The density of the stone is 5 because the formula is mass/volume and the volume is 5 and the mass is 25
Answer:
please the answer Is false
It is E. because the predators will have slightly more meals with their prey slightly increasing population.
Answer:
The porpoise would hear its echo first because sound travels faster in water than in air. The bat would hear its echo first because the amplitude of sound waves ...
Explanation:
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.