K(eq) = concentration of products/concentration of reactant = [Cu+2] / [Ag+]^2
Activity of pure solid and liquid is taken as 1.
Hence last option is correct.
Hope this helps, have a great day ahead!
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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:
iron sulfate
Explanation:
Electrostatic attraction as bonds between ions is characteristic of the electrovalent bond or the ionic bond
In this type of bond, there is electron transfer from one atom to another. The atom that looses electrons become positively charged while the atom that gains electrons becomes negatively charged.
In iron sulfate, there is electrostatic attraction between Fe II ions and sulphate ions, making iron sulfate an ionic compound.
Explanation:
pls, refer to the above picture, i hope you will find it helpful.