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ad-work [718]
3 years ago
9

A living cell with a tonicity (solute concentration) equivalent to 0.9% NaCl is placed in a solution containing 2% NaCl. Assume

that aquaporins are present and that the membrane therefore is permeable to water.
Chemistry
1 answer:
Stells [14]3 years ago
4 0

Answer:

This question is incomplete

Explanation:

This illustration refers to an hypertonic solution. Hypertonic solution is a solution in which the surrounding solution has a higher solute concentration (2% of NaCl) than the cell's cytosol (0.9% of NaCl). In hypertonic solution, the solution outside the cell (with higher concentration) pulls the water from the cell's cytosol (via osmosis) causing the cell to shrink.

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Thus your answer is OPTION B
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The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.11 g of water boils at atmospheric pressure?
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Answer:

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Explanation:

A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.

To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:

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the molar mass of water is:

H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole

So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?

moles of water=\frac{3.11 grams*1 mole}{18 gramos}

moles of water= 0.1728

Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?

heat=\frac{0.1728 moles*40.66 kJ}{1 mole}

heat= 7.026 kJ

<u><em>The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.</em></u>

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