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Who wants a medal? How much of a 60% acid solution must be mixed with 50 gallons of a 24% acid solution to obtain a solution that is 50% acid? a. 130 gal c. 13 gal b. 125 gal d. None of these

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I would use the idea of a mass balance of the acid.
For example, the amount of acid in the 50 gallons of a 24% acid solution would be 50*.24=12 gallons of acid.
Make sense so far?

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Other answers:

Sweet. Since we don't have the volume of the 60% acid solution, I will call that x.
Can you write an expression that shows the volume of acid of the 60% acid solution?
0.5 = 0.6x + (50*0.24) I have a feeling that this isn't right.
That is close. The addition side looks good, but what about the volume of the combined solution?
sorry I had to step a way for a bit
No worries :)
(0.5*50) = 0.6x + (50*0.24) Is this what you meant?
Closer. :) The volume of the 24% solution is 50 gals and the 60% acid solution has volume x gals, so wouldn't the combined 50% acid solution have more volume that 50 gals?
So we would have to use two variables since we don't know the number of gallons need of the 60% solution nor do we know the number of gallons of the final solution?
We could, but we don't have to. Since we know we are adding two volumes together and one volume is 50 gallons and we defined the second volume to be x gallons. So the combined volume is 50+x gallons.
(x+50) = 0.6 + (50*0.24) It's still wrong isn't it?
Last time you put (0.5*50) = 0.6x + (50*0.24) which was closer. The only thing you needed to change was (0.5*[50+x]) = 0.6x + (50*0.24)
The 50 to a 50+x
Do you follow me?
Ooohhh! I see now. :D
Can you solve for x?
(0.5*[50+x]) = 0.6x + (50*0.24) 25 + 0.5x = 0.6x + 12 13 + 0.5x = 0.6x 13 = 0.1x 130 = x right?
Awesome! :)
Thank you so much. :)
No worries

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