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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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a sample of sulfur dioxide occupies a volume of 652 mL at 40.0 C and 0.75 atm. What volume will the sulfur dioxide occupy at STP
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The volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles. Details about volume can be found below.

<h3>How to calculate volume?</h3>

The volume of a gas can be calculated using the following formula:

PV = nRT

  • P = pressure
  • V = volume
  • n = number of moles
  • R = gas law constant
  • T = temperature

0.75 × 0.652 = n × 0.0821 × 313

0.489 = 25.69n

n = 0.489/25.69

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Therefore, the volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles.

Learn more about volume at: brainly.com/question/1578538

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