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Oksana_A [137]
3 years ago
8

Two 0.9mL aliquots of the same specimen are placed in test tubes. To one is added 0.1 mL of water (spec A). To the other is adde

d 0.1 mL of 200 mg/dl urea (spec B). They are analyzed for their results: Spec A: 25 mg/dl, Spec B: 43 mg/dl. What is the percent recovery of this method?
Chemistry
1 answer:
Tanya [424]3 years ago
8 0
We assume that the method that made use of urea was able to recover all of the recoverable substance. The method in question is the method that makes use of water.
The total amount of substance is 43 mg/dl. The recovered amount is 25 mg/dl. The percent recovery is
(25 mg/dl / 43 mg/dl) * 100 = 58.14%
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CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

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Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

 

Therefore the final concentration of Cl- in the solution mixture is:

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