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When 71.43 mL of 0.855 M HNO3 is diluted with 20.57 mL of water, the molarity of the nitric acid becomes

  • chemistry -

    The dilution is
    c1v2 = c2v2 but I prefer to reason it saying M has been diluted by how much?
    0.855M x (71.43 mL/(71.43+20.57) = ?
    By the way, this assumes that the volumes are additive which is not always the case.

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