Homework Help is Here – Start Your Trial Now!
Science
Chemistry
How do you find the standard deviation and average deviation using the data table above?

How do you find the standard deviation and average deviation using the data table above?

Chemistry
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...
See similar textbooks
icon
Related questions
Question
How do you find the standard deviation and average deviation using the data table above?
T₁ = trig11 Tq: Trialy T2 = trial 2 Record with appropriate significant figures! That means TWO decimal places for volume! Trial 1 Trial 2 Trial 4 0.25 76 Mass of KHP decimal places) Moles KHP Final ml NaOH Initial mL NaOH Moles NaOH That Reacted Volume NaOH Used (2 decimal places!) M NaOH 0.29 Dlo3 moles 343L 0.22 0.00103 moles 0.00108 moles 41.75mL 28.32 ml 28.32mL 15.75ml 0.00103 moles 0.00108 moles 13.43m2 12.57mL 0.0860m Trial 3 mistake 0.0767 m Average NaOH M 0.080 m (remember, trial one may be omitted for the average if it was sloppy) Standard deviation Average deviation M Na CH Calculations: 1 Show a sample calculation for one of the trials below: 13.43m4 IL = 0.01343L 11000 ML - 0.0766939687 mass Naot 0.00 122 moles ML C.00 ML 15.75mL 0.00122 males 15.75mL 0.0 775 m 2.04 0.0767m+ 0.0860m + 0.0775m 0.0800666667 Moles Calculations TH 0.25 KP. I mol KHP 204.22kMP = 0.00122417 moles aba
Transcribed Image Text:T₁ = trig11 Tq: Trialy T2 = trial 2 Record with appropriate significant figures! That means TWO decimal places for volume! Trial 1 Trial 2 Trial 4 0.25 76 Mass of KHP decimal places) Moles KHP Final ml NaOH Initial mL NaOH Moles NaOH That Reacted Volume NaOH Used (2 decimal places!) M NaOH 0.29 Dlo3 moles 343L 0.22 0.00103 moles 0.00108 moles 41.75mL 28.32 ml 28.32mL 15.75ml 0.00103 moles 0.00108 moles 13.43m2 12.57mL 0.0860m Trial 3 mistake 0.0767 m Average NaOH M 0.080 m (remember, trial one may be omitted for the average if it was sloppy) Standard deviation Average deviation M Na CH Calculations: 1 Show a sample calculation for one of the trials below: 13.43m4 IL = 0.01343L 11000 ML - 0.0766939687 mass Naot 0.00 122 moles ML C.00 ML 15.75mL 0.00122 males 15.75mL 0.0 775 m 2.04 0.0767m+ 0.0860m + 0.0775m 0.0800666667 Moles Calculations TH 0.25 KP. I mol KHP 204.22kMP = 0.00122417 moles aba
Expert Solution
steps

Step by step

Solved in 3 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Statistics and Analytical Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
SEE MORE TEXTBOOKS