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Grace [21]
2 years ago
8

Se requieren preparar 1500 ml de una solución al 55% en masa de hidróxido de calcio, sabiendo que la densidad de dicha solución

es de 1.17 g/ml. Calcula la masa de hidróxido para preparar dicha solución.
Chemistry
1 answer:
Masja [62]2 years ago
6 0

Respuesta:

968 g Ca(OH)₂

Explicación:

Paso 1: Calcular la masa de solución

Tenemos 1500 mL de una solución cuya densidad es 1.17 g/mL, es decir, 1 mL de solución tiene una masa de 1.17 g.

1500 mL × 1.17 g/mL = 1.76 × 10³ g

Paso 2: Calcular la masa de hidróxido de calcio en 1.76 × 10³ g de solución

La solución tiene una concentración de 55% en masa de hidróxido de calcio, es decir, cada 100 gramos de solución hay 55 gramos de hidróxido de calcio.

1.76 × 10³ g Solución × 55 g Ca(OH)₂/100 g Solución = 968 g Ca(OH)₂

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Free-energy change, ΔG∘, is related to cell potential, E∘, by the equationΔG∘=−nFE∘where n is the number of moles of electrons t
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a)\Delta G=372490 J

b)\Delta G=-568614 J

Explanation:

a) The reaction:

Mg(s) +Fe^{2+}(aq) \longrightarrow Mg^{2+}(aq) + Fe(s)

The free-energy expression:

\Delta G=-n*F*E

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The element wich is reduced is the Fe and the one that oxidates is the Mg:

-0.44V=E_{red}-(-2.37V)=1.93V

The electrons transfered (n) in this reaction are 2, so:

\Delta G=-2mol*96500 C/mol * 1.93 V

\Delta G=-372490 J

b) If you have values of enthalpy and enthropy you can calculate the free-energy by:

\Delta G=\Delta H - T* \Delta S

with T in Kelvin

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7 0
2 years ago
A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f
a_sh-v [17]

Answer:

m_{H_2O}=39.0g

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g

Best regards.

8 0
3 years ago
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