Show work please **Educational Content Transcription: Reaction Completion**At 25°C, the second-order reaction \( \text{NOCl (g)} \rightarrow \text{NO (g)} + \frac{1}{2} \text{Cl}_2 \text{(g)} \) is 50% complete after 5.82 hours when the initial concentration of \(\text{NOCl}\) is 4.46 mol/L. How long will it take for the reaction to be 75% complete?**Multiple Choice:**- 23.0 hr- 11.6 hr- 8.22 hr- 10.5 hr**Explanation:**This question involves a second-order chemical reaction with respect to the decomposition of NOCl. Students are asked to determine the time required for the reaction to reach 75% completion using the provided data on the time taken to reach 50% completion.**Note if Graphs or Diagrams Were Present:**Currently, there are no graphs or diagrams associated with this question. If a diagram of a typical second-order reaction curve were provided, it would show the concentration of \(\text{NOCl}\) decreasing over time, illustrating how the concentration declines more sharply at the beginning and levels out as the reaction progresses.The graph of concentration (\(\text{NOCl}\)) versus time would typically display a curve that begins steep and becomes more gradual as it approaches completion, characteristic of second-order reactions. Understanding this would provide learners with a visual representation of how reaction rates change over time. The image displays a section of a multiple-choice question, presumably as part of an educational quiz or test. The question itself is not visible, but several answer options are provided as follows:1. 23.0 hr2. 11.6 hr3. 8.22 hr4. 15.5 hr5. 17.5 hrEach option is preceded by a radio button, allowing the user to select one answer. There are no graphs or diagrams accompanying the question. The bottom of the interface shows "Prev" and "Next" buttons, suggesting that this is part of a sequence of questions and this particular item is number 9 out of 18.

The time required for the given second order reaction to complete 75% is to be determined from the…

At 25°C, the second-order reaction NOC(g) NO(g) + 2Cl2(9)is 50% complete after 5.82 hours when the initial concentration of NOCI is 4.46 mol/L. How long will it take for the reaction to be 75% complete? Multiple Choice 23.0 hr 116 hr

At 25°C, the second-order reaction NOC(g) NO(g) + 2Cl2(9)is 50% complete after 5.82 hours when the initial concentration of NOCI is 4.46 mol/L. How long will it take for the reaction to be 75% complete? Multiple Choice 23.0 hr 116 hr

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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AHMAD, ASHNA 5 of 35 1 4. 6. 7 A student has two rubber balls of different masses to perform an experiment. Which experiment can the student do to objects? O Roll the rubber balls on thick carpet, and observe if either ball produces a spark. O Move a magnet near each rubber ball, and observe if either ball is attracted to the magnet. O Place the rubber balls ona rough surface, apply a small push, and record the time for each ball to move 30 cm. O Drop the rubber balls at the same time from 5 m, and record the time for each ball to hit the ground. P Type here to search 5.
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VISUALIZATION Heating Curves I 1.0 g Water, H₂O O 1.0 g Benzene, C6H6 Heat Sample Reset Recheck 8th attempt Incorrect Temperature (°C) Submit Answer 50 25- 0 -25 Generate the heating graph for H₂O. Determine the quantity of energy transfer required to heat 1.0 g solid H₂O from -50 °C to the melting point, 0 °C. * J Next) (1 of 6) [Review Topics] -50 0 50 100 150 200 250 300 350 400 450 500 550 600 650 Quantity of energy transferred (J) Retry Entire Group [References] Incorrect 9 more group attempts remaining Previous Next>
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AutoSave Off H 2. U Document1 - Word P Search retroget R File Home Insert Design Layout References Mailings Review View Help A Share P Comments O Find - E Replace Times New Roma v 12 - A A Aa v A AaBbCcDc AaBbCcDc AaBbC AaBbCcC AaB 1 Normal Paste BIU v ab x, x A ev A v A - I No Spac. Heading 1 Heading 2 Dictate Editor Reuse Title A Select Files Clipboard Font Paragraph Styles Editing Editor Reuse Files Voice Consider a reversible reaction in equilibrium, CO (g) + 3 H2 (g) = CH4 (g) + H20 (g). How the equilibrium of the reaction is disturbed by adding (i) CO, (ii) H2, (iii) CH4, (iv) and H2O2 in the reaction? 한 (Ctri)- Page 1 of 1 W English (Canada) D. Focus 100% 40 words 8:06 AM W 26°C Mostly sunny 4) a O 08/06/2021
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Let C(t) be the carbon dioxide level in parts per million in the atmosphere where t is the time in years since 2000. Under two possible models the derivative functions are 1. C'(t)=0.45 +0.03 11. C'(t)=0.45e0.025 If the carbon dioxide level was 370 ppm in 2000, find C(t) for each model. Then find the carbon dioxide level in 2030 for each model. Using Model I, C(t) Using Model II., C(t) and the carbon dioxide level in 2030 is C(30) - ppm. and the carbon dioxide level in 2030 is C(30)-pp ppm.
ľ m 5.06 Home Insert Draw Design Layout A^ A AutoSave OFF Paste Page 6 of 8 Times New... v 12 B I U 1693 words ✓ ab R 2 V x² A V References Aa ✓ English (United States) A A ✓ V Mailings Review V View V V Tell me ↓ Time (hr) 0 0.5 1 2 4 6 10 15 20 25 50 75 PHA 371_Q3.docx AaBb CcDdF Heading 1 c. 0.05/hr d. 0.05 mm/hr e. 1.175 mm/hr c. 6.065 hours d. 13.86 hours e. 69.3 hours [A] mM 10 Recovered Compatibility Mode 9.975 9.95 9.9 9.8 9.7 9.5 9.25 9 8.75 7.5 6.25 c. 6.065 hours d. 7.5 AaBbCcDdEe Normal (Question 18, 19, 20, 21) Drug A and B undergoes degradation reaction in solution. Investigation of this reaction led to the availability of [A] and [B] versus time data, which is represented in the table shown below. Please use this information in answering the various parts of this problem 3 AaBbCcDdEe No Spacing [B] mM 10 9.950 9.900 9.802 9.608 9.418 9.048 8.607 8.187 7.788 19. The kinetic constant for the degradation of B is about a. 0.01 mm/hr b. 0.01 /hr 6.065 4.724 18. Which of the…
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