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2. In Part A of this experiment, you prepare five FeSCN solutions (one that is just a blank) according to the reaction below Fe (aq) SCN (aq) > FeSCN (a) SCN formed We assume that the starting SCN determines the concentration of Fe (because Fe is in excess and SCN is limiting). Calculate the concentration ot FeSCN2 that forms for each of the solutions (Beakers 1-4) and fill out the table below. Show your calculations beneath the table. Concentration of Beaker 0.200 M0.0020 M Number Fe(NOs)s KSCN (mL) H2O (mL) FeSCN M) mL 5.0 5.0 5.0 5.0 5 (Blank)5.0 5.0 4.0 3.0 2.0 0.0 40.0 41.0 42.0 43.0 45.0
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Answer #1

Anowa 2 As frescn²t] is proportional to [SEN] So, [feseNe can be calculated asi- Presen?] . molarity of K KSCN X volume of Ks

Answery Pe3t + SEN & Presenzer ke rescN21] [Fe3+] [sent] from absorbance data 0.118 . m = 0.118 -0.0483 - 10.000d 3 M7 2417.5

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