Answer:
A reversible reaction is one where <u><em>B) there is little change in the net free energy between substrate and product.</em></u>
Explanation:
A reversible reaction is one that reagents are transformed into products and at the same time products are transformed into reagents. That is to say that as the products appear in the reaction, they can react with each other by regenerating the reagents again. It is represented by a double arrow, indicating that the reaction can be carried out both in one direction and the other way around.
At the start of the reaction, there is a large amount of reagents. As time goes by, that amount decreases and speed too.
On the other hand, at the beginning of the reaction there are no products. As the reaction happens, the products are being formed and their speed will increase to match the speed of the reagents. When the rates of products and reagents are equal and constant, it is possible to say that the reaction is in chemical equilibrium. At this point, both reactions continue to happen, but the total concentrations of reagents and products no longer change.
The Gibbs free enthalpy or free energy of a system is a measure of the amount of usable energy (energy that a job can perform) in that system.
When a reaction system is in chemical equilibrium, it is in the lowest possible energy state (it has the lowest possible free energy). If a reaction is not in equilibrium, it will move spontaneously towards it because that allows it to reach a state of lower and more stable energy. Then when the reaction moves towards equilibrium, the free energy of the system decreases more and more.
Finally, <u><em>a reversible reaction is one where B) there is little change in the net free energy between substrate and product.</em></u>
B boiling point https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/03%3A_Matter_and_Energy/3.05%3A_Differences_in_Matter%3A_Physical_and_Chemical_Properties#Summary
Answer :
AgBr should precipitate first.
The concentration of 
when CuBr just begins to precipitate is, 
Percent of remains is, 0.0018 %
Explanation :
for CuBr is 
for AgBr is 
As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.
Now we have to calculate the concentration of bromide ion.
The solubility equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Cu^+][Br^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BBr%5E-%5D)
![4.2\times 10^{-8}=0.073\times [Br^-]](https://tex.z-dn.net/?f=4.2%5Ctimes%2010%5E%7B-8%7D%3D0.073%5Ctimes%20%5BBr%5E-%5D)
![[Br^-]=5.75\times 10^{-7}M](https://tex.z-dn.net/?f=%5BBr%5E-%5D%3D5.75%5Ctimes%2010%5E%7B-7%7DM)
Now we have to calculate the concentration of silver ion.
The solubility equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ag^+][Br^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%2B%5D%5BBr%5E-%5D)
![7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M](https://tex.z-dn.net/?f=7.7%5Ctimes%2010%5E%7B-13%7D%3D%5BAg%5E%2B%5D%5Ctimes%205.75%5Ctimes%2010%5E%7B-7%7DM)
![[Ag^+]=1.34\times 10^{-6}M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%3D1.34%5Ctimes%2010%5E%7B-6%7DM)
Now we have to calculate the percent of remains in solution at this point.
Percent of remains = 
Percent of remains = 0.0018 %
The degree to which a specified material conducts electricity, calculated as the ratio of the card density in the material to the electric field that causes the flow of current. It is the reciprocal of the resistivity.
