Answer:
pH of the solution is 10.37
Explanation:
![pOH=pkb+log\frac{[salt]}{[base]}](https://tex.z-dn.net/?f=pOH%3Dpkb%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D)
kb = 
pkb = -log kb
= 
= 3.35
salt is methylammonium bromide and methylamine is base
Substitute the values in the above expression as follows:
![pOH=pkb+log\frac{[salt]}{[base]} \\=3.35+log\frac{0.35}{0.18} \\=3.35+0.28\\=3.63](https://tex.z-dn.net/?f=pOH%3Dpkb%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%20%5C%5C%3D3.35%2Blog%5Cfrac%7B0.35%7D%7B0.18%7D%20%5C%5C%3D3.35%2B0.28%5C%5C%3D3.63)
pH = 14 - pOH
= 14 - 3.63
= 10.37
pH of the solution is 10.37
Explanation:
pH is used to measure the concentration of H+ ions ([H+]) and therefore, whether a substance is acidic or basic (alkaline). Solutions with a pH of less than seven are acidic, while those with a pH greater than seven are basic (alkaline).
Answer:
1) Separate the aqueous layer from the organic layer using the separation funnel.
2) Treat the aqueous layer to obtain compound A.
3) Distilated the organic layer to obtain compound B.
Explanation:
When <u>NaOH is added</u> to the mixture the acid groups will react to produce a salt and increases the polarity of the compounds due to the net charges generated. (Figure 1).
Therefore, the salt produced by compound A will move to the <u>aqueosus layer</u>. Compound B dont react due to the lack of <u>acid groups</u>. So, this molecule will stay in the <u>organic layer</u>.
When the aqueous layer is separated from the organic layer using the separation funnel we will have a <u>separation</u>. The compound will remain in the aqueous layer and compound b will remain in the organic layer.
Then we aqueous layer can be <u>treated with HCl</u> in order to obtain the initial A molecule, in other words: Undo the ionic form of compound A.
The organic layer can be removed by <u>distillation</u> in order to obtain the pure form of compound B.
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