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3 years ago

How many molecules of water are released during the polymerization of a 20 monomer-long cellulose molecule?

Chemistry

1 answer:

Alborosie3 years ago

8 0

Answer:

During the polymerization of a 20 monomer-long cellulose molecule, 19 molecules of water are released.

Explanation:

In simple terms, cellulose is the biopolymer formed by many glucose units. This is cellulose is the polymer and glucose is the monomer.

To have a 20 monomer-long cellulose molecule, 20 monomers have been chemically bonded by reacting 19 times, as it is explained in the next paragrpahs, and so 19 molecules of water have been released.

You can imaging the polymerization process as a step-by-step reaction in which the first step is the condensation reaction of one glucose molelecule to produce a 2 monomer-long chain, with the release of one molecule of water: the second step would be the condensation reaction between the 2 monomer-long chain with another glucose molecule, with the release of an additional molecule of water, and so on, until 19 condensation reactions happen, to obtain the 20 monomer-long cellulose molecule.

Condensation is the loss of water in a chemical reaction.

When two glucose molecules react together, condensation occurs. One OH group from each glucose molecule come together, the OH from one glucose molecule combines with the H part of the OH from the other glucose molecule, to form H₂O (water that is released).

The two glucose molecules (monomers) will form one bigger molecule where the two glucose monomers are bonded through the oxygen atom that did not form part of the water molecule released.

Then, a 20-monomer chain means 19 condenstation reactions, with the release of 19 molecules of water.

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Dihydrogen dioxide decomposes into water and oxygen gas. Calculate the amounts requested if 1.34 moles of dihydrogen dioxide rea

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Answers:

a. Moles of oxygen formed: 0.670 mol

b. Moles of water formed: 1.34 mol

c. Mass of water formed: 24.1 g

d. Mass of oxygen formed: 21.4 g

Explanation:

Dihdyrogen dioxide is the chemical name for a compound made of two hydrogen atoms and two oxide atoms, i.e. H₂O₂, which is also known as hydrogen peroxide or oxygenated water.

The decomposition reaction of dihydrogen dioxide into water and oxygen gas is represented by the balanced chemical equation:

$2H_2O_2(l)\rightarrow 2H_2O(l)+O_2(g)$

The mole ratios derived from that balanced chemical equation are:

2 mol H₂O₂ : 2 mol H₂O : 1 mol O₂

a. Moles of oxygen formed

Set the proportion using the theoretical mole ratio of H₂O₂ to O₂ and the amount of moles of dyhydrogen dioxide that react:

$2\text{ mol }H_2O_2/1\text{ mol }O_2=1.34\text{ mol }H_2O_2/x$

When you solve for x, you get:

x = 1.34 mol H₂O₂ × 1 mol O₂ / 2 mol H₂O₂ = 0.670 mol O₂

b. Moles of water formed

Set the proportion using the theoretical mole ratio of H₂O₂ to H₂O and the amount of moles of dyhydrogen dioxide that react:

$2\text{ mol }H_2O_2/2\text{ mol }H_2O=1.34\text{ mol }H_2O_2/x$

When you solve for x, you get:

x = 1.34 mol H₂O₂ × 2 mol H₂O / 2 mol H₂O₂ = 1.34 mol H₂O

c. Mass of water formed

Using the number of moles of water calculated in the part b., you calculate the mass of water formed, in grams, using the molar mass of water:

Molar mass of water = 18.015 g/mol

Number of moles = mass in grams / molar mass

⇒ mass in grams = number of moles × molar mass

⇒ mass in grams = 1.34 mol × 18.015 g/mol = 24.1 g

d. Mass of oxygen formed

Using the number of moles of oxygen determined in the part a., you calculate the mass in grams using the molar mass of O₂.

Molar mass of O₂ = 32.00 g/mol

mass = molar mass × number of moles

mass = 32.00 g/mol × 0.670 mol = 21.4 g.

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