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...
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc65448d-ddaf-46ec-8ff7-ab22850b6815%2F44bd0892-6fcd-477e-aa15-b17c0ad3aa33%2F6qjoeme_processed.jpeg&w=3840&q=75)
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
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