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egoroff_w [7]
2 years ago
13

The volume of an air bubble at the bottom of a lake is 1.35 mL. If the pressure at the bottom of the lake is 125 kPa and the air

pressure at the top of the lake is 105 kPa, what will the volume of the bubble be when it rises to the surface?
Chemistry
1 answer:
nikitadnepr [17]2 years ago
8 0

Answer:

1.61 mL

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Initial volume (V1) = 1.35 mL

Initial pressure (P1) = 125 kPa

Final pressure (P2) = 105 kPa

Final volume (V2) =..?

Step 2:

Determination of the volume of the air bubble at the surface of the lake.

This can be obtained by use the Boyle's law equation as follow:

P1V1 = P2V2

125 x 1.35 = 105 x V2

Divide both side by 105

V2 = (125 x 1.35)/105

V2 = 1.61 mL

Therefore, the volume of the air bubble at the surface of the lake is 1.61 mL

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The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add
JulijaS [17]

Answer:

a)The Ksp was found to be equal to 13.69

Explanation:

Terminology

Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.

Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.

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Step-by-step solution:

To solve this: 

#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.

#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.

#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

a) The equation of solubility equilibrium for KCL is thus;

KCL_(s) ---> K+(aq) + Cl- (aq)

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Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

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In pure water KCl

Ksp =13.69 KCl =[K+][Cl-]

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Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio

37M moles/L

The Ksp was found to be equal to 14.

4.0 M HCl = KCl =[K+][Cl-]

Let y= molar solubility :. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

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Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

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