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2 years ago

What is difference between vapourization and boiling point?

Chemistry

2 answers:

Pepsi [2]2 years ago

6 0

Answer:

Vaporization is basically just evaporation which does not require nearly as much heat and happens naturally however when it comes to boiling point this is the point where water has been heated enough to start bubbling and changing state slowly.

Explanation:

aev [14]2 years ago

3 0

Vaporization is a phase transition from the liquid phase to vapour while boiling point is when the temperature is high enough for a phase transition from liquid to gas.

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2 years ago

Iron fluoride (FeF2) dissociates according to the following equation:

melomori [17]

Answer:

S = 0.788 g/L

Explanation:

The solubility product (Kps) is an equilibrium solubization constant, which can be calculated by the equation:

$Kps = \frac{[product]^x}{[reagent]^y}$

Where x and y are the stoichiometric coefficients of the product and the reagent, respectively. Because of the aggregation form, the concentration of solids is always equal to 1 for use in this equation.

Analyzing the equation, we see that for 1 mol of $Fe^{+2}$ is necessary 2 mols of $F^-$ , so if we call "x" the molar concentration of $Fe^2$ , for $F^-$ we will have 2x, so:

$Kps = [Fe^{+2}].[F^-]^2\\\\2.36x10^{-6} = x(2x)^2\\\\2.36x10^{-6} = 4x^3\\\\x^3 = 5.9x10^{-7}\\\\x = \sqrt[3]{5.9x10^{-7}} \\\\x = 8.4x10^{-3} mol/L$

So, to calculate the solubility (S) of FeF2, which is in g/L, we multiply this concentration by the molar mass of FeF2, which is:

Fe = 55.8 g/mol

F = 19 g/mol

FeF2 = Fe + 2xF = 55.8 + 2x19 = 93.8 g/mol

So,

[tex]S = 8.4x10^{-3}x93.8

S = 0.788 g/L

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2 years ago

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2 years ago

What is difference between vapourization and boiling point?​

What is difference between vapourization and boiling point?