gammagirl

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Everything posted by gammagirl

  1. How is a correlation coefficient of 0.999992 better than a correlation coefficient of 0.99996 even though both analyses resulted in the same molar absorptivity for the analyte using spectrophotometric analysis? Does 0.999992 result in a higher correlation coefficient than 0.99996? If so, why?
  2. Ksp

    Determine the mass in grams of lead (II) iodide that will dissolve in 500.0 mL of a solution containing 1.03 grams of lead (II) nitrate. Ksp of lead (II) iodide is 1.4 x 10-8. 1.03 g x mol = .00622 mol ------ ------ ------------- 0.5 L 331.2 L PbI2 (s) ==> PbI2(aq) ==>Pb2+ + 2I- .00622 2x^2 (.00622)(2x)^2=1.4 x 10^-8 x = 7.5 x 10^-4 M 7.5 x 10^-4 m/L x .5 L x 461.01 g/mole = .173 g
  3. Ksp

    yes should if be (x +.00622)(4x^2)? Idk how the solution to that math.
  4. Keq puzzle

    1.Consider a reaction mixture that has initial concentrations of Fe3+ = 0.0050 M and SCN– = 0.0050 M. Without doing any calculations, which of the following values do you know? a) The equilibrium concentrations of Fe3+, SCN– , and FeSCN2+ b) The sum of the equilibrium concentrations of Fe3+ , SCN– , and FeSCN2+ c) The product of the equilibrium concentrations of Fe3+ , SCN– , and FeSCN2+ d) The ratio of equilibrium concentrations of products to reactants: [FeSCN2+]/[Fe3+][SCN– ] e) The ratio of equilibrium concentrations of reactants to products: [Fe3+][SCN– ]/[FeSCN2+] Explain your answer, Earlier in the experiment, a calibration curve measuring absorbance on the y-axis and concentration on the x-axis was generated from a set of 3 standard solutions. So, I am thinking perhaps the d and e are known due to a concentration/absorbance ratio that graph? Next is this question, Consider a reaction mixture that has an initial concentration of FeSCN2+ = 0.0050 M, no Fe3+ or SCN. Without doing any measurements or calculations, which of the following two values do you know? a) The equilibrium concentrations of Fe3+, SCN– , and FeSCN2+ b) The sum of the equilibrium concentrations of Fe3+, SCN– , and FeSCN2+ c) The product of the equilibrium concentrations of Fe3+, SCN– , and FeSCN2+ d) The ratio of equilibrium concentrations of products to reactants: [FeSCN2+]/[Fe3+][SCN– ] e) The ratio of equilibrium concentrations of reactants to products: [Fe3+][SCN– ]/[FeSCN2+] Explain your answer. By the way the equation for both is , the equation is Fe3+ (aq) + HSCN (aq) <-----> FeSCN2+ (aq) and everything is 1:1. Using the ice table, Fe3+ is 0.0050M -x , SCN-, is 0.0050M- x. So, we know a and b?
  5. Keq puzzle

    Absorbance of FeScn2+ And the link IS the experiment And that is the point the reaction is in equilibrium but the questions are confusing. If The Fe3+ was in excess, then Scn-=FeScn2+. But for the first question question, Fe 3+=Scn - is the same amount (as in the lab). It is something intuitive and easy.
  6. What do you mean by slightly different noise?
  7. Is the bottom line that both values are nearly perfect straight lines making the differences between the numbers negligible?
  8. I used 3 mL of 95% ethanol? 3mL x 0.789/ml x mol/46.068 g = 0.0513gm OR 95gm/100ml =x/3ml, x= 2.85 gm ethanol x mol/46.068 g= 0.06187 gm
  9. IR puzzle

    I was thinking "A shoulder band usually appears on the lower wavenumber side in primary and secondary liquid amines arising from the overtone of the N–H bending band: this can confuse interpretation."
  10. IR puzzle

