How do you find the standard deviation and average deviation using the data table above? T₁ = trig11Tq: TrialyT2 = trial 2Record with appropriate significant figures! That means TWO decimal places for volume!Trial 1Trial 2Trial 40.2576Mass of KHPdecimalplaces)Moles KHPFinal ml NaOHInitial mLNaOHMoles NaOHThat ReactedVolume NaOHUsed(2 decimalplaces!)M NaOH0.29Dlo3 moles343L0.220.00103 moles0.00108 moles41.75mL28.32 ml28.32mL15.75ml0.00103 moles 0.00108 moles13.43m2 12.57mL0.0860mTrial 3mistake0.0767 mAverage NaOH M 0.080 m(remember, trial one may be omitted for the average if it was sloppy)Standard deviationAverage deviationM Na CH Calculations: 1Show a sample calculation for one of the trials below:13.43m4IL= 0.01343L11000 ML- 0.0766939687mass Naot0.00 122 molesML C.00 ML15.75mL0.00122 males15.75mL0.0 775 m2.040.0767m+ 0.0860m + 0.0775m0.0800666667Moles CalculationsTH0.25 KP.I mol KHP204.22kMP= 0.00122417 moles aba

Given, Molarity of NaOH=0.0767 M, 0.0860 M, 0.0775 M Average,x¯= 0.080 M n=3 To calculate, Average…

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

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter1: Chemical Foundations

Section: Chapter Questions

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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

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