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Pie
1 year ago
7

Would you expect a heavily crosslinked polyethylene that has a glass-transition temperature of 0°c to be an elastomer, a thermos

et (nonelastomer), or a thermoplastic polymer at room temperature?
Chemistry
1 answer:
musickatia [10]1 year ago
4 0

The heavily crosslinked polyethylene that has a glass-transition temperature of 0°c to be an <u>thermoset (non-elastomer).</u>

A cross-linked kind of polyethylene has been known as cross-linked polyethylene, sometimes known by the acronyms PEX, XPE, or XLPE. It must be primarily utilized in residential water piping, high tension electrical cable insulation, hydronic radiant heating as well as cooling systems, construction management pipework systems, and infant play mats.

Tanks composed of linear polyethylene as well as XLPE are both constructed using heated resins to produce cured plastic. But because of the variations in how they are produced, the polyethylenes they produce have extremely varied structural strengths. Simply put, cross-linking is just the creation of links between polymer chains.

Therefore, the correct answer will be option (a).

To know more about polyethylene

brainly.com/question/15913091

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Give the direction of the reaction, if K &gt;&gt; 1. Give the direction of the reaction, if K &gt;&gt; 1. The forward reaction i
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A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

K=\frac{Products}{reagents}

In any case  

aA +Bb  equilibrium Cd +dD

where K is:

K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})

Where: H is reaction enthalpy, R is the gas constant and T temperature.  

Clearing the equation for K_{2} we get:

K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}

Here we can study two cases: when delta H^{o} is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so K_{2} is greater that K_{1} and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so K_{2} will be less than K_{1} the forward reactions is not favored.  

{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}

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3 years ago
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