Answer:
9839 is your answer mate .
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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:
The molar mass of dry air = 28.94
Density = 1.29
Explanation:
Mm of N2= 28, Mm of O2= 32, Mm Ar= 39.9
78% of N2 by mole give = 0.78×28= 21.84
21% of O2 by mole give = 0.21× 32= 6.72
1% of At by mole give = 0.01× 39= 0.39
Hence overall molecular weight of air = 21.84+6.72+0.39= 28.94
Density = m/ v,
1- mole of air occupy 22.4dm^3 volume
Density= 28.94/22.4= 1.29
D. Zinc metal (zn)
Zinc is a chemical element with atomic number 30.
Answer:
24.28 kPa
Step-by-step explanation:
You want to convert atmospheres to kilopascals.
The conversion factor is 101.325 kPa = 1 atm
p = 0.2396 atm × (101.325 kPa/1 atm)
p = 24.28 kPa
