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SashulF [63]
3 years ago
6

Studying the decay of radioactive isotopes in dead organisms helps scientists to identify fossilized remains. The ratio of C-12

to C-14 in the atmosphere is 1 x 1012. The table shows this ratio inside the body of three organisms.
C12/C14 Ratio
Organism C12/C14 Ratio
X 8 x 1012
Y 2 x 1012
Z 1 x 1012


What can most likely be concluded from the above information?
Only Organism Z is alive, and Organism X has been dead longer than Organism Y.
Only Organism Z is alive, and Organism Y has been dead longer than Organism X.
Only Organism X is alive, and Organism Y has been dead longer than Organism Z.
Only Organism X is alive, and Organism Z has been dead longer than Organism Y.
Chemistry
1 answer:
satela [25.4K]3 years ago
7 0

Answer:

Only Organism Z is alive, and Organism X has been dead longer than Organism Y.

Explanation:

After the death of an organism, there is radioactive disintegration of C-14 allotrope of carbon, which increases the ratio of C-12 to C-14 in a dead organism as compared to a living organism.

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What substance is added during hydrolysis in order to break down food macromolecules?
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So, water (in terms of the word) is added to help split and breakdown macromolecules. 

I hope this helps!
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3 years ago
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Which component is required for the activation of some enzymes?
shepuryov [24]
The best answer is letter A. 
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Determine the LIMITING reactant in the following balanced equation:
padilas [110]

Answer:

KBr is limiting reactant.

Explanation:

Given data:

Mass of  KBr =4g

Mass of Cl₂ = 6 g

Limiting reactant = ?

Solution:

Chemical equation:

2KBr + Cl₂      →    2KCl + Br₂

Number of moles of KBr:

Number of moles = mass/molar mass

Number of moles = 4 g/ 119 gmol

Number of moles = 0.03 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 6 g/ 70 gmol

Number of moles = 0.09 mol

Now we will compare the moles of reactant with product.

              KBr            :            KCl

                2              :              2

            0.03            :            0.03

             KBr            :              Br₂

                2             :               1

             0.03           :          1/2×0.03= 0.015

               Cl₂             :            KCl

                 1              :              2

            0.09            :           2/1×0.09 = 0.18

               Cl₂             :              Br₂

                1              :               1

             0.09           :            0.09

Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂  is present in excess.

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Are pure substances homogeneous or heterogeneous
Svetllana [295]

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The grouping element, ul, contains the definition associated with a term from a description list. True False
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False is your answer

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