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Oksana_A [137]
4 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]4 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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Answer: D

Explanation:

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307 g of an unknown mineral is heated to 98.7 °C and placed into a calorimeter that contains 72.4 g of water at 23.6 °C. The hea
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Answer:

0.138 J/g.K

Explanation:

According to the law of conservation of energy, the sum of the heat released bt he mineral and the heats absorbed by the water and the calorimeter is equal to zero.

Qm + Qw + Qc = 0

Qm = - Qw - Qc   [1]

The heat absorbed by the calorimeter can be calculated using the following expression.

Q = C . ΔT

where,

C is the heat capacity

ΔT is the change in temperature

The heat released by the mineral and the one absorbed by the water can be calculated using the following expression.

Q = c . m . ΔT

where,

c is the specific heat capacity

m is the mass

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Replacing in [1],

Qm = -Qw - Qc\\c_{m}.m_{m}.\Delta T_{m}=-c_{w}.m_{w}.\Delta T_{w}-C_{c}.\Delta T_{c}\\c_{m}=\frac{-c_{w}.m_{w}.\Delta T_{w}-C_{c}.\Delta T_{c}}{m_{m}.\Delta T_{m}} \\c_{m}=\frac{-(4.18J/g.K).(72.4g).(305.4K-296.6K)-(15.7J/K).(305.4K-296.6K) }{(307g).(305.4K-371.7K)} \\c_{m}=0.138J/g.K

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