Answer:
Imma just say acids have more acidity and bases have lower acidity...
Letter d, because they are both alkali metals (group one)
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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Yes you are correct hope you do well
Answer:
Option B = 60,600 mg (correct option)
Explanation:
First of all we will have an idea which numbers are consider as significant.
1 = All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.
2= Leading zeros are not consider as a significant figures. e.g. 0.02 in this number only one significant figure present which is 2.
3= Zero between the non zero digits are consider significant like 105 consist of three significant figures.
4= The zeros at the right side e.g 3400 are also significant. There are four significant figures are present.
In given options, Option A 60.6 mg have 3 significant figures.
Option B have 5 significant figures.
Option C have 4 significant figures.
Option D have 3 significant figures.
Thus option b is correct option which have more significant figures.