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Vedmedyk [2.9K]
3 years ago
9

Aqueous solutions of isopropyl alcohol are commonly sold as rubbing alcohol. The boiling point of isopropyl alcohol is 82.4 °C.

It evaporates readily from the surface of the skin. Ethylene glycol (CH2OHCH2OH) is sold as antifreeze/antiboil. The boiling point of ethylene glycol is 198 °C. The molecular masses of isopropyl alcohol and ethylene glycol are similar. Provide an explanation for the observed difference in their boiling points.
Chemistry
1 answer:
Monica [59]3 years ago
8 0

Answer:

This is due to more hydrogen bonding in ethylene glycol than it is in isopropyl alcohol

Explanation:

The boiling point of isopropyl alcohol is 82.4 °C it contains only a single OH group, hence intermolecular hydrogen bonding is solely responsible for it's boiling point, whereas Ethylene glycol (CH2OHCH2OH) contains 2-OH group and both intermolecular and intramolecular hydrogen bonding are responsible for the higher boiling point of ethylene glycol at 198 °C.

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There are three other structural isomers of 1-butanol: 2-butanol (sec-butyl alcohol), 2-methyl-1-propanol (isobutyl alcohol), and 2-methyl-2-propanol (tert-butyl alcohol). 2-Butanol, or sec-butanol, or sec-butyl alcohol, or s-butyl alcohol, is a four-carbon chain, with the OH group on the second carbon.

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Hope this helps

7 0
3 years ago
Consider the following unbalanced reaction: P4(s) + F2(g) → PF3(g) What mass of fluorine gas is needed to produce 120. g of PF3
Studentka2010 [4]

Answer:

44.28 grams.

Explanation:

Let us write the balanced reaction:

P_{4}+6F_{2}-->4PF_{3}

As per balanced equation, six moles of fluorine gas will give four moles of PF₃.

The mass of PF₃ required = 120 g

The molar mass of PF₃ = 88g/mol

Moles of PF₃ required =\frac{mass}{molarmass}=\frac{120}{88}=1.364mol

The moles of fluorine gas required = \frac{4X1.364}{6}=0.91

the mass of fluorine gas required = moles X molar mass = 0.91x38 = 34.58g

Now this much mass will be required if the reaction is of 100% yield

But as given that the yield of reaction is only 78.1%

The mass of fluorine required = \frac{massX100}{78.1} =\frac{34.58X100}{78.1} =44.28g

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