Answer:
standard entropy of vaporization of ethanol = 142.105 J/K-mol
Explanation:
given data
enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × 
J/mol
entropy of vaporization of ethanol boiling point = 285 K
to find out
standard entropy of vaporization of ethanol
solution
we get here standard entropy of vaporization of ethanol that is expess as
standard entropy of vaporization of ethanol ΔS = 
.............1
here ΔH is enthalpy of vaporization of ethanol and T is temperature
put value in equation 1
standard entropy of vaporization of ethanol ΔS = 
standard entropy of vaporization of ethanol = 142.105 J/K-mol
Answer: (i) F = 2
(ii) F = 3
(iii) F = 2
Explanation:
We would be applying the famous Gibbs Phase Rule to explaining this problem;
By applying the formula;
F+P = C +2
Where P = this represent the phase
F = this is called the degree of freedom
C = this represent the component in the system
Ok let us begin;
(i). from this we can see that there are 2 components i.e. (water + ethanol) and the phase in question is a vapor phase + liquid phase.
So from the formula;
F = C-P+2
F = 2 – 2 + 2 = 2
Therefore, F = 2.
(ii). Also, from the statement, we can figure there are 3 components, while the phases are two like the previous one above, i.e. liquid + vapor
F = 3 – 2 + 2 = 5 – 2 = 3
F = 3
(iii). From this statement, we can figure there are 3 components, and the phases are 3 i.e. (2 liquid phases + 1 vapor phase)
From the formula;
F = 3 – 3 + 2 = 0 + 2
F = 2
Answer:
it would be an element because its an element
Explanation:
For a single atom, the charge is the number of protons minus the number of electrons
