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yKpoI14uk [10]
3 years ago
8

What happens during the second trimester

Biology
2 answers:
netineya [11]3 years ago
3 0

The babys organs will start to develop, and the baby will be able to sleep and hear other people outside the womb, the bby starts to grow more essential things it will need to live outside of the mothers womb.


Lemur [1.5K]3 years ago
3 0

The vocal cords development

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Which accurately describes the relationship between nutrients and life-forms in an estuary?
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O A. As nutrients increase, the number of life-forms increases.

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The bones from an animal found at an archaeological dig have a C614 activity of 0.10 Bq per gram of carbon. The half-life of C61
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C14 is an isotope used in radiocarbon dating techniques to date organic matter remains. The age of these bones is approximately<u> 6890 years.</u>

<h3>What is Carbon 14?</h3>

Carbon 14, also known as radiocarbon, is a radioactive carbon isotope.

Isotopes are the atoms of the same element -carbon- that vary in neutrons and, hence, in their massic number. They are alternative forms of the same element.

The radioactive C14 nucleus contains 6 protons and 8 neutrons and has a half-life of 5730 years.

The term half-life is a reference. It means that an organism that has been dead for 5730 years has half the C14 amount or concentration than the same organism had when it was alive.

Knowing the half-life of an element is useful to determine the age of the dead matter.

C14 is used in radiocarbon dating techniques or methods to estimate the age of fossils. This is a reliable technique used for dating organic samples that are less than 50,000 years old.

<u>Available data</u>:

  • The half-life of C14 is 5730 years
  • Bones activity of 0.10 Bq per gram of carbon

To answer this question we can make use of the following equation

Ln (C14T₁/C14 T₀) = - λ T₁

Where,

  • C14 T₀ ⇒ Amount of carbon in a living body. We know, by bibliography, that living organism activity is 0.23 Bq per gram of carbon. So, C14 T₀ = 0.23 Bq/g
  • C14T₁ ⇒ Amount of carbon in the dead body. C14T₁ = 0.1 Bq/g
  • λ ⇒ radioactive decay constant = (Ln2)/T₀,₅
  • T₀,₅ ⇒ The half-life of carbon 14 = 5730 years
  • T₀ = Time when the organism was alive
  • T₁ = Age of bones

Let us first calculate the radioactive decay constant.

λ = (Ln2)/T₀,₅

λ = 0.693/5730

<u>λ = 0.0001209</u>

Now, let us calculate the first term in the equation

Ln (C14T₁/C14 T₀) = Ln (0.1/0.23) = Ln 0.4347 =<u> - 0.833</u>

Finally, let us replace the terms, clear the equation, and calculate the value of T₁.

Ln (C14T₁/C14 T₀) = - λ T₁

- 0.833 = - 0.0001209 x T₁

T₁ = - 0.833 / - 0.0001209

T₁ =  6889.99 ≅ <u>6890 years</u>

The bones are approximately<u> 6890 years.</u>

You can learn more about dating organic matter with carbon14 at

brainly.com/question/4149380

#SPJ1

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