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 molarity is 0.56
Explanation:
In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes. Then, the molarity or molar concentration is the number of moles of solute per liter of solution.
In other words, molarity is the number of moles of solute that are dissolved in a given volume.
The Molarity of a solution is determined by:

Molarity is expressed in units (
).
Then you must know the number of moles of Cu(NO₂)₂. For that it is necessary to know the molar mass. Being:
-
Cu: 63.54 g/mol
- N: 14 g/mol
- O: 16 g/mol
the molar mass of Cu(NO₂)₂ is:
Cu(NO₂)₂= 63.54 g/mol + 2*(14 g/mol + 2* 16 g/mol)= 155.54 g/mol
Now the following rule of three applies: if 155.54 g are in 1 mole of the compound, 225 g in how many moles are they?

moles= 1.45
So you know:
- number of moles of solute= 1.45 moles
- volume=2.59 L
Replacing in the definition of molarity:

Molarity= 0.56
<u><em>The molarity is 0.56</em></u>
<u><em></em></u>
Answer:
Explanation:
Molarity = number of moles / volume
If 550 mL of a 3.50 M KCl solution are set aside and allowed to evaporate until the volume of the solution is 275 mL, which is half of 550 mL, the molarity of the solution with the same number of moles of KCl is 3.5 * 2 = 7.00 M
Answer: I believe the answer is 12
Explanation: