1. A 100-fold dilution (1/100-dilution; 1 part sample: 99 parts diluent, most of the time water) of a 15 µM solution results in a final concentration of : uM or nM. 2. A 20-fold dilution (1/20) of a 3 µM results in a final concentration of µM or 3MM 6.15 MM =_ nm nM. 20 R, A solution çontaining 0.3 umoles/ml of a solute has a concentration also expressed as:

1. A 100-fold dilution (1/100-dilution; 1 part sample: 99 parts diluent, most of the time water) of a 15 µM solution results in a final concentration of : uM or nM. 2. A 20-fold dilution (1/20) of a 3 µM results in a final concentration of µM or 3MM 6.15 MM =_ nm nM. 20 R, A solution çontaining 0.3 umoles/ml of a solute has a concentration also expressed as:

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter27: Molecular Fluorescence Spectroscopy

Section: Chapter Questions

Problem 27.13QAP

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Quinine in a 1.664 g antimalarial tablet was dissolved in sufficient 0.10 M HCl to give 500 mL of solution. A 15.00 mL aliquot was then diluted to 100.0 mL with the acid. The fluorescense intensity for the diluted sample at 347.5 nm provided a reading of 288 on an arbitrary scale. A standard 100 ppm quinine solution registered 180 when measured under conditions identical to those for the diluted sample. Calculate the mass in milligrams of quinine in the tablet.
Solution. Vol. of A solution (mL) [methyl red], M Absorbance 1 10.00 _1.64 x10^-16_ _0.256_ 2 15.00 _2.45 x10^-16_. _0.373_ 3 20.00 _3.27 x10^-16_. _0.486_ 4 25.00 _4.09 x10^-16_. _0.620_ Plot a graph of the absorbance (y axis) as a function of the concentration of methyl red (x axis). What is the slope of the graph?
analyte concentration(C)(mg/ml) injection volume (ul) elution time (time) peak DAD signal(mAU) caffeine 1 1 4.67 302.85 aspartame 5 1 7.53 15.83 benzoic acid 1 1 8.14 89.98 saccharin 1 1 1.91 84.86 mixture(add everything above with 1:1:1:1 ratio) 1 4.47 69.58 How to get the concentration of the mixture in this case?
explain the results of this experimet Table 2 Absorbance vs CoCl2 concentration Test Tube Number Cobalt Chloride Concentration (mol/mL) Absorbance at 510 nm 1 0.000 0.000 2 0.009 0.024 3 0.018 0.055 4 0.027 0.085 5 6 0.036 0.045 0.112 0.138 7 (Unknown) 0.088
5.00 mL of stock solution is diluted to 25.00 mL, producing solution ALPHA. 10.00 mL of solution ALPHA is diluted to 25.00 mL, resulting in solution BETA. 10.00 mL of solution BETA is then diluted to 25.00 mL, producing solution GAMMA. dilution factor for ALPHA from stock solution = 0.167 dilution factor for BETA from ALPHA solution = 0.0476 part c and d?
A 0.0200 gram blood sample was decomposed by a microwave digestion technique followed by dilution to 100.0 mL in a volumetric flask. Aliquots of the sample solution were treated with a lead complexing reagent and water as follows: Solution 1: 10.0 ml blood sample + 20.0 mL complexing agent + 30.0 mL H20. Solution 2: 10.0 ml blood sample + 20.0 mL complexing agent + 26.0 mL H20 + 4.00 mL of 78 ppb Pb2+ standard. The resulting solutions were analyzed by UV/Vis at 375 nm. Absorbance for solution 1 = 0.155 and for solution 2 = 0.216. Calculate the concentration of lead (ppb) in the original sample.
A toluene solution (1.50 mg / L) shows an absorbance of 0.30 in a cuvette with a 0.025 mm optical path. Calculate the limit of detection (in ppb) for toluene if the noise with the solvent spectrum is 0.002 absorbance units in a 0.050 mm cuvette.
Method used was Volhard Method Sample preparation: 5ml sample diluted in 100 ml. 20ml aliquot was used for analysis. Concentration of AgNO3: 0.0609M Calculate mass of NaCl and % NaCl (w/v)
Based on the data, calculate the concentration of analyte in the ORIGINAL sample in ppm. Report to 1 decimal place Sample: A 10.00 mL aliquot of the unknown was placed in a 100.0 mL volumetric flask and made up to the mark will distilled water. The absorbance of this solution is 0.185. Sample + Spike: A 10.00 mL aliquot of the unknown and a 13.00 mL aliquot of the 46.00 ppm stock were both placed in the same 100.0 mL volumetric flask and diluted to the mark with water. The absorbance of this solution is 0.326. ( Please type answer note write by hend)
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)
Quinine in a 1.553-g antimalarial tablet was dissolved in sufficient 0.10 M HCl to give 250 mL of solution.A 10.00-mL aliquot was then diluted to 50.00 mL with the acid. The fluorescence intensity for the dilutedsample at 347.5 nm provided a reading of 196 on an arbitrary scale. A standard 100-ppm quinine solutionregistered 125 when measured under conditions identical to those for the diluted sample. Calculate the massin milligrams of quinine in the tablet.
Frankie determined the trendline for their Calibration Curve for the relationship between Absorbance and [FeSCN2+] was y = 3823.3 x − 0.0127 During Part B, Frankie measured the absorbance of Equilibrium Solution 3 as 0.292 a.u. (absorbance units). Question 1: What is the equilibrium concentration of the [FeSCN2+]? Anjelica used the absorbance of 0.449 a.u. to determine the equilibrium concentration of [FeSCN2+] for Run 5 was 1.17 x 10-4 M. Previously, Anjelica calculated the initial concentrations of [Fe3+] = 1.00 x 10-3 M and [SCN-] = 1.00 x 10-3 M. Question 2: What are the equilibrium concentrations of [Fe3+] and [SCN-]? Please answer completely all questions will give rating surely