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Using the data sheet, identify the rate order of reactant B.
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- If 5 µM enzyme was used to obtain the data in the summary plot below, V-max = ________________ µM sec-1 and Km = ____________ µM for this enzyme (Enter numeric values to the nearest integer; Do NOT write units.)NH4+ {aq) + NO2(aq) -> N2(g) +2H2O{l} Data Initial [NH4+] Initial [NO2-] rate 1 0.0100 0.200 5.4 x10-7 2 0.0200 0.200 10.8x10-7 3 0.0400 0.200 21.5x10-7 4 0.200 0.0202 10.8x10-7 5 0.200 0.0404 21.6x10-7 6 0.200 0.0808 43.3x10-7 Find x,y,kWhich of the following would result in an increase in the calculated ngas 10, assuming the rest of the parameters are held constant? I. Inserting 23.0 mg instead of the prescribed 15.0mg of folded Mg ribbon into the eudiometer. II. Performing the experiment in a different laboratory with a cooler temperature. III. A shorter equilibration time prior to volume measurement. IV. Acid used was 12.0 M HNO3 instead of 6.0 M HCI.
- Which of the following would result in an increase in the calculated ngas in the Chemistry Experiment, assuming the rest of the parameters are held constant? I. Inserting 23.0 mg instead of the prescribed 15.0mg of folded Mg ribbon into the eudiometer. II. Performing the experiment in a different laboratory with a cooler temperature. III. A shorter equilibration time prior to volume measurement. IV. Acid used was 12.0 M HNO3 instead of 6.0 M HCl. a. II and III b. I and II c. I only d. II, III, and IVConcentration=2.5 X 10^3 μM OD400nm=0.859 1cm cuvette is used what is the extinction coefficient (μM^-1 cm^-1)You 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?
- What is ΔHsys for a reaction at 16.9 °C with ΔSsurr = -159 J mol-1 K-1 ? Express your answer in kJ mol-1 to at least two significant figures. (Please type answer no write by hend)3. a. Briefly explain why internal standardization method is useful inanalytical chemistry?b. Why does a response factor of an instrument’s detector need to becalculated?If the enzyme lactase has a Vo of 0.111111111111 mM per minute when [S] = 1.0 mM, and a Vo of 0.20 mM per minute when [S] = 5.0 mM, what is its Km? Calculate Vmax of the above enzyme (lactase). Calculate the slope on a Lineweaver-Burk plot (Km/Kmax) for the above enzyme (lactase).
- Initial temp: 19.06 C trial 1: 0.5 M concentration trial 2 : 1.0 M concentration 200 mL of HCl and 200 mL of NaOH are combined in an insulated container. HCl = 0.5 mol/ml x 200 x 10^-3 ml = 0.1 mol NaOH = 0.5 mol/ml x 200 x 10^-3 ml= 0.1 mol H+(aq) + OH-(aq) - - -> H2O (l) ∆H H++ = 0, ∆H OH−− = -230 kJ/mol, ∆H H2O = -286 kJ/mol ∆H reactants = ∆H H++ + ∆H OH−− = -230 + 0 = -230 kJ/mol ∆H reactants = -286 kJ/mol - ( -230 kJ/mol) = -56 kJ/mol For each trial, calculate how much energy is released during this reaction.Chemistry The levels of an organic pollutant (P) in the groundwater at the perimeter of a plant were a cause for concern. A 10 mL sample of the water was taken and the pollutant was extracted with 95% efficiency using 25 mL of diethyl ether. GC was used to analyse the concentration of P in diethyl ether. A calibration curve was plotted for a series of standards of P which yielded the following results: Peak Area Toluene Conc. (µg/ml) 12,000 2.6 23,700 5.0 35,500 7.7 46,800 9.9 31,250 Sample Determine the concentration of P in ppb in the initial groundwater sample.Before determining the amount of Na2CO3 in an unknown sample, a student decides to check her procedure by analyzing a sample known to contain 98.76% w/w Na2CO3. Five replicate determinations of the %w/w Na2CO3 in the standard were made with the following results 98.71% 98.59% 98.62% 98.44% 98.58% Is the mean for these five trials significantly different from the accepted value at the 95% confidence level (α = 0.05)