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Mazyrski [523]
3 years ago
15

You are given that kc = 10-1 kg eq-1 min-1, ku = 10-3 kg2 eq-2 min-1 and [A]0 = 10 eq kg-1, where kc is the rate constant for a

catalyzed step-growth polymerization, ku is the rate constant for an uncatalyzed step-growth polymerization, and [A]0 is the initial concentration of monomer A in the step-growth polymerization. Assume that the stoichiometry of monomer A to monomer B is 1:1, that the reactive groups on A only react with the reactive groups on B, and that the reactivity of A and B is independent of the chain length.
A. Calculate the time in minutes for conversion (p) to reach values of 0.2, 0.4, 0.6, 0.8, and 0.99 for both catalyzed and uncatalyzed polymerizations.B. Next, calculate the number-average degree of polymerization at each of these times.C. Assume that you can recover all of the catalyst (i.e., the cost of the catalyst is not a factor), which polymerization route is more economical from an industrial perspective?
Engineering
1 answer:
victus00 [196]3 years ago
7 0

Answer:

-ajkdbvójadh`toug511

Explanation:

b

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Answer:

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int main()

{

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#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

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try{

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  {hour = hour + 12;

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  cin.clear();

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catch (int c) { cout << "Invalid hour input!";}

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  cin.clear();

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catch (int e) { cout << "Invalid minute input!" << endl;}

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int convertTime::invalidSec(int sec)

{

try{

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  cin.clear();

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catch (int t) { cout << "Invalid second input!" << endl;}

}

void convertTime::printMilTime()

{

cout << "Your time converted: " << hour << ":" << min << ":" << sec;

}

Explanation:

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