This dilution problem uses the equation
M
a
V
a
=
M
b
V
b
M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
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This represents a primary amine. An amine has a nitrogen group that is connected to three substituents via single bonds. The number of carbon-based substitutents determines whether it is primary, secondary, or tertiary. In this case, since 2 substitutents are just hydrogen atoms, and only one has a carbon-based skeleton, this is a primary amine.
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