According to Grahams law the rate of effusion of a gas is inversely proportional to the square root of it's molecular weight. The rate of diffusion is the measure of rate at which two gases mix. From this law we can say that for the two gases carbon monoxide and carbon dioxide, the rate of effusion of carbon monoxide is greater than that of carbon dioxide, this is because carbon monoxide is lighter (28 g) compared to carbon dioxide (44 g).
Explanation:
attribute of a person that often cannot be measured directly but can be assessed using numbers of indicators or manifest variables
Answer:
c or b
Explanation:
the story is entertaining and it shares the teachers personal experiences with the teachers students
Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
