SOLUTION: Ok i have missed 18 days of school. So im completely lost on this. Im trying to get work done today since its the last day before we go back to school from holiday break. It says a
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Question 699209: Ok i have missed 18 days of school. So im completely lost on this. Im trying to get work done today since its the last day before we go back to school from holiday break. It says a Chemist has a beaker of a 3% acid solution and a beaker of 7% acid solution. He needs to make 75 mL of a 4% acid solution. He wants me to make a table makeing a equation. B) says use the information in the table to write a system of equations. C) says solve the system of equations to find how much he will use from each beaker Answer by solver91311(24713) (Show Source):
I don't see where making a table is going to help much.
Let represent the number of mL of 3% solution and let represent the number of mL of 7% solution. The total amount of solution in the end is 75 mL, so:
Three percent of the 3% solution is pure acid. Seven percent of the 7% solution is pure acid. And four percent of the final 75 mL is pure acid, hence there are of pure acid in the final mixture. Since the only place you get the pure acid in the final mix is the amount that is in the 3% solution plus the amount that is in the 7% solution, you can say:
Solve the system of equations for . Substitution works well.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
