C = 4 mol/l
v = 0.5 l
n(NaCl)=cv
n(NaCl) = 4 mol/l · 0.5 l = 2 mol
2 moles of NaCl must be dissolved
Answer: 1.4x10-3 g N2O4
Explanation: First convert molecules of N2O4 to moles using Avogadro's Number. Then convert moles to mass using the molar mass of N2O4.
9.2x10^18 molecules N2O4 x 1 mole N2O4 / 6.022x10²³ molecules N2O4
= 1.53x10-5 moles N2O4
1.53x10-5 moles N2O4 x 92 g N2O4/ 1 mole N2O4
= 1.4x10-3 g N2O4
I found this is the order of activity in Internet
Mn>Zn>Cr>Fe>Cd>Co>Ni>Pb
You can see that Fe is between Cr and Cd
Cr is more active than Fe and Fe is more active than Cd.
The Fe will displace Cd but not Cr.
Answer:Fe
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.