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3 years ago

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A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba 2 . When the concentration of F - exceeds __________ M,

BaF 2 will precipitate. Neglect volume changes. For BaF 2,

1 answer:

Stella [2.4K]3 years ago

5 0

Answer:

When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate

Explanation:

Ksp of BaF₂ is:

BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

Ksp = 1.7x10⁻⁶ = [Ba²⁺] [F⁻]²

The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.

As the concentration of [Ba²⁺] is 0.0144M, the product [Ba²⁺] [F⁻]² will be equal to ksp just when:

1.7x10⁻⁶ = [Ba²⁺] [F⁻]²

1.7x10⁻⁶ = [0.0144M] [F⁻]²

1.18x10⁻⁴ = [F⁻]²

0.0109M = [F⁻]

That means, when [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate

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18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

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3 years ago

How many cm 3 are in 0.14 m 3?

nlexa [21]

Answer:

There are 14 cm in 0.14 m

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3 years ago

Compare and contrast the ease at which molecules are able to move relative to each other in the three states of matter

ANTONII [103]

The arrangement of molecules within the 3 phases of matter are shown in the picture.

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3 years ago

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Genrish500 [490]

Answer:

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Explanation:

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$moles = \frac{Mass\ taken}{Molar\ mass}$

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$4Al+3O_2\rightarrow 2Al_2O_3$

4 moles of aluminum react with 3 moles of oxygen gas

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Pressure = 1 atm

Temperature = 273.15 K

Using ideal gas equation as:

$PV=nRT$

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T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K

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Answer:

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Explanation:

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