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
The answer is 200 J
The kinetic energy (KE) is:
KE = 1/2 m * v²
m - mass of the object
v - velocity of the object
We have:
m = 4 kg
v = 10 m/s
KE = 1/2 * 4 kg * 10² m/s
KE = 200 J
I found these four statements for that question:
Each molecule contains four different elements.
Each molecule contains three atoms.
Each molecule contains seven different bonds.
Each molecule contains six oxygen atoms.
The last one is true. Each molecule contains six oxygen atoms.
The number to the right of O and of (NO3) ares subscripts.
The chemical formula uses subscripts to indicate the number of atoms.
The subscript 2 in (NO3)2 means that there are two NO3 radicals.
And the subscript 3 to the right of O means that each NO3 radical has three atoms of O.
Then, the number of atoms of O is 2 * 3 = 6.
So, the true statement is the last one: each molecule of Ba (NO3)2 has six atoms of O.
From that molecule you can also tell:
- Each molecule contains one atom of barium
- Each molecule contains two atoms of nitrogen
- Each molecule contains two NO3 radicals
Answer:
B
Explanation:
I am not sure but I think it is B. Out of all the answers that one makes sense.
