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zhenek [66]
2 years ago
15

A una mezcla de 300g, formada con 60% P/P de Hierro y 40% P/P de Arena, se le adicionan 135g de Cobre y 2,77g de Aluminio. ¿Cuál

es la concentración final P/P de cada uno de los componentes?
Chemistry
1 answer:
Svetllana [295]2 years ago
5 0

Answer:

\%P/P_{hierro}=41.1\%\\\\\%P/P_{arena}=24.4\%\\\\\%P/P_{cobre}=30.8\%\\\\\%P/P_{aluminio}=0.6\%

Explanation:

¡Hola!

En este caso, dado que estamos tratando con problem sobre porcentaje peso/peso de hierro, arena, cobre y aluminio, primero debemos calcular la masa inicial de estos dos primeros en la mezcla original de acuerdo con:

m_{hierro}=300g*0.60=180g\\\\m_{arena}=300*0.40=120g

Ahora si podemos calcular la masa de la mezcla final como la suma de las masas de todos los constituyentes de la mezcla:

m_T=180g+120g+135g+2.77g=437.77g

Finalmente, podemos calcular los porcentajes P/P como se muestra a continuación:

\%P/P_{hierro}=\frac{180g}{437.77g} *100\%=41.1\%\\\\\%P/P_{arena}=\frac{120g}{437.77g} *100\%=24.4\%\\\\\%P/P_{cobre}=\frac{135g}{437.77g} *100\%=30.8\%\\\\\%P/P_{aluminio}=\frac{g}{437.77g} *100\%=0.6\%

¡Saludos!

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