What is the result when you total the individual percent abundances?

100 %

%Abundance = part / total *100
%Abundance_1 = parts_1 / total * 100
%Abundance_2 = parts_2 / total * 100
%Abundance_3 = parts_3 / total * 100

 %Abundace_1 + %Abundace_2+%Abundace_3 = parts_1/total *100 + parts_2/total*100 + parts_3/total*100

Factor the right side:

%Abundace_1 + %Abundace_2+%Abundace_3 = [parts_1+parts_2+part2_3]/total *100

Given the sum of the parts = total 

 %Abundace_1 + %Abundace_2+%Abundace_3 = [total]/total *100 = 100

So, the sum of the individual % of abundaces is 100 %.

The basic equation for diluting solutions is:

(M1)(V1) = (M2)(V2)

M1 and V1 belong with the stock solution, the stronger one.

The mL has to be turned into liters, so convert first. 175 mL = .175 L

So, 1.6 M represents M1 and .175 L represents V1. 1.0 is V2 because it's asking for the new concentration, which is M2. Therefore, M2 is the variable you're looking for.

Plugging it in:

(1.6M)(.175) = (M2)(1)

Simplify and Solve:

.28 = M2