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dem82 [27]
3 years ago
9

Which solution has the highest vapor pressure?

Chemistry
1 answer:
nata0808 [166]3 years ago
7 0

Answer:

the highest vapor pressure is the solution (b) P = 23.552 torr

∴ sln (a): P = 23.33 torr

∴ sln (c): P = 23.370 torr

Explanation:

vapor pressure:

  • ΔP = - (P*a)(Xb,l) = P - P*a

∴ a: solvent (water)

∴ b: solute

assuming T=25°C:

∴ P*a (25°C) = 23.8 torr

molecular weight:

∴ Mw H2O = 18 g/mol

∴ Mw glucose: 180.156 g/mol

∴ Mw sucrose = 342.3 g/mol

∴ Mw potassium acetate = 98.15 g/mol

∴ mol H2O = (100 mL)(1 g/mL)(mol/18 g) = 5.55 mol H2O

a) moles glucose = (20.0 g)(mol/180.156 g) = 0.111 mol

⇒ Xb,l = (0.111 mol C6H12O6)/(5.55 + 0.111) = 0.0196

⇒ ΔP = - (23.8 torr)(0.0196) = - 0.466 torr = P - P*a

⇒ P = 23.8 torr - 0.466 torr = 23.33 torr

b) moles sucrose = (20 g)(mol/342.3 g) = 0.0584 mol

⇒Xb,l = (0.0584)/(5.55 + 0.0584) = 0.0104

⇒ ΔP = - (23.8 torr)(0.0104) = - 0.248 torr

⇒ P = 23.8 torr - 0.248 torr = 23.552 torr

c) moles KC2H3O2 = (10 g)(mol/98.15 g) = 0.102 mol

⇒ Xb,l = (0.102)/(5.55 + 0.102) = 0.0180

⇒ ΔP = - (23.8 torr)(0.0180) = - 0.429 torr

⇒ P = 23.8 torr - 0.429 torr = 23.370 torr

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sertanlavr [38]
In the periodic table, elements of the same group are characterized by having the same similar properties.
So, first we will check the elements that lie within the same group as <span>beryllium  and then we will attempt to choose the elements with atomic mass higher than 130.

So, elements in the same group as </span>beryllium are: magnesium, calcium, strontium, barium and radium.
Among these elements, we will find that:
radium has atomic mass of 226 amu
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Based on this, the two elements would be barium and radium.

3 0
3 years ago
Detecta cuál de las siguientes fórmulas estructurales no cumple con la tetravalencia del carbono (formar cuatro enlaces)
o-na [289]

El ejercicio completo con las formulas es el siguiente:

cuál de las siguientes fórmulas no cumple con la tetravalencia del átomo de carbono.

a. CH3-CH2-CH2-OH

b. CH3-CH=CH2

c. CH3-CH2=CH2

d. CH3-CH2-CH3

Answer:

La respuesta correcta es la opción C

Explanation:

En la formula C, podemos ver claramente que no se cumple con la tetravalencia del carbono pues al sumar el total de las valencias del carbono nos da 5

CH3 - CH2  =   CH2 = Total de valencias del carbono 5

         1 + 2+  2   =  5

Para que la formula sea correcta debe eliminarse un hidrogeno quedando la formula de la siguiente manera:

                      CH3   -  CH   =   CH2

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8 0
3 years ago
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.

Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).

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)

The solubility of KCl given is 3.7 M.

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

The Ksp was found to be equal to 14.

In pure water KCl

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

Let x= molar solubility [K+],/[Cl-] :. × , x

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)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

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

Let b= molar solubility :. B, b+8

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

17.1 moles/L

Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.

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