Answer:
B. 3'
Explanation:
Polymerization is the process of forming large molecules by joining together many small molecules. The small molecules are known as the monomers, and the macromolecules are called polymers.
There are two(2) types of polymerization.
a. addition polymerization
b. condensation polymerization
Polymerization is characterized by eliminating small molecules such as water, carbon(IV)oxide, methanol e.t.c.
In complex compounds, polymerization starts at the 3' end of the template.
Answer:
the empirical (lowest raios) is
C2H4Cl
Explanation:
A compound is known to consist solely of carbon, hydrogen, and chlorine. Through elemental analysis, it was determined that the compound is composed of 24.27% carbon.
What is the empirical formula of this compound?
the compound has ONLY C, H, and Cl
the % Cl = 100% - 24.27% -4.03% = 71.7%
in 100 gm, there are 71.7 gm Cl, 24.27 gm C, and 4.03 gm H
the number of moles are Cl=71.7/70.91 =1.01= ~ 1
C = 24.27/12.0 = 2.02 =~ 2
H = 403/1.01 = 3.97 =~ 4
so the empirical (lowest raios) is
C2H4Cl
Answer:
1.2 liters.
Explanation:
Focus on the 4th digit: that's the ones column. The 3rd digit is the decimal place, just be sure to round up.
Question:
A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.
Answer:
The answer to the question is as follows
The mass of solution the student should use is 23.42 g.
Explanation:
To solve the question we note the following
A solution containing 42.7 % w/w of isopropenylbenzene in acetone has 42.7 g of isopropenylbenzene in 100 grams of the solution
Therefore we have 10 g of isopropenylbenzene contained in
100 g * 10 g/ 42.7 g = 23.42 g of solution
Available solution = 120 g
Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.
Given:
Half life(t^ 1/2) :30 years
A0( initial mass of the substance): 200 mg.
Now we know that
A= A0/ [2 ^ (t/√t)]
Where A is the mass that remains after t years.
A0 is the initial mass
t is the time
t^1/2 is the half life
Substituting the given values in the above equation we get
A= [200/ 2^(t/30) ] mg
Thus the mass remaining after t years is [200/ 2^(t/30) ] mg
