The IUPAC rules are
a) Find out the longest chain of carbon in the given organic compound
b) We will name the longest chain.
c) We will identify the main functional group and will assign a suffix to the compound.
d) We will number the carbons in the longest chain selected so that the attached groups attain lowest numeral as substituent
e) We will name the side groups or chains.
Hello!
The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.
We have the following data:
mo (initial mass) = 53.3 mg
m (final mass after time T) = ? (in mg)
x (number of periods elapsed) = ?
P (Half-life) = 10.0 minutes
T (Elapsed time for sample reduction) = 25.9 minutes
Let's find the number of periods elapsed (x), let us see:
Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:
I Hope this helps, greetings ... DexteR! =)
Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
Explanation : Given,
Mass of oxygen in sulfur dioxide = 3.49 g
Mass of sulfur in sulfur dioxide = 3.50 g
Mass of oxygen in sulfur trioxide = 9.00 g
Mass of sulfur in sulfur trioxide = 6.00 g
Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.
Mass of oxygen per gram of sulfur for sulfur dioxide =
Mass of oxygen per gram of sulfur for sulfur dioxide =
and,
Mass of oxygen per gram of sulfur for sulfur trioxide =
Mass of oxygen per gram of sulfur for sulfur trioxide =
Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.