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lozanna [386]
3 years ago
6

Si desea preparar 300 gramos de una solución de azúcar en agua que tenga una concentración de 20% m/m en cuanto gramos de soluto

de azúcar se necesita?
Chemistry
1 answer:
Digiron [165]3 years ago
5 0

Answer:

m_{soluto}=60g

Explanation:

¡Hola!

En este caso, al considerar la unidad de concentración de porcentaje masa/masa, podemos escribir su fórmula como:

\%m/m=\frac{m_{soluto}}{m_{solucion}}*100\%

Podemos identificar la masa de soluto (azúcar) como la incógnita, y resolverla como se muestra a continuación:

m_{soluto}=\frac{\%m/m*m_{solucion}}{100\%}\\\\m_{soluto}=\frac{20\%*300g}{100\%}\\\\m_{soluto}=60g

¡Saludos!

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Solid aluminum and gaseous oxygen read in a combination reaction to produce aluminum oxide. 4Al(s) + 3O_2(g) rightarrow 2Al_2O_3
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Answer:

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Explanation:

4Al(s) + 3O₂ (g) ⇄ 2Al₂O₃ (s)

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Ratio is 4:3. 4 moles of Al react with 3 moles of O₂

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P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

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R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

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