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2 years ago

Which of the following compounds has the lowest boiling point?

Chemistry

1 answer:

sergejj [24]2 years ago

6 0

Answer : The correct answer is option A : Diethyl ether

Explanation :

The boiling point of a compound depends on the intermolecular forces (IMF) of attractions present among its molecules.

Stronger the IMF, more difficult it is to separate the molecules. And hence the compound shows higher boiling point.

The most common types of IMF are

a) Hydrogen bonding : Hydrogen bonding occurs when a compound has hydrogen atom directly attached to strongly electro-negative atoms like O, N and F

Hydrogen bondings are the strongest IMF. Therefore compounds that have hydrogen bondings show higher boiling point.

In the given examples, 2-butanol and 4-octanol both have -OH group where H is directly attached to highly electro-negative O atom. As a result hydrogen bonding is present in these compounds.

Therefore they show higher boiling point.

b) Dipole interactions : These are seen in case of polar compounds.

c) London dispersion forces : These are present in all the compounds but they are predominant in case of non polar compounds.

Both diethyl ether and diphenyl ether predominantly show London dispersion forces. Since these forces are weaker as compared to other IMF, the molecules having london dispersion tend to have lower boiling points.

But the magnitude of dispersion forces increases as the molecular weight of the compound increases.

Therefore diphenyl ether which has a molecular weight of 170 g/mol has much stronger london dispersion forces as compared to diethyl ether which has a molecular weight of 74 g/mol

From above discussion, we can conclude that diethyl ether has the weakest intermolecular forces of attraction. Hence it has the lowest boiling point.

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Answer:

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2 years ago

A silver sphere has a mass of 5.492 g and a diameter of 10.0 mm. What is the density of silver metal in grams per cubic centimet

Blizzard [7]

Answer:

Explanat $\rho=10.5\frac{g}{cm^3}$ ion:

Hello,

In this case, considering the given diameter which is related to a radius of 5.0 mm and the formula for the calculation of the volume of the sphere, its volume in cubic centimeters (5.00 mm = 0.5 cm) is then:

$V=\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*(0.5cm)^3\\\\V=0.524cm^3$

In such a way, the density turns out:

$\rho =\frac{m}{V} =\frac{5.492g}{0.048m^3} \\\\\rho=10.5\frac{g}{cm^3}$

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8 0

1 year ago

You decide to establish a new temperature scale on which the melting point of ammonia (-77.75 ∘c) is 0∘a, and the boiling point

Hoochie [10]

First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:

-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
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The boiling point of water in ∘a would be 400.33∘a.

7 0

1 year ago

Read 2 more answers

Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank

marysya [2.9K]

The rate of Formation of Carbocation mainly depends on two factors'

1) Stability of Carbocation:
The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.

2) Ease of detaching of Leaving Group:
The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

R-I > R-Cl > R-F

B > C > A

3 0

1 year ago

A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a pist

REY [17]

Answer:

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

Explanation:

Determination of sign of q

Temperature of the water bath before the reaction = 25 °C

Temperature of the water bath after the completion of the reaction = 28 °C

After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.

If heat is absorbed by the system, then q is indicated by positive sign and if heat is released by the system, then q is indicated by negative sign.

As in the given case, heat is released by the system, so sign of q is negative, or q < 0

Determination of sign of w

After the completion of the reaction, piston moved downward, that means volume of the system decreases or compression occur. During the compression, work is done on the system.

if work is done on the system, sign of w is positive.

If work is done by the system, sign of w is negative.

In the given case, work is done on the system, therefore sign of w is positive, or w > 0

Determination of sign of ΔE

Relationship between ΔE, q and w is given by first law of thermodynamics:

ΔE = q + w

In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.

So, among the given option, option c is correct.

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

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1 year ago