<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine,
, the reaction will be
and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4
And
![kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}](https://tex.z-dn.net/?f=kb%20%3D%20%20%5Cfrac%7B%28%20%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%5D%2A%20%20OH%5E%7B-%7D%20%29%7D%7B%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%5D%7D%20)
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
Charge and uncharged particles
From the calculations, the heat of fusion of the substance is 0.73 kJ
<h3>What is is the heat of freezing?</h3>
The heat of freezing is the energy released when the substance is converted from liquid to solid.
Now we know that the molar mass of the substance is 82.9 g/mol hence the number of moles of the substance is; 13.3 g /82.9 g/mol = 0.16 moles
Now the heat of fusion shall be;
H = 4.60 kj/mol * 0.16 moles
H = 0.73 kJ
Learn more about freezing:brainly.com/question/3121416
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Covalent since they are both non metals, a metal and a non metal make an ionic bond
Answer:
Explanation:
Step 1. Calculate the pOH
pOH =-log[OH⁻]
pOH =-log(1.0 × 10⁻⁹)
pOH = 9.00
Step 2. Calculate the pH
pH + pOH = 14.00
pH + 9.00 = 14.00
