I need help atleast an equation to get started please
A lab assisstant wants to make five liters of 27.8% acid solution. If solutions of 50% and 13% are in stock, how many liters of each must be mixed to prepare the solution.
.5x + .13(5-x) = .278*5
x=2
so, use 2L of 50% and 3L of 13%
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1 n 4
To find out how many liters of each solution should be mixed, we can use a system of equations.
Let's denote the volume of the 50% acid solution as x and the volume of the 13% acid solution as y.
Since we want to make a 5-liter solution, we can create the first equation:
x + y = 5
Now, let's calculate the amount of acid in each solution.
For the 50% acid solution, the amount of acid is 50% of the volume (x liters) or 0.5x liters.
For the 13% acid solution, the amount of acid is 13% of the volume (y liters) or 0.13y liters.
The total amount of acid in the final 5-liter solution is 27.8% of 5 liters or 0.278 * 5 liters = 1.39 liters.
So, the second equation we can create is:
0.5x + 0.13y = 1.39
Now we have a system of equations:
x + y = 5
0.5x + 0.13y = 1.39
To solve this system, we can use the method of substitution or elimination.
Let's solve it using the method of substitution:
From the first equation, we can rewrite y in terms of x: y = 5 - x.
Now we substitute this expression for y into the second equation:
0.5x + 0.13(5 - x) = 1.39
Let's simplify:
0.5x + 0.65 - 0.13x = 1.39
Combine like terms:
0.37x + 0.65 = 1.39
Subtract 0.65 from both sides:
0.37x = 0.74
Divide by 0.37:
x = 0.74 / 0.37
x = 2
Now, substitute this value of x back into the first equation to find y:
2 + y = 5
y = 5 - 2
y = 3
Therefore, you would need to mix 2 liters of the 50% acid solution with 3 liters of the 13% acid solution to make 5 liters of a 27.8% acid solution.