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Strike441 [17]
3 years ago
9

old blooded animals modulate the fatty acid composition of their membranes as a function of temperature in order to A) ensure co

nsistent membrane fluidity. B) maximize available fatty acids for metabolic use. C) adjust the membrane thickness and increase thermal insulation. D) compensate for decreasing cholesterol solubility. E) all of the above
Chemistry
1 answer:
horsena [70]3 years ago
6 0

Answer:

<em>In order to ensure consistent membrane fluidity</em>

Explanation:

<h3>Cold Blooded Animals</h3>

Cold-blooded animals rely on the temperature of the surrounding environment to maintain its internal temperature, their blood is not cold.  Their body temperature fluctuates,  based on the external temperature of the environment. If it  is 30 °F outside, their body temperature  will eventually normalize to  30 °F, as well. If it eventually rises to 120 °F, their body temperature will follow the same pattern to 120 °F.

<h3>Membrane fluidity</h3>

The cell membrane of cold-blooded animals contains cholesterol, which acts as a shield for the membrane. Membrane fluidity is enhanced by temperature, as the temperature increases membrane fluidity increases and it decreases when the temperature goes down.

<em>Most cold-blooded animals adjust their feeding habit to contain  more </em><em>unsaturated fats</em><em> from plants. This helps them to maintain their </em><em>motor coordination and membrane </em><em> fluidity during the long winter</em>

Cold-blooded animals modulate the fatty acid composition of their membrane to stabilize their membrane fluidity

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Answer: The percent yield of this combination reaction is 41.3 %

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First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

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\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

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Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

2Na+Cl_2\rightarrow 2NaCl

From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

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So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

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Molar mass of NaCl = 58.5 g/mole

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Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

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