    When obtaining an IR spectrum, a secondary amine should exhibit only one N-H stretch, however, when a student ran his sample, two N-H stretches appeared in the spectrum. What could account for this?
  11. Both are good points. Thank-you
  12. IR spectroscopy

    A secondary amine should only exhibit only one N-H stretch. However, 2 N-H stretches appeared in the spectrum. What could account for this?
  13. Henderson-Hasselbalch

    Thank-you
  14. Henderson-Hasselbalch

    When extracting diphenhydramine from hexane and calculating the concentration of diphenhydramine using the Beer-Lambert law, the ratios of NH3/NH4+ don't match the expected ratios based on the Henderson-Hasselbalch equation. Why?
  15. Henderson-Hasselbalch

    The answers mentioned above are correct, but I thoroughly understand the experiment at a higher level. Therefore, the Henderson-Hasselbach case is closed.
  16. Determine the solution to the following differential equation: dA/dt = -0.3A and A(2) = 400
  17. Can you help with the Henderson-Hasselbalch question, please? 

  18. Henderson-Hasselbalch

    Part 1: (1) Made an unbuffered stock solution of DPH-diphenhydramine (100mg/100ml) in water (2) Made ph4 buffered stock solution of DPH (3) Made ph7 buffered stock solution of DPH (4) Made a ph10 buffered stock solution of DPH Part 2: Obtain absorbance at 252 of solutions in Part 1 Part 3: Add 5ml of DPH stock solutions to 5 ml hexanes and obtain absorbances. question: even though a buffered solution at pH7 should have a 100:1 ratio of ammonium: amine, why does the absorbance of the aqueous solution after extraction have such a large difference when compared to the unextracted stock solution of DPH of ph7? question: Assuming molar absorptivity of DPH is 388L/mole cm (a) calculate the approximate conc of DPH in each stock solution, before and after extraction with hexane. (did it) (b) Why don't these values match the expected ratios based on Henderson-Hasselbach equation? (my answer: HH is valid when it contains equilibrium concentrations of an acid and conjugate base. In this lab, changing pH increases the amount of DPH moving to the organic layer from the aqueous layer, changing the expected ratios of conjugated base and acid.)
  19. Henderson-Hasselbalch

    I have another question. Even though a buffered solution at pH 7 should have a 100:1 ratio of ammonium ion: amine, why does the absorbance of the aqueous solution after extraction with hexane have such a large difference when compared to the unextracted stock solution of the amine a ph 7?
  20. Henderson-Hasselbalch

    Step 1: Creating pH 4, 7, 10 buffered stock solutions. Step 2: Obtain absorbance by spectroscopy of buffered solutions. Step 3: Obtain absorbance of stock solutions after extraction with hexanes. Step 4: calculate concentration with Beer's Law Question: Why don't these values match the expected ratios based on the Henderson-Hasselbalch equation? The difference is due to the buffer solutions are able to withstand most changes in pH and maintain a reasonably stable pH?
  21. Recrystallization

    Only someone really smart could give an answer like that, especially since no solubility data was given in cold versus hot to hint at that issue as the main point. That explanation is the entire basis of recrystallization.
  22. Recrystallization

    A student performed successive recrystallizations on an impure mixture where there was a 10% by weight impurity. After one recrystallization, she obtained 80% of the original weight of crystals back. After a second recrystallization, she obtained 60% of the original weight of crystals back. After a third recrystallization, she obtained 40% of the original weight of crystals back. If the original amount of contaminant was only 10%, why did she only obtain 40% of the final pure product? This is what I understand so far, 10g impure so 90 gm pure.........so now there is 80 gm. However, the 10% impurity may mean that part of the 90 % is contaminated. Can someone push me along?
  23. Recrystallization

    The property you are exploiting is removal of the impurities. And why is always the same......is the puzzle to me. There are impurities that are soluble and insoluble in the mother filtrate along with the pure product.
  24. Recrystallization

    Dear hypervalent_iodine, if you start off with a 100gm mass 10gm is impurity 90 gm is pure product. So during the second crystallization, at most 80 gm is the maximum yield. Then, ......