Answer:
If you want to separate black grapes from the mixture of black and green grapes, then you will simply pick black grapes using your hands from the mixture. In this way you are actually using handpicking separation method.
Explanation:
German scientist Döbereiner was one responsible for grouping elements into triads based on most notably atomic mass, many of which can be found in the periodic table to be in a pattern (for example <span><span>Iron </span><span>Cobalt </span><span>Nickel, elements 26, 27, 28)</span></span>
Answer:
The answer to your question: 0.7 M
Explanation:
Data
V of KOH = 90 ml
[KOH] = ?
V H2SO4 = 21.2 ml
[H2SO4] = 1.5 M
2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)
Molarity = moles / volume
moles of H₂SO₄ = (1.5) (21.2)
= 31.8
2 moles of KOH -------------- 1 mol of H₂SO₄
x -------------- 31.8 mol of H₂SO₄
x = (31.8)(2) / 1
x = 63.8 moles of KOH
Molarity = 63.8 / 90
= 0.7 M
Answer:
0!
Explanation:
- You need to search your pKa values for Asn (2.14, 8.75), Gly (2.35, 9.78) and Leu(2.33, 9.74), the first value corresponding to -COOH, the second to -NH3 (a third value would correspond to an R group, but in this case that does not apply), and we'll build a table to find the charges for your possible dissociated groups at indicated pH (7), we need to remember that having a pKa lower than the pH will give us a negative charge, having a pKa bigger than pH will give us a positive charge:
-COOH -NH3
pH 7------------------------------------------------------
Asn - +
Gly - +
Leu - +
- Now that we have our table we'll sketch our peptide's structure:
<em>HN-Asn-Gly-Leu-COOH</em>
This will allow us to see what groups will be free to react to the pH's value, and which groups are not reacting to pH because are forming the bond between amino acids. In this particular example only -NH group in Ans and -COOH in Leu are exposed to pH, we'll look for these charges in the table and add them to find the net charge:
+1 (HN-Asn)
-1 (Leu-COOH)
=0
The net charge is 0!
I hope you find this information useful and interesting! Good luck!
