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Mrrafil [7]
3 years ago
5

0.398 mol AgNO3 and 2.73 mol NH3 were added to enough water to give 1.00 L of solution at 25 °C. Calculate the equilibrium conce

ntration of free silver ion , [Ag+], in this solution. The formation constant for Ag(NH3)2+ (Kf) = 1.50x107 at 25 °C.
Chemistry
1 answer:
scoundrel [369]3 years ago
8 0

Answer:

[Ag⁺] = 1x10⁻⁸M

Explanation:

In the reaction:

Ag⁺ + 2NH₃ ⇄ Ag(NH₃)₂⁺

Kf = 1.50x10⁷ =  [Ag(NH₃)₂⁺] / [NH₃]²[Ag⁺]

If in the beginning you add 0.398 moles of Ag⁺ and 2.73 moles of NH₃. Concentrations in equilibrium are:

[Ag(NH₃)₂⁺]: X

[NH₃]: 2.73mol - 2X

[Ag⁺]: 0.398mol - X

Replacing:

1.50x10⁷ =  [X] / [2.73-X]²[0.398-X]

1.50x10⁷ =  [X] / (7.4529-5.46X+X²)(0.398-X)

1.50x10⁷ =  [X] / 2.9662542-9.62598X+5.858X²-X³

-1.50x10⁷X³+8.787x10⁷X²-1.443897x10⁸X + 4.4493813x10⁷ = X

-1.50x10⁷X³+8.787x10⁷X²-1.443897x10⁸X + 4.4493813x10⁷ = 0

Solving for X:

X = 0.39799999

Thus, equilibrium concentration of Ag⁺, [Ag⁺] is 0.398mol - X. Replacing:

[Ag⁺] = 0.398mol - 0.39799999 = 1x10⁻⁸M

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frutty [35]

Answer:

Gold and copper are example of alloy

Explanation:

8 0
2 years ago
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
BigorU [14]
Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).

The first step is to find the number of moles of Mg in 4.03g of Mg.  You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg.  Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg.  To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂.  From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP.  Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.

I hope this helps. Let me know in the comments if anything is unclear.
6 0
3 years ago
Read 2 more answers
Fill in the Blank
Arisa [49]

Answer: Quantitative data

Explanation:

3 0
3 years ago
Help PLEASEEE !!!!!! will give brainliest !
BabaBlast [244]

Answer:

6. O₂ + Cu —> CuO

7. H₂ + Fe₂O₃ —> H₂O + Fe

8. O₂ + H₂ — > H₂O

9. H₂S + NaOH —> Na₂S + H₂O

10. Al + HCl —> H₂ + AlCl₃

Explanation:

6. Oxygen gas react with solid copper metal to form copper(II) oxide

Oxygen gas => O₂

Copper => Cu

copper(II) oxide => CuO

The equation is:

O₂ + Cu —> CuO

7. hydrogen gas and iron(III) oxide powder react to form liquid water and solid iron power

hydrogen gas => H₂

Iron(III) oxide => Fe₂O₃

Water => H₂O

Iron => Fe

The equation is:

H₂ + Fe₂O₃ —> H₂O + Fe

8. Oxygen gas react with hydrogen gas to form liquid water

Oxygen gas => O₂

hydrogen gas => H₂

Water => H₂O

The equation is:

O₂ + H₂ — > H₂O

9. Hydrogen sulphide gas is bubbled through a sodium hydroxide solution to produce sodium sulphide and liquid water

hydrogen sulphide => H₂S

sodium hydroxide => NaOH

Sodium sulphide => Na₂S

Water => H₂O

The equation is:

H₂S + NaOH —> Na₂S + H₂O

10. Hydrogen gas and aluminum chloride solutions are produced when solid aluminum react with hydrochloric acid

Aluminum => Al

Hydrochloric acid => HCl

hydrogen gas => H₂

Aluminum chloride => AlCl₃

The equation is:

Al + HCl —> H₂ + AlCl₃

5 0
2 years ago
An atom of aluminum in the ground state and an atom of gallium in the ground state have the same
Kobotan [32]
(4) total number of valence electrons, because they exist in the same group.
3 0
3 years ago
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