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A 46.2-mL, 0.568 M calcium nitrate [Ca(NO3)2] solution is mixed with 80.5 mL of1.396 M calcium nitrate solution. calculate the concentration of the final solution

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3 answers
  1. Not 100% sure if this is right, but here's what I think it is.

    M = mol/L
    .568 = mol/.0462 L
    = .0262416 mol

    1.396 = mol/.0805 L
    = .112378 mol

    (.0262416 mol + .112378 mol) / (.0462 L + .0805 L) = M

    M = 1.0094077348

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  2. Thanks, the answer's 1.09M

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  3. oops!

    (.0262416 mol + .112378 mol) / (.0462 L + .0805 L) = M

    M = 1.0941 and not 1.0094077

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