Answer:
900 J/mol
Explanation:
Data provided:
Enthalpy of the pure liquid at 75° C = 100 J/mol
Enthalpy of the pure vapor at 75° C = 1000 J/mol
Now,
the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.
Thus, mathematically,
The heat of vaporization at 75° C
= Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C
on substituting the values, we get
The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol
or
The heat of vaporization at 75° C = 900 J/mol
Answer:
is reduced in the reaction
Explanation:
The given reaction is
The oxidation number of 
is changed from 
And The oxidation number of 
is changed from 
Hence, 
is oxidized and is reduced
Seven elements make up the compound
First, find out how many grams are in one mole of CO2(the two oxygen atoms means you need to multiply oxygen’s amu by 2,then add whatever carbon’s amu is to that). Then divide 26 grams by that number and that will be your moles. There are only two significant figures, so round your answer correctly.
