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valentinak56 [21]
3 years ago
7

Explain how the attractive forces between the particles in a liquid are related to the equilibirum vapour pressure of that liqui

d.
Chemistry
1 answer:
drek231 [11]3 years ago
5 0

Answer:

Attractive forces between particles are inversely proportional to vapour pressure.

Explanation:

Inside a liquid, molecules undergo random motion (thermal motion), but also interact with one another via electromagnetic forces of different kinds, like Van der Waals forces, ion-dipole interactions, hydrogen bonds, etc. These forces keep the liquid together, giving it a definite volume, in distinction to gases, which take the volume of the vessel that contains them.

Now, some molecules in a liquid can attain a high velocity as a random outcome of thermal motion, if this molecule is at the liquid's surface, it might actually escape! actually, many molecules might do that, and form a vapour over the liquid's surface.

Now, we know that liquids exist, therefore this process has to reach an equilibrium, that means, once the vapour becomes <em>dense </em>(or <em>concentrated</em>)<em> </em>enough, it would be as likely for a vapour molecule to re-enter the liquid as it is likely for a liquid molecule to leave the liquid and enter into the vapour.

This is called vapour-liquid equilibrium.  

How can we measure how "concentrated" the vapour is? by measuring the pressure above the liquid. We know by the ideal gas law that the number of molecules in a gas is proportional to pressure at constant volume and temperature.

But how does vapour pressure relate to intermolecular forces?

Simply, the stronger the intermolecular forces, the less likely a molecule at the liquid's boundary will be to shoot of into the vapour phase! and viceversa, if intermolecular forces are very weak, the molecules won't hold together much and many molecules will leave the liquid.

As an extreme case imagine a solid, for which intermolecular forces are the strongest, what's the vapour pressure of a solid? Do solids evaporate into the air?  The answer is no, solids (with few exceptions) don't evaporate, and their vapour pressure is extremely small.

Cheers!

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3 years ago
What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?
mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

C=\frac{n}{V} .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is K_{2}SO_{4} thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass

Putting the values,

n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol

Thus, number of moles of  K_{2}SO_{4} will be 0.1061\times 0.5=0.053 mol.

The volume of solution is 225 mL, converting this into L,

1 mL=10^{-3}L

Thus,

225 mL=0.225 L

Putting the values in equation (1),

C=\frac{(0.053 mol}{0.225 L}=0.236 M

Therefore, concentration of potassium sulfate solution is 0.236 M.


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3 years ago
At what height the acceleration due to gravity of the earth become half ?​
Serggg [28]

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Acceleration due to gravity is reduced to half its value on the earth's surface at an altitude of 2.65×106 m

3 0
3 years ago
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A
liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

p^{OH}=14-56.17

      =8.823

The volume of the NH_{3} = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated NH_{3} =0.10 M

The volume of the NH_{4}Cl= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated NH_{4}Cl= 0.10 M

The volume of the H_{2}So_{4}= 30 ml

convert into the liter= 0.030L  

The value of concentrated H_2So_4=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of NH_3 = \frac{0.10\times 0.040}{0.120}= 0.33 \ M

calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

5 0
3 years ago
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