55. Volume of Titram (ml) 24.70 24.80 24.90 25.00 25.10 25.20 25.30 Potential (mV) (A) 24.96 mL (B) 25.00 mL (C) 25.04 mL (D) 25.14 mL (E) 25.50 ml. 210 222 240 360 600 616 625 Potential Change pèr 0.1 mL Volume Change 12 18 120 240 16 9 The table above contains potential readings near the equivalence point of a potentiometric titration. The volume of titrant needed to reach the equivalence point is

55. Volume of Titram (ml) 24.70 24.80 24.90 25.00 25.10 25.20 25.30 Potential (mV) (A) 24.96 mL (B) 25.00 mL (C) 25.04 mL (D) 25.14 mL (E) 25.50 ml. 210 222 240 360 600 616 625 Potential Change pèr 0.1 mL Volume Change 12 18 120 240 16 9 The table above contains potential readings near the equivalence point of a potentiometric titration. The volume of titrant needed to reach the equivalence point is

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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If may calculations po, please show all steps leading to the final answer po. Here’s the link in accordance with the topic po: https://drive.google.com/file/d/1_FnDtXCrFKSol3RNWIG_9tNQ7IxgxD6t/view?usp=drivesdk
Is data correct?
Show all steps leading to the final answer po. Here’s a pdf file in accordance with the topic po: https://drive.google.com/file/d/1_FnDtXCrFKSol3RNWIG_9tNQ7IxgxD6t/view?usp=drivesdk
y= mx+ b Slope () = -4E-09 y-intercept () = 0.4021 and OD= 0.540 , Calculate the concentration (x)
Determine the value of the slope (m), y-intercept (b), and the coefficient of determination (r²) A) m=0.1128; b= -4.0 x 105; r² = 0.9762 B) m = 0.1128; b=3.0 x 10-3 r2 = 0.9762 C) m = 0.1128; b=3.0 x 103, r² = 0.9880 D) m=0.1128; b = -4.0 x 105; r2 = 0.9880
how would u justify this question? my answer is correct i just do not understand the justification. this is a non graded practice worksheet
A researcher wants to calculate the flux of palm oil (PO) in a simulated body fluid (SBF). To perform the experiment, he designed a 30 cm capillary tube. At the very beginning of the tube, he designed a partition with a separator as shown in the figure below. He filled the partition with a stock solution prepared by solving 10 g of PO in a liter of SBF and the capillary tube with pure SBF. Then he removed the separator. The viscosity of the BSF is 3.6x10-3 Pa.s and molecular weight of PO is 850.92.a. Calculate the flux of PO in kg mol/s.m2 at steady state. (Use Polson equation for diffusivity.) (Assume the temperature is maintained at 37oC during whole experiment.)b. Concentration of PO at the half of the tube.