A standard sample was prepared containing 20.00 ppm of an analyte and 10.00 ppm internal standard....
A standard sample was prepared containing 20.00 ppm of an analyte and 10.00 ppm internal standard. The signal readings for these were 0.115 units and 0.321 units respectively. Sufficient internal standard was added to an unknown sample solution to make it contain 15.00 ppm internal standard. This solution gave readings for the analyte and internal standard of 0.266 and 0.275 units respectively. Determine the concentration of the analyte in the unknown sample.
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A standard sample was prepared containing 20.00 ppm of an
analyte and 10.00 ppm internal standard....
A standard sample was prepared containing 20.00 ppm of an analyte and 10.00 ppm internal standard....
A standard sample was prepared containing 20.00 ppm of an analyte and 10.00 ppm internal standard. The signal readings for these were 0.115 units and 0.321 units respectively. Sufficient internal standard was added to an unknown sample solution to make it contain 15.00 ppm internal standard. This solution gave readings for the analyte and internal standard of 0.266 and 0.275 units respectively. Determine the concentration of the analyte in the unknown sample. Select one: O A. 1.71 ppm OB. None...
Mn was used as an internal standard for measuring Fe by atomic absorption. A standard mixture...
Mn was used as an internal standard for measuring Fe by atomic absorption. A standard mixture containing 2.00 Hg/mL Mn and 2.50 Hg/mL Fe gave a quotient (Fe signal/Mn signal) 1.05/1.00. A solution was prepared by mixing 10.00 mL of unknown Fe solution with 10.00 mL of standard containing 8.24 ug/mL Mn, and diluting to 50.00 mL; the measured absorbance signal at the Fe wavelength was 0.197, and at the Mn wavelength the absorbance was 0.116. Calculate the molar concentration...
5-29. A solution containing 3.47 mM X (analyte) and 1.72 mM S (standard) gave peak areas...
5-29. A solution containing 3.47 mM X (analyte) and 1.72 mM S (standard) gave peak areas of 3 473 and 10 222, respectively, in a chromatographic analysis. Then 1.00 mL of 8.47 mM S was added to 5.00 mL of unknown X, and the mixture was diluted to 10.0 mL This solution gave peak areas of 5 428 and 4 431 for X and S respectively (a) Calculate the response factor for the analyte (b) Find the concentration of S...
A solution containing 3.47 mM X (analyte) and 1.72 mM S (Standard) gave peak areas of...
A solution containing 3.47 mM X (analyte) and 1.72 mM S (Standard) gave peak areas of 3.473 and 10222, respectively, in a chromatographic analysis. Then 1mL of 8.47 mM S was added to 5mL of unknown X, and the mixture was diluted to 10mL. This solution gave a peak areas of 5,428 and 4,431 for X and S, respectively. a.) Calculate the response factor for the analyte. b.) Find the concentration of S (mM) in the 10mL mixture. c.) Find...
A 10.00-g sample that contains an analyte is transferred to a 250-mL volumetric flask and diluted...
A 10.00-g sample that contains an analyte is transferred to a 250-mL volumetric flask and diluted to volume. When a 10.00 mL aliquot of the resulting solution is diluted to 25 ml it gives a signal of 0.235 (arbitrary units). A second 10.00 mL portion of the solution is spiked with 10.00 mL of a 1.00-ppm standard solution of the analyze and diluted to 25.00 mL. The signal for the spiked sample is 0.502. Calculate the weight percent of analyte...
Consider a solution containing 4.69 mM of an analyte, X, and 1.22 mM of a standard,...
Consider a solution containing 4.69 mM of an analyte, X, and 1.22 mM of a standard, S. Upon chromatographic separation of the solution peak areas for X and S are 3793 and 10323, respectively. Determine the response factor for X relative to S. F = To determine the concentration of X in an unknown solution, 1.00 mL of 8.70 mM S was added to 3.00 mL of the unknown X solution and the mixture was diluted to 10.0 mL. After...
A standard solution with a concentration of 12.5 ppm gave a fluorescence emission intensity of 5.227...
A standard solution with a concentration of 12.5 ppm gave a fluorescence emission intensity of 5.227 x 104 counts at 457 nm. An unknown sample was diltued by placing 5 mL of the sample into a 250 mL volumetric flask and diluted to the mark. The resulting solution gave an emission intensity of 3.921 x 104 counts at 457 nm. Determine the concentration of analyte in the unknown sample in ppm.
A standard solution containing 5.6 10-8 M iodoacetone and 2.5 10-7 M p-dichlorobenzene (an internal standard)...
