SOLUTION: I submitted a problem earlier today and the response definitely helped, but I am stumped on a different problem. My professor assigned 2 mixture problems and I could not set up th
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Question 98306This question is from textbook Intermediate Algebra through Applications
: I submitted a problem earlier today and the response definitely helped, but I am stumped on a different problem. My professor assigned 2 mixture problems and I could not set up the chart for either. The problem is:
A chemist has a 50 ml solution that is 25% acid. How much water should she add to make a 10% solution?
I tried to set up the chart like this:
Solution Acid Water Total
________________________________________
1st 2.5 x 50
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2nd 1 x 50
You can put this solution on YOUR website!
Let x=amount of water that needs to be added
Now we know that the amount of pure acid in the 25% solution (0.25(50)) plus the amount of pure acid in the water that's added (0) has to equal the amount of pure acid in the final mixture (0.10(50+x)). So our equation to solve is:
0.25*50=0.10(50+x) get rid of parens
12.5=5+0.10x subtract 5 from both sides
12.5-5=5-5+0.10x combine like terms
7.5=0.10x divide both sides by 0.10
x=75 ml------------------------------amount of water needed
CK
0.25*50=0.10*125
12.5=12.5
Hope thsi helps---ptaylor
