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leonid [27]
3 years ago
11

What concentration of the lead ion, pb2+, must be exceeded to precipitate pbf2 from a solution that is 1.00×10−2 m in the fluori

de ion, f−? ksp for lead(ii) fluoride is 3.3×10−8 ?
Chemistry
1 answer:
makkiz [27]3 years ago
5 0

Answer is: concentration of Pb²⁺ must be exceeded is 3.3·10⁻⁴ M.

<span> Chemical reaction : </span>Pb²⁺(aq) + 2F⁻(aq) → PbF₂(s).

<span> Ksp(PbF</span>₂) = 3.3·10⁻⁸.<span>
c(F</span>⁻) = 0.01 M.<span>
Ksp(PbF</span>₂) = c(Pb²⁺) · c(F⁻)².<span>
c(Pb²</span>⁺) = Ksp(PbF₂) ÷ c(Cl⁻)².<span>
c(Pb²</span>⁺) = 3.3·10⁻⁸ ÷ (0.01 M)².<span>
c(Pb²</span>⁺) = 0.000000033 M³ ÷ 0.0001 M².<span>
c(Pb²</span>⁺) = 0.00033 M = 3.3·10⁻⁴ M.

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A student makes two mixtures, both made of the same substances. What must be true of the two mixtures? A Both mixtures have the
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• Molecular mass of Iron (III) tetraoxide

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deff fn [24]

Answer:

Explanation:

This is a limiting reactant problem.

Mg(s)

+

2HCl(aq)

→

MgCl

2

(

aq

)

+ H

2

(

g

)

Determine Moles of Magnesium

Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).

4.86

g Mg

×

1

mol Mg

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g Mg

=

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Determine Moles of 2M Hydrochloric Acid

Convert

100 cm

3

to

100 mL

and then to

0.1 L

.

1 dm

3

=

1 L

Convert

2.00 mol/dm

3

to

2.00 mol/L

Multiply

0.1

L

times

2.00 mol/L

.

100

cm

3

×

1

mL

1

cm

3

×

1

L

1000

mL

=

0.1 L HCl

2.00 mol/dm

3

=

2.00 mol/L

0.1

L

×

2.00

mol

1

L

=

0.200 mol HCl

Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,

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×

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mol H

2

1

mol Mg

×

2.01588

g H

2

1

mol H

2

=

0.403 g H

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0.200

mol HCl

×

1

mol H

2

2

mol HCl

×

2.01588

g H

2

1

mol H

2

=

0.202 g H

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The limiting reactant is

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0.202 g H

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under the stated conditions.

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