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DEQ:
A fungal specie of cordyceps turns humanity into a zombie-like state. The rate of growth of a single
cordyceps was observed and was determined to be proportional to itself. If between noon and 2:00
PM, the population triples, at what time, no controls being exerted, should the cordyceps becomes 100
times what it was in noon?
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Solved in 4 steps with 4 images
- A reaction increases its rate four times when temperature increases from 250 K to 300 K. What is the activation energy of the reaction? Round off your answer to 3 significant figures and with correct units:The decomposition of aqueous sucrose to form the isomers glucose and fructose is a common organic reaction, which requires a strong catalyst: C12H22O11(aq) + H2O(l) → 2C6H12O6(aq). The following data were collected during the process: Time (min) [C12H22O11] (mol/L) 0 0.316 39 0.274 80 0.238 140 0.190 210 0.146 Use the following grids to plot three separate graphs of [C12H22O11], ln[C12H22O11], and 1/[C12H22O11] against time to determine the reaction order. (a)Write the rate law expression in its general form (water should not be included). (b)Determine the rate constant with units. (c)Determine the half-life. (d)If we performed a new trial with an initial concentration of sucrose of 0.400 mol/L, what concentration would remain after 4.0 h has passed?A chemistry grad student measures the performance of the new pump in his lab. The result is: Z=19.1 kPa·mm3·s-1 conver Z to mJ·s-1
- A chemical reaction is shown below for oxidation of benzene (C6H6) to carbo (CO2) and water (H2O) using oxygen (O2). Using the theoretical oxygen (ThOD) approach, calculate how many pounds per day of oxygen (lb/day O2) will be ri to completely degrade benzene given the following conditions. Report your al the nearest unit value (no decimal). • Flow (Q) = 6 mgd Benzene concentration (C) = 107 mg/L Oxygen transfer efficiency (n) = 85 % C6H6+ 7.5 O2 —› 6 CO2 + 3 H20You are trying to come up with a drug to inhibit the activity of an enzyme thought to have a role in liver disease. In the laboratory the enzyme was shown to have a Km of 1.0 x 10-6 M and Vmax of 0.1 micromoles/min.mg measured at room temperature. You developed an uncompetitive inhibitor. In the presence of 5.0 x 10-5 M inhibitor, the apparent Vmax was determined to be 0.02 micromoles/min.mg. What is the Ki of the inhibitor?Determine the average rate of change of BB from ?=0 st=0 s to ?=272 s.t=272 s. A⟶2BA⟶2B Time (s) Concentration of A (M) 0 0.7300.730 136136 0.4450.445 272272 0.1600.160 rateB= __________M/s
- حد يدبرلنا طريقة حل فرع C A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction HAPPY ⇌ SAD. The researchers begin to characterize the enzyme. (a) In the first experiment, with [Et] at 4 nM, they find that the Vmax is 1.6 μM s−1. Based on this experiment, what is the kcat for happyase*? (Include appropriate units.) (b) In another experiment, with [Et] at 1 nM and [HAPPY] at 30 μM, the researchers find that V0 = 300 nM s−1. What is the measured Km of happyase* for its substrate HAPPY? (Include appropriate units.) (c) Further research shows that the purified happyase used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase preparation and the two experiments repeated, the measured Vmax in (a) is increased to 4.8 μM s−1, and the measured Km in (b) is now 15 μM. For the inhibitor ANGER, calculate the values of α and α′. . (d)…As discussed by Adeleye et al. (2015), what are the rationale for the following tablet characteristics when tablet compression is increased: Decrease in tablet thickness Increase in tablet’s mechanical strength Decrease in tablet’s friability No effect on drug release rate and kineticsThe decomposition of aqueous sucrose to form the isomers glucose and fructose is a common organic reaction, which requires a strong catalyst: C12H22O11(aq) + H2O(l) → 2C6H12O6(aq). The following data were collected during the process: Time (min) [C12H22O11] (mol/L) 0 0.316 39 0.274 80 0.238 140 0.190 210 0.146 (A) If we performed a new trial with an initial concentration of sucrose of 0.400 mol/L, what concentration would remain after 4.0 h has passed?
- The following growth equation was derived for color-dyed rock candy: (R - 1.8 * 10^-3 mm/min)/(3 * 10^-3 mm/min - R) = 1/(AC^m) where R is growth rate in units [mm/min] and C is concentration of dye in parts per million (ppm). Growth rate data was collected for the following concentrations: C (ppm) 50 75 100 125 150 R (mm/min) 0.0025 0.0022 0.00204 0.00195 0.0019 What are the numerical values and units of A and m, which are both constants in the equation? For a dye concentration of 400ppm, what is the estimated growth rate? Are you confident in this estimate, why or why not?A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction The researchers begin to characterize the enzyme. a)In the first experiment, with [Et] at 4 nM, they find that the Vmax is 1.6 μM s-1. Based on this experiment, what is the kcat for happyase*? (Include appropriate units.) b)In another experiment, with [Et] at 1 nM and [HAPPY] at 30 μM, the researchers find that V0300 nM s-1. What is the measured Km of happyase* for its substrate HAPPY? (Include appropriate units.) c)Further research shows that the purified happyase* used in the first two experiments wasactually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase* preparation, and the two experiments repeated, the measured Vmax in (a) is increased to 4.8 μM s-1, and the measured Km in (b) is now 15 μM. For the inhibitor ANGER, calculate the values of αand α’.A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction HAPPY ⇌ SAD. The researchers begin to characterize the enzyme. (a) In the first experiment, with [Et] at 4 nM, they find that the Vmax is 1.6 μM s−1. Based on this experiment, what is the kcat for happyase*? (Include appropriate units.) (b) In another experiment, with [Et] at 1 nM and [HAPPY] at 30 μM, the researchers find that V0 = 300 nM s−1. What is the measured Km of happyase* for its substrate HAPPY? (Include appropriate units.) (c) Further research shows that the purified happyase used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase preparation and the two experiments repeated, the measured Vmax in (a) is increased to 4.8 μM s−1, and the measured Km in (b) is now 15 μM. For the inhibitor ANGER, calculate the values of α and α′. . (d) Based on the information given…