Getting a solution that contains 40% iodine

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The discussion focuses on calculating the volumes of two iodine solutions needed to create a 50-liter mixture containing 40% iodine. The first solution is 30% iodine, and the second is 80% iodine. The equation derived from the problem is 50 * 0.4 = 0.3x + 0.8y, where x is the volume of the 30% solution and y is the volume of the 80% solution. The final solution confirms that 40 liters of the 80% iodine solution is required to achieve the desired concentration.

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Richay
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For solving this problem?

Two containers are on the shelf. The first one contains a 30% iodine solution and the other contains an 80% iodine solution. How much of the 30% solution should be used to get 50 liters of a solution that is 40% iodine?

Thanks for any help.
 
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you should really attepmt the problem before posting here.. but-


okay... how many litres of the 80% stuff is there.. that's the limiting, otherwise you might as well use 25 litres of it diluted with water to 50 litres


or if you can't dilute with distilled water, like any reasonable person would... then i suppose i would set the thing to equal 50*.4 then the "thing" as i ahve described it would be .3*x and .8*y

...so 50*.4=.3*x plus .8*y

then you get some limiting somehting or other prob in the equation...
 
Last edited:
I attempted the problem but my freaken fomula kept being weird.. uh weird results

Ohkay I get it thanks.

The answer is 40 ^__^
 

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