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statuscvo [17]
2 years ago
8

Explain how electrostatic attraction affects the van’t Hoff factor. Why is the effect always to lower the van’t Hoff factor from

the simple assumption that dissolved ionic substances always form whole number values of ions?
Chemistry
1 answer:
Olin [163]2 years ago
4 0

Electrostatic attraction affects the van’t Hoff factor as the solute concentration increases, thereby decreasing the van't Hoff factor.

<h3>What is van't Hoff factor?</h3>

The van't Hoff factor is therefore becomes a measure of a deviation from an ideal behavior. In order words, the lower the van't Hoff factor, the greater the deviation.

Learn more about dissolution of ionic substances:

brainly.com/question/6198647

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Determine the empirical formula of a
k0ka [10]

Answer:

AuCl

Explanation:

Given parameters:

Mass of Gold  = 2.6444g

Mass of Chlorine  = 0.476g

Unknown:

Empirical formula  = ?

Solution:

Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.

Elements                                     Au                                             Cl

Mass                                         2.6444                                     0.476

Molar mass                                 197                                          35.5

Number of moles                  2.6444/197                                 0.476/35.5

                                                 0.013                                           0.013

Divide by the

smallest                                 0.013/0.013                                 0.013/0.013

                                                       1                                                   1

The empirical formula of the compound is AuCl

3 0
2 years ago
What mass of helium is in a 2.00 L balloon at STP? pls help me lol
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3 0
2 years ago
What particle joins or bonds atoms to form a compound or molecule
agasfer [191]

<h3>I hope it is helpful for you ....</h3>

6 0
3 years ago
Calculate the mass of a solid metal cylinder with a density of 2.6 g/cm", a
worty [1.4K]

Answer:

V= π ×r² × h

V = 3.14 × (0.9)² × 4

V = 3.14 × 0.81 × 4

V = 10.1736

Mass = Volume × density

M = 10.1736 × 2.6

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8 0
3 years ago
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten c
gtnhenbr [62]

Answer:

1.13 × 10⁶ g

Explanation:

Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.

Al³⁺ + 3 e⁻ → Al

We can establish the following relations:

  • 1 Ampere = 1 Coulomb / second
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  • 1 mole of Al is produced when 3 moles of electrons circulate
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The mass of aluminum produced under these conditions is:

90.0 s \times \frac{1s}{620c} \times \frac{96,468c}{1mole^{-} } \times \frac{3mole^{-}}{1molAl} \times \frac{26.98gAl}{1molAl} =1.13 \times 10^{6} g Al

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3 years ago
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