The phosphorus in a 4.258 g sample of a plant food was converted to PO, and precipitated as AGPO, through the addition of 50.00 mL of 0.0820 M A£NO3. The excess AGNO, was back-titrated with 4.06 mL of 0.0625 M KSCN. Express the results of the analysis in terms of %P,Og. The chemical reactions are: P2O5 + 9H2O → 2PO,"+ 6H3O* 2PO. + 6Ag" → 2Ag‚PO46) Ag* + SCN' → AgSCN)
The phosphorus in a 4.258 g sample of a plant food was converted to PO, and precipitated as AGPO, through the addition of 50.00 mL of 0.0820 M A£NO3. The excess AGNO, was back-titrated with 4.06 mL of 0.0625 M KSCN. Express the results of the analysis in terms of %P,Og. The chemical reactions are: P2O5 + 9H2O → 2PO,"+ 6H3O* 2PO. + 6Ag" → 2Ag‚PO46) Ag* + SCN' → AgSCN)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![The phosphorus in a 4.258 g sample of a plant food was converted to PO,* and
precipitated as AgPO, through the addition of 50.00 mL of 0.0820 M AgNO3. The
excess AgNO3 was back-titrated with 4.06 mL of 0.0625 M KSCN. Express the results
of the analysis in terms of %P,O5.
The chemical reactions are:
P2O5 + 9H20 → 2PO,*-+ 6H;O*
2PO, + 6Ag' → 2Ag;PO43)
Ag* + SCN' → A£SCN9)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F67501ed6-15f9-4c8d-b221-3bde038769c2%2F2089a455-a920-4b5c-a20a-9fdf62dda971%2F4s80z2_processed.png&w=3840&q=75)
Transcribed Image Text:The phosphorus in a 4.258 g sample of a plant food was converted to PO,* and
precipitated as AgPO, through the addition of 50.00 mL of 0.0820 M AgNO3. The
excess AgNO3 was back-titrated with 4.06 mL of 0.0625 M KSCN. Express the results
of the analysis in terms of %P,O5.
The chemical reactions are:
P2O5 + 9H20 → 2PO,*-+ 6H;O*
2PO, + 6Ag' → 2Ag;PO43)
Ag* + SCN' → A£SCN9)
![e. In a GC analysis for the determination of limonene in orange peel, separate solutions
of 100 ppm limonene standard and 100 ppm propanol internal standard (IS) gave
peaks of 1273 and 598 respectively. A sample of the extract (50.0 mL) was spiked
with 10.0 mL of 1000 ppm propanol and the mixture diluted to 100 mL. This
solution gave peak areas of 723 and 519 for limonene and propanol respectively.
What is the concentration of limonene in the extract?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F67501ed6-15f9-4c8d-b221-3bde038769c2%2F2089a455-a920-4b5c-a20a-9fdf62dda971%2F5lgeudp_processed.png&w=3840&q=75)
Transcribed Image Text:e. In a GC analysis for the determination of limonene in orange peel, separate solutions
of 100 ppm limonene standard and 100 ppm propanol internal standard (IS) gave
peaks of 1273 and 598 respectively. A sample of the extract (50.0 mL) was spiked
with 10.0 mL of 1000 ppm propanol and the mixture diluted to 100 mL. This
solution gave peak areas of 723 and 519 for limonene and propanol respectively.
What is the concentration of limonene in the extract?
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