ucose-cocp-2000+100g ge Find Cyoy the value of the [HgCl2] (M) [C2042-1 (M) | Rate (Mis) 0.10 0.10 1.3.10-7 0.10 0.20 5.2.10-7 0.20 0.20 1.0-10-6 A) 1.4 x 10-8 1/M2-s B) 1.3 x 10-7 1/M2-s C) 1.4 x 10-5 1/M2-s D) 1.3 × 10-4 1/M2-s
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![ucose-cocp-2000+100g ge
Find Cyoy
the value of the
[HgCl2] (M) [C2042-1 (M) | Rate (Mis)
0.10
0.10
1.3.10-7
0.10
0.20
5.2.10-7
0.20
0.20
1.0-10-6
A) 1.4 x 10-8 1/M2-s
B) 1.3 x 10-7 1/M2-s
C) 1.4 x 10-5 1/M2-s
D) 1.3 × 10-4 1/M2-s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9dcc4d2-674c-4a07-a038-b6299ea22f35%2Ffda6a390-846a-45b2-bd01-9f37dd63ebd2%2Fmtw26zb_processed.jpeg&w=3840&q=75)
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- The kinetics of an enzyme are measured as a function of [S] in the presence and absence of 2 mM I. Compute Km and Vmax in the absence and presence of I S] (uM) V (uM/min) without I with I 3 10.4 4.1 5 14.5 6.4 10 22.5 11.3 30 33.8 22.6 90 40.5 33.8Using the data in the table, calculate the rate constant of this reaction. A+B⟶C+D Trial [?] (?) [?] (?) Rate (M/s) 1 0.400 0.340 0.0126 2 0.400 0.952 0.0988 3 0.680 0.340 0.0214Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D Trial [?] (?) [?] (?) Rate (M/s) 1 0.340 0.210 0.0204 2 0.340 0.420 0.0204 3 0.680 0.210 0.0816 ?=
- 6 I- + BrO3- + 6 H+ ----- 3 I2 + Br- + 3 H2O Rate= k[I-]m[BrO3-]n[H+]p rate= (∆[I2])/(3*∆time) Trial [I-] (M) [BrO3-] (M) [H+] (M) 1 1.67*10-3 6.67*10-3 1.67*10-2 2 3.33*10-3 6.67*10-3 1.67*10-2 3 1.67*10-3 1.33*10-2 1.67*10-2 4 1.67*10-3 6.67*10-3 3.33*10-2 Reaction time: Trial 1 - 301.90 secs Trial 2 - 213.28 secs Trial 3 - 214.92 secs Trial 4 - 86.45 secs Calculate the rate for each trial and Using the information above determine the rate law (m, n and p) for the reaction. m_____________ n_________________ p ____________ Rate= k[I-]m[BrO3-]n[H+]pUsing the data in the table, calculate the rate constant of this reaction. A+B⟶C+D Trial [?] (?) [?] (?)[) Rate (M/s) 1 0.370 0.390 0.0234 2 0.370 0.936 0.135 3 0.592 0.390 0.0374 ?= ? units= ?An enzyme with a single active site has a Michaelis-Mentin constant (KM ) of 25 mM and aturnover number of 4.0 x 107 s –1 .(a) Calculate the initial rate of the reaction in μM/s if the total enzyme concentration is0.016 μM and the initial substrate concentration is 4.32 μM.(b) Calculate the value of Rmax
- A dye degradation experiment was carried out and the rate law was determined to be: rate=k[A]2. The initial absorbance, [A]0, was 1.00 and the absorbance at t=60.0 s, [A]t, was 0.00125. What is the value of the rate constant, k for the experiment?A group of students compiled the data shown in data table 1 below. What is the units for the average rate constant value (k)? Exp. # [IO3-]0 (M) [I-]0 (M) [H+]0 (M) Time (s) 1 0.005 0.05 2 x 10-5 22.12 2 0.010 0.05 2 x 10-5 86.84 3 0.005 0.10 2 x 10-5 5.35 4 0.005 0.05 4 x 10-5 2.65 Question 8 options: M-2 s-1 M-3 s-2 M2 s-2 M-4 s-2 M3 s3 M-4 s-3 M-2 s-2 M2 s2 M s M s-1 M-3 s-1 M-1 s-1 M-3 s-3An enzyme has a Km equal to 1.0 x 10-4 M (molar) for a certain substrate S. When [S] = 1.0 x 10-4 M, 2% of the substrate is transformed into product in 1 minute. Which of the following statements are correct? Hint: use the Michaelis-Menten equation; Remember that a reaction rate can be expressed in molar units / min (M / min). Select one or more than one: a) When [S] = 1.0 x 10-4 M, the initial reaction rate, V0, is 2.0 x 10-7 M / min b) For this enzyme Vmax = 4.0 x 10-7 M / min c) For this enzyme Vmax is equal to 2.002 x 10 -7 M / min. d) When [S] 0 = 1.0 x 10-3 M, the initial speed is 3.64 x 10-7 M / min. e) When [S] = 0.01 M, the initial speed is 4.0 x 10-3 M / min
- A + 3B + 2C --> D + 2E Determine the rate law and rate constant using the experimental data below [A] (M) [B] (M) [C] (M) Rate (M/sec) Exp. 1 0.100 5.00 x 10-4 1.00 x 10-2 0.137 Exp. 2 0.100 1.00 x 10-3 1.00 x 10-2 0.268 Exp. 3 0.200 1.00 x 10-3 1.00 x 10-2 0.542 Exp. 4 0.400 1.00 x 10-3 2.00 x 10-2 1.084Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D Trial [?] (?) [?] (?) Rate (M/s) 1 0.360 0.290 0.0144 2 0.360 0.580 0.0144 3 0.720 0.290 0.0576 k=An enzyme-catalyzed reaction has a Km of 1 mM and a Vmax of 4 nmole/L sec-1. The reaction velocity (nmole/L sec-1) when the substrate concentration is 0.25 mM is: A. 1.25 B. 10.0 C. 5.0 D. 0.50 E. 1.0 F. 100 G. 150 H. 55 I. 75
