Conductivity Table 1: Conductivity (mS/cm) of KCI Solutions of Various Concentrations (% w/v). Concentration (% w/v) Conductivity (mS/cm) 0.50 1.0 2.5 5.0 3.903 4.767 13.76 30.79 Data Analysis Using your data from your stock and dilute solutions, create a graph of Conductivity (mS/cm) vs. Molarity Concentration (M). Make sure your graph is attached to this report, and include the following: Proper title for graph, and proper labels for x and y axis (with proper units) • Include a line of best fit that represents the trend observed in your data Use your trend line from your graph to complete the following statement: The conductivity of a KCI solution that has a concentration of 0.75 % (w/v) is mS/cm.

Conductivity Table 1: Conductivity (mS/cm) of KCI Solutions of Various Concentrations (% w/v). Concentration (% w/v) Conductivity (mS/cm) 0.50 1.0 2.5 5.0 3.903 4.767 13.76 30.79 Data Analysis Using your data from your stock and dilute solutions, create a graph of Conductivity (mS/cm) vs. Molarity Concentration (M). Make sure your graph is attached to this report, and include the following: Proper title for graph, and proper labels for x and y axis (with proper units) • Include a line of best fit that represents the trend observed in your data Use your trend line from your graph to complete the following statement: The conductivity of a KCI solution that has a concentration of 0.75 % (w/v) is mS/cm.

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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ts/431885/variants/431885/take/15/ TEXT ANSWER Question 18 Calculate the molarity of a sodium hydroxide solution if 50.0 mL of this solution was completely neutralized by 46.6 mL of 0.750 M sulfuric acid. Balanced equation: H₂SO4+2NaOH → Na₂SO4 + 2H₂O Show your work. Use the following formula: X [NaOH] = 2x mL H₂SO4 x [H₂SO4] mL NaOH X₂ X² Normal AA 2 x 46.6= 93.2 1 Ďõõŏ õõ õõŏõõčõõčŏo O 20 of 20 Total Questions Answered 2 90 C N NJ H I III 123 B I US | √x ΩΡΑ Ix ((( All Changes Saved V ✰ ✰ 294 words left 7:56 P 5/2/202 ^ ☎ ☎ ‹ b