The chemistry lab at the university of hardwoods keeps two acid

solutions on hand. one is 20% acid and the other is 35% acid.

how much 20% acid solution and how much 35% acid solution should

be used to prepare 25 liters of a 26% acid solution?

[note: your answer should look something

like this: 45, 10]

fill in the blanks and list your answers it in same order as the

blanks.

x = liters of 20% solution and liters of the 35%

solution.

  [15, 10]

Step-by-step explanation:

Let x represent the quantity of 35% solution. Then the amount of acid in the mix is ...

  (25 -x)×20% + (x)×35% = (25)×26%

  5 -0.2x + 0.35x = 6.5 . . . . . simplify

  0.15x = 1.5 . . . . . . . . . . . . . . .subtract 5

  x = 1.5/0.15 = 10 . . . . . . . . . . divide by 0.15

Then the amount of 20% solution is 25-10 = 15 liters.

  [20% solution, 35% solution] = [15, 10] . . . liters

You can also consider the ratio of one solution to the other. That ratio will be ...

  high concentration : low concentration = (mix% -low%) : (high% -mix%)

  = (26 -20) : (35 -26) = 6 : 9 = 2 : 3

The total of these ratio units is 2+3 = 5, and the total number of liters of mixed solution is 25 liters, so each ratio unit must stand for 25/5 = 5 liters. Then you have ...

  high concentration : low concentration = (2·5 liters) : (3·5 liters) = 10 : 15 liters

That is, ...

  [15, 10] liters of [20%, 35%] solution must be mixed to get 26% solution.

This method of solving mixture problems is so simple it can often be done mentally.

a variable is any characteristics, number, or quantity that can be measured or counted. a variable may also be called a data item. hope i !

where is the graph, sweetheart?

step-by-step explanation: