Answer:
Explanation: See images below for explanation
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!
The answer is b. 4.2 mole. The balanced reaction formula is 2LiOH + H2SO4 -->Li2SO4 + 2H2O. And the ratio of mole number of the reactants is the same as the ratio of coefficients.
The reaction between NaOH and H₂SO₄ is as follows;
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
Therefore molarity of NaOH - 0.352 M
Answer:
Answer choice A
Explanation:
When an electron is transferred to another atom, both atoms involved become ions.
