A test for drug use involves addition of 5.00mL of colorless reagent to 10.00 mL urine.If...
A test for drug use involves addition of 5.00mL of colorless reagent to 10.00 mL urine.If drug residue is present in the sample the solution turns blue(λmax = 605 nm, ε = 18,200M-1cm-1). Because other things in urine may give slightly positive results, only tests that measure a minimum concentration of 5.0 x 10-5 M drug residue in the original urine sample are considered positive for illegal drugs. When you measure the sample using the test described above and get an Abs of 0.28 at 605 nm (b = 1 cm). Can you conclude that this person has been using illegal drugs? Explain.
Show your work please.
Homework Answers
The relation between absorption (A), Concentration (C), Path
length of sample (b) and Molar extinction coefficient (
) can be given
as below
A = 0.28 for sample, = 18200
M-1 cm-1 , b = 1 cm
C= 0.28 / 18200 x1 =1.5348 x 10 -5
To conclude that a person is using drug, the minimum concentration of drug should be 5 x 10 -5 M. But, from the absorption data the concentration found is less (1.5348 x 10 -5 < 5 x 10 -5 M).
So, we can conclude that the person is not using illegal drugs in the recent times.
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A test for drug use involves addition of 5.00mL
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4. A test for drug use involves addition of 10.00 mL of colorless reagent to 10.00 mL urine. If drug residue is present in the sample the solution turns blue ( 605 nm, <= 18,200 M-'cm'). Because other things in urine may give slightly positive results only tests that measure a minimum concentration of 5.0x10° drug residue in the original urine sample are considered positive for illegal drugs. When you measure the sample using the test described above and get...
F17 UTA-401 4. A test for drug use involves ad residue is present in the sample the solution Because other things in ur...
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conclusion
F17 MATERIALS 0.12 g [Fe(phen)3]Cl2 (student's red crystals from the previous experiment) 0.5 mL unknown [Fe(phen)3]Cl2 solution 175 mL DI water A. Series of Dilutions: Concentration of stock solution 3.15x103M 105gx mol 667.354 IL 0.05004 Concentrations of the series of dilutions Solution Concentration 0.2/25 2.52x10m 0.4/25 5.04×105m 0.6/25 7.56 X10 sm 0.8/25 1.0088 10" 1.0/25 1.26x10 "m Abs atmas 0.20_ Unknown 514nm 0.52 1.42 Solution 0.2/25 0.4/25...
4. A test for drug use involves addition of 10.00 mL of colorless reagent to 10.00 mL urine. If drug residue is present in the sample the solution turns blue ( 605 nm, <= 18,200 M-'cm'). Because other things in urine may give slightly positive results only tests that measure a minimum concentration of 5.0x10° drug residue in the original urine sample are considered positive for illegal drugs. When you measure the sample using the test described above and get...
F17 UTA-401 4. A test for drug use involves ad residue is present in the sample the solution Because other things in urine may give sig concentration of 5.0x10 M drug residue in drug use involves addition of 10.00 mL of colorless reagent to 10.00 mL urine. If drug it in the sample the solution turns blue (A max = 605 nm, ε = 18,200 M'cm). ngs in urine may give slightly positive results, only tests that measure a minimum...
Percent by mass of dye in Fruit Loops cereal 11.56*100 QUESTIONS: 1. If the molar extinction coefficient (E) of a compound is 2.5 x 108 moli'cmat $28 nm and the Abs Cat 528 nm) 0.254 Dind the concentration of that solution. (assume a path length of 1.0 cm) 2. If a student measures a sample that gives an Abs = 1.85 at the 2max, is this data reliable? If not, then what might they do to improve their data? 3....
I
need help doing just the questions, not the graph or the
conclusion
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