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3 years ago

7

Decane (C10H22) used to produce poly(ethene). what conditions are needed?

1 answer:

Strike441 [17]3 years ago

6 0

<span>Cracking of hydrocarbons involves thermal decomposition.
This means that large hydrocarbon molecules break into
smaller molecules when they are heated. The <span>hydrocarbons
</span>are boiled and the <span>hydrocarbon gases </span>are either
mixed with steam and heated to a very high temperature or
passed over a hot <span>powdered </span>aluminium oxide catalyst.
The catalyst works by providing the <span>hydrocarbon gases
</span>with a convenient surface for the cracking to take place.</span>

<span>For example, decane (an alkane with 10 carbons)
can be cracked to produce octane and ethene.</span>

decane<span> octane + ethene.
<span>C10H22</span>(g) <span>C8H18</span>(g) + C2<span>H4</span>(g)</span>

<span>Octane is used as petrol.
Ethene is used in the manufacture of polymers.</span>

<span>Cracking an alkane produces a smaller alkane plus an alkene.
If you add up the number of hydrogen atoms in the
above reaction, you will see that there are 22 on each side.
An alkene is produced because the original alkane does
<span>not </span>have enough hydrogen atoms to produce two more<span> alkanes</span>.</span>

<span>
</span>

<span>hope this helps </span>

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An example of polytamic anoin would be a. Ca+2 b. NH4+ c. F d. PO4-3

ella [17]

$PO_4^{-3}$ and $NH^{4+}$ are polyatomic anions composed of two or more atoms.

<h3>What is a polyatomic anion?</h3>

A polyatomic ion is an ion composed of two or more atoms. Examples: The hydroxide cation ( $OH^-$ ) and the phosphate cation ( $PO_4^{3-}$ ) are both polyatomic ions.

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2 years ago

When atoms from Column I (Group 1) combine with atoms from Column VII (Group 17): A nearly 100% covalent bond forms. The bond wi

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The answer you're looking for is B. The bond will be ionic.

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3 years ago

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Consider an exceptionally weak acid, HA, with a Ka = 1x10 -20 . You make a 0.1M solution of the salt Na

Vedmedyk [2.9K]

Answer:

$pH=13$

Explanation:

Hello,

In this case, given the acid, we can suppose a simple dissociation as:

$HA\rightleftharpoons H^+ + A^-$

Which occurs in aqueous phase, therefore, the law of mass action is written by:

$Ka=\frac{[H^+][A^-]}{[HA]}$

That in terms of the change $x$ due to the reaction's extent we can write:

$1x10^{-20}=\frac{x*x}{0.1M-x}$

But we prefer to compute the Kb due to its exceptional weakness:

$Kb=\frac{Kw}{Ka}=\frac{1x10^{-14}}{1x10^{-20}} =1x10^{-6}$

Next, the acid dissociation in the presence of the base we have:

$Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}$

Whose solution is $x=0.0999M$ which equals the concentration of hydroxyl in the solution, thus we compute the pOH:

$pOH=-log([OH^-])=-log(0.0999)=1$

Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:

$pH+pOH=14\\\\pH=14-pOH=14-1\\\\pH=13$

Regards.

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4 years ago

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3 years ago

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Not sure what you are asking. I have two possible answers though...

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Again, I don't know what you were asking, but one of these answers may be correct.

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