You have measured the blank for a determination of arsenic in food samples by hydride-generation atomic fluorescence spectrometry. The blank values are: 0.23 ppb, 0.14 ppb, 0.16 ppb, 0.28 ppb, 0.18 ppb, 0.09 ppb, 0.10 ppb, 0.20 ppb, 0.15 ppb, 0.21 ppb As. What is the LOD?
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You have measured the blank for a determination of arsenic in food samples by hydride-generation atomic fluorescence spectrometry. The blank values are: 0.23 ppb, 0.14 ppb, 0.16 ppb, 0.28 ppb, 0.18 ppb, 0.09 ppb, 0.10 ppb, 0.20 ppb, 0.15 ppb, 0.21 ppb As. What is the LOD? |
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Given,
Sr. No. (N) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Blank values(ppb) | 0.23 | 0.14 | 0.16 | 0.28 | 0.18 | 0.09 | 0.10 | 0.20 | 0.15 | 0.21 |
Limit of detection (LOD)=?
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- 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.Exactly 5.00 mL aliquots of a solution containing analyte X were transferred into 50.00-mL volumetric flasks and the pH of the solution is adjusted to 9.0. The following volumes of a standard solution containing 2.00 µg/mL of X were then added into each flask and the mixture was diluted to volume: 0.000, 0.500, 1.00, 1.50 and 2.00 mL. The fluorescence of each of these solutions was measured with a fluorometer, and the following values were obtained: 3.26, 4.80, 6.42, 8.02 and 9.56, respectively. ii. Using relevant functions in Excel, derive a least-squares equation for the data, and use the parameters of this equation to find the concentration of the phenobarbital in the unknown solution.An organic chloro-compound has the molecular formula: C₁4H22Cl2. Give the ratio of the% (M+2)* to M* peak intensity for the above compound to two decimal places. The masses in Daltons of selected isotopes and their abundances are given in the table below: Isotope abundance and mass data Element Mass (Da) 1.00783 12.0000 14.0031 15.9949 18.9984 27.9769 30.9738 31.9721 34.9689 78.9183 126.9045 H C N F Si P S CI Br I Mt Isotope H-1 C-12 N-14 0-16 F-19 Si-28 P-31 S-32 Cl-35 Br-79 I-127 ANS - (M+2)/M* (%) = (M+1) Isotope H-2 C-13 N-15 0-17 F-20 Si-29 P-33 S-33 CL-36 Br-80 I-128 Abundance 0.015 1.08 0.37 0.04 0 5.1 0 0.8 0 0 0 (report the % to 2 decimal places) (M+2)* Isotope Abundance H-3 0 C-14 0 N-16 0 O-18 0.2 F-21 0 Si-30 3.4 P-34 0 S-34 4.40 CI-37 32.5 Br-81 98.0 I-129 0
- During a spectroscopic evaluation of the several solutions of C6H5CH=CHCCl₂ in ethanol using a 1.0 cm spectrophotometric cell, different values of absorbance (A) were obtained, which are given in the following table: 105 x conc. mol dm-3 A time min A 0.446 0.080 0 0.560 C When the reaction of C6H5CH=CHCCl₂ with C₂H5O in an ethanolic medium is followed spectroscopically at a convenient wavelength (2), the following results are found: 0.812 0.145 40 0.283 1.335 1.711 0.240 0.305 80 0.217 120 2.105 0.378 0.168 140 0.149 The values of ε at λ and the rate constant for that reaction are: a. ε = 1.79x10-6 dm³ mol-¹ cm-¹ and k = 0.034 dm³ mol-¹ min-¹ b. ε = 1.79×104 dm³ mol-¹ cm-¹ and k = 0.034 dm³ mol-¹ min-¹ c. ε = 1.79x10-6 dm³ mol-¹ cm-1 and k = 8.9x10-³ dm³ mol-¹ min-¹ d. ε = 1.79×104 dm³ mol-¹ cm-¹ and k = 8.9x10-³ dm³ mol-¹ min¯¹The Na* concentration in a blood serum sample was determined using the method of standard addition and atomic spectroscopy. 