how many liters of a 10%solution of acid should be added to 20 liters of 60%solution of acid to obtain a 50%solution?
This looks like a chemistry assignment.
Assuming the density doesn't change and the solutions mix with no volume change, the following should do it.
L = liters we need to add.
L(10) + 20(60) = (L+20)*50
Solve for L.
Frank=
You should show the subject matter in the heading. In this case the heading should have been chem or math. Had it not been for Ms Sue, I wouldn't have looked at anything called journal assignment
how many liters of a 10%solution of acid should be added to 20 liters of 60%solution of acid to obtain a 50%solution
To solve this problem, we can use the concept of the concentration of a solution, which is defined as the amount of solute (acid) dissolved in the solvent (water).
Let's assume that x liters of a 10% solution of acid will be added to 20 liters of a 60% solution of acid. This means that there will be (20 + x) liters of the final solution.
To determine the amount of acid in each solution, we can calculate it by multiplying the volume of the solution by its concentration:
For the 10% solution: 10% of x liters is 0.1x liters of acid.
For the 60% solution: 60% of 20 liters is 0.6 * 20 = 12 liters of acid.
We want to obtain a 50% solution, which means that the final solution will have the same amount of acid as the sum of the acid in the initial solutions.
Therefore, we can set up the equation:
0.1x + 12 = 0.5(20 + x)
Now, we can solve this equation to find the value of x:
0.1x + 12 = 10 + 0.5x
12 - 10 = 0.5x - 0.1x
2 = 0.4x
x = 2 / 0.4
x = 5
So, to obtain a 50% solution, 5 liters of a 10% acid solution should be added to 20 liters of a 60% acid solution.