Respuesta :
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 "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.
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.
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 "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.
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.
Pheromones are the hormones secreted by females to attract male species. The molecular formula of pheromone is [tex]\text C_{19} \text H_{38} \text O[/tex] also known as 2-nonadecannone. The pheromone secreted by females is usually 1.8 10 -12 grams. The number of molecules in this quantity will be [tex]9.45\times10^{37}[/tex].
Given that,
Molecular formula of pheromone is [tex]\text C_{19} \text H_{38} \text O[/tex]
- Molar Mass of C = 12.01 g/mol
- Molar Mass of H = 1.01 g/mol
- Molar Mass of O = 16 g/mol
Also,
There are 19 molecules of Carbon or C, it will be = 19 X 12.01 = 228.19 g/mol
There are 38 molecules of Hydrogen or H, it will be = 38 X 1.01 = 38.38 g/mol
There are 1 molecule of Oxygen or O, it will be = 1 X 16 = 16 g/mol
Molecular mass of [tex]\text C_{19} \text H_{38} \text O[/tex] = 282.57 g/mol
Now, given that female produces 1.8 10-12 g of pheromones, then it will be:
[tex]\frac{1.8\times 10 ^{-12}}{282.57 \text{g/mol}} = 6.35\times 10^{-15}\;\text {moles}[/tex]
The value of Avogadro's number of molecules in 1 mol is [tex]6.02\times 10^{23}[/tex].
Therefore, the number of molecules in pheromone will be:
[tex]\frac{6.02\times10^{23}}{6.37\times10^{-15}}&=9.45\times10^{37} \text{molecules}[/tex]
Thus, the number of molecules in [tex]\text C_{19} \text H_{38} \text O[/tex] will be [tex]9.45\times10^{37} \text{molecules}[/tex].
To know more about pheromones, refer to the following link:
https://brainly.com/question/11391766?referrer=searchResults
