Can you help me find the mixtures and the average on the second page. The first page has the experimental data on it Conclusion:KMixture #1133Mixture #2Mixture #3Mixture #4Mixture #5Average|||| Vol. of 2.00 x 10³ MVol. of 2.00 x 10:3 MVol. of H20MixtureFe(NO;)3 (in mL)KSCN (in mL)(in mL)Absorbance[FESCN*2] = A/5x10315.001.004.00.115.115/5x103 = 2.3x10-5M5.002.003.00.234.34035.003.002.0045.004.001.00.430.5255.005.000.00Mixture 1 Calculation:Fe*3SCN-1FESCN+2Start (2x10-³M)(5ml) = 1x10-3M(10ml)-X(2x10-3M)(1ml) = 2×104M((10ml)-XChangeX+Equil. 1x10-3М-X2×104M -X9.77х104M1.77x10-M2.3x10-5MK = [FESCN+21[Fe**][ SCN]= 133

We are given Initial molarity of iron(III) nitrate = 2×10-³ M Initial molarity of KSCN = 2×10-⁴ M…

Answered: Mixture #2 Mixture #3 Mixture #4…

Science

Chemistry

Mixture #2 Mixture #3 Mixture #4 Mixture #5 Average

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter26: Molecular Absorption Spectrometry

Section: Chapter Questions

Problem 26.21QAP

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Can you help me find the mixtures and the average on the second page. The first page has the experimental data on it

Conclusion:
K
Mixture #1
133
Mixture #2
Mixture #3
Mixture #4
Mixture #5
Average
|
|||

Vol. of 2.00 x 10³ M
Vol. of 2.00 x 10:3 M
Vol. of H20
Mixture
Fe(NO;)3 (in mL)
KSCN (in mL)
(in mL)
Absorbance
[FESCN*2] = A/5x103
1
5.00
1.00
4.00
.115
.115/5x103 = 2.3x10-5M
5.00
2.00
3.00
.234
.340
3
5.00
3.00
2.00
4
5.00
4.00
1.00
.430
.525
5.00
5.00
0.00
Mixture 1 Calculation:
Fe*3
SCN-1
FESCN+2
Start (2x10-³M)(5ml) = 1x10-3M
(10ml)
-X
(2x10-3M)(1ml) = 2×104M
((10ml)
-X
Change
X+
Equil. 1x10-3М-X
2×104M -X
9.77х104M
1.77x10-M
2.3x10-5M
K = [FESCN+21
[Fe**][ SCN]
= 133

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