The two algae seen here are classified as green algae. That means they make their own food by photosynthesis. The algae on the l
eft is one-celled, or unicellular. The algae on the right is many-celled, or multicellular.
How are these two types of algae different?
Question 5 options:
Only the the multicellular algae contains a chloroplast.
They are different only in how many cells make up each algae.
Only the multicellular algae has chlorophyll for photosynthesis.
The multicellular algae cel has many more parts than the unicellular algae.
How are these two types of algae different?
Question 5 options:
Only the the multicellular algae contains a chloroplast.
They are different only in how many cells make up each algae.
Only the multicellular algae has chlorophyll for photosynthesis.
The multicellular algae cel has many more parts than the unicellular algae.
2 answers:
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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-]
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>
For triehtylamine,
and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4And
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7