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Aleksandr-060686 [28]
3 years ago
12

If body fats has a density of 0.94 g/mL and 3.0 liters of fat are removed, how many pounds of fat were removed from the patient?

Chemistry
1 answer:
-Dominant- [34]3 years ago
4 0

Answer:

6.217 pounds

Explanation:

We are given;

  • Density of body fats 0.94 g/mL
  • Volume of fats removed = 3.0 L

We are required to determine the mass of fats removed in pounds.

We need to know that;

Density = Mass ÷ volume

1 L = 1000 mL, thus, volume is 3000 mL

Rearranging the formula;

Mass = Density × Volume

         = 0.94 g/mL × 3000 mL

         = 2,820 g

but, 1 pound = 453.592 g

Therefore;

Mass = 2,820 g ÷ 453.592 g per pound

         = 6.217 pounds

Thus, the amount of fats removed is 6.217 pounds

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42.9 liters of hydrogen are collected over water at 76.0 °C and has a pressure of 294 kPa. What would the pressure (in kPa) of t
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Answer:

494.1 kPa

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

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According to the information provided in this question,

P1 = 294 kPa

P2 = ?

V1 = 42.9 liters

V2 = 22.8 liters

T1 = 76.0°C = 76 + 273 = 349K

T2 = 38.7°C = 38.7 + 273 = 311.7K

294 × 42.9/349 = P2 × 22.8/311.7

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36.14 = 22.8P2/311.7

Cross multiply

36.14 × 311.7 = 22.8P2

11264.605 = 22.8P2

P2 = 11264.605 ÷ 22.8

P2 = 494.1 kPa

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