25.00 mL aliquots of serum were pipetted into five 50.00 mL volumetric flasks into which 2.640 M NaCl standard was added as shown in the table below. After dilution to volume, the emission of each solution was measured and used to plot a calibration graph. Flask 1 23 4 5 Vol. of standard (mL) 0 1.00 2.00 3.00 4.00 Emission signal 3.13 5.40 7.89 10.30 12.48 Analytical signal (emission) 14 42 10 12 4 2 0 0 0.05 y = 44.697x + 3.12 R² = 0.9995 0.1 0.15 Concentration of added analyte (M) 0.2 a) Using the plotted calibration graph, calculate the Na* concentration in the blood serum sample. b) Provide a possible reason why the serum sample might have been analysed using this analytical procedure. 0.25The quenching of tryptophan fluorescence by dissolved O₂ gas was monitored by measuring the emission lifetimes at 348 nm in aqueous solutions. [0₂]/(10-² mol dm-³) T/(10-⁹ s) 0 2.3 5.5 8 10.8 2.6 1.5 0.92 0.71 0.57 A. 0.95 E-2 M B. 4.3 E-2 M C. 1.8 E-2 M D. 3.0 E-2 M E. None of the choices From the data above, predict the value of [0₂] required to decrease the intensity of tryptophan emission to 50% of the unquenched value.
- A sample solution containing quinine was analysed by fluorescence spectroscopy. 5 mL of the sample solution was diluted with 0.1 M HCl to a final volume of 100 mL. The fluorescence of the solution was measured and gave a value of 54. A reference solution containing 0.086 mg/mL of quinine gave a fluorescence reading of 38. A blank solution gave a fluorescence reading of 21. Calculate the quinine content of the sample solution in mg/mL.Please interpreting IR of DEETIn the HQS-Mg fluorescence part of Lab 6, a calibration curve is made of fluorescence intensity vs. MgCl2 concentration (M). The standard solutions for the calibration curve were made by mixing a small amount of 0.01 M MgCl2 with some HQS solution. Calculate the concentration of MgCl2 (M) in each of the following standards. Report your concentrations to 3 sig figs. Fluorescence Volume of 0.01 M Volume of 1 mM Concentration of MgCl2 (M) Intensity MgCl2 (mL) HQS (mL) 0.031250902 0.8 100 0.049490754 0.8 50 0.059803507 2 100 0.085834029 3.6 100 Hint: The total volume of each solution is volume of 0.01 M MgCl2 + volume of 1 mM HQS.
- An atomic absorption method for the determination of the amount of iron present in used jet engine oil was found from pooling 30 triplicate analyses to have a standard deviation s = 3.6 mg Fe/mL. If s is a good estimate of σ, calculate the 95 and 99% confidence intervals for the result 18.5 mg Fe/mL if it was based on (a) a single analysis, (b) the mean of two analyses, and (c) the mean of four analyses.An atomic absorption method for the determination of the amount of iron present in used jet engine oil was found from pooling 30 triplicate analyses to have a standard deviation s = 3.1 µg Fe/mL. If s is a good estimate of o, calculate the 95 and 99% confidence intervals for the result 21.7 µg Fe/mL if it was based on the following criteria. (a) a single analysis 95% confidence interval= 21.7 + 99% confidence interval = 21.7 + (b) the mean of two analyses 95% confidence interval = 21.7 + 99% confidence interval= (c) the mean of six analyses 21.7 ± 95% confidence interval = 21.7 + 99% confidence interval= 21.7 + μg Fe/mL μg Fe/mL μg Fe/mL μg Fe/mL μg Fe/mL μg Fe/mLA UVIVIS molecular absorption spectrometric determination of analyte X in aqueous solution gave the following calibration data (absorbances corrected for the blank). Estimate the concentration of Xin an aqueous sample that yielded an instrument response of 0.238 (A 1.00- mL cuvette was used for all absorbance measurement s) OA. 7.7 ppm OB. 8.0 ppm, 148 O c. 23.8 ppm O D.0.0308 ppb OE. 0.8 ppm Concentration (ppm X) 1.0 2.0 5.0 7.0 8.5 10.0 Absorbance 0.031 0.068 0.150 0.221 0.262 0.308 3:49 PM
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