A. Fewer valence electrons
Valence electrons are electrons that are held most loosely in the outermost shell.
Answer:
The pH of the final solution is 7.15
Explanation:
50 mL of 2.0 M of
and 25 mL of 2.0 M of
were mixed to make a solution
Final volume of the solution after dilution = 200 mL
Final concentration of ![K_2HPO_4, [K_2HPO_4] = \frac{50 mL\times 2 M}{200 mL} = 0.5 M](https://tex.z-dn.net/?f=K_2HPO_4%2C%20%5BK_2HPO_4%5D%20%3D%20%5Cfrac%7B50%20mL%5Ctimes%202%20M%7D%7B200%20mL%7D%20%3D%200.5%20M)
Final concentration of![KH_2PO_4, [KH_2PO_4] = \frac{25 mL\times 2 M}{200 mL} = 0.25 M](https://tex.z-dn.net/?f=KH_2PO_4%2C%20%5BKH_2PO_4%5D%20%3D%20%5Cfrac%7B25%20mL%5Ctimes%202%20M%7D%7B200%20mL%7D%20%3D%200.25%20M)
We use Hasselbach- Henderson equation:
![pH = pK_a+ log \frac{[salt]}{[acid]}pka of KH_2PO_4 = 6.85](https://tex.z-dn.net/?f=pH%20%3D%20pK_a%2B%20log%20%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7Dpka%20of%20KH_2PO_4%20%3D%206.85)
Substituting the values:

Therfore, the pH of the final solution is 7.15
Answer:
-1043,8 KJ
Explanation:
Hello!
It is necessary to accommodate the reactions to get the reaction of my interest.
Reaction A reverses what modifies its energy in its sign (+72.6 KJ).
Reaction B doubles, which also doubles its energy (-514.4 KJ x 2).
Reaction C reverses what modifies its energy in its sign (-87.6 KJ).
By sum of the energies I get the energy of the global reaction.
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