The following two solutions were combined and mixed well: 150.00 mL of a 3.500 M iron(¡¡¡) nitrate solution and 760.00mL of a 1.600M magnesium nitrate solution. 20.00mL of the final solution was transferred with a pipette into an empty 250.00mL volumetric flask. It was made up to the mark with distilled water. Calculate the nitrate concentration in the final solution.

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  1. millimols Fe(NO3)3 = mL x M = 150 x 3.5 = 525
    mmols NO3^- from Fe(NO3)3 = 3*525 = approx 1500 but you need it more accurately.
    mmols Mg(NO3)2 = 760 x 1.6 = 1216
    mmols NO3^- from Mg(NO3)2 = 2*1216 = approx 2400
    Total NO3^- = approx 2400 + 1500 = approx 3900 mmols.
    Total volume assuming volumes are additive = 150 + 760 = 910 mL
    You took 20 of that so you have how many millimols NO3^- in that 20?
    That's approx 3900 x (20/910) = approx 86. So now you have 85 mmols NO3^- in 250 mL and concn is M = millimols/mL = ?

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