Next the student conducts two different runs: In RUN 1, 10.0 mL of 5.00 M NaOH is combined with 10.0 mL of 0.010 M CO and 80.0 mL of wate In RUN 2, 20.0 mL of 5.00 M NAOH is combined with 10.0 mL of 0.010 M CO and 70.0 mL of wate The following plots were obtained for each of the runs: (crystal orange] vs time (RUN 1) In[crystal orange] vs time (RUN 1) 0.0012 0.001 y= -0.0002x + 0.0007 R-0.6565 y--0.4463x - 7.3334 R=0.893 0.0008 0.0006 -6 0.0004 -8 0.0002 -10 2. 0.0002 -12 time (hr) time (hr) 1/[crystal orange] vs time (RUN 1) 1/[crystal orange] vs time (RUN 2) 12000 25000 10000 20000 8000 15000 6000 y = 2000x + 1000 R=1 10000 y= 4000x + 1000 4000 5000 2000 time (hr) time (hr) Figure 4. Kinetic traces for the reaction between crystal orange and sodium hydroxide. What is the order with respect to CO? I Select) What is the order with respect to NaOH? | Select ) What is the value of the averaged rate constant for the reaction above? Give your answer to 3 sig. figs. 1/[crystal orange] [M (crystal orange] (M) 1/(crystal orange] (M In(crystal orange] (M)

Next the student conducts two different runs: In RUN 1, 10.0 mL of 5.00 M NaOH is combined with 10.0 mL of 0.010 M CO and 80.0 mL of wate In RUN 2, 20.0 mL of 5.00 M NAOH is combined with 10.0 mL of 0.010 M CO and 70.0 mL of wate The following plots were obtained for each of the runs: (crystal orange] vs time (RUN 1) In[crystal orange] vs time (RUN 1) 0.0012 0.001 y= -0.0002x + 0.0007 R-0.6565 y--0.4463x - 7.3334 R=0.893 0.0008 0.0006 -6 0.0004 -8 0.0002 -10 2. 0.0002 -12 time (hr) time (hr) 1/[crystal orange] vs time (RUN 1) 1/[crystal orange] vs time (RUN 2) 12000 25000 10000 20000 8000 15000 6000 y = 2000x + 1000 R=1 10000 y= 4000x + 1000 4000 5000 2000 time (hr) time (hr) Figure 4. Kinetic traces for the reaction between crystal orange and sodium hydroxide. What is the order with respect to CO? I Select) What is the order with respect to NaOH? | Select ) What is the value of the averaged rate constant for the reaction above? Give your answer to 3 sig. figs. 1/[crystal orange] [M (crystal orange] (M) 1/(crystal orange] (M In(crystal orange] (M)

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...

See similar textbooks

Similar questions

The pKb of ammonia is 4.74. Ksp of Mg(OH)2 is 5.6 x 10-12 If one 100 mL 0.1 M MgCl2 to 100 mL 0.1M NH3, would a precipitate of Mg(OH)2 form? Show using calculations why this would or would not happen.
A 0.5131-g sample containing KBr is dissolved in 50 mL distilled water. Titrating with 0.04614 M AgNO3 requires 25.13 mL to reach the Mohr endpoint. A blank titration requires 0.65 mL to reach the same endpoint. Report the % (w/w) KBr in the sample. [Ans. 26.19 % (w /w)]
How would you use an experimentally determined Ksp value to calculate that Ag2CrO4 should precipitate when 5ml of 0.0040 M AgNO3 are added to 5ml of 0.0024 M K2CrO4. You can come up with your own experimentally Ksp value. I just want to see steps in solving
Perform these calculations for nickel(II) carbonate. (a) With what volume of water must a precipitate containing NiCO3 be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with NiCO3 (Ksp = 1.36 × 10–7).(b) If the NiCO3 were a contaminant in a sample of CoCO3 (Ksp = 1.0 × 10–12), what mass of CoCO3 would have been lost? Keep in mind that both NiCO3 and CoCO3 dissolve in the same solution.
Using the two images attached please write one paragraph about the ANALYSIS Please please please answer as fast as possible
14. A 130.00 mL solution of 0.00335 M A3B4 is added to a 160.00 mL solution of 0.00105 M C3D4. What is pQsp for A3D4?
39. Three more samples of KHP (GFW=204.23g/mole) weighing 0.8001g, 0.7001, and 0.7666g were titrated with the same sodium hydroxide solution. The volumes for the titrations were 21.55mL, 17.44, and 18.37mL respectively. A.) Calculate three more values for the molarity =, then B.) Calculate the mean for all four measurements. And calculate the average for all for measureme ement ave. mean.
I have seen other "experts" answer this question but i'm not understanding the process. Please write out the steps and why. And please explain why we use 0.100-x and x.
I made some titrations and the calculated molarity of HCI was 0.096, 0.097 and 0.098. The nomical concentrsrion of HCl is 0.1. How can I explain my results by comparing it to the nominal concentration if my confidence intervals don't include the nominal concentration?
8) 25.00 mL of a halide solution, which contains approximately 0.025 M iodide and approximately 0.05 M chloride, is titrated with 0.0500 M AgNO3. Ksp (Agl) = 8.3 x 10-17 and Ksp (AgCl) = 1.8 x 10-10./25.00 mL van 'n haliedoplossing, wat ongeveer 0.025 M jodied en ongeveer 0.05 M chloried bevat, word getitreer met 0.0500 M AgNO3. Ksp (Agl) = 8.3 x 10-17 en Ksp (AgCl) = 1.8 × 10-10. a) Sketch a rough pAg* vs volume AgNO3 added curve of the titration of 25 mL of the halide solution with the AgNO3 solution. Your sketch should be consistent with the information provided, including that the pAg* scale should be an approximate representation of the true value at different points of the titration. Your curve should also include realistic estimates of the equivalence points. / Skets 'n rowwe benoemde pAg* vs volume AgNO3 bygevoeg kurwe van die titrasie van 25 mL van die halied oplossing met die AgNO3 oplossing. Jou skets moet ooreenstem met die inligting wat verskaf word, insluitend dat die…
Find the solubility of AgI in 3.3 M NH3. (Ksp of AgI = 8.3 10-17 and Kf of Ag(NH3)2+ = 1.7 107)
It is found that 24.68 mL of 0.1165 M NaOH is needed to titrate 0.2931 g of an unknown acid to the phenolphthalein end point. Calculate the equivalent mass of the acid.