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:
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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I hope this is the answer that you are looking for .
Explanation:
1
Number of nucleon =
Molarmassofnucleon
Massofatom
=
1.6726×10
−24
g/nucleon
3.32×10
−23
g
=19.8=20(approximately)
It is given that element comprises of 2 atoms
Hence,number of nucleon = 2×20=40
2
You have 4.70 mol H2O
There are two H atoms in 1 molecule H2O.
Therefore, there must be 2*4.70 = 9.40 mols H in 4.70 mols H2O.
How many mols O in 4.70 mols H2O? That's 4.70 mols, of course.
Said another way, you have 2 mols H for every 1 mol H2O and 1 mol O for every 1 mol H2O.
So for 50 mols H2O you have 100 mols H and 50 mol O.