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: d) 164.9 g
Explanation:
To calculate the moles :
The balanced chemical reaction is:
According to stoichiometry :
2 moles of
require = 25 moles of
Thus 0.412 moles of
will require=
of
Mass of
Thus 164.9 g of oxygen is consumed.
Particle, Relative Mass<span>, Actual </span>Mass<span> (g) </span>Electron<span>- 1 , 9.11*10 power -28 g. ... Calculate: What is the diff </span>expressed in kg between the mass<span> of a </span>proton and the mass<span> ... Best Answer: </span>difference<span>in </span>mass<span> = [1.673 x 10^(-24) - 9.11 x ...</span>
Answer:
<u>= 14.24g of </u>
<u> is required.</u>
Explanation:
Reaction equation:
→ 
Mole ratio of ammonium carbonate to carbon dioxide is 1:1
1 mole of CO2 - 44g
?? mole of CO2 - 6.52g
= 6.52/44 = 0.148 moles was produced from this experiment.
Therefore, if 1 mole of
- 96.09 g
0.148 mol of
-- ?? g
=0.148 × 96.09
<u>= 14.24g of </u>
<u> is required.</u>