Wednesday, May 29, 2013

how much pure acid should be mixed with 3 gallons of 20% acid solution to get a 40% acid solution?


unknown amt: x
20% acid: 3(0.2)
new mixture: (x+3)(0.4)
%0D%0Ax%2B%283%29%280.2%29=%28x%2B3%29%280.4%29%0D%0A

x=1




:)
Posted by at
Labels: , ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)