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AURORKA [14]
2 years ago
8

How do sodium and potassium ions transfer in and out of an axon?

Chemistry
1 answer:
aliya0001 [1]2 years ago
7 0

Answer:

Option A= through gates that,s open up

Explanation:

The sodium and potassium ions transfer in and out the axon through electrical process. This process is called depolarization and re-polarization.

Depolarization:

Depolarization is occur when stimulus is given to the resting neuron. In this process gates of sodium ions on the membrane become open and sodium ions that are present out side. inter into the cell. because of this process the charge of the nerve changes (-70 mv to -55 mV).

Re-polarization:

when the re-polarization occur, potassium gates are open and the potassium ions goes outside the membrane. During this process electrical potential becomes negative inside the cell until the potential of -70 mV is re-attain i.e, resting potential.

In short we can say that depolarization allow sodium ions to inter into the nerve membrane and re-polarization allow potassium ions to moves out side the membrane.

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Answer:-

(a) 3.5

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Explanation:-

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onsider the reversible dissolution of lead(II) chloride. P b C l 2 ( s ) − ⇀ ↽ − P b 2 + ( a q ) + 2 C l − ( a q ) PbClX2(s)↽−−⇀
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Answer:

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Explanation:

Step 1:

The balanced equation for the reaction.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

Step 2:

Data obtained from the question:

Mass of PbCl2 = 0.2393 g

Volume = 50mL

concentration of Pb^2+, [Pb^2+] = 0.0159 M

Concentration of Cl^-, [Cl^-] = 0.0318 M

Equilibrium constant, Kc =?

Step 3:

Determination of the number of mole PbCl2.

The number of mole of PbCl2 can be obtained as follow:

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Number of mole =Mass /Molar Mass

Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

Step 4:

Determination of Molarity of PbCl2.

At this stage we shall obtain the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.61x10^-4 mole

Volume = 50mL = 50/1000 = 0.05L

Molarity of PbCl2 =?

Molarity = mole /Volume

Molarity of PbCl2 = 8.61x10^-4/0.05

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Step 5:

Determination of the equilibrium constant Kc.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

The equilibrium constant Kc for the equation above is given by:

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

[Pb^2+] = 0.0159 M

[Cl^-] = 0.0318 M

[PbCl2] = 0.01722 M

Kc =?

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2/ 0.01722

Kc = 9.34x10^-4

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