Molar mass of C is 12.01g/mol.
Molar mass of H is 1.01g/mol.
Molar mass of O is 16.00g/mol.
If we have 19 atoms of C, 38 atoms of H, and 1 atom of O, then we have:
19*12.01+38*1.01+1*16.00=282.57g/mol
282.57g/mol is the molar mass of the pheromone molecule.
Supposing that "<span>1.8 10-12 g" is 1.8*10^(-12)g then, this quantity of pheromones, shoud be divided by the molar mass of the pheromone molecule:
1.8*10^(-12)/282.57g/mol=6.37*10^(-15)moles
Using now Avogadro's number of molecules in 1mol (6.02*10^(23)) it is possible to know how many molecules are there in this pheromone quantity.
</span>6.02*10^(23)/6.37*10^(-15)=9.45*10^(37)molecules
So, in this quantity of the pheromone there are approximately 9.45*10^(37)molecules of it.
The biggest difference<span>, however, is that the line in the </span>logistic growth<span> graph changes direction and begins to level off as it nears the carrying capacity. That means that the main </span>difference between exponential<span> and </span>logistic growth<span> is that </span>logistic growth takes<span> into account carrying capacity. hope that helped</span>
If organisms that lived there before had exceeded the carrying capacity then the ecosystem would no longer be able to sustain life
Answer:
Option c (
) is the correct alternative.
Explanation:
Given:
Common ancestor,
= 50 million years ago
Base differences,
= 92
Now,
The predicted mutation rate will be:
= 
= 
= 
or,
=
Thus the above solution is the right one.
