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Sonbull [250]
2 years ago
5

What ion movement changes the membrane potential in a contractile cell from negative to positive during the rapid depolarization

phase?
Biology
1 answer:
Lesechka [4]2 years ago
8 0

Answer:

The influx of calcium ions.

Explanation:

The influx of calcium ions at a certain point produces phase of the action potential also called depolarization which results in the ion movement changes the membrane potential from negative to positive inside the cell.

In action potential, calcium ions may control gene transcription, cell excitability, and neurotransmitter release. In chemical synapses, the influx of calcium ions leads to vesicles filled with neurotransmitters move to the surface of the cell, and into the synaptic cleft release their contents.

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Iron-55 has a half-life of 3 years. How much of a 40 gram sample will remain after 12 years? A. 0 grams B. 4 grams C. 2.5 grams
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Answer:

\boxed {\boxed {\sf C. \ 2.5 \ grams }}

Explanation:

We are asked to find how much of a 40 gram sample remains after 12 years.

Iron-55 has a half-life of 3 years. Therefore, after 12 years, 4 half-lives have been completed.

  • 12 years/3 years = 4 half-lives

Every time a half-life is completed, half of the sample's mass decays. Remember we start with a 40 gram sample.

  • 1 half- life: 40 g / 2 = 20 g
  • 2 half-lives: 20 g / 2= 10 g
  • 3 half-lives: 10 g / 2 = 5 g
  • 4 half-lives: 5 g / 2 = 2.5 g

There is also a formula that can be used to solve this problem.

A= A_o(\frac {1}{2})^{\frac{t}{hl}

Where A₀ is the initial amount, t is the time, and hl is the half-life.

We know 40 grams is the inital amount, 12 years is the time, and 3 years is the halflife.

  • A₀= 40 g
  • t= 12
  • hl= 3

A= 40 \ g (\frac{1}{2})^\frac{12}{3}

A= 40 \ g (\frac{1}{2})^4

A= 40 \ g * 0.0625

\bold {A= 2.5 \ g}

After 12 years, <u>2.5 grams </u> of Iron-55 will remain.

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