Some info Fe2+ + 3 phen - Fe(phen)?+ Imax = 502 nm Colorless Orange amp; = 0.840 mM-1cm-1 phenanthroline 2.5 g/L (180.21 g/mol) -gt; 13.87 mM

*please ignore the writing, it is not mine so I am not sure how to go about this question!

Some infoFe2+ + 3 phen - Fe(phen)?+Imax = 502 nmColorlessOrange& = 0.840 mM-1cm-1phenanthroline 2.5 g/L (180.21 g/mol) -> 13.87 mMFe(NHA)2(SOA)2 . 6H20 2.81 g/L (392.14 g/mol) -> 7.17 mM* Titrate 10 ml of the Fe soln & monitor spectrophotometrically. *Questions & CalculationsgivemICMV. What is the equiv pt volume Ve?product. Complete the following tableA=EbcXFe =nFe* . Plot A VS . XFeDV+10mLnFe +nphenfractionVtotBeer'sA Lann = combined moles of free andVolV (mL)[product]bound analytesAbs0.25Ve3.875 0, 00129XFeXphen = \0.50V.7.750.75Ve11.GZSVe15.51.25Ve 19.37S1.50Ve 33.08