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olganol [36]
2 years ago
15

Calcule la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) que se preparó al mezclar, en un recipiente aforado, 4 mo

les del ácido con suficiente agua hasta completar 8 litros de solución. *
0.4
0.6
0.3
0.5
Chemistry
1 answer:
oee [108]2 years ago
8 0

Considerando la definición de molaridad, la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) es 0.5 \frac{moles}{litro}.

La molaridad es una medida de la concentración de un soluto en una disolución que se define como el número de moles de soluto que están disueltos en un determinado volumen.

La molaridad de una solución se calcula dividiendo los moles del soluto por el volumen de la solución:

Molaridad=\frac{numero de moles de soluto}{volumen}

La Molaridad se expresa en las unidades \frac{moles}{litro}.

En este caso, sabes que una solución acuosa se preparó al mezclar 4 moles del ácido con suficiente agua hasta completar 8 litros de solución. Entonces, sabes que:

  • número de moles de soluto= 4 moles
  • volumen= 8 litros

Reemplazando en la definición de molaridad:

Molaridad=\frac{4 moles}{8 litros}

Resolviendo:

Molaridad= 0.5 \frac{moles}{litro}

Finalmente, la molaridad de una solución acuosa de ácido sulfúrico (H₂SO₄) es 0.5 \frac{moles}{litro}.

<em>Aprende más</em>:

  • brainly.com/question/17647411?referrer=searchResults
  • brainly.com/question/21276846?referrer=searchResults

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Answer:

The answer to your question is 576 grams of FeO

Explanation:

Balanced chemical reaction

                      2Fe  +  O₂  ⇒  2 FeO

       Reactant             Elements         Products

             2                         Fe                     2

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1.- Convert moles of Fe to grams

Atomic mass = 56 g

                         56 g of Fe ------------------ 1 mol

                           x               ------------------ 8 moles

                           x = (8 x 56) / 1

                           x = 448 g of Fe

2.- Use the coefficients of the balanced reaction to find the grams of FeO.

Molecular mass of FeO = 2[ 56 + 16]

                                       = 144 g

               2(56) g of Fe ----------------------- 144 g of FeO

               448 g             -----------------------  x

                x = (448 x 144) / 2(56)

                x = 64512 / 112

               x = 576 grams of FeO

                       

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