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klemol [59]
2 years ago
9

Calculating and using the molar mass of diatomic elements

Chemistry
1 answer:
beks73 [17]2 years ago
4 0
Moles(grams/moles)
F2=19.00(2)=38g

0.075(38g/mol) =2.9
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The enthalpy of vaporization of Bromine is 15.4 kJ/mol. What is the energy change when 80.2 g of Br2 condenses to a liquid at 59
Aleksandr-060686 [28]

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:

Q = n \Delta H_{\text{vapo.}

where,

Q = Energy change

n = number of moles

\Delta H_{\text{Vapo.}} = Molar enthalpy of vaporization

Now find the number of moles

Number of moles (n) = \frac{\text{Given Mass}}{\text{Molar mass}}

                                   = \frac{80.2\ g }{159.8\ g/mol}

                                   = 0.5 mol

Now put the values in above formula we get

Q = - n \Delta H_{\text{vapo.}         [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

#SPJ1

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its not a question but i wanted to tell you guys that if you guyz like my answers then please don't forget to give like and foll
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thank you for this oibnki

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