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ludmilkaskok [199]
3 years ago
13

Ou have a 5 ml sample of a protein in 0.5 m nacl. you place the protein/salt sample inside dialysis tubing (see fig. 2-14) and p

lace the bag in a large beaker of distilled water. if your goal is to remove as much nacl from the sample as possible, which would be more effective: (1) placing the dialysis bag in 4 l of distilled wa- ter for 12 h, or (2) placing the bag in 1 l of distilled water for 6 h and then in another 1 l of fresh distilled water for another 6 h
Chemistry
1 answer:
Oduvanchick [21]3 years ago
4 0

Answer:

Procedure (2)  

Explanation:

Assume the dialyses come to equilibrium in the allotted times.

Procedure (1)

If you are dialyzing 5 mL of sample against 4 L of water, the concentration of NaCl will be decreased by a factor of

\dfrac{5}{4000} = \dfrac{1}{800}

Procedure (2)

For the first dialysis, the factor is

\dfrac{5}{1000} = \dfrac{1}{200}

After a second dialysis, the original concentration of NaCl will be reduced by a factor of  

\dfrac{1}{200} \times \dfrac{1}{200} = \dfrac{1}{40000}

Procedure (2) is more efficient by a factor of  

\dfrac{40000}{800} = \mathbf{50}

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Strong acids are assumed 100% dissociated in water. True As a solution becomes more basic, the pOH of the solution increases. Fa
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Answer:

Strong acids are assumed 100% dissociated in water- True

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A mixture of nacl and sucrose (c12h22o11) of combined mass 10.2 g is dissolved in enough water to make up a 250 ml solution. the
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<span>% NaCl = (1.5971 / 10.2)*100 = 15.66%</span>
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