Can you draw the peptide structure at ph 7.0 (including peptide bonds)
peptide bond sequence : tryptophan - serine - asparagine - glycine - cysteine - histidine

See image below

Explanation:

The image is labeled according to the sequence N'-Trp-Ser-Asg-Gly-Cys-His-COOH' which means that in the main chain of the peptide, the amino group of the Tryptophan and the carboxylic group of the Histidine are free and thus its charge depends on the pH; other groups that rely on the pH are the side groups of the Cysteine and the Histidine.

Overall, ionizable groups in this peptide are:

Then, the amino group of Trp and SH group of Cys are protonated since the peptide is at a pH below the pKa. The secondary amine of the Histidine is deprotonated because the pH is greater than the pKa, as well with its carboxylic acid group.

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: