The amine here is the easiest to spot since there’s only one structure that has a nitrogen atom, which would be the first (the first structure is a primary amine).
The distinguishing functional group of an alcohol is the hydroxy group (—OH). Both the second and third structures have an —OH group, but the —OH in the third structure is part of a carboxyl group (—COOH or —C(=O)OH). A carboxyl group takes priority over hydroxy group. Thus, the second structure would be an alcohol and the third structure would be a carboxylic acid.
That leaves us with the fourth structure, a hydrocarbon with a halogen substitutent, or, aptly named, a halocarbon.
Answer:
1.67 M
Explanation:
75.0 g *1 mol/180 g = (75/180) mol of glucose
250 mL = 0.250 L
(75/180)/0.250 = 1.67 mol/L =1.67 M
If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.
In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.
So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC
1.17×6×1023
Wich equals 7.02×1023
Your final answer is 7.02×1023
35.25 is the answer of the probem
