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Marta_Voda [28]
3 years ago
8

A student claims that if she wanted to make a solution quickly, she should use small pellets instead of powder along with heatin

g and stirring. Do you agree or disagree with the student's claim? Answer using CER format. PLS HELP I’m a little confused and don’t know what to put.
Chemistry
1 answer:
torisob [31]3 years ago
4 0

Answer:

<em>I disagree with the student's claim</em>

<em></em>

Explanation:

<em>From basic chemistry, the rate of chemical reaction increases with an increased surface area.</em>

<em>This simply means that a powdered substance will react faster than a small pellet of the same substance.</em>

Let us break this down further. The pellet has a smaller exposed surface per grain to the solvent, here is how. Imagine the powdered substance bunched up together to make the small pellet, grains at the center of the pellet is blocked by other grains around it, and will take time before it comes in contact with the solvent. For the powdered grains, each grain has a surface independently exposed to the solvent and can easily come in contact with the solvent, taking less time to react with the solvent.

Technically, a powdered substance has more exposed surface area and logically will take a shorter time for the reaction to be complete.

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Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
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Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

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