Answer:the pH is 12
Explanation:
First We need to understand the structure of trimethylamine
Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline
C3H9N+ H2O --> C3H9NH + OH-
![k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BC3H9NH%5D%2A%5BOH-%5D%7D%7B%5BC3H9N%5D%7D)
Then:
The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine
[OH-]=
[OH-]=0.01
pH=14-(-log[OH-])
pH=12
The student's test average is : 0.92
Total score obtained = 369
Number of tests = 4
Since Each test is exactly 100 points ;
The total score obtainable is (100 × 4) = 400
Average = score obtained / total score obtainable
Average = 369 / 400
Average = 0.9225
Average = 0.92
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Answer:
14.3mL you require to reach the half-equivalence point
Explanation:
A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:
CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻
As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.
Initial moles of CH₃CH₂NH₂ are:
20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =
0.0106moles CH₃CH₂NH₂
To reach the half-equivalence point you require:
0.0106moles ÷ 2 = 0.005304 moles HClO₄
As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:
0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =
<h3> 14.3mL you require to reach the half-equivalence point</h3>
