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sergiy2304 [10]
3 years ago
13

Electrochemical gradients Because ions carry a charge (positive or negative) their transport across a membrane is governed not o

nly by concentration gradients across the membrane but also by differences in charge across the membrane (also referred to as membrane potontia) Together, the conoentration (chemical) gradient and the charge difforence (electrical gradient) across the plasma membrane make up the electrochemical gradient. consider the plasma membrane of an animal cell that oontains a sodium potassium pump as wel as two non-gated (always open) ion channelse a Nat channel and a K channel. The effect of the sodium potassium pump on the concentrations of Na and K as well as the distribution of charge across the plasma membrane is indicated in the figure below. Outside cell channel (Nat) high K 1 high inside cel which of the following statements correctly describes) the driving forces for diffusion of Na and K ions through their respective channels? a. The diffusion of Na ions into the cell is facilitated by the Na concentration gradient across the plasma membrane. b. The diffusion of Na" ions into the cell is impeded by the electrical gradiont across the plasma membrane c. The diffusion of K ions out of the cel is impeded by the KT concentration gradient across the plasma membrane.
Chemistry
1 answer:
Vlad1618 [11]3 years ago
4 0

Answer:

<h2> a. The diffusion of Na ions into the cell is facilitated by the Na concentration gradient across the plasma membrane.</h2>

Explanation:

Cells differ in the concentration of Na+  and many other chemicals inside and out side of the cell, so diffusion of Na+ ions into the cell is facilitated by the Na+ concentration gradient across the membrane.

The diffusion of K+ ions out of the cell is also prevented by the electrical gradient across the plasma membrane.

In the cell, the electro chemical gradient is larger for Na+ than for K+ and many other substances.

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Answer:

A = -213.09°C

B = 15014.85 °C

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Explanation:

Given data:

Initial volume of gas = 5.00 L

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Final volume = 1100 mL, 280 L, 87.5 mL

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Solution:

Formula:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

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T₂ = Final temperature

Conversion of mL into L.

Final volume = 1100 mL/1000 = 1.1 L

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Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 1.1 L × 273 K / 5.00 L

T₂ = 300.3 L.K / 5.00 K

T₂ = 60.06 K

60.06 K - 273 = -213.09°C

2)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 280 L × 273 K / 5.00 L

T₂ = 76440 L.K / 5.00 K

T₂ = 15288 K

15288 K - 273 = 15014.85 °C

3)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 0.0875 L × 273 K / 5.00 L

T₂ = 23.8875 L.K / 5.00 K

T₂ = 4.78 K

4.78 K - 273 = -268.37°C

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Answer:

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