CHM 111/112 LAB MANUAL >C<
CHM 111/112 LAB MANUAL >C<
11th Edition
ISBN: 9781337310956
Author: SLOWINSKI
Publisher: CENGAGE LEARNING (CUSTOM)
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Chapter 24, Problem 2ASA

In an acid-base titration, 21.16 mL of an NaOH solution are needed to neutralize 20.04 mL of a 0.0997 M HCl solution. To find the molarity of the NaOH solution, we can use the following procedure:

  1. First note the value of MH+ in the HCl solution. ____________M
  2. Find MOH- in the NaOH solution. (Use Eq.3.) ____________M
  3. Obtain MNaOH from MOH. ____________M

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Provide the IUPAC name of the following molecule. Don't forget to include the proper stereochemistry where appropriate.
3. 2. 1. On the graph below, plot the volume of rain in milliliters versus its height in centimeters for the 400 mL beaker. Draw a straight line through the points and label it "400 mL beaker." Volume (mL) 400 350 300 250 200 150 750 mL Florence Volume Versus Height of Water 400 mL beaker 100 50 0 0 2 3 4 5 Height (cm) 6 7 8 9 10 Explain why the data points for the beaker lie roughly on a straight line. What kind of relationship is this? How do you know? (see page 276 text) the design of the beaker is a uniform cylinder the volume of liquid increases evenly with its height resulting in a linear relationship. What volume would you predict for 10.0 cm of water? Explain how you arrived at your answer. Use the data table and the graph to assist you in answering the question. 4. Plot the volume of rain in milliliters versus its height in centimeters for the 250 mL Florence flask on the same graph. Draw a best-fit curve through the points and label it "250 mL Florence flask." oke came
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Acid-Base Titration | Acids, Bases & Alkalis | Chemistry | FuseSchool; Author: FuseSchool - Global Education;https://www.youtube.com/watch?v=yFqx6_Y6c2M;License: Standard YouTube License, CC-BY