Answer:
81.59%
Explanation:
First we <u>convert 107.50 g of NH₃ into moles</u>, using its <em>molar mass</em>:
- 107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃
Now we <u>calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃</u>:
- 6.32 mol NH₃ *
= 6.32 mol NO
Then we <u>convert 6.32 moles of NO to grams</u>, using its <em>molar mass</em>:
- 6.32 mol NO * 30 g/mol = 189.60 g NO
Finally we <u>calculate the percent yield</u>:
- 154.70 g / 189.60 g * 100% = 81.59%
Answer: It is exhibiting retrogade motion.
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!
Answer:
droughts in California have become more common in California
Explanation:
Climate is weather over time which means that the first and the third one are just isolated incidents and the last answer choice shows climate but it doesn't show a change in climate
Answer:
Fluorine
Explanation:
These particles stick in the atoms and make them radioactive.