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irinina [24]
3 years ago
7

Van der Waals forces can best be described as the attractive or repulsive forces that are formed

Chemistry
2 answers:
blsea [12.9K]3 years ago
5 0

Vander Waals forces can best be described as the attractive or repulsive forces that are formed during dipole-dipole interactions between molecules.

Vander Waals forces arises from interaction between uncharged atoms or molecules and these are formed during dipole-dipole interactions between molecules. Vander waal forces are weak forces that vanishes when molecules moves apart.


dem82 [27]3 years ago
3 0

Correct answer: by a random, short-lived redistribution of electrons.

van der Waal's forces are the weak intermolecular forces of attraction or repulsions that are seen in all molecules, when they come close to each other. These forces of attraction or repulsion are due to the momentary charges that develop between the electron clouds of one atom and the nuclei of the other atom. Therefore, the van der Waal's forces are due to the instantaneous dipoles that are produced between two atoms close to each other, due to the redistribution of electron clouds.

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Object A has a mass of 3,600 kilograms. Object B has a mass of 600 kilograms. The weight of Object B will be ________ times the
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How many moles in 4.65 g of Helium?
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3 years ago
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

7 0
3 years ago
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