Quiz 6 - Totalscore: 110/140=78.5714%...

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Total score: 110/140 = 78.5714% Total score adjusted by 0.0  Maximum possible score: 140  1. The compound RX 3  decomposes according to the equation:  RX 3   (1/3)R + (1/3)R 2 X 3  + X 2 In an experiment, the following data were collected for the decomposition at 100°C. What is the average rate of reaction  over the first ten seconds of the experiment? t (s) [RX 3 ] (M) 0 0.0534 4 0.0432 6 0.0222 10 0.0135 12 0.0089 16 0.0082  Student  Response Value Correct Answer Feedback 1.   2.0 x 10 -2  M/s 2.   2.0 x 10 -3  M/s 3.   4.0 x 10 -3  M/s 100%  4.  0.040 M/s 5.  0.020 M/s General Feedback: average rate = -Δ[RX 3 ]/Δt We are considering the first ten seconds so, average rate = -(0.0135 M - 0.0534 M)/10. s = 0.0040 M/s Score: 10/10 
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Comments: 2. "Glycemic index" is a popular, though controversial measurement of how quickly glucose is released from starchy foods  into the blood during digestion.  Slow release (low glycemic index)is proposed to be better for preventing type II diabetes,  while fast release (high glycemic index) would be preferable for endurance athletes during competition.  Given starch  extracted from a variety of foods, what colour of staining with I 2  would indicate the lowest glycemic index?  Student  Response Value Correct Answer Feedback 1.   red 2.   blue 3.   purple 0%  4.   green General Feedback: -amylose stains blue with I α 2 . Because it is a linear, non-branched starch molecule, it will  be hydrolyzed more slowly than amylopectin, which stains red with I 2 , or a mixture, which  would stain purple. Score: 0/10  Comments: 3. Nitric oxide (NO) reacts with hydrogen (H 2 ) according to the equation: 
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2 NO(g) + 2 H 2 (g) ---> N 2 (g) + 2 H 2 O(g)  The following initial rates of reaction have been measured for the given reactant concentrations. Exp. # [NO] / M  [H 2 ] / M Rate / (M/h) 0.010 0.020 0.020 2 0.015 0.020 0.030 3 0.010 0.010 0.005 Which of the following is the rate law (rate equation) for this reaction?  Student  Response Value Correct Answer Feedback 1.  rate = k[NO][H 2 ] 2.   rate = k[NO][H 2 ] 4 3.   rate = k[NO] 2 [H 2 ] 4.  rate =  k[NO] 1/2 [H 2 ] 1/4 5.   rate = k[NO][H 2 ] 2 100%  General Feedback: Consider exp. 1 and exp. 2. ([H 2 ] is the same in both of these experiments, so any change in rate must  be due to the change in [NO].) Changing [NO] by a factor of (0.015 M/0.010 M) = 1.5 results in an  increase in rate of (0.030 M/h / 0.020 M/h) = 1.5 = (1.5) 1 , so the reaction is  1 st-order in NO. Now, consider exp. 1 and exp. 3. ([NO] is the same in both of these experiments, so any change in 
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This note was uploaded on 04/13/2011 for the course CHEM 1AA3 taught by Professor H during the Spring '08 term at McMaster University.

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Quiz 6 - Totalscore: 110/140=78.5714%...

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