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3 years ago

The first step in the free radical mechanism for the preparation of polyethylene is:

Chemistry

1 answer:

kherson [118]3 years ago

3 0

Answer:

The first step in the free radical mechanism for the preparation of polyethylene is:

b. heating an organic peroxide to break the O-O bond

Explanation:

The preparation of ethylene occur by free radical mechanism , which occurs in three steps :

1. Chain Initiation

2. Chain propagation

3. Chain termination.

The first step i.e chain initiation require a radical. The compound are generally peroxide(O-O linkage) that produce radical is :

Benzoyl Peroxide

This compound when heated (with U.V radiation) produce two species of radicals .

(see the attached image)

These RADICALS are highly unstable and very reactive in nature.

They readily react with ethene and initiate the polymerisation.

Thus,

The first step in the free radical mechanism for the preparation of polyethylene is:

b. heating an organic peroxide to break the O-O bond

<u><em>The sequence of Polymerisations are:</em></u>

b. heating an organic peroxide to break the O-O bond

d. propagation of the free radicals

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3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

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3 years ago