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if I have this given to me how do I find the concentration of FeSCN2+ at...
If this data is given how do I fill in the table Beaker [FeSCN2+), M 1.60E-04 Absorbance 1 SDY 0.618 1.20E-04 N 6.11E-02 SDX 9.08E- 06 6.81E- 06 4.54E- 06 2.27E- 06 0.492 3 8.00E-05 4.86E-02 0.346 4 4.00E-05 3.42E-02 0.157 1.55E-02 From Prelab Y=mx+b Y = absorbance @ 455.9 m = slope = 3822.5 b = (Y-intercept) - 0.021 +252.3 20.02764 X= Concentration 0.8 270 characters English Part I: Absorbance reading for Standard Solutions <1.5 marks> Beaker [FeSCN?) Absorbance...
How do I find the concentration of KSCN & FeSCN2+ from the following data? Solution Beaker 1 Beaker 2 Beaker 3 Beaker 4 Beaker 5 Volume of 0.200 M Fe(NO3)3. mL 2.5 mL 2.5 mL 2.5 mL 2.5 mL 2.5 mL Volume of 0.0020 M KSCN, mL .50 mL .75 mL 1.0 mL 1.25 mL 1.5 mL Volume of 0.50 M HNO3, mL 7 mL 6.75 mL 6.5 mL 6.25 mL 6 mL Total volume, mL 10 mL 10 mL...
please do 1 through 4. Thank you. 0 1. Fe 1.SCN Procedure B: 1. Prepare ICE tables for beakers 2 - 6 using the example below as a guideline (you will have 5 different ICE tables). Table 3. Sample "ICE" table. Fe(aq)*3 + SCN(aq) = FeSCN2 Initial: **M calculated **M calculated using your Table 2 using your Table 2 volumes volumes Change: - 1x - 1x Equilibrium: M 1x 1x = concentration calculated from Procedure A slope-intercept equation +1x M...
How do you find the initial concentration if you are only given the initial volume and final volume? I have tried using the V(i) x C(i) =V(f) x C(f) equation but it needs at least 3 figures so I'm a bit lost right now. DILUTION WORKSHEET Solution #1 Concentration = 0.0232 0% Solution in volumetric flask before water is added to the line Solution in volumetric flask after water has been added to the line Diluted Solutions V #3 18.00mL...