time (s) 135 265 404 544 692 838 995 Experiment # 1 Total Moles Consumed 2.00E-04 4.00E-04 6.00E-04 8.00E-04 1.00E-03 1.20E-03 1.40E-03 [S20x²1 0.050 -0.05M IN 0.050 Slope 1.35E-06 Rate (M/s) 1.35E-05 Solution 1. Initial [S,O,2] = 0.05 M; initial [1] = 0.05 M. Sample calculation of [1]: (25mL)(0.2M) (100 mL is the total final volume) 100ml
time (s) 135 265 404 544 692 838 995 Experiment # 1 Total Moles Consumed 2.00E-04 4.00E-04 6.00E-04 8.00E-04 1.00E-03 1.20E-03 1.40E-03 [S20x²1 0.050 -0.05M IN 0.050 Slope 1.35E-06 Rate (M/s) 1.35E-05 Solution 1. Initial [S,O,2] = 0.05 M; initial [1] = 0.05 M. Sample calculation of [1]: (25mL)(0.2M) (100 mL is the total final volume) 100ml
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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Question
![6x 104
8 x 10
10 x 104
12 x 10-4
14 x 104
3:17
time (s)
135
265
404
544
692
838
995
Task 1: Find Rate
Solution 1
Experiment #
Experiment 3 De...ion (1) (1) (1) (1) 2
1
Present
6
7
Total Moles
Consumed
2.00E-04
4.00E-04
6.00E-04
8.00E-04
1.00E-03
1.20E-03
1.40E-03
-0.05M
[S20x²]
0.050
LOG
Edit
A
SO
0.050
& TU
6 x 104
8 x 104
10 x 104
12 x 104
14 x 104
Rates of Chemical Reactions A Clock Reaction
Slope
1.35E-06
Solution 1. Initial [S₂0,2] = 0.05 M; initial [1] = 0.05 M. Sample calculation of [1]:
(25mL)(0.2M)
(100 ml, is the total final volume)
100ml
a
Find
PAR
www
Rate (M/s)
1.35E-05
Share](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7d6dfbf-bc8a-4631-8738-d4572dcdcbd9%2F34fce9a2-9be7-4d74-9eeb-f1f1ff1a85f7%2Ftf3o2ce_processed.jpeg&w=3840&q=75)
Transcribed Image Text:6x 104
8 x 10
10 x 104
12 x 10-4
14 x 104
3:17
time (s)
135
265
404
544
692
838
995
Task 1: Find Rate
Solution 1
Experiment #
Experiment 3 De...ion (1) (1) (1) (1) 2
1
Present
6
7
Total Moles
Consumed
2.00E-04
4.00E-04
6.00E-04
8.00E-04
1.00E-03
1.20E-03
1.40E-03
-0.05M
[S20x²]
0.050
LOG
Edit
A
SO
0.050
& TU
6 x 104
8 x 104
10 x 104
12 x 104
14 x 104
Rates of Chemical Reactions A Clock Reaction
Slope
1.35E-06
Solution 1. Initial [S₂0,2] = 0.05 M; initial [1] = 0.05 M. Sample calculation of [1]:
(25mL)(0.2M)
(100 ml, is the total final volume)
100ml
a
Find
PAR
www
Rate (M/s)
1.35E-05
Share
![Name
Section
Aliquot #
1 (already in sol.)
2 (first one added)
3
4
5
Rates of Chemical Reactions:
Solution 1. Initial [S₂O82] = 0.05 M; initial [I] = 0.05 M. Sample calculation of [I]:
(25mL)(0.2M)
(100 mL is the total final volume)
100mL
6
7
= 0.05M
t, sec
135
265
404
544
692
838
995
t, sec
REPORT SHEET
Moles S₂02
consumed
Date
2 x 10-4
4 x 10-4
6 x 10-4
8 x 10-4
10 × 10-4
12 x 10-4
14 x 10-4
Solution 2. Initial [S₂082] = 0.10 M; Initial [I] = 0.05 M.
Moles S₂0,²2
Aliquot #
consum
ΤΑ
[S₂082-1"
log [S₂0]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7d6dfbf-bc8a-4631-8738-d4572dcdcbd9%2F34fce9a2-9be7-4d74-9eeb-f1f1ff1a85f7%2Fl58m0fj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Name
Section
Aliquot #
1 (already in sol.)
2 (first one added)
3
4
5
Rates of Chemical Reactions:
Solution 1. Initial [S₂O82] = 0.05 M; initial [I] = 0.05 M. Sample calculation of [I]:
(25mL)(0.2M)
(100 mL is the total final volume)
100mL
6
7
= 0.05M
t, sec
135
265
404
544
692
838
995
t, sec
REPORT SHEET
Moles S₂02
consumed
Date
2 x 10-4
4 x 10-4
6 x 10-4
8 x 10-4
10 × 10-4
12 x 10-4
14 x 10-4
Solution 2. Initial [S₂082] = 0.10 M; Initial [I] = 0.05 M.
Moles S₂0,²2
Aliquot #
consum
ΤΑ
[S₂082-1"
log [S₂0]
Expert Solution
Step 1
Solutions-
Here we have to find the Rate of chemical reaction with the help of given values of time and moles of [S2O82-].
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