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:
C. Shield Volcano
Explanation:
Flow after flow pours out in all directions from a central summit vent, or group of vents, building a broad, gently sloping cone of flat, domical shape, with a profile much like that of a warrior's shield.
Answer:
V₂ = 648.53 mL
Explanation:
Given data:
Initial volume of gas = 490. mL
Initial temperature = -35°C (-35 + 273 = 238 k)
Final temperature = 42°C = (42+273 = 315 k)
Final volume = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 490 mL × 315 K / 238 k
V₂ = 154350 mL.K / 238 K
V₂ = 648.53 mL
The second volume : V₂= 0.922 L
<h3>
Further explanation
</h3><h3>Given
</h3>
7.03 Liters at 31 C and 111 Torr
Required
The second volume
Solution
T₁ = 31 + 273 = 304 K
P₁ = 111 torr = 0,146 atm
V₁ = 7.03 L
At STP :
P₂ = 1 atm
T₂ = 273 K
Use combine gas law :
P₁V₁/T₁ = P₂V₂/T₂
Input the value :
0.146 x 7.03 / 304 = 1 x V₂/273
V₂= 0.922 L
Answer:
<h2>ignore your body being cold</h2>
Hope it helps
