Reviee Censtants I Becieds Tatie The integrated rate laws for zero-, first, and second-order reaction may be arranged such that hey resemble the equation for a straight line. y= mz+b. • Part A Order Integrated Rate Law Graph Slope JA JA| vs. t kt + JA. kt + In|A. In[A] vs. -k The reactant concentration in a zero-order reaction was 7.00x102 Mater 170 s and 3.00x10 M ater 360 s. What is the rate constant for this reaction? In A Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. View Available Hints) k 211x10 M Previous Answers v Correct Comect answer is shown. Your answer 2.1-10- was ether rounded diferenty or used a diferent number of significant figures than required for this part Important: fyou use this answer in later parts, use the full unrounded value in your caloulations. • Part B What was the initial reactant concentration for the reaction described in Part A? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. View Availlable Hintis) |Ajp = 0.0307 Enter your answer using units of chemical concentration. No credit lost. Try again. Submit Previous Answers

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<Chapter 14 - Attempt 1
Item 7
7 of 16
>
I Review | Constants | Periodic Table
The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that
they resemble the equation for a straight line, y = mI +b.
Part A
Order
Integrated Rate Law
Graph
Slope
[A]=
=- kt + [Alp
[A] vs. t
k
The reactant concentration in a zero-order reaction was 7.00x10-2 M after 170 s and 3.00x10-2 M after 360 s. What is the rate constant for this reaction?
1
In[A]:
kt + In[A]o In[A] vs. t
Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.
- kt + TANo
[A]
A vs. t
2
k
> View Available Hint(s)
koth = 2.11x10-4
Submit
Previous Answers
v Correct
Correct answer is shown. Your answer 2.1-10-4 was either rounded differently or used a different number of significant figures than required for this part.
Important: If you use this answer in later parts, use the full unrounded value in your calculations.
Part B
What was the initial reactant concentration for the reaction described in Part A?
Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.
> View Available Hint(s)
[A]o = |0.0307
-1
S
Enter your answer using units of chemical concentration.
No credit lost. Try again.
Submit
Previous Answers
Transcribed Image Text:<Chapter 14 - Attempt 1 Item 7 7 of 16 > I Review | Constants | Periodic Table The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line, y = mI +b. Part A Order Integrated Rate Law Graph Slope [A]= =- kt + [Alp [A] vs. t k The reactant concentration in a zero-order reaction was 7.00x10-2 M after 170 s and 3.00x10-2 M after 360 s. What is the rate constant for this reaction? 1 In[A]: kt + In[A]o In[A] vs. t Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. - kt + TANo [A] A vs. t 2 k > View Available Hint(s) koth = 2.11x10-4 Submit Previous Answers v Correct Correct answer is shown. Your answer 2.1-10-4 was either rounded differently or used a different number of significant figures than required for this part. Important: If you use this answer in later parts, use the full unrounded value in your calculations. Part B What was the initial reactant concentration for the reaction described in Part A? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. > View Available Hint(s) [A]o = |0.0307 -1 S Enter your answer using units of chemical concentration. No credit lost. Try again. Submit Previous Answers
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