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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Answer:
fitness. the ability of an organism to survive and reproduce in its environment. adaptation. the inherited characteristic that increases an organism's chance of survival.
Answer:
The final temperature will be "12.37°".
Explanation:
The given values are:
mass,
m = 0.125 kg
Initial temperature,
c = 22.0°C
Time,
Δt = 4.5 min
As we know,
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
Physical because you can see it evaporate have a great día
Answer:
(i)The mole fractions are :
(ii)
(iii)ΔG = 1.974kJ
Explanation:
The given equation is :
⇄
Let
be the number of moles dissociated per mole of 
Thus ,
<em>The initial number of moles of :</em>
+
⇄
+ 
And finally the number of moles of ![C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930](https://tex.z-dn.net/?f=C%5Btex%5D%20is%200.9%3C%2Fp%3E%3Cp%3EThus%20%2C%3C%2Fp%3E%3Cp%3E%5Btex%5D3%5Calpha%3D0.9%5C%5C%5Calpha%3D0.3%5Btex%5D%3C%2Fp%3E%3Cp%3E%3Cem%3E%3Cstrong%3EThe%20final%20number%20of%20moles%20of%3A%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fp%3E%3Cul%3E%3Cli%3E%3Cem%3E%3Cstrong%3E%5Btex%5DA%20%3D%201-2%5Calpha%3D1-2%2A0.3%3D0.4mol%5Btex%5D%20%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fli%3E%3C%2Ful%3E%3Cul%3E%3Cli%3E%3Cem%3E%3Cstrong%3E%5Btex%5DB%3D2%281-%5Calpha%29%3D2%281-0.3%29%3D1.4mol%5Btex%5D%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fli%3E%3C%2Ful%3E%3Cul%3E%3Cli%3E%3Cem%3E%3Cstrong%3E%5Btex%5DD%3D1%2B2%5Calpha%3D1%2B2%2A0.3%3D1.6mol%5Btex%5D%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fli%3E%3C%2Ful%3E%3Cp%3EThus%20%2C%20total%20number%20of%20moles%20are%20%3A%200.4%2B1.4%2B0.9%2B1.6%3D4.3%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%28i%29The%20mole%20fractions%20are%20%3A%20%3C%2Fstrong%3E%3C%2Fp%3E%3Cul%3E%3Cli%3E%3Cstrong%3E%5Btex%5DA%3D%5Cfrac%7B0.4%7D%7B4.3%7D%20%5C%5C%3D0.0930)
(ii)

Where ,
are the partial pressures of A,B,C,D respectively.
Total pressure = 1 bar .
∴
<em>
</em>
<em>
</em>
<em>
</em>
<em>
</em>

(iii)
Δ
ΔG = 