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!
Option A
using verbal and nonverbal communication to show a person is listening is a critical part of listening
<u>Explanation:</u>
An adequate hearer obligation listens and recognizes the language sounds focused approaching them, know the information of the sounds, critically estimate or estimate that information, retrieve what’s implied stated, and acknowledge either orally or nonverbally to erudition they’ve earned.
Answering occasions imparting literal and nonverbal intelligence that show attentiveness and perception or a shortage thereof. We convey verbatim and nonverbal feedback while the added character is speaking and later they are ingested. Assess the trustworthiness, completeness, and quality of an information ere responding with lexical and nonverbal beacons.
Answer:
976.8 g
Explanation:
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂0 ------------(1)
number of moles of reaction 1 can be shown as;
1 : 8 → 5 : 6
thus, 5 moles of CO₂ will form from the combustion of 1 mole of C₅H₁₂
molecular weight of CO₂ = 44 g/mol
molecular weight of C₅H₁₂ = 72 g/mol
mass of C₅H₁₂ = 320 g
number of moles of C₅H₁₂ = mass ÷ molecular weight = 320 / 72 = 4.44 moles
thus, number of moles of CO₂ formed = 5 * 4.44 = 22.2 moles
mass of CO₂ formed = 22.2* 44 = 976.8 g
A) Compounds because its the other word of the formation of new substances!