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Given the following data in enzyme-catalyzed reaction, what is the Vm, Km and type of inhibition of Experiment B?
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- 2I– (aq) + H2O2 (aq) + 2H3O + (aq) → I2 (aq) + 4H2O (l) (slow) C6H8O6 (aq) + 2H2O (l) + I2 (aq) → C6H6O6 (aq) + 2H3O + (aq) + 2I– (aq) (very fast) I2 (aq) + I– (aq) ⇌ I3- (aq) I3- (aq) + starch → blue I3- · starch complex (aq) (fast) A 0.100 L solution is prepared with initial concentrations of 4.0 × 10−3 M iodine I2 , 8.0×10−3 M iodide I– , and 5.0×10−3 M ascorbic acid C6H8O6 . After the second reaction goes to completion, what will the molar concentrations of iodide and ascorbic acid in the solution be?0% 25% 50% 75% 100% Depth of H2O2 Solution (d) 2.1 cm 2.1 cm 2.1 cm 2.1 cm 2.1 cm Trial 1 Time 180 sec 84.61 sec 43.52 sec 36.90sec 25.90 sec Trial 2 Time 180 sec 92.25 sec 38.16 sec 34.36 sec 23.57sec Trial 3 Time 32.53 sec 18.82 sec Average Time (t) 180 sec 88.43 sec 40.84 Sec 34.5 sec 22.76 sec Rate of the Reaction(R = d/t) 0.012 cm/sec 0.02 cm/sec 0.051 cm/sec 0.061 cm/sec 0.0923cm/sec , graph rate of reaction on the y-axis and percent concentration of enzyme on the x-axis. If the points are linear, draw a “best-fit” straight line through or near all of the data points. Based on the information in the data table and your graph, explain the relationship between percent concentration of catalase and rate of reaction. Did your actual results match your hypotheses? If not, why?Q) water hardness of each trial and average ppm with calculation, please. Hard Water Trial 1 Hard Water Trial 2 Hard Water Trial 3 Initial Syringe Reading 1.0ml 1.0ml 1.0ml Final Syringe Reading 0.88ml 0.84ml 0.85ml Volume of EDTA Consumed 0.12ml 0.16ml 0.15ml Water Hardness ppm CaCO3 Average ppm
- analyte concentration(C)(mg/ml) injection volume (ul) elution time (time) peak DAD signal(mAU) caffeine 1 1 4.67 302.85 aspartame 5 1 7.53 15.83 benzoic acid 1 1 8.14 89.98 saccharin 1 1 1.91 84.86 mixture(add everything above with 1:1:1:1 ratio) 1 4.47 69.58 How to get the concentration of the mixture in this case?2I– (aq) + H2O2 (aq) + 2H3O + (aq) → I2 (aq) + 4H2O (l) (slow) C6H8O6 (aq) + 2H2O (l) + I2 (aq) → C6H6O6 (aq) + 2H3O + (aq) + 2I– (aq) (very fast) I2 (aq) + I– (aq) ⇌ I – 3 (aq) I3- (aq) + starch → blue I3- · starch complex (aq) (fast) A different 0.100 L reaction mixture is prepared with initial concentrations of 1.0×10−2 M iodide I– , 2.0 × 10−3 M ascorbic acid C6H8O6 , and 0.135 M hydrogen peroxide H2O2 . If the first two reactions above procede at the same rate (they’re limited by the first one!), which of these reactants will be completely consumed first?2I– (aq) + H2O2 (aq) + 2H3O + (aq) → I2 (aq) + 4H2O (l) (slow) C6H8O6 (aq) + 2H2O (l) + I2 (aq) → C6H6O6 (aq) + 2H3O + (aq) + 2I– (aq) (very fast) I2 (aq) + I– (aq) ⇌ I – 3 (aq) I3- (aq) + starch → blue I3- · starch complex (aq) (fast) (a) A 0.100 L solution is prepared with initial concentrations of 4.0 × 10−3 M iodine I2 , 8.0×10−3 M iodide I– , and 5.0×10−3 M ascorbic acid C6H8O6 . After the second reaction goes to completion, what will the molar concentrations of iodide and ascorbic acid in the solution be?
- Which of the following describes a gradient elution process? stationary phase has varying concentration mobile phase has varying concentration stationary phase has static concentration mobile phase has static concentrationYou finish doing an experiemnt with Benzoin. These are the results you get:Boiling start point: 137 ceciusBoiling end point: 130 ceciusVile while empty (with cap on): 15.348Vile with crude prod (with cap on: 15.748 Your starting weight: 1.00115.748 - 15.348 = 0.400 / 1.001 = 0.3996 or 39.96% YieldAnswer the following questions with this information:Initial Mass of impure sample:Mass of recrystallized benzoin:Percent reccovery of Benzoin:Melting pot range of purified benzoin:Literature melting point for benzoin:Please write a paragraph of the objective stated using the two images attached below Please please answer as fast as possible thank you
- if you perform QC for sodium on a chemistry analyzer and the value is greater than 2SD above the mean. This is the second day that this level of control has been >2SD for that analyte. You mixed the vial and repeated. QC was still "out". It is time to troubleshoot. What do you do?If some of the analytes were found to be out of range after doing the Quality Control and the machine repeatedly gives invalid result. What are the possible problem and best solution as a Medical Technologist?18. Jet is an undergraduate chemistry student, he’s out in the laboratory trying to determine the volatile organic compounds as well as overall protein content of the leaf and stem of a malunggay (Moringa oleifera). He subjected the leaf and stem in a separate digestion reaction (treatment of sulfuric acid), afterwards he subjected the products to high temperature induction to get a dry ash like substance. Which type/s of chemical analysis did Jet employ to reach his objective?I. Qualitative AnalysisII. Quantitative Analysis III. Instrumental Chemical AnalysisIV. Wet Chemical AnalysisA. II & IV onlyB. I & IV onlyC. I, II & IV onlyD. I, II, III, IV