So what do you want me to do
Explanation
Answer:
1. pH = 1.23.
2. 
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on 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)
Whereas the pKa is:

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

Which is also shown in net ionic notation.
Best regards!
Answer:
The center of the atom contains the nucleus so yes
Explanation:
Answer:
Percentage abundance of 121 Sb is = 57.2 %
Percentage abundance of 123 Sb is = 42.8 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope, 121 Sb :
% = x %
Mass = 120.9038 u
For second isotope, 123 Sb:
% = 100 - x
Mass = 122.9042 u
Given, Average Mass = 121.7601 u
Thus,

Solving for x, we get that:
x = 57.2 %
<u>Thus, percentage abundance of 121 Sb is = 57.2 %
</u>
<u>percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %</u>