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How many milliliters of 60% carbonic acid must be mixed with how many milliliters of 15% carbonic acid to make 650 milliliters of a 38% carbonic acid solution

Sagot :

maacastrobr

Answer:

348.9 mL of the 60% solution and 251.1 mL of the 15% solution.

Explanation:

First, we calculate how many mililiters of pure carbonic acid are there in 650 mL of a 38% solution:

  • 650 mL * 38/100 = 247 mL

Then we can express the sum of both initial solutions as:

  • 1)  x * 60/100 + y * 15/100 = 247

for the volume of carbonic acid; and

  • 2)  x + y = 600 mL

For the volume of the solutions.

We now have a system of two equations and two unknowns (x is the volume of the 60% solution and y is the volume of the 15% solution).

We express x in terms of y in equation 2):

  • x = 600 - y

And replace x in equation 1):

  • (600 - y) * 60/100 + y * 15/100 = 247
  • 360 - 0.6y + 0.15y = 247
  • -0.45y = -113
  • y = 251. 1 mL

Finally we calculate x using equation 2):

  • x + 251.1 = 600
  • x = 348.9 mL

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