Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

Thus, the concentration of ethylamine in solution is:
![[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M](https://tex.z-dn.net/?f=%5Bethylamine%5D%3D%5Cfrac%7B0.0816mol%7D%7B0.1850L%2B0.1144L%7D%3D0.2725M)
Now, we can also infer that some salt is formed, and has the following concentration:
![[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D%5Cfrac%7B0.0549mol%7D%7B0.1850L%2B0.1144L%7D%3D0.1834M)
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
![pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%20%29%5C%5C%5C%5CpOH%3D3.19%2Blog%28%5Cfrac%7B0.1834M%7D%7B0.2725M%7D%29%5C%5C%5C%5CpOH%3D3.0)
Finally, the pH turns out to be:

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
Best regards!
Answer:
B
Explanation:
Iodine has a negative charge, so it will gain electrons.
Answer:
the enthalpy change for this reaction is -57.7 kJ/mol
Explanation:
Given:
HB₂O₃ = -1272.8 kJ/mol
HCOCl₂ = -218.8 kJ/mol
HBCl₃ = -403.8 kJ/mol
HCO₂ = -393.5 kJ/mol
Those are all standard enthalpies
Question: Calculate the enthalpy change for this reaction, ΔHreaction = ?
The enthalpy of the reaction is calculated using the standard enthalpies of formation of both products and reagents. To understand better, the reaction is as follows
B₂O₃ + 3COCl₂ → 2BCl₃ + 3CO₂
Where the compounds on the left are the reactants and the compounds on the right are the products
ΔHreaction = ∑ΔHproducts - ∑ΔHreactants

Please be careful with the signs.
Answer:
Thank you
I didn't wait because there is someone who will take it and there is someone who answer a wrong answer
Explanation: