Q=mc(delta)T
118=0.005×c×15
c=118/(0.005×15)
c= 1573 J/kg/C
The C=O stretch appears as a very sharp and intense peak in an IR spectrum. Since, C=O is a double bond, it appears in the "double bond" region of the IR spectrum, which is typically in the 1500-2000 cm-1 range. More specifically, C=O generally falls in the range of 1650-1850 cm-1. The reason for the range is that there are many types of functional groups that contain a carbonyl (C=O), such as a ketone, aldehyde, amide, or ester. Each of these will have a slightly different value as each stretch will have a different energy due to various factors such as conjugation.
B. 20 that’s the answer to your question
For seven, simply divide 321 joules by 4.184 to get 76.72 calories.
Fencepost method:
321 joules | 1 calorie
_________________ = 76.72
| 4.184 J
Because it doesn't have a 3' -OH group and that is the place where the next nucleotide would attach on to. So since it's not there, it can't attach, and therefore terminates.
