1. What is meant by the term gaseous exchange?

Growing willow for biofuel is sustainable development because it is a source of fuel which potentially will never run out. For every willow tree that is cut down, another can be planted in its place. This means that the source of fuel is potentially limitless and subsequently sustainable.

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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 6890 years.

<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.

Available data:

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

λ = 0.0001209

Now, let us calculate the first term in the equation

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

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 ≅ 6890 years

The bones are approximately 6890 years.

You can learn more about dating organic matter with carbon14 at

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6 0

2 years ago

Which method is most likely to reduce soil loss?

jenyasd209 [6]

Answer:

B

Explanation:

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3 years ago

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