Answered: 4. Which of the functions satisfy the…

4. Which of the functions satisfy the boundary conditions for a particle in the one-dimensional box shown below? V(w) O COS 2X a. cos b. sin RX 2a c. x (a-x) d. All are correct

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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Trial I Trial II Final reading HCI 26. 40 ml 27.10 ml Initial reading HCI 1.2 ml 1.8 ml Volume of HCl used ? ? Final reading NaOH 33.7 ml 32.8 ml Initial reading NaOH 0.8 mL 0.5 ml Volume of NaOH used ? ? For trial I: Normality of NaOH : 0.25 N Vol. of HCl used = final volume - initial volume Solve for the average Normality of HCI : ? NHCI x VHCI = NNAOH x VNAOH Vol. of NaOH used = final volume - initial volume NNAOH x VNAOH For trial II: NHCI = Vol. of HCl used = final volume - initial volume VHCI Activa Vol. of NaOH used = final volume - initial volume Go to S
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ASSAY Calibration solutions Calibration solutions of naproxen in the range 5 – 25 μg/mL were prepared. Sample preparation 20 tablets weighing 12.3819 g were crushed to a fine powder. A portion of the powder (145.4 mg) was shaken with approximately 150 mL of acetic acid (0.05 M) for 5 min and then made up to volume in a 250 mL volumetric flask (stock solution). Approximately 50 mL of the stock solution was filtered and a 25 mL aliquot was diluted to 100 mL in a volumetric flask. 10 mL of the resulting solution was further diluted to 100 mL with acetic acid (0.05 M). Analysis The standards and sample solutions were analyzed by HPLC under the following conditions: Column: octadecylsilyl (ODS), 4.6 mmx150 mm, mobile phase: acetonitrile : 0.05 M acetic acid (85:15), flow rate: 1 ml/min, UV detection at 243 nm. Results: A calibration curve of concentration versus peak area was constructed for the standard solutions and gave the straight-line equation: y = 3555.6x + 85, r = 0.9999 The area…