A chemist is trying to classify an unknown substance as either a metal or nonmetal. What question should the chemist use to help
1 answer:
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Answer:
pH 9,8 is likely to work best for this separation
Explanation:
Ion exchange chromatography is a chemical process where molecules are separated by affinity to an ion exchange resin. To separate different aminoacids you must use the isoelectric point (That is the pH where the aminoacid will be in its neutral form).
For lysine, PI is:
9,8
For arginine:
10,75
At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.
Thus, <em>pH 9,8 is likely to work best for this separation</em>
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I hope it helps!
Answer:
The concentration of KOH is 0.186 M
Explanation:
First things first, we need too write out the balanced equation between HBr and KOH.
This is given as;
KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)
From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.
We use the acid base formular in calculating unknown concentrations. This is given as;
where;
Ca = Concentration of acid
Va = Volume of acid
Cb = Concentration of base
Vb = Volume of base
na = Number of moles of acid
nb = Number of moles of base
KOH is the base and HBr is acid.
Hence;
Ca = 0.225
Va = 35
Cb = ?
Vb = 42.3
na = 1
nb = 1
Making Cb subject of formular we have;
Cb = (0.225 * 35 * 1) / (42.3 * 1)
Cb = 0.186 M