Question

Partner: Data Sheet & Calculations. Part I: For this part of the experiment assume (SCN"]= FeSCN21 Table 3: Trial % Transmittance [SCN); [Fe(SCN)?) Absorbance 2 89.9 0.00004 0.0462 0.00004 3 79.1 0.00008 0-1018 0.00012 4 66.1 10.17918 000012 5 58.3 0.00016 10.2343 o.ooola 0.00020 6 51.0 0.2924 0.00 02 apb00% Calculate the [SCN Ji and [Fe3+]; accounting for dilution using MeV. - MV., for each solution in Table 1. Show one sample calculation for absorbance, (SCN) and [FeSCN?"). Enter all calculated values into Table 3 above. Absorbance: Use Eqn. #2 (2.000 - log (%) Dilution Formula [SCN-) Trial 2 A=2-09 (89.9) =0.0462 [FeSCN2") _- _[SCN-) [Fe] >> [SCN']; equilibrium in ean. 4 is far to the right, using up essentially all of the SCN ions For every one mole of SCN reacted, one mole of FeSCN is produced. Thus, equilibrium moles and molarity of FeSCN are assumed to be equal to initial moles SCN, Graph: Plot Absorbance vs. [FeSCN2") from the data from Table 3 above & fit a trendline. Print a well-labeled graph with a complete Data Table, showing original data and calculated and graphed values. Report the trendline equation along with the R2 value.

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question: in.....typed & image format...

$\Rightarrow$Answer:

As shown in the Explanation, section below...

$\Rightarrow$Explanation:

• Step -1:

We have been given the following data:  ( table prepared with MS Excel, presented in..image format...)

Trial Transmittance (SCN^-Jinit, M (mol/L) 2 فيما 4 89.9 79.1 66.1 58.3 51 4.00E-051 8.00E-05 1.20E-041 1.60E-04 2.00E-04 5 6

$\Rightarrow$ Following is the graph , of  [SCN^-]_i vs. %Transmittance , using the above data: prepared by using MS Excel, pesented here in...image format...

[SCN-linit vs. % Transmittance graph 2.50E-04 2.00E-04 1.50E-04 [SCN-Jinit, M (mol/L) 1.00E-04 [SCN-init vs. %Transmittance graph Linear (SCN-linit vs. % Transmittance graph) y = -4E-06% +0.000 5.00E-05 0.00E+00 0 20 40 60 80 100 %Transmittance

From the above graph, we get the followinng information:

$\Rightarrow$ Equation of Linear Trendline:    y = -4 x 10^-6  x ( x ) + 0.00 ------------------Eq.(1)

Where:

1. y = [SCN^-]_i =   Initial molar concentration of SCN^- [=] M (mol/L)
2. x = %Transmittance
3. slope = -4.0 x 10^-6

Therefore, using Eq.(1), we can convert:  %Transmittance into [SCN^-]_i, i.e. initial molar concentration of SCN^- ...

• Step - 2:

​​​​​​​We know the following relationship:

$\Rightarrow$A = 2.0 - log₁₀ ( %T ) -------------------------------------Eq. (2)

We are given with the following information, in table format, and using Eq.(2) and the given information, we can calculate the following: Absorbances , using MS Excel, presented here in....image format....

Trial A B %Transmittance Absorbance 88.1 0.0550240921 82.1 0.085656843 77.7 0.109578981 72.3 0.140861703 68.8 0.162411562) C D E

$\Rightarrow$ Above is the Table: ( 4 ), we have obtained... ( Answer )

• Step - 3:

​​​​​​​We know:

$\Rightarrow$ [SCN^-]_init = [Fe³⁺]_i   

Therefore:

Following will be the desired values of [SCN^-]_ivalues and [Fe³⁺]_i  values , obtained using Eq. ( 1 ) and MS Excel...

in...image format...

Trial %Transmittance [SCN^-/init [Fe^3+jinit A 88.1 -0.0003524 B 82.1 -0.0003284 C 77.7 -0.0003108 72.3 -0.00028921 E 68.8 -0.0002752 -0.0003524 -0.0003284 -0.00031081 -0.0002892 -0.00027521 A

$\Rightarrow$ Above is the desired:   Table: (5) .......i.e. Answer....

------------------------------------------------------------------------------------------------------------------

​​​​​​​