A standard solution containing 5.6 10-8 M iodoacetone and 2.5 10-7 M p-dichlorobenzene (an internal standard) gave peak areas of 306 and 877, respectively, in a gas chromatogram. A 3.00 mL unknown solution of iodoacetone was treated with 0.100 mL of 1.0 10-5 M p-dichlorobenzene and the mixture was diluted to 10.00 mL. Gas chromatography gave peak areas of 685 and 329 for iodoacetone and p-dichlorobenzene, respectively. Find the concentration of iodoacetone in the 3.00 mL of original unknown.
please show steps. A solution was prepared by mixing 10.00 mL of unknown (X) with 5.00 mL of standard (S) containing...
please show steps.
A solution was prepared by mixing 10.00 mL of unknown (X) with 5.00 mL of standard (S) containing 8.24 μg/mL and diluting the mixture to 50.0 mL. The measured signal quotient was (signal due to X/signal due to S) 1.690/1.000. what is the value of X in
A solution was prepared by mixing 10.00 mL of unknown (X) with 5.00 mL of standard (S) containing 8.24 μg/mL and diluting the mixture to 50.0 mL. The measured signal...
Atomic emission spectroscopy and the method of standard addition are used to determine the Nat co...
Atomic emission spectroscopy and the method of standard addition are used to determine the Nat concentration and its uncertainty in an unknown sample. Increasing amounts of a 1.75 μg/m standard Na+ solution are added to a series of volumetric flasks containing 15.00 mL of the unknown 200.0 mL. Atomic emission spectroscopy is then used to measure the emission intensity of each solution. solution. The volumetric flasks are then diluted to a final volume of Unknown volume Standard volume Final volume...
A standard sample was prepared containing 20.00 ppm of an analyte and 10.00 ppm internal standard. The signal readings for these were 0.115 units and 0.321 units respectively. Sufficient internal standard was added to an unknown sample solution to make it contain 15.00 ppm internal standard. This solution gave readings for the analyte and internal standard of 0.266 and 0.275 units respectively. Determine the concentration of the analyte in the unknown sample. Select one: O A. 1.71 ppm OB. None...
Mn was used as an internal standard for measuring Fe by atomic absorption. A standard mixture containing 2.00 Hg/mL Mn and 2.50 Hg/mL Fe gave a quotient (Fe signal/Mn signal) 1.05/1.00. A solution was prepared by mixing 10.00 mL of unknown Fe solution with 10.00 mL of standard containing 8.24 ug/mL Mn, and diluting to 50.00 mL; the measured absorbance signal at the Fe wavelength was 0.197, and at the Mn wavelength the absorbance was 0.116. Calculate the molar concentration...
5-29. A solution containing 3.47 mM X (analyte) and 1.72 mM S (standard) gave peak areas of 3 473 and 10 222, respectively, in a chromatographic analysis. Then 1.00 mL of 8.47 mM S was added to 5.00 mL of unknown X, and the mixture was diluted to 10.0 mL This solution gave peak areas of 5 428 and 4 431 for X and S respectively (a) Calculate the response factor for the analyte (b) Find the concentration of S...
Consider a solution containing 4.69 mM of an analyte, X, and 1.22 mM of a standard, S. Upon chromatographic separation of the solution peak areas for X and S are 3793 and 10323, respectively. Determine the response factor for X relative to S. F = To determine the concentration of X in an unknown solution, 1.00 mL of 8.70 mM S was added to 3.00 mL of the unknown X solution and the mixture was diluted to 10.0 mL. After...
please show steps.
A solution was prepared by mixing 10.00 mL of unknown (X) with 5.00 mL of standard (S) containing 8.24 μg/mL and diluting the mixture to 50.0 mL. The measured signal quotient was (signal due to X/signal due to S) 1.690/1.000. what is the value of X in
A solution was prepared by mixing 10.00 mL of unknown (X) with 5.00 mL of standard (S) containing 8.24 μg/mL and diluting the mixture to 50.0 mL. The measured signal...
Atomic emission spectroscopy and the method of standard addition are used to determine the Nat concentration and its uncertainty in an unknown sample. Increasing amounts of a 1.75 μg/m standard Na+ solution are added to a series of volumetric flasks containing 15.00 mL of the unknown 200.0 mL. Atomic emission spectroscopy is then used to measure the emission intensity of each solution. solution. The volumetric flasks are then diluted to a final volume of Unknown volume Standard volume Final volume...
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