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lora16 [44]
2 years ago
11

When an aqueous solution of sodium phosphate (Na3PO4) and an aqueous solution of magnesium sulfate (MgSO4) are mixed, a precipit

ate forms. Write the chemical formula of the precipitate.
Chemistry
1 answer:
Akimi4 [234]2 years ago
8 0

Answer: The chemical formula of this precipitate is Mg_{3}(PO_{4})_{2}.

Explanation:

An equation which depicts the chemical reaction of substances in the form of chemical formulas is called a chemical equation.

For example, chemical equation for an aqueous solution of sodium phosphate (Na_{3}PO_{4}) and an aqueous solution of magnesium sulfate (MgSO_{4}) are mixed, a precipitate forms is as follows.

3MgSO_{4}(aq) + 2Na_{3}PO_{4}(aq) \rightarrow Mg_{3}(PO_{4})_{2}(s) + 3Na_{2}SO_{4}(aq)

Here, trimagnesium phosphate is the precipitate. The chemical formula of this precipitate is Mg_{3}(PO_{4})_{2}.

Thus, we can conclude that the chemical formula of this precipitate is Mg_{3}(PO_{4})_{2}.

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How do catalysts affect energy of reactions
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Explanation:

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3 years ago
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onsider the reversible dissolution of lead(II) chloride. P b C l 2 ( s ) − ⇀ ↽ − P b 2 + ( a q ) + 2 C l − ( a q ) PbClX2(s)↽−−⇀
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Answer:

9.34x10^-4

Explanation:

Step 1:

The balanced equation for the reaction.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

Step 2:

Data obtained from the question:

Mass of PbCl2 = 0.2393 g

Volume = 50mL

concentration of Pb^2+, [Pb^2+] = 0.0159 M

Concentration of Cl^-, [Cl^-] = 0.0318 M

Equilibrium constant, Kc =?

Step 3:

Determination of the number of mole PbCl2.

The number of mole of PbCl2 can be obtained as follow:

Molar Mass of PbCl2 = 207 + (35.5x2) = 278g/mol

Mass of PbCl2 = 0.2393 g

Number of mole =Mass /Molar Mass

Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

Step 4:

Determination of Molarity of PbCl2.

At this stage we shall obtain the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.61x10^-4 mole

Volume = 50mL = 50/1000 = 0.05L

Molarity of PbCl2 =?

Molarity = mole /Volume

Molarity of PbCl2 = 8.61x10^-4/0.05

Molarity of PbCl2 = 0.01722 M

Step 5:

Determination of the equilibrium constant Kc.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

The equilibrium constant Kc for the equation above is given by:

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

[Pb^2+] = 0.0159 M

[Cl^-] = 0.0318 M

[PbCl2] = 0.01722 M

Kc =?

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2/ 0.01722

Kc = 9.34x10^-4

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3 years ago
EXPLAIN how to identify the reducing agent in a reaction of magnesium with oxygen
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Explanation:

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Mg(2+)and O(2-)

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8 0
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Answer:

The graph of this equation is shown in Figure 1. As you can see this is a straight line with negative slope and does not intersect the y-axis. So the ...

Explanation:

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