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Inessa05 [86]
3 years ago
11

Osmosis is the process responsible for carrying nutrients and water from groundwater supplies to the upper parts of trees. The o

smotic pressures required for this process can be as high as 20.1 atm . What would the molar concentration of the tree sap have to be to achieve this pressure on a day when the temperature is 29 ∘C?
Chemistry
1 answer:
ad-work [718]3 years ago
7 0

Answer:

The molar concentration would have to be 0,81 M.

Explanation:

The osmotic pressure equation is:

                                               \pi = M R T

where:

\pi: osmotic pressure [atm]

M: molar concentration [M]

R: gas constant 0,08205 [atm.L/mol.°K]

T: absolute temperature [°K]

To solve the problem, we just clear M from the osmotic pressure equation and then replace our data using the appropiate units. Clearing the variable M we have:

M = \frac{\pi }{RT}

We have to use temperature as absolute temperature (in kelvins), T=29+273=302 °K. Now we can replace our values in the equation:

M = \frac{20,1 atm}{0,08205 \frac{atm.L}{mol.K} * 302 K }

As we can see, all units will be simplified and we'll have the molar concentration in mol/L.

M = 0,81 \frac{mol}{L} = 0, 81 M

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The equilibrium membrane potential is 41.9 mV.

To calculate the membrane potential, we use the <em>Nernst Equation</em>:

<em>V</em>_Na = (<em>RT</em>)/(<em>zF</em>) ln{[Na]_o/[Na]_ i}

where

• <em>V</em>_Na = the equilibrium membrane potential due to the sodium ions

• <em>R</em> = the universal gas constant [8.314 J·K^(-1)mol^(-1)]

• <em>T</em> = the Kelvin temperature

• <em>z</em> = the charge on the ion (+1)

• <em>F </em>= the Faraday constant [96 485  C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]

• [Na]_o = the concentration of Na^(+) outside the cell

• [Na]_i = the concentration of Na^(+) inside the cell

∴ <em>V</em>_Na =

[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV

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3 years ago
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Determine the equilibrium constant, K, for the equilibrium below given the
meriva

Answer:

or reactions that are not at equilibrium, we can write a similar expression called the reaction quotient QQQ, which is equal to K_\text cK

c

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Explanation:

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Explanation:

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raketka [301]

Answer:

This question is incomplete

Explanation:

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