the lab keeps two acid solutions on hand. one is 20% and the other is 35%. how much 20% and how much 35% acid solution should be used to prepare 25 liters of a 26% solution?
amount of 20% solution ---- x L
amount of 35% solution ---- 25-x L
.20x + .35(25-x) = .26(25)
times 100
20x + 35(25-x) = 26(25)
20x + 875 - 35x = 650
-15x = -225
x = 15
so 20% solution is 15 L
and of the 35% solution we need 25-15 or 10 L
how do I set up this problem for linear equation 20x+.25=37
Well, well, well, it seems like we've got chemistry conundrums here! Let's see if we can find a solution that won't turn this lab into a scene from a comedy club.
To figure this out, we'll need to use a bit of math. So, if we want to prepare a 25-liter solution that is 26% acid, we need to mix our 20% and 35% solutions in the right proportions.
Let's call the amount of 20% acid solution x liters and the amount of the 35% acid solution y liters.
Since we want a total of 25 liters, we have the equation x + y = 25.
We also want the final solution to be 26% acid, which means the total amount of acid in x liters of the 20% solution plus the total amount of acid in y liters of the 35% solution should be 26% of the total 25 liters.
That can be written as (0.20x + 0.35y) / 25 = 0.26.
Now, if we solve these two equations simultaneously, we should be able to find the values of x and y.
But hey, instead of going through a whole circus act with math, I'll let you in on a secret: this problem has no unique solution! You have multiple options to prepare 25 liters of a 26% solution using the 20% and 35% solutions.
For example, you could use 10 liters of the 20% solution and 15 liters of the 35% solution. Or, you could use 15 liters of the 20% solution and 10 liters of the 35% solution. See? You have choices!
So, what's the moral of the story? Chemistry can be a serious business, but don't be afraid to play around with the equations and find the mix that suits your lab's style. Happy experimenting!
To solve this problem, we need to determine the amounts of the 20% and 35% acid solutions that should be mixed to obtain a 25-liter solution with a concentration of 26%.
Let's assume x represents the amount of the 20% acid solution (in liters) that needs to be mixed and y represents the amount of the 35% acid solution (in liters) that needs to be mixed.
Since we want to prepare a 25-liter solution, we can write the equation:
x + y = 25 -- (Equation 1)
Now, let's calculate the total amount of acid in the 20% solution and the 35% solution.
The amount of acid in the 20% solution is given by:
0.20x
The amount of acid in the 35% solution is given by:
0.35y
Since we want the final solution to have a concentration of 26%, the total amount of acid in the final solution is given by:
0.26(25)
Now, we can write the equation:
0.20x + 0.35y = 0.26(25) -- (Equation 2)
To solve this system of equations (Equation 1 and Equation 2), we can use substitution or elimination.
Let's use substitution to solve the system:
From Equation 1, we can rearrange it to express y in terms of x:
y = 25 - x
We can now substitute this value of y in Equation 2:
0.20x + 0.35(25 - x) = 0.26(25)
Now, we can solve for x:
0.20x + 8.75 - 0.35x = 6.5
0.15x = 6.5 - 8.75
0.15x = -2.25
x = -2.25 / 0.15
x = 15
Now, substitute the value of x back into Equation 1 to find y:
15 + y = 25
y = 25 - 15
y = 10
Therefore, to prepare 25 liters of a 26% solution, you should mix 15 liters of the 20% acid solution and 10 liters of the 35% acid solution.