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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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Answer:

k=10^{3}

Explanation:

k stand for equilibrium constants in terms of reaction

The higher the value of an equilibrium constant the faster the equilibrium reaction comes to completion.

Consider the example below:

R_{1}\ + R_{2}⇄P_{1}\ + P_{2}

where

The\ equilibrium\ constant\ K = \frac{P_{1}P_{2}}{R_{1}R_{2}}

For a faster reaction the numerator i.e. the right hand side of the equation have to be higher than the left hand side (the denominator). therefore the higher the numerator, the higher the value of the equilibrium constant and the faster the reaction get to completion thus option c is correct.

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Osmium is the most dense element we know of. A 22 g sample of Osmium has a volume of 100 cL. Calculate the density of Osmium in
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Answer:

a. V = 1000 mL

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Explanation:

a.

First we need to convert the volume of the Osmium into mL. For that purpose we are given the conversion unit as:

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Hence, the given volume of Osmium will be:

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<u>V = 1000 mL</u>

b.

The density of Osmium is given by the following formula:

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HCOOH \rightleftharpoons HCOO^{-} + H^{+}

The acid dissociation constant of formic acid, k_a is:

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Rearranging the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

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