4) If the classification of the chromatography run is a reverse-phase chromatography, discuss the relative polarities of caffeine, aspartame and benzoate
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4) If the classification of the chromatography run is a reverse-phase chromatography, discuss the relative polarities of caffeine, aspartame and benzoate
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- A solution of tryptophan has an absorbance of 0.64 at 280 nm. Given with of 6.04 x 103 Msolution of tryptophan and path length of 0.5 cm. What is the molar absorptivity oftryptophan?Please create a caption for this table. Solution NaCl Conc. (%) Osmolality (mOsm) % transmittance Absorbance % hemolysis % crenation C distilled 0 0 0.001029 4.987584625 100 0.03354 1 0.177179111 54.61 0.001551 4.809388202 96.42720001 0.05837 2 0.297126222 91.58 0.01012 3.994819487 80.09527231 0.08444 3 0.442542222 136.4 3.849 1.414652089 28.3634704 0.134 4 0.590164444 181.9 64.8 0.188424994 3.777880643 0.2125 5 0.74752 230.4 95.64 0.019360433 0.388172513 0.3368 6 0.89644 276.3 99.56 0.001915112 0.038397585 0.5336 7 1.095648889 337.7 99.98 8.68676E-05 0.001741676 0.9834 8 1.336711111 412 100 0 0 2.1 9 1.755568889 541.1 100 0 0 7.9 10 2.674395556 824.3 100 0 0 57.83 11 4.490211111 1384 100 0 0 99.72As part of an investigation on Pb in biological ecosystems, some grasses that grew along the road and were exposed to emissions from motor gasoline were collected. 6.2501 g of grass were calcined to destroy all organic matter, the inorganic residue was properly treated and finally made up to 100 mL. An aliquot of 50 mL of this solution was read in an atomic absorption spectrophotometer registering an absorbance of 0.1250; then 0.10 mL of a standard solution of 55 µg of Pb / mL were added to the other 50 mL, with this fortified sample, another absorbance measurement was made, which was 0.1798. Calculate the µg of Pb / g of grass
- Attatched is UV-Vis Spectrum (Isopropanol, c= 1.65 * 10-4 M) Include the UV-Vis data in a table with wavelength, absorbance. and molar extinction coefficient Wavelength Absorbance Molar Extinction Coefficient 300-700 nmThe absorbance values at 250nm of 5 standard solutions, and sample solution of a drug are given below: Conc. (ug/ml) A 250 nm10 0.16820. 0.32930 0.50840. 0.66050 0.846Sample. 0.661Calculate the concentration of the sampleCalculate the concentration of a colored solution if the molar absorptivity is 39.80 M-1•cm-1 and the absorbance is 0.777. The path length of the sample is 1.00 cm.
- A calibration plot of absorbance vs concentration (ppm) was obatined with standard known Red 40 dye solution. - the slope of the best-fit straight line of the plot is 0.057 ppm^-1 - The absorbance of the dilute unknown sports drink was 0.88 What is the concentration of this dilute unknown sports drink?______ the answer is 15.4 ppm ( can you please show me how to calculate this problem)Sally obtains a standard calibration curve for their assigned food dye by plotting absorbance versus concentration (in M) of known solutions and finds the following slope and intercept: y = 1.646x + 0.026 Note: Copying and/or posting this or other questions without the express written permission from Dr. Burke and the Department of Chemistry & Biochemistry at the University of Delaware is a violation of intellectual property copyright law. Sally finds that her original Kool-aid sample is too concentrated, so they dilute it by transferring 13.00 mL of the original Kool-aid sample into a new container and diluting to a total volume of 37.00 mL. If the absorbance of the dilute Kool-aid sample at the appropriate wavelength for their assigned dye is 0.950, determine the concentration (in M) of the assigned food dye in the original Kool-aid sample. Report your answer with three places after the decimal.Compound X was dissolved in a 5-mL volumetric flask. A 1.00-mL aliquot of the compound was placed in a 10-mL volumetric flask and diluted to the mark. The solution was read in a UV-Vis spectrophotometer and an absorbance of 0.427 in a 1.000-cm cuvette was observed at 340 nm. Solve the following question. Data for Compound X:Molecular mass = 292.16 g/molMolar absorptivity (ɛ) at 340 nm= 6130 M -1 cm -1 a. Calculate the concentration of compound X in the cuvette. b. How many milligrams of compound X were used to make the 5-mL solution?
- Compound X was dissolved in a 5-mL volumetric flask. A 1.00-mL aliquot of the compound was placed in a 10-mL volumetric flask and diluted to the mark. The solution was read in a UV-Vis spectrophotometer and an absorbance of 0.427 in a 1.000-cm cuvette was observed at 340 nm. Solve the following question. Data for Compound X:Molecular mass = 292.16 g/molMolar absorptivity (ɛ) at 340 nm= 6130 M -1 cm -1 1. What was the concentration of compound X in the 5-mL flask?Calculate the concentration of a solution of peptide AWSME, in mM if the absorbance at 280 nm for the solution is 0.6 absorbance units, and a 0.2-cm cuvette is used. For the extinction coefficient, take into account that e280(Trp) = 5690 M-1 cm-1 and e280(Tyr) = 1280 M-1 cm-1. Round your answer to two decimals.A solution of Tryptophan has an absorbance at 280 nm of 0.54 in a 0.5 cm length cuvette. Given the absorbance coefficient of trp is 6.4 x 103 Lmol-1cm-1. What is the concentration of solution? The options are as follows A) 0.17 x10-3 mol/L B)2.56x10-3mol/L C)0.17x10+3mol/L D)2.56x10+3mol/L