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C1 * V1 = C2 * V2 is probably what you need to calculate this
you always use algebra to rearrange equations to solve for the missing piece
\(= V_2\), but yes
C's are the two concentrations. I use \(M_1*V_1 = M_2*V_2\), but the idea is exactly the same
why would you put an unknown in the denominator? solve for \(V_2\) with the UNITS, not the NUMBERS
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since\[M_1*V_1 = M_2*V_2\]then\[V_2 = \frac{M_1*V_1}{M_2}\]
yes
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