1.661x10 negative 24 squared
Answer:
endoplasmic reticulum (ER)
<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
Answer:
- <u><em>The change in the temperature of the iron is a decrease of 14.0ºC.</em></u>
Explanation:
See the file attached, since it was not possible to show the complete answer in the canvas.
V = 75 mL = 0,075 L = 0,075 dm³
C = 2.1M
n = ?
---------------
C = n/V
n = C×V
n = 2.1×0,075
n = 0,1575 mol
--------
mKCl: 39+35.5 = 74,5 g/mol
74,5g --------- 1 mol
Xg ------------- 0,1575 mol
X = 74,5×0,1575
X = 11,73375g KCl
:•)