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Crazy boy [7]
3 years ago
9

What is the percent composition by mass of sulfur in the compound mgso4

Chemistry
1 answer:
Margaret [11]3 years ago
6 0

Answer:

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

Explanation:

Molar mass of compound = 120 g/mol

Number of sulfur atom = 1

Atomic mass of sulfur = 32 g/mol

Percentage of element in compound :

=\frac{\text{Number of atoms}\times \text{Atomic mass}}{\text{molar mas of compound}}\times 100

Sulfur :

=\frac{1\times 32 g/mol}{120 g/mol}\times 100=26.7\%

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

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Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
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Answer:

[Pb^{2+}]=3.9 \times 10^{-2}M

this is the concentration required to initiate precipitation

Explanation:

PbCl_2  ⇄ Pb^{2+}+2Cl^-

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

Ksp=2.4\times 10^{-4}

lets solubility=S

[Pb^{2+}] = S

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Ksp=[Pb^{2+}]\times [Cl^-]^2

Ksp=S \times (2S)^2

Ksp=4S^3

S=\sqrt[3]{\frac{Ksp}{4} }

S=3.9\times 10^{-2}

[Pb^{2+}]=3.9 \times 10^{-2}M this is the concentration required to initiate precipitation

4 0
3 years ago
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