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serious [3.7K]
2 years ago
14

When an ionic compound dissolves in water, the individual ions first leave the crystal lattice and then each ion becomes surroun

ded by a cluster of polar water molecules. What is the sign of the entropy change for each of these two steps in the dissolving process
Chemistry
1 answer:
vichka [17]2 years ago
3 0

The entropy of the process in which the  individual ions first leave the crystal lattice is positive while the entropy of the process whereby the each ion becomes surrounded by a cluster of polar water molecules is negative.

<h3>What is entropy?</h3>

The term entropy has to do with the degree of disorder in a system. The higher the entropy of the system, the more the disorderliness of the system.

Now, the entropy of the process in which the  individual ions first leave the crystal lattice is positive while the entropy of the process whereby the each ion becomes surrounded by a cluster of polar water molecules is negative.

Learn more about entropy: brainly.com/question/13146879

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A loaf of bread has a green substance growing on each slice. Which of the following types of spoilage could have occurred?
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What is a semiconductor?
makkiz [27]
It is c: a conductor that operates only at low temperatures
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3 years ago
How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?
Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

C_{6}H_{8}O_{6}

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g

To know the percent of each element, we need to to the following:

C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

5 0
3 years ago
A compound possessing a carboxylic acid produces a yellow solution when dissolved in diethyl ether. This yellow solution is tran
Serga [27]

Answer:

A. The top layer will be diethyl ether, and the top layer will be yellow.

Explanation:

The purpose of the addition of the saturated aqueous solution of polar solvents like sodium chloride in the liquid-liquid extraction techniques is to remove as well as separate any kind of water which may be dissolved in the ether. Water and sodium chloride are both polar and thus, they forms the bottom layer and only ether forms the top layer. The compound being organic and is colored is in the top layer with the ether.

Hence, answer - A. The top layer will be diethyl ether, and the top layer will be yellow.

6 0
3 years ago
The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

Ea = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

Now put all the given values in this formula, we get

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

Therefore, the value of the rate constant at 775.0 K is 4.3\times 10^{-13}s^{-1}

5 0
3 years ago
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