I need help with this problem:
if you want to reduce the strength of 100 mL of 10 N NaOH to 3N, how much water would you need to add?
Thanks in advance.

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asked by John
  1. mL1 x N1 = mL2 x N2

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    posted by DrBob222
  2. Oh yeah! Thanks, so you would substitute:
    (10)(100) = (3)(x) to get:
    x = 333.33 mL, right?

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    posted by John

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