A chemist has a 9% acid solution and a 4% acid solution. He wishes to make 3 liters of a 7% acid solution. ANSWERS ASAP! There are four parts to it: A)There's a chart with 3 sections labeled "liters of solution", "percent acid", and "liters of acid".* B)Write an equation to find how many liters of each solution the chemist must mix together to make 3 liters of a 7% acid solution. C)Solve your equation. D)How many liters of each solution must the chemist mix together? *Three rows going down say "9% acid solution", "4% acid solution", and "7% acid solution". THANK YOU!

Expert Answers

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Let us say the chemist mix V1 volume from 9% acid solution  and V2 from 4% acid solution.

Moles in 9% acid solution `n1 = (9V1)/100`

Moles in 4% acid solution `n2 = (4V2)/100`


It is given that in the final solution the acid concentration is 7%.

We have 3 L of this solution.

Moles in 7% acid solution` n = (7*3)/100`


But we know;

`n1+n2 = n`

`V1+V2 = 3`


`n1+n2 = n`

`(9V1)/100+(4V2)/100 = (3*7)/100`

`9V1+4V2 = 21 -----(1)`


`V1+V2 = 3 -----(2)`



 `9V2-4V2 = 27-21`

          `V2 = 1.2L`


From (2)

    `V1 = 3-1.2 = 1.8L`


So the chemist needs to mix 1.8L from the 9% acid solution and 1.2L from the 4% acid solution to get the final acid solution.


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