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4 years ago

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A hospital needs to dilute a 50% boric acid solution to a 10% solution. If it needs 25 liters of the 10% solution, how much of t

he 50% solution and how much water should be used?

2 answers:

Leona [35]4 years ago

7 0

You need one equation of this problem to balance the amount of a boric acid.

The amount of boric acid in 50% solution is expressed as 0.50*x

Next, the amount of water is taken from 25 liters and x (based on 50% solution), that is 25 - x. However, there is NO amount of BORIC ACID in water, and that is 0.

For the output solution, the amount of boric acid will be 0.10*2.5 = 2.5

Then, the balance equation is
0.50*x + 0*(25 - x) = 2.5
Solving for x,
0.50*x + 0 = 2.5
x = 2.5/0.5 = 5

Therefore the amount of liters in 50% solution is
5*0.50 = 2.5 liters
The amount of water is 25 - 5 = 20 liters

stiv31 [10]4 years ago

5 0

Answer:

20 liters of water and 5 liters of 50% solution

Step-by-step explanation:

Balance the amount of acid:

amount of 50% solution: x liters

amount of acid in 50% solution: .5x liters (concentration * volume)

amount of water: 25 - x liters (since 25 altogether, and x is 50% solution)

amount of acid in water: 0

final solution: 25 liters

amount of acid in final 10% solution: .10 * 25 = 2.5

balance the "goes into" and the "comes outta"

.50x + 0 = 2.5

x = 2.5 / .5 = 5

therefore 5 liters of 50% solution (for 2.5 liters of acid)

mixed with 25 - 5 = 20 liters of water

will yield 25 liters of a 10% solution (2.5 liters of acid)

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Step-by-step explanation:

5x + y = 8

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Step 2: Subtract 120 from both sides.

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Now that we know the value of x, we can solve for y in any of the equations, but let's use the equation "y = −5x + 8"

$\mathrm{So\:it\:would\:look\:like\:this:\ y = -5 \left( \dfrac{ 53 }{ 30 } \right) +8}$

$\mathrm{Now\:lets\:solve\:for\:"y"\:then}$

$y = -5 \left( \dfrac{ 53 }{ 30 } \right) +8}$

$\mathrm{Express\: -5 \times \dfrac{ 53 }{ 30 }\:as\:a\:single\:fraction}$

$y = \dfrac{ -5 \times 53 }{ 30 } +8$

$\mathrm{Multiply\:-5 \:and\:53\:to\:get\:-265 }$

$y = \dfrac{ -265 }{ 30 } +8$

$\mathrm{Simplify\: \dfrac{ -265 }{ 30 } \:,by\:dividing\:both\:-265\:and\:30\:by\:5} }$

$y = \dfrac{ -265 \div 5 }{ 30 \div 5 } +8$

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$\mathrm{Turn\:8\:into\:a\:fraction\:that\:has\:the\:same\:denominator\:as\: - \dfrac{ 53 }{ 6 }}$

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$\mathrm{Convert\:8\:to\:fraction\:\dfrac{ 48 }{ 6 }}$

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So there is only one solution to the equation.

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