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For a college chemistry experiment, students need to prepare a solution containing ethylene glycol. They are each given 2020 fluid ounces of a solution containing 2%2\% ethylene glycol, to which they will add another solution that contains 12%12\% ethylene glycol. How much of the 12%12\% solution should they add to obtain a 5%5\% ethylene glycol solution? \newlineWrite your answer as a whole number or as a decimal rounded to the nearest tenth.\newline\newline____\_\_\_\_ fluid ounces\newline

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Q. For a college chemistry experiment, students need to prepare a solution containing ethylene glycol. They are each given 2020 fluid ounces of a solution containing 2%2\% ethylene glycol, to which they will add another solution that contains 12%12\% ethylene glycol. How much of the 12%12\% solution should they add to obtain a 5%5\% ethylene glycol solution? \newlineWrite your answer as a whole number or as a decimal rounded to the nearest tenth.\newline\newline____\_\_\_\_ fluid ounces\newline
  1. Denote 1212\% Solution: Let's denote the amount of the 1212\% solution to be added as xx fluid ounces.
  2. Calculate Ethylene Glycol: The amount of ethylene glycol in the original 2020 fluid ounces of 2%2\% solution is 20×0.0220 \times 0.02.\newlineCalculation: 20×0.02=0.420 \times 0.02 = 0.4 fluid ounces of ethylene glycol.
  3. Calculate Ethylene Glycol: The amount of ethylene glycol in the xx fluid ounces of 12%12\% solution is x×0.12x \times 0.12.\newlineCalculation: x×0.12=0.12xx \times 0.12 = 0.12x fluid ounces of ethylene glycol.
  4. Total Volume After Mixing: After mixing the solutions, the total volume of the solution will be 20+x20 + x fluid ounces, and the total amount of ethylene glycol will be 0.4+0.12x0.4 + 0.12x fluid ounces.
  5. Equation for 55% Solution: To have a final solution of 55% ethylene glycol, the equation (0.4+0.12x)/(20+x)=0.05(0.4 + 0.12x) / (20 + x) = 0.05 must be true.
  6. Multiply Both Sides: Multiply both sides of the equation by (20+x)(20 + x) to get rid of the denominator.\newlineCalculation: (0.4+0.12x)=0.05×(20+x)(0.4 + 0.12x) = 0.05 \times (20 + x)
  7. Distribute 0.050.05: Distribute 0.050.05 on the right side of the equation.\newlineCalculation: 0.4+0.12x=1+0.05x0.4 + 0.12x = 1 + 0.05x
  8. Rearrange Equation: Rearrange the equation to isolate xx on one side.\newlineCalculation: 0.12x−0.05x=1−0.40.12x - 0.05x = 1 - 0.4
  9. Combine Like Terms: Combine like terms.\newlineCalculation: 0.07x=0.60.07x = 0.6
  10. Solve for x: Solve for x by dividing both sides by 0.070.07.\newlineCalculation: x=0.60.07x = \frac{0.6}{0.07}
  11. Perform Division: Perform the division to find the value of xx.\newlineCalculation: x=8.57142857x = 8.57142857
  12. Round to Nearest Tenth: Round the result to the nearest tenth.\newlineCalculation: x≈8.6x \approx 8.6 fluid ounces

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