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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
- 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+?A 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-13) 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 A 1.00 mL aliquot of this was withdrawn, placed into a 10.00 mL volumetric flask and diluted to the mark.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)
- 4. 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.The absorption coefficient of rhodopsin at 496 nm is 40,000 M-1cm-1. If you have a 1M solution of rhodopsin, how many times will you dilute the solution so that the diluted solution will give you an absorption reading at 496 nm that is in the range of 0.1-1.0. And what will be the absorption by the diluted solution?Unrounded Rounded ε∗,L/μmolε∗,L/μmol 0.0259825 0.0260 heres data: TZ # Concentration Absorbance at 430 nmnm 1 33.6480 0.931 2 25.2360 0.757 3 16.8420 0.210 4 8.41200 0.137 5 4.20600 0.122
- Describe the mesomeric or resonance effect and differentiate between types +E or +M and -R or -M.If a radiation intensity l0 = 2.5x1010 fotones s-1 cm2 causes a dissolución and an absorbance of 0.95 will be recorded. How much incident radiation is absorbed by the music screen?What will be the corresponding concentration for a solution having an absorbance of 0.350 and a molar absorptivity of 420 L mol-1 cm-1. (path length = 1.0 cm)
