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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The population decreases when the birth rate is less than the death rate.
Answer: alkaline earth metals (group-IIA)
Explanation:
The element which donates the electron is known as electropositive element and forms a positively charged ion called as cation. The element which accepts the electrons is known as electronegative element and forms a negatively charged ion called as anion.
Alkaline earth metals donate 2 valence electrons to acquire noble gas configuration.
For example: Berrylium is the first alkaline earth metal with atomic number of 4 and thus has 4 electrons
Electronic configuration of berrylium:
![[Be]:4:1s^22s^2](https://tex.z-dn.net/?f=%5BBe%5D%3A4%3A1s%5E22s%5E2)
Berrylium atom will loose two electrons to gain noble gas configuration and form berrylium cation with +2 charge.
![[Be^{2+}]:2:1s^2](https://tex.z-dn.net/?f=%5BBe%5E%7B2%2B%7D%5D%3A2%3A1s%5E2)
Thus Elements donate 2 electron to produce a cation with a 2+ charge are alkaline earth metals.
Answer:
0.8988
Explanation:
To calculate the fractional saturation of hemoglobin , the formula used is
YO_2=
now putting the values
YO_2= 
= 
=0.8988
Therefore, fractional saturation of hemoglobin= 0.8988
where
YO_2= fractional saturation of hemoglobin
pO_2= partial pressure of oxygen
p50= is the pO_2 at which hemoglobin is 50% saturated
Answer:
b. The molarity of the solution increases
Explanation:
The correct answer is option b, that is the molarity of the solution increases.
Because the molarity is the concentration of the solution and it is explained as the amount of solute in amount of solution.
Solution: is the solute dissolved in solvent.
So if we increases the amount of solute in solvent the concentration in terms of molarity of solution increases and if we increase amount of solvent or water then the concentration or molarity increases.
Suppose we have form a sugar solution of 1 L by adding 4 mole of sugar then what happen
Use the Molarity formula
Molarity = no. of moles / 1 L of solution
put values in the formula
Molarity = 4/ 1 L of solution = 4 M
So the molarity of solution is 4 now if we add 2 mole more sugar to the same amount of sugar and amount of solution remain the same
now the no. of moles of sugar = 6 mole
So,
Use the Molarity formula
Molarity = no. of moles / 1 L of solution
put values in the formula
Molarity = 6 mol / 1 L of solution = 6 M
So the correct option is b.
