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2 years ago

What explanation can you give for why the sodium-potassium pump does not run out of ions to move in or out of the cell

1 answer:

6 0

The sodium-potassium pump does not run out of ions since ion exchange is essential for the action potential to take place and to maintain homeostasis.

The cell has variable concentrations of different substances compared to the environment that surrounds it, with significant differences with sodium and potassium.

The main function of the sodium-potassium pump is to maintain homeostasis of the intracellular medium, controlling the concentrations of these two ions.

In order to carry out the adequate exchange of sodium and potassium ions in the extra and intracellular medium, the cells need an active transport process that is carried out thanks to the sodium potassium pump.

This process is needed for the maintenance and functioning of cells, and it is essential for the action potential to be executed, necessary for the transmission of electrical impulses from neuron to neuron.

Therefore, we can conclude that the sodium potassium pump produces an exchange of potassium ions for sodium ions which keeps the cellular system functioning properly.

Learn more here: brainly.com/question/24336764

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First, we have to see how K2O behaves when it is dissolved in water:

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3 years ago

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The equilibrium constant, Kc, for the following reaction is 5.10×10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) Calculate the equilibri

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Answer: The equilibrium concentration of HCl is $2.26\times 10^{-3}M$

Explanation:

We are given:

Moles of $NH_4Cl(s)$ = 0.564 moles

Volume of vessel = 1.00 L

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$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Molarity of $NH_4Cl=\frac{0.564}{1}=0.564M$

The given chemical equation follows:

$NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)$

Initial: 0.564

At eqllm: 0.564-x x x

The expression of $K_c$ for above equation follows:

$K_c=[NH_3][HCl]$

The concentration of pure solid and pure liquid is taken as 1.

We are given:

$K_c=5.10\times 10^{-6}$

Putting values in above equation, we get:

$5.10\times 10^{-6}=x\times x\\\\x=2.26\times 10^{-3}M,-2.26\times 10^{-3}M$

Negative sign is neglected because concentration cannot be negative.

So, $[HCl]=2.26\times 10^{-3}M$

Hence, the equilibrium concentration of HCl is $2.26\times 10^{-3}M$

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