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ratelena [41]
3 years ago
14

Give an example of a polymer formed by addition polymerization.

Chemistry
1 answer:
miv72 [106K]3 years ago
8 0

Answer:

During the addition polymerization, all monomers are consumed and no byproducts are formed. Common examples of addition polymerization are polyethylene, polyvinyl chloride (PVC), acrylics, polystyrene, polytetrafluoroethylene, and polyoxymethylene (acetal).

What is meant by addition polymer?

In polymer chemistry, an addition polymer is a polymer that forms by simple linking of monomers without the co-generation of other products. Addition polymerization differs from condensation polymerization, which does co-generate a product, usually water.

Explanation:

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What is the % yield when 140.0 grams of Ethylene gas (C2H4) reacts with excess chlorine to form 280.0 grams of 1,2-Dichloro Etha
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Answer:

percent yield = 56.6 %

Explanation:

Data given:

mass of  Ethylene gas (C₂H₄) = 140 g

actual yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)= 280 g

percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂) = ?

Reaction Given:

                        C₂H₄ + Cl₂ -------> C₂H₄Cl₂

Solution:

First we have to find theoretical yield.

So,

Look at the reaction

                       C₂H₄ + Cl₂ -----—> C₂H₄Cl₂

                      1 mol                         1 mol

As 1 mole of C give 1 mole of CH₄

Convert moles to mass

molar mass of C₂H₄ = 2(12) + 4(1)

molar mass of C₂H₄ = 24 + 4

  • molar mass of C₂H₄ = 28 g/mol

molar mass of C₂H₄Cl₂ = 2(12) + 4(1) + 2(35.5)

molar mass of C₂H₄Cl₂ = 24 + 4 + 71

  • molar mass of C₂H₄Cl₂ = 99 g/mol

Now

                       C₂H₄     +       Cl₂    -----—>     C₂H₄Cl₂

                1 mol (28 g/mol)                        1 mol (99 g/mol)

                          28 g                                          99 g

28 grams of Ethylene gas (C₂H₄) produce 99 grams of C₂H₄Cl₂

So

if 28 g of C₂H₄ produce 99 g of C₂H₄Cl₂ so how many grams of C₂H₄Cl₂ will be produced by 140 g of C₂H₄.

Apply Unity Formula

                        28 g of C₂H₄ ≅ 99 g of C₂H₄Cl₂

                        140 g of C₂H₄≅ X of C₂H₄Cl₂

Do cross multiply

                      mass of C₂H₄Cl₂ = 99 g x 140 g / 28 g

                      mass of C₂H₄Cl₂ = 495 g

So the Theoretical yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)= 495 g

Now Find the percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)

Formula Used

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

                percent yield = 280g / 495 g x 100 %

                 percent yield = 56.6 %

percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂) = 56.6 %

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