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beks73 [17]
3 years ago
8

"The occurrence of the all-or-none ________________ results from depolarization in response to rapid entry of ____________ ions

into the neuron A. resting potential; sodium B.action potential; sodium C.resting potential; potassium D.action potential; potassium E.hormonal response; potassium F. hormonal response; sodium"
Chemistry
2 answers:
Leno4ka [110]3 years ago
8 0

Answer:

The correct answer is

B.action potential; sodium

Explanation:

Action potential

Nerve action potentials are initiated and propagated in an all or nothing fashion. The all-or nothing action potential is initiated when the membrane potential threshold value is reached

Sodium

Voltage-gated Na⁺ ions membrane channels open enabling an influx of sodium ions to rapidly enter the cell, raising the membrane potential up to +40 mV

Depolarization  in the neuron is based ion exchange channels opening to allow the rapid influx of sodium ions (Na⁺) and the efflux of potassium ions (K⁺)

The neuronal responds rapidly and for short periods to the activation of an ion channel receptor by a drug or natural neurotransmitter is rapid rel

Anestetic [448]3 years ago
4 0

Answer:

B. action potential; sodium.

Explanation:

Action potential occurs when the membrane potential of a specific cell location rapidly rises and falls.

These are generated by special types of voltage-gated ion channels embedded in a cell's plasma membrane. They are shut when the membrane potential is near the resting potential of the cell, but they open rapidly if the membrane potential increases to a defined threshold voltage, depolarising the membrane potential, this then allows the inward flow of sodium ions, which changes the electrochemical gradient.

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3 years ago
A buffer is prepared containing 0.75 M NH3 and 0.20 M NH4 . Calculate the pH of the buffer using the Kb for NH3. g
denis-greek [22]

Answer:

pH=8.676

Explanation:

Given:

0.75 M NH_{3}

0.20 M NH_{4}

The objective is to calculate the pH of the buffer using the kb for NH_3

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pOH=pka+log\frac{[salt]}{[base]}\\

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pOH=-log(1.77*10^{-5})+log\frac{0.75}{0.20}\\        =4.75+0.5740\\        =5.324

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pH=8.676

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