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<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
<span>B) phosphodiester </span> is the correct answer
Answer:
0.0498 mol
Explanation:
Number of moles = concentration in mol/L × volume in L
Concentration = 1 M = 1 mol/L
Volume = 49.8 mL = 49.8/1000 = 0.0498 L
Number of moles = 1×0.0498 = 0.0498 mol
1. Convert the number of moles to number of particles (divide 6.41 by 6.022 x 10^23)
2. K2O dissociates into three ions (2 potassium ions and 1 oxide ion), so multiply the result from step 1 by 3 to get your answer