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vampirchik [111]
3 years ago
8

You cannot use Carbon dating to find the age of something if it is...

Biology
1 answer:
Sveta_85 [38]3 years ago
4 0
My best guess is C, when it is not made of carbon, but I’m only around 70% sure.
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Would you except to find muscle tissue in a plant?Why or why not?
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Answer:

No, Because muscle tissue is on found inside humans and animals.

5 0
3 years ago
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
erastova [34]

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

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6 0
2 years ago
A species is represented by individuals that form a population:________.
Setler [38]

AnsweR

D.

Explanation:

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Submit your project explaining your choices for each of the five scenarios presented.
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Hellp asap i have 5 mins
marissa [1.9K]

When reading information on websites you should look for all of the aforementioned except: A. number of hits.

<h3>What is a website?</h3>

A website can be defined as a collective name which connotes a series of webpages that are interconnected and linked together with the same domain name, so as to provide certain information to end users over the Internet or an active network connection.

<h3>What is a web server?</h3>

A web server can be defined as a type of computer that is designed and developed to run websites and distribute webpages as they are being requested over the Internet by end users (clients).

<h3>What is a hit?</h3>

In Computer technology, a hit can be defined as the number of digital files that are downloaded from a website and registered on the web server over a specific period of

In this context, we can infer and logically deduce that when reading information on websites you should look for all of the following:

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Read more on websites here: brainly.com/question/3733655

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Complete Question:

When reading information on websites you should look for all of the following except:

A. Number of hits.

B. Author's name and credentials.

C. Website sponsors.

D. Date last updated.

6 0
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