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murzikaleks [220]
3 years ago
5

Can someone please help me with question 1

Biology
2 answers:
marshall27 [118]3 years ago
4 0
Graduating from high school this year makes me feel a bit sad. Biology was such a fun but crazy class.

Enough of my sentiments; I'm here to help, :-).

This answer is dependent upon what ecosystem you choose your project over.

So, abiotic factors are nonliving things in an ecosystem. Biotic factors are living organisms and anything alive.

For instance, if you choose a pond for your ecosystem then the water, soil, maybe trash would be abiotic factors. Some could argue there is living material within the water and soil but those abiotic factors are not alive. I mean water is not alive but it could be a home to biotic factors like microorganisms such as protozoans and worms. The grass and trees and animals around a pond would be biotic factors.

Hope you get the idea. if there's any questions then leave a comment and I'll help. Good luck!
Tpy6a [65]3 years ago
4 0
My answer is the same as her (SORRY) ( I need some points to ask my question)
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svetlana [45]

Answer:

Explanation:

From the information given:

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By using Goldman's equation:

E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}}      \Big )

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and the two species be Na⁺ and Cl⁻

E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}

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z = ionic charge on the species = + 1

F = faraday constant

∴

100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In}  \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\  53.57 = \dfrac{4}{[K^+]_{in}}

[K^+]_{in} = \dfrac{4}{53.57}

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For [Cl⁻]:

100 \times 10^{-3} = -0.0257 \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

-3.981 =  \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

0.01867 =  \dfrac{120}{[Cl^-]_{in}}

[Cl^-]_{in} = \dfrac{120}{0.01867}

[Cl^-]_{in} =6427.4

For [Na⁺]:

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

53.57= \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

[Na^+]_{in}= 2.70

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