John wanted to determine the protein content of his blood. To do so, he obtained 1.00 mLof blood, and diluted it to 10.00 mL. From that solution, he subjected 1.00 mL to Kjeldahl analysis. After digestion, the liberated NH3 was distilled in a flask containing exactly 25.00 mL of 0.0150 M H3BO3. Titration requires 5.85 mL of standard 0.0150 M NaOH to reach the desired endpoint. If a typical blood sample contains 15.45 g N per 100 g protein, calculatethe protein content (in g/mL) in the sample of John’s blood. 6. John wanted to determine the protein content of his blood. To do so, he obtained 1.00 mLof blood, and diluted it to 10.00 mL. From that solution, he subiected 1.00 mL to Kjeldahlanalysis. After digestion, the liberated NH3 was distilled in a flask containing exactly 25.00mL of 0.0150 M H;BO3. Titration requires 5.85 mL of standard 0.0150 M NaOH to reach thedesired endpoint. If a typical blood sample contains 15.45 g N per 100 g protein, calculatethe protein content (in g/mL) in the sample of John's blood.

Given data, Volume of sample = 1.0mL Molarity of H3BO3= 0.0150M Volume of H3BO3= 25.0mL = 0.025L…

Answered: John wanted to determine the protein…

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter19: Transition Metals And Coordination Chemistry

Section: Chapter Questions

Problem 11E: Iron (II) can be oxidized to iron (III) by dichromate ion, which is reduced to chromium (III) in...

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John wanted to determine the protein content of his blood. To do so, he obtained 1.00 mL
of blood, and diluted it to 10.00 mL. From that solution, he subjected 1.00 mL to Kjeldahl analysis. After digestion, the liberated NH3 was distilled in a flask containing exactly 25.00 mL of 0.0150 M H3BO3. Titration requires 5.85 mL of standard 0.0150 M NaOH to reach the desired endpoint. If a typical blood sample contains 15.45 g N per 100 g protein, calculate
the protein content (in g/mL) in the sample of John’s blood.

6. John wanted to determine the protein content of his blood. To do so, he obtained 1.00 mL
of blood, and diluted it to 10.00 mL. From that solution, he subiected 1.00 mL to Kjeldahl
analysis. After digestion, the liberated NH3 was distilled in a flask containing exactly 25.00
mL of 0.0150 M H;BO3. Titration requires 5.85 mL of standard 0.0150 M NaOH to reach the
desired endpoint. If a typical blood sample contains 15.45 g N per 100 g protein, calculate
the protein content (in g/mL) in the sample of John's blood.

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