This is a 2 part question Carbon monoxide (CO) is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does oxygen (O₂), as indicated by these approximate standard tree-energy changes in bloodreaction A: Hb+0₂HbO₂.AG-70 kJ/molreaction B: Hb+CO HbCO, AG= -80 kJ/molPart AEstimate the equilibrium constant Kat 298 K for the equilibriumExpress your answer numerically using one significant figure.>View Available Hint(s)K-SubmitVAEPrevious Answers?H6O, + CÓ HACO | Chemical energy is released or absorbed from reactions in various forms. Themost easily measurable form of energy comes in the form of heat, or enthalpyThesenthalpy of a reaction can be calculated from the heats of formation of thesubstances involved in the reaction:AHAH; (products)-AH; (reactants)Entropy change, AS", is a measure of the number of energetically equivalentmicrostates Introduced into the system during the reaction. The degree ofspontaneity of a reaction is represented by the Gibbs free energy, AG. TheGibbs free energy depends on both the enthalpy and entropy changes that takeplace during the reaction:AGAH-TAS"where is standard temperature, 298 K.Part ACalculate the standard enthalpy change for the reactionwhere the heats of formation are given in the following table:SubstanceABCDAH;(kJ/mol)All--273381200-511Express your answer in kilojoules.▸ View Available Hint(s)197) ΑΣΦ$?kJ2A+B 2C+2D

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Carbon monoxide (CO) is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does oxygen (O₂), as indicated by these approximate standard tree-energy changes in blood reaction A: Hb+0₂ -HbO₂. AG-70 kJ/mol reaction B: Hb+CO AG= -80 kJ/mol Part A Estimate the equilibrium constant Kat 298 K for the equilibrium Express your answer numerically using one significant figure. > View Available Hint(s) K- Submit VAE Previous Answers ? HbCO, ньо, + со ньсо+02

Carbon monoxide (CO) is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does oxygen (O₂), as indicated by these approximate standard tree-energy changes in blood reaction A: Hb+0₂ -HbO₂. AG-70 kJ/mol reaction B: Hb+CO AG= -80 kJ/mol Part A Estimate the equilibrium constant Kat 298 K for the equilibrium Express your answer numerically using one significant figure. > View Available Hint(s) K- Submit VAE Previous Answers ? HbCO, ньо, + со ньсо+02

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Part c
Name: Part I: Calibration of the Plastic Pipette Number of Drops Total Mass (g) Mass of 10 Drops (g) Volume of 10 Drops (mL) 23.3124 10 23.8216 20 24.2999 30 24.8245 1. Temperature of deionized water (°C) 24.7 2. Density of water at this temperature from (g/mL) Density calculator https://antoine.frostburg.edu/chem/senese/javascript/water-density.html 3. Average volume of 10 drops (mL) show calculation: Part II: Determination of Standard Solutions mL of 0.200M CuSO4'5 H20 added M CuSO4 diluted Absorbance Tube # mL H20 added 0.200 1.346 about 3 0.945 3 4.00 1.00 0.715 4 3.00 2.00 0.469 2.00 3.00 0.281 6. 1.00 4.00 show calculation for Tube #3 diluted concentration:
Station 2: Density column In the graduated cylinder, you have molasses, water (blue), corn oil, and ethanol (red). They have the following densities: Substance Molasses Water Corn Oil Ethanol Density 1.45 g/mL 1.0 g/mL 0.92 g/mL 0.79 g/mL Polarity Polar Polar Nonpolar Polar Viscosity High Low Low Low 1. Explain why the column is arranged in the order it is according to density.
3. + 3HCI B H. H. H. と-
Suppose a normally distributed set of data with 8900 observations has a mean of 123 and a standard deviation of 16. Use the 68-95-99.7 Rule to determine the number of observations in the data set expected to be above a value of 107. Round your answer to the nearest whole value. Hint: This problem is asking for how many observations not the percent. Answer Tip: Don't round any probabilities or percentages in your calculations. Keep all decimal places and round at the END of the problem.
Trial I Trial II Final reading HCI 26. 40 ml 27.10 ml Initial reading HCI 1.2 ml 1.8 ml Volume of HCl used ? ? Final reading NaOH 33.7 ml 32.8 ml Initial reading NaOH 0.8 mL 0.5 ml Volume of NaOH used ? ? For trial I: Normality of NaOH : 0.25 N Vol. of HCl used = final volume - initial volume Solve for the average Normality of HCI : ? NHCI x VHCI = NNAOH x VNAOH Vol. of NaOH used = final volume - initial volume NNAOH x VNAOH For trial II: NHCI = Vol. of HCl used = final volume - initial volume VHCI Activa Vol. of NaOH used = final volume - initial volume Go to S
10. Explain the step in filtration technique.
In a particular trial, a student got a value for R of 0.083. Calculate the percent deviation of this result. Enter a number without the % sign. The percent deviation should be reported as a positive value.
Express Surgery Center's (ESC) preferred stock, which has a par value equal to $160 per share, pays an annual dividend equal to 10 percent of the par value. If investors require a 14 percent return to purchase ESC's preferred stock, what is the stock's market value? Do not round intermediate calculations. Round your answer to the nearest cent.
Part D [OH ] = 4.0x10-11 M Express your answer using two decimal places. ? pH = Submit Request Answer Part E [H3O*] = 6.5x10-8 M Express your answer using two decimal places. V ΑΣφ pH %3D Suhmit Request Answer
The answer is supposed to be C though, it was a provincial question in the Chem 12 august 2005 exam. Why is it C?
7. Don believes that fewer than one-third of the students at his college hold part-time jobs while in school. To investigate this, he randomly surveyed 116 students and found that 32 students held part-time jobs. a. ( - Provide a 90% confidence interval for the population proportion of students that hold a part-time job. b. 1 (a). ;Give the correct interpretation of the interval in part c. Extra Credit: Does this interval support Don's claim? Why?