The volume of the 39% acid solution that Roger needs to use to make a mixture of 86 mL of a 36% acid solution is 74.27 mL.
The mixture of the acids solutions is given by:
[tex] CV = C_{1}V_{1} + C_{2}V_{2} [/tex] (1)
Where:
C: is the concentration if the mixture = 36%
V: is the total volume of the mixture = 86 mL
C₁: is the concentration of acid 1 = 39%
V₁: is the volume if acid 1 =?
C₂: is the concentration of acid 2 = 17%
V₂: is the volume of acid 2
The sum of V₁ and V₂ must be equal to V, so:
[tex] V = V_{1} + V_{2} [/tex]
[tex] V_{2} = V - V_{1} [/tex] (2)
By entering equation (2) into (1), we have:
[tex] CV = C_{1}V_{1} + C_{2}(V - V_{1}) [/tex]
[tex]36\%*86 mL = 39\%*V_{1} + 17\%(86 mL - V_{1})[/tex]
Changing the percent values to decimal ones:
[tex] 0.36*86 mL = 0.39*V_{1} + 0.17(86 mL - V_{1}) [/tex]
Now, by solving the above equation for V₁:
[tex] V_{1} = 74.27 mL [/tex]
Therefore, the volume of the 39% acid solution is 74.27 mL.
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We have that the volume of the 39% acid solution that Roger needs to use to make the mixture.
x=74mL
From the question we are told that
Roger needs to make a mixture of 86 mL of a 36% acid solution from a 39% acid solution and a 17% acid solution.
Hence
86 - x of the 17\% acid solution is taken by Rogers
Generally the equation for the mixture concentration is mathematically given as
C_m=\frac{39% solution concentration x volume of solution x 17% solution conc x volume of solution taken}{volume of solution mixture}
36 =\frac{ (39 * x + 17 * (86-x)) }{ 86}
x=74mL
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