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Nady [450]
3 years ago
5

Which interactions and processes contribute to the dissolution of ionic compounds in water?

Chemistry
1 answer:
artcher [175]3 years ago
6 0

Answer:

Ion-dipole attraction and solvation of the ions  

Step-by-step explanation:

Assume you have an ionic compound MX.

Ion-dipole interactions:

Water is a highly polar molecule.  

The partially negative O atom is strongly attracted to the M⁺ cations at the surface of the solid. The partially positive H atoms are strongly attracted to the X⁻ ions.

If the water molecules hit the surface of the solid with the right orientation, the ion-dipole attractions will help dislodge the ions from the surface.

Solvation:

Once the ions are in solution, the ions are stabilized by solvation. The strong ion-dipole attractions cause a shell of water ions to form around the ionss.

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1. What pressure would be exerted by 46.0grams of hydrogen gas placed into a 3.00 L container at
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Answer:

187.34 atm

Explanation:

From the question,

PV = nRT.................. Equation 1

Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.

make P the subject of the equation

P = nRT/V.............. Equation 2

n = mass(m)/molar mass(m')

n = m/m'............... Equation 3

Substitute equation 3 into equation 2

P = (m/m')RT/V............ Equation 4

Given: m = 46 g, T = 25°C = (25+273) = 298 K, V = 3.00 L

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2 years ago
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Answer:

The elements mentioned in the series are Na, Mg, Al, Si, and P. It can be seen that all these elements are located in the same period. The atomic number of the mentioned elements are Na-11, Mg-12, Al-13, Si-14, and P-15.

a) There will be an increase in the ionization energy with the increase in the element's atomic number across a period. More energy is needed to withdraw an electron from a completely occupied shell in comparison to an incompletely occupied shell.

The atomic number of Na is 11. When one electron is withdrawn from Na, it gets converted into the inert gas configuration of Ne. Thus, it will require more energy to withdraw the second electron from Na. Hence, Na exhibits the lowest second ionization energy.

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c) The metallic nature of the elements reduces from left to right across a period in the periodic table. Thus, P is the least metallic element.

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Explanation:

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