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10 December, 20:26

How many liters of a 10% silver iodide solution must be mixed with 3 L of a 4% silver iodine solution to get s 6% solution?

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  1. 10 December, 21:05
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    To answer this problem, you should just substitute some figures in order to get the required amount.

    So, let a be the amount of 10% iodide solution

    So ...

    4 (3) + 10a = 6 (3+a)

    [4 represents the silver iodine solution, the 3 is the liters, while we are looking for the 6% solution]

    12 + 10a = 18 + 6a

    10a - 6a = 18 - 12

    4a = 6

    a = 6/4

    a = 3/2

    So you will need 3/2 or 1.5 liters of 10% iodide solution.
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