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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Answer:
He had the most potential energy going down hill
Explanation:
because he picked up he went faster
Answer:
False
Explanation:
Not all of it but some of it. Hope this helps! (ノ◕ヮ◕)ノ*:・゚✧
Answer:
650 mmol.
Explanation:
The equation for the fermentation of one mole of glucose is:
C₆H₁₂O₆ + 2 NAD⁺ + 2 ADP + 2 P i + 2 NADH → 2 EtOH + 2 ATP + 2 NADH + 2 NAD⁺
Since NAD⁺/NADH is used and regenerated, we can eliminate it from the equation:
C₆H₁₂O₆ + 2 ADP + 2 P i → 2 EtOH + 2 ATP
With the equation, we calculate the maximum amount of ethanol that could be obtained theoretically:
1000 mmol C₆H₁₂O₆ ------------ 2000 mmol EtOH
325 mmol C₆H₁₂O₆ ------------- x= 650 mmol EtOH
Therefore, the maximum amount of ethanol that could be produced is 650 mmol.
Answer:
the energy should be potential