapor mixture may obey to the equation:
, where P is the total pressure and y is the fraction in the vapor phase, so:
%
The fractions of B can be calculated according to the fact that the sum of the molar fractions is equal to 1.
