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.
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Adding a catalyst as this would speed up the reaction and the rest would slow it down
Answer:
E
Explanation:
This is because all steps from A-D are important to obtain an accurate result
Answer:
Explanation:
Discussion
When Pressure increases equilibrium shifts to the side with the smallest number of moles. But which side is that?
N2(g) + 3H2(g) ⇌ 2NH3(g)
The left side has 1 mol of nitrogen (N2) and 3 moles of Hydrogen = 4 mols
on the left side.
The right side has 2 mols of NH3 = 2 mols on the right.
Conclusion: You tell the number of mols by the Balance numbers to the left of each chemical in an equation.
Since the left side N2 + 3H2 = 4 mols, the equilibrium does NOT shift left.
2NH3 is only two mols.
The equilibrium shifts Right
Answer
D