Answer:
d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2
Explanation:
Hello,
In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:
![pH=pKa+log(\frac{[Base]}{[Acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%20%29)
We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:
a. ![\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.497M%7D%7B0.365M%7D%3D1.36)
b. ![\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.217M%7D%7B0.521M%7D%3D0.417)
c. ![\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.713M%7D%7B0.821M%7D%3D0.868)
d. ![\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.116M%7D%7B0.121M%7D%3D0.959)
Therefore, the d. solution has the best buffering capacity.
Regards.
Answer:
bottom
Explanation:
the ones with the greatest density will sink. if they have a lighter density than water, they will float. if they have about the same, they will stay in the middle. and if they have a higher density they will sink.
Answer:
reee, here is your answer.
Explanation: CH3OH(l) + 3O2(g) rightarrow CO2(g) + 3H2O(g) CH3OH(l) + O2(g) rightarrow CO2(g) + 2H20(g) CH3OH(l) + 2O2(g) rightarrow 2CO2(g) + 4H20(g) 2CH3OH(l) + 3O2(g) rightarrow 2CO2(g) + 4H20(g) Correct Calculate Delta H degree rxn at 25 degree C.
