Apolymer sample combines five different molecular-weight fractions, each of equal weight. the molecular weights of these fractions increase from 20,000 to 100,000 in increments of 20,000. calculate mn, mw, and mz. based upon these results, explain whether this sample has a broad or narrow molecular-weight distribution compared to typical commercial polymer samples.

Mn = 43,783

Mw = 60,000

Mz = 73,333

narrow distribution = 1.37

Explanation:

The molecular weights of these fractions increase from 20,000 to 100,000 in increments of 20,000. This means their Mi is respectively: (The molar weight (Mi) of the fractions)

Fraction 1 : Mi = 20  *10^3

Fraction 2: Mi = 40 *10^3

Fraction 3 : Mi = 60 *10^3

Fraction 4: Mi = 80 *10^3

Fraction 5 : Mi = 100  *10^3

The ΣMi = 300*10^-3

The Wi (mass of the fractions is for all the fractions the same, let's say 1)

So Wi = 1+1+1+1+1 = 5

Since number of moles = mass / Molar mass

The number of moles is respectively: ni = Wi/Mi (x10^5)

Fraction 1 : ni = Wi/Mi = 1/20000 = 5

Fraction 2: ni = 1/40000 = 2.5

Fraction 3 : ni =1/60000 = 1.67

Fraction 4: ni = 1/80000 = 1.25

Fraction 5 : ni= 1/100000 = 1

The Σni = 11.42

Mn = ΣWi/ni = 5/11.42*10^-5 = 43,783

Mw = (ΣWi * Mi)/ΣWi  = 300,000 /5 = 60,000

Mz = (ΣWi * Mi²)/ΣWi *Mi = (4*10^8 +16*10^8 +36*10^8 +64*10^8 +100*10^8) /300,000  =73,333

Mz/Mn = narrow distribution =60,000/43,783 = 1.37

c is the answer because when something is left after something evaporates that shows the chemical change.

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.