# gammagirl

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1. ## Determine the solution to the following differential equation: dA/dt = -0.3A and A(2) = 400

I appreciate you.
2. ## Determine the solution to the following differential equation: dA/dt = -0.3A and A(2) = 400

Determine the solution to the following differential equation: dA/dt = -0.3A and A(2) = 400
3. ## gammagirl    hypervalent_iodine

Can you help with the Henderson-Hasselbalch question, please?

4. ## 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.)
5. ## 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?
6. ## 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?
7. ## 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?
8. ## 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.
9. ## 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.
10. ## 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, ......
11. ## Recrystallization

Gentlemen, The vague answer, possibly, is that some crystals are left behind in the solvent during each recrystallization so this causes a decrease in recovery. In addition, successive recrystallizations result in soluble impurities contaminating the filtrate, which reduces the yield of pure crystals. The question is," Is there some math that goes along with this 10 % impurity that results in successive 20% decrease in yield with each subsequent crystallization?"
12. ## Recrystallization

My thoughts exactly. That the solvent amount is reducing, but how does that account for the 20% drop? 1st recrystallization 20 gm impure 80 pure. 2nd recrystallization 40gm impure 60 gm pure. 3rd recrystallization 60 impure and 40 gm pure. The solvent is not mentioned in the problem. What is the math behind this?
13. ## 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?
14. ## Rank the following molecules from most to least acidic?

You always come through. Thanks for confirming. I appreciate you.
15. ## Rank the following molecules from most to least acidic?

Fellow companions, I am following the rules of charge, atoms, resonance, inductive effects, and orbitals. For (c) I do not understand why molecule 1 is ranked first. I would have thought molecule 3 would be the most acidic because of atoms and the ortho arrangement. Please enlighten me as soon as possible. Why would the first molecule win? Anyways, that is what the answer key says.