The french broad? idk sorry
Complete question is;
When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have
a. a pH of 7 at the equivalence point.
b. two equivalence points below 7.
c. no equivalence point.
d. one equivalence point.
e. two distinct equivalence points
Answer:
Option E - Two Distinct Equivalence points
Explanation:
I've attached a sample diprotic acid titration curve.
In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.
Thus, it is clear that there are two distinct equivalence points.
<u>Answer:</u> The enthalpy of the formation of
is coming out to be -65.3 kJ/mol
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SiC(s))})+(2\times \Delta H^o_f_{(CO(g))})]-[(1\times \Delta H^o_f_{(SiO_2(s))})+(3\times \Delta H^o_f_{(C(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SiC%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SiO_2%28s%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![624.6=[(1\times \Delta H^o_f_{(SiC(s))})+(2\times (-110.5))]-[(1\times (-910.9))+(3\times (0))]\\\\\Delta H^o_f_{(SiC(s))}=-65.3kJ/mol](https://tex.z-dn.net/?f=624.6%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SiC%28s%29%29%7D%29%2B%282%5Ctimes%20%28-110.5%29%29%5D-%5B%281%5Ctimes%20%28-910.9%29%29%2B%283%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SiC%28s%29%29%7D%3D-65.3kJ%2Fmol)
Hence, the enthalpy of the formation of
is coming out to be -65.3 kJ/mol.