A 100.0 mL solution containing 0.100 M NaBr was titrated with 0.100 M AGNO3 and monitored with a S.C.E. What voltage readingwould be observed after 65.0 mL? Ksp AgBr = 5.0 x 10-13 E° Ag*/Ag3 0.558 V%3DA 0.072 VBnone of the choices(c)- 0.081 VD) 0.081 V(E)-0.072 V A 20.00 mL urine sample was treated with molybdenum blue reagents to produce a species absorbing at 820 nm, after which thesample was diluted to 100.00 mL. A 25.00 mL aliquot gave an absorbance of 0.428 a.u.. Addition of 1.00 mL of a solution containing0.0500 mg of phosphate to second 25.00-mL aliquot gave an absorbance of 0.517 a.u. Calculate for the concentration of phosphatein mg per mL of the sample.0.390 mg/mL0.427 mg/mLc) 0.0780 mg/mLD) none of the choices0.0429 mg/mL

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A 100.0 ml solution containing 0.100 M NaBr was titrated with 0.100 M AgNOz and monitored with a S.C.E. What voltage reading would be observed after 65.0 mL? Ksp AgBr = 5.0 x 10-13 E° Ag*/Ag= 0.558 V A 0.072 V none of the choices C) - 0.081 V 0.081 V E- 0.072 V

A 100.0 ml solution containing 0.100 M NaBr was titrated with 0.100 M AgNOz and monitored with a S.C.E. What voltage reading would be observed after 65.0 mL? Ksp AgBr = 5.0 x 10-13 E° Ag*/Ag= 0.558 V A 0.072 V none of the choices C) - 0.081 V 0.081 V E- 0.072 V

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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Question

A 100.0 mL solution containing 0.100 M NaBr was titrated with 0.100 M AGNO3 and monitored with a S.C.E. What voltage reading
would be observed after 65.0 mL? Ksp AgBr = 5.0 x 10-13 E° Ag*/Ag3 0.558 V
%3D
A 0.072 V
B
none of the choices
(c)
- 0.081 V
D) 0.081 V
(E)-0.072 V

A 20.00 mL urine sample was treated with molybdenum blue reagents to produce a species absorbing at 820 nm, after which the
sample was diluted to 100.00 mL. A 25.00 mL aliquot gave an absorbance of 0.428 a.u.. Addition of 1.00 mL of a solution containing
0.0500 mg of phosphate to second 25.00-mL aliquot gave an absorbance of 0.517 a.u. Calculate for the concentration of phosphate
in mg per mL of the sample.
0.390 mg/mL
0.427 mg/mL
c) 0.0780 mg/mL
D) none of the choices
0.0429 mg/mL

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