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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<u>Answer:</u> The value of
for the net reaction is 
<u>Explanation:</u>
The given chemical equations follows:
<u>Equation 1:</u> 
<u>Equation 2:</u> 
The net equation follows:
As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.
The value of equilibrium constant for net reaction is:

We are given:


Putting values in above equation, we get:

Hence, the value of
for the net reaction is 
We have to complete all the given reactions.
1. Fe(s) + CuCl₂ → Cu + FeCl₂
2. Cu(s) + FeCl₂(aq) → NR (no reaction takes place)
3. K(s) + NiBr2(aq) → NR (no reaction takes place)
4. Ni(s) + KBr(aq) → K + NiBr₂
5. Zn(s) + Ca(NO₃)₂(aq) → NR (no reaction)
6. Ca(s) + Zn(NO₃)₂(aq) → Zn(s) + Ca(NO₃)₂(aq)
Answer:
V2 = 2.88L
Explanation:
P1= 78atm, V1= 2L, T1= 900K, P2= 45atm, V2=? T2= 750K
Applying the general gas equation
P1V1/T1 = P2V2/T2
Substitute the above
(78*2)/900= (45*V2)/750
V2= (78*2×750)/(900*45)
V2= 2.88L