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...
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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)`
`(1)*9-(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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