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faltersainse [42]
2 years ago
10

The solubility product for chromium(iii) fluoride is ksp = 6.6 × 10–11. what is the molar solubility of chromium(iii) fluoride?

Chemistry
2 answers:
Vlad [161]2 years ago
6 0

Answer:

x=1.3x10^{-3}M

Explanation:

Hello,

in this case, one considers the dissolution as an ionic reaction:

CrF_3Cr^{+3}+3F^{-}

In such a way, we write the equilibrium equation based on the change x due to the slight rate of dissolution of the considered salt as:

Ksp=[Cr^{+3} ]_{eq} [F^{-} ]_{eq} ^{3} \\Ksp=x(3x)^{3}  \\x=\sqrt[4]{\frac{Ksp}{27}} =\sqrt[4]{\frac{6.6x10^{-11} }{27}}\\x=1.3x10^{-3}M

Such x accounts for the molar solubility of the chromium (III) fluoride.

Best regards.

uysha [10]2 years ago
3 0
Following is the dissociation reaction of CrF3

CrF3              →            Cr3+            +             3F-

Now, solubility product (Ksp) = [Cr3+] [F-]^3

Let [Cr3+]  =   [F-]   =   x

∴Ksp = x(x)^3 
         = x^4

But Given: Ksp = 6.6 X 10^-11

6.6 X 10^-11 = x^4
x = 2.85 x 10^-3 M

Thus, <span>the molar solubility of chromium(iii) fluoride is 2.83 X 10^-3 M</span>
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Answer:

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6 0
2 years ago
When one atom loses an electron and another atom accepts that electron a(n) bond between the two atoms results?
morpeh [17]
When an element losses its electron its called a cation. When an element accepted that electron it called anion. This is called an ionic bond.
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2 years ago
If the actual yield of the reaction was 75% instead of 100%, how many molecules of no would be present after the reaction was ov
artcher [175]
To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:

N₂ + O₂ → 2 NO

Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.

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8 0
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Read 2 more answers
Suppose that the microwave radiation has a wavelength of 12.4 cm . How many photons are required to heat 255 mL of coffee from 2
vivado [14]

Answer:

Explanation:

wavelength λ = 12.4 x 10⁻² m .

energy of one photon = h c / λ

= 6.6 x 10⁻³⁴ x 3 x 10⁸ /  12.4 x 10⁻²

= 1.6 x 10⁻²⁴ J .

Let density of coffee be equal to density of water .

mass of coffee = 255 x 1 = 255 g

heat required to heat up coffee = mass x specific heat x rise in temp

= 255 x 4.18 x ( 62-25 )

= 39438.3 J  .

No of photons required = heat energy required / energy of one photon

= 39438.3 / 1.6 x 10⁻²⁴

= 24649 x 10²⁴

= 24.65 x 10²⁷ .

5 0
2 years ago
The empirical formula of a gaseous fluorocarbon is CF2. At a certain temperature and pressure, a 1-L volume holds 8.93 g of this
dimaraw [331]

Answer:

C₄F₈

Explanation:

Using their mole ratio to compute their mass

molar mass of carbon = 12.0107 g/mol

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let x = mass of carbon

given mass of fluorine = 1.70 g

x / 12.01067 = 1.70 / 37.99687

cross multiply

x = ( 1.70 × 12) / 37.99687 = 20.4 / 37.99687 = 0.53688 g

mass of one mole of CF₂ = 0.53688 + 1.70 = 2.23688 g

number of mole of CF₂ = 8.93 g / 2.23688 = 3.992 approx 4

molecular formula of CF₂ = 4 (CF₂) = C₄F₈

3 0
2 years ago
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