0001 M HCl is the same as saying that 1 *10-4 moles of H+ ions have been added to solution. The -log[. 0001] =4, so the pH of the solution =4.
<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:
a) 2-bromopyrrole
Explanation:
Our options for this questions are:
a) 2-bromopyrrole
b) 2,3-dibromopyrrole
c) N-bromopyrrole
d) 3-bromopyrrole
To understand how the reaction works we have to start with the <u>resonance structures</u>. (Figure 1), on these structures, we will obtain a n<u>egative charge on carbon 2</u> in the pyrrole ring, therefore on this carbon we can generate an attack to an electrophile.
The second step is to check how the mechanism take place. An <u>electrophile is generated</u> by the
and
. This electrophile can be <u>attacked</u> by the negative charge on carbon 2 producing the 2-bromopyrrole. (See figure 2).
I hope it helps!
The correct answer is that chemical changes form new compounds. Chemical change is a change where the substance changes in identity or form new substances after undergoing a process. On the other hand, a physical change is a change in the properties of matter that does not change the identity of the substance.