Answer:
1. pH = 1.23.
2. 
Explanation:
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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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The answer is the second choice.
Photosynthesis: The process of making something ( carbohydrates ) with light
CO+H2O &Sunlight=O2+C6H6O12
Physical properties include: appearance, texture, color, boiling point, melting point, ect.
Answer:
The answer is quartet 2.40 ppm.
Note: Kindly find an attached image below for the part of the solution to this question
Sources: The image was researched from Course hero platform
Explanation:
Solution
Multiplicity or (n+1) rule:
It helps in determination of multiplicity of an individual proton or individual types of proton which are available in the molecule.
Multiplicity =(n+1)
Thus
The non equivalent protons which are attached from adjacent atom is denoted by n.
Now because there are three non-equivalent protons are present at adjacent carbon of methylene group, hence the multiplicity of methylene hydrogen is given as follows:
The multiplicity will be the same for the two hydrogen's. thus we compute multiplicity only for one hydrogen atom stated below:
Non- equivalent = 3
Multiplicity = (3 +1)
= 4
= Quartet for 2H
A quartet for 2H indicates that the hydrogen atoms attached from the carbon, which is attached one side from a methyl group and the other side form an atom that have no any hydrogens.
Now due +I effect of carbonyl group, chemical shift value is high for these two hydrogens which is exactly at 2.40 ppm or 2.40 Quartet.