The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
<h3>What is Enthalpy of Vaporization ?</h3>
The amount of enthalpy or energy that must be added to a liquid substance into gas substance is called Enthalpy of Vaporization. It is also known as Latent heat of vaporization.
<h3>How to find the energy change from enthalpy of vaporization ?</h3>
To calculate the energy use this expression:

where,
Q = Energy change
n = number of moles
= Molar enthalpy of vaporization
Now find the number of moles
Number of moles (n) = 
= 
= 0.5 mol
Now put the values in above formula we get
[Negative sign is used because Br₂ condensed here]
= - (0.5 mol × 15.4 kJ/mol)
= - 7.7 kJ
Thus from the above conclusion we can say that The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
Learn more about the Enthalpy of Vaporization here: brainly.com/question/13776849
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Hi! Chemistry is complicated, right?
A hydroxyl group is a group that consists of
oxygen and hydrogen bonds. When they combine, ethanol may be released.
I hope I helped!
Answer:
A. Theoretical yield = 3.51g
B. %yield = 75%
Explanation:
The balanced equation for the reaction is given below:
C2H4 + 3O2 —> 2CO2 + 2H2O
Molar Mass of O2 = 16 x 2 = 32g/mol
Mass of O2 from the balanced equation = 3 x 32 = 96
Molar Mass of H20 = (2x1) + 16 = 18g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36g
A. From the equation,
96g of O2 produced 36g of H2O
Therefore, 9.35g of O2 will produce = (9.35 x 36)/96 = 3.51g of H2O
Therefore,theoretical yield of water (H2O) = 3.51g
B. Theoretical yield = 3.51
Actual yield = 2.63g
%yield =?
%yield = Actual yield/Theoretical yield x 100
%yield = 2.63/3.51
%yield = 75%