The ones that are able to survive pass on their genes to the next generation. Bye doing that they pass on certain traits to the next generation causing them to inherit those traits and develop new ones and that prosses repeats for thousands and millions of years and that's how they are able to evolve.
Answer:
848.1 g/mol
Explanation:
Data given:
Standard of atomic mass of Ca = 50 amu
molar mass of Aluminium Acetate = ?
Solution:
Relative atomic mass represented by Ar. It is ratio of actual mass with respect to the 1/12th mass of C-12 but here Ca-50 is standard.
Formula of Aluminium Acetate = Al(CH₃COO)₃
In order to calculate Ar
first we will calculate 1/50 of Ca
As we know
mass of 1/12 of C-12 = 1.993 x 10⁻²⁶/12 = 1.661 x 10⁻²⁷ Kg
So, for Ca-50
mass of 1/50 of Ca-50 = 1.993 x 10⁻²⁶/50 = 3.986 x 10⁻²⁸ Kg
Now
Relative Atomic mass for element X = rest mass of "X"/ 3.986 x 10⁻²⁸ Kg . . . . . . (1)
First we have to know the relative atomic masses of Aluminium, carbon, hydrogen, oxygen atoms involve in Aluminium Acetate formula with respect to new standard Ca-50
By using equation-1 we can calculate Ar for which we have reported rest masses of atoms as below
Rest mass of Aluminium = 4.48 x 10⁻²⁶ Kg
Rest mass of carbon = 1.993 x 10⁻²⁶ Kg
Rest mass of hydrogen = 1.608 x 10⁻²⁷ Kg
Rest mass of oxygen = 2.657x10⁻²⁶ Kg
Now put values in equation 1 for each atom
Ar for Aluminium = 112.5 amu
Ar for Carbon= 50 amu
Ar for hydrogen = 4 amu
Ar for Oxygen = 66.6 amu
Now find the molar mass ofAl(CH₃COO)₃
molar mass of Aluminium Acetate = Al(CH₃COO)₃
Al(CH₃COO)₃ = 112.5 + 3 (50 + 3(4) + 50 + 66.6 +66.6)
Al(CH₃COO)₃ = 112.5 + 3 (50 + 12 + 50 + 66.6 +66.6)
Al(CH₃COO)₃ = 112.5 + 3 (245.2)
Al(CH₃COO)₃ = 112.5 + 735.6
Al(CH₃COO)₃ = 848.1 g/mol
molar mass of Aluminium Acetate = 848.1 g/mol
A) Mr. Greenjeans needs a control group for comparison.
B) Mr. Greenjeans should use corn and radish seeds as a comparison.
C) Mr. Greenjeans should use three different types of pumpkin seeds.
D) Mr. Greenjeans probably should monitor the pumpkins for a longer period of time.
Mr. Greenjeans needs a control group for comparison.
How do we know that the fertilizer had any impact on growth since Mr. Greenjeans has no control group for comparison? He must know how the pumpkins will grow without any fertilizer as well. The control group gives us a way to compare the experimental groups to help form more valid conclusions.