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Mandarinka [93]
3 years ago
8

What does all wetlands have in common?

Biology
2 answers:
sveticcg [70]3 years ago
7 0

Answer: A.They contain plants that purify water.

vodomira [7]3 years ago
4 0
They’re marshy swamps with tall grasses and the water is very still
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Please help
Arada [10]

25% is the approximate production efficiency of this animal

Explanation:

Given ,

A ladybug eats 200 J of plant material

100 J is eliminated as feces

75 J are used in cellular respiration,

Now,

The approximate production efficiency of this animal :

200 - (100 +75) = 25 %

7 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

#SPJ1

6 0
2 years ago
What is important in Hubble's discovery about the red shift in the spectra of galaxies? a. It proves the Big Bang theory. b. It
dedylja [7]

Answer:

c

Explanation:

Hubble's brilliant observation was that the red shift of galaxies was directly proportional to the distance of the galaxy from earth. That meant that things farther away from Earth were moving away faster. In other words, the universe must be expanding.

5 0
2 years ago
Which statement best explains why cells were observed in more detail using a compound microscope than a simple microscope?
Reil [10]
The statement best that explains why cells were observed in more detail using compound microscope than a simple microscope is : compound microscope has greater magnification ability than the simple microscope .

This will allow us to observe the subject with a much smaller size, making it easier for us to observe the detail

hope this helps
3 0
3 years ago
Read 2 more answers
GIVING BRAINLIEST AND 20 POINTS PLS HELP lol
dolphi86 [110]

Answer:

Explanation:

4

8 0
3 years ago
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