The enthalpy of vaporization for water is 40.67 kJ/mol and the enthalpy of fusion for water is

Chemistry

1 answer:

S_A_V [24]2 years ago

6 0

Answer:

$\Delta H_{dep}=-46.68kJ/mol$

Explanation:

Hello!

In this case, since vaporization is the processes by which a liquid goes to the gas phase, fusion by which a solid goes to liquid and deposition by which a gas goes to solid; we infer that the following set up relates the enthalpies associated to each process:

$\Delta H_{fus}+\Delta H_{vap}=-\Delta H_{dep}$

Because deposition goes from a state with more energy to a state with less energy, therefore it is negative; in such a way, by plugging in we obtain:

$\Delta H_{dep}=-(\Delta H_{fus}+\Delta H_{vap})\\\\\Delta H_{dep}=-(6.01kJ/mol+40.67kJ/mol)\\\\\Delta H_{dep}=-46.68kJ/mol$

Best regards!

You might be interested in

Which item is made from a synthetic material?

seropon [69]

First one because Common synthetic materials are nylon, acrylic, polyester, carbon fiber, rayon and spandex. Synthetic materials are made from chemicals and are usually based on polymers. They are stronger than natural and regenerated materials.

7 0

2 years ago

Read 2 more answers

Consider the reaction below. HI + H2O mc014-1.jpg H3O+ + I– Which is an acid-conjugate base pair? HI and H2O H2O and H3O+ H3O+ a

lesya692 [45]

Answer: Option (d) is the correct answer.

Explanation:

According to Bronsted-Lowry, species which donate a proton are known as acid. The species which accept a proton are known as a base.

In the given reaction, acids and bases are as follows.

HI + $H_{2}O$ $\rightarrow$ $H_{3}O^{+}$ + $I^{-}$

Acid Base Conjugate acid Conjugate base

Therefore, the acid HI loses a proton to form a conjugate base that is $I^{-}$ .

Thus, we can conclude that HI and $I^{-}$ is an acid conjugate base pair.

4 0

3 years ago

Read 2 more answers

What organic product would be formed from the reaction of 1-bromo-2-methylpropane (isobutyl bromide), (ch3)2chch2br, with mg, et

miskamm [114]

The reaction of 1-bromo-2-methylpropane (isobutyl bromide), $(CH_3)_2CHCH_2Br$ , with $Mg$ , and $Et_2O$ results in the formation of an adduct of formula $(CH_3)_2CHCH_2MgBr$ known as grignard reagent, an organometallic reagent, which on reacting with $H_3O^{+}$ results in the formation of isobutane.