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Explain the types of electromeric effects +E and -E.
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- Three compounds (A, B and C) are mixed together ( they are present in the same container). Molar absorptivity coefficients of these compounds at 575nm -1 -1 are: 2500, 3750 and 5250Lcm mol, respectively. Concentrations are: 2 × 10M for A, 5× 105 M for B and -6 1.5×10 M for C. What is the absorbance measured at 575mm, in 1.0cm cuvette of the solution containing all three compounds? Show your calculations.A 2.0 gram of a certain substance with a molecular weight of 100 g/mol is dissolved in a 500 ml solution. The length of the cuvette is 2.0 cm and absorbance of 0.301. What is the extinction coefficient in cm-1 M-1?7) The correct structure that corresponds to the spectroscopic data given below is [IR 2720, 1710 1 point cm*1. 1H NMR 9.80(s, 1H), 7.50(dd, J = 8,0Hz, 2Hz, 1H), 7.40(d, J = 2.0Hz, 1H), 6.90(d, J = 8.0Hz, 1H), 3.90(s, 3H), 3.80(s, 3H)] COOH OMe Me CHO OMe OMe CHO OMe OMe COOH OMe Me
- Calculate and report the energy, in units of joules, associated with each absorbance maximum.the absorbance of an equilibrium mixture of Fescn2+ was measured at 447nm and found to be 0.319. What is the equilibrium concentration of FeSCN2+?Explain the role of timescale of the experiment and the type of information you get from NMR.
- Please help me with this, I am trying to prepare for my final exam next week and I am just completely lostA 2.78 x 10-4 M solution of a compound has an absorbance of 0.469 at 520 nm in a 1.00 cm cell. The solvent's absorbance under the same conditions is 0.024. (a) What is the molar absorptivity of the unknown compound? 4.0 1687 X M-¹cm-1Avobenzone is used in sunscreens. A UV spectrum of a 3.0 mL sample of avobenzone (310.4 g/mol) in a 1.0 cm cell was obtained. The molar extinction coefficient for avobenzone is 9120 M¹¹cm ¹. The UV spectrum revealed a λ(max) of 300 nm with an absorbance of 0.80. Calculate the mass of avobenzone in the solution. O
- 3) A compound, X, with a molecular weight of 292.16 g/mol was dissolved in a 5.00 mL volumetric flask and diluted to the mark. A 1.00 mL aliquot of this was withdrawn, placed into a 10.00 mL volumetric flask and diluted to the mark. The absorbance at 340 nm was 0.427 in a 1.000 cm cuvette. The molar absorptivity of X at 340 nm is 6130 M-¹ cm-¹ a) Using Beer's Law, what is the [X] (concentration in molarity) for the diluted solution in the cuvet ? mmmmm b) Use the dilution equation (McVc = MdVa) to calculate [X] in the original 5.00 mL flask. c) How many milligrams of X was used in the original solution ? (Find total moles X using original concentration and original volume - and then use the molar mass to go from moles to grams and then milligrams)Identify, synthesized and show the mechanism show in the spectroscopy4. The total absorbance of a solution is the sum of the absorbances of all different materials present. AT = Aa + Ap + Ac .. etc. At 427 nm, the molar absorbtivity of Yellow #5 is 2.73×10°M-1-cm1. At 427 nm the molar absorptivity of Red 40 is 7.49x103 M-1.cm-1. Calculate the total absorbance of a solution in which the concentration of Yellow #5 = 2.00x10-5 M and the concentration of Red 40 = 4.80x10-5 M, when measured at 427 nm in a 1.00 cm cell.
