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katrin [286]
3 years ago
14

Si una solucion 0.0150 M tiene un porcentaje de transmitancia de 35% a una determinada longitud de onda, cual sera el porcentaje

de transmitancia para una solucion 0.0300M de la misma muestra?
Chemistry
1 answer:
tangare [24]3 years ago
4 0

Answer:

17.5% es la transmitancia de la solución problema

Explanation:

Basados en la ley de Beer que dice que:

La transmitancia de una solución es inversamente proporcional a su concentración.

Podemos hallar la transmitancia de una concentración problema sabiendo la concentración de una solución y su transmitancia:

Si una solución 0.0150M tiene una transmitancia de 35%

Y ahora la solución duplica su concentración, su transmitancia se disminuirá en la mitad, esto es:

35% / 2 =

<h3>17.5% es la transmitancia de la solución problema</h3>
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If calcium carbonate (caco3) decomposes, what would the product of the reaction be? caco3 → co2
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If calcium carbonate decomposes, the product of the reaction would be calcium oxide and carbon dioxide.

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CaCO₃ is calcium carbonate.

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Chemical decomposition is the separation of a single chemical compound (in this example calcium carbonate) into its two or more simpler compounds (in this example calcium oxide and carbon dioxide).

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A solution is made by dissolving 4.35 g of glucose (C6H1206) in 25.0 mL of water at 25 °C.Calculate the molality of glucose in t
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The molality is unchanged (0.96 molal)

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<u>Step 1: </u>Data given

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volume of water = 25.0 mL

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Molar mass of glucose = 180.156 g/mol

<u>Step 2:</u> Calculate number of moles

moles of glucose = mass of glucose / Molar mass of glucose

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<u>Step 3:</u> Calculate mass of water

mass = density * volume

mass of water = 1.00 g/mL * 25.0 mL

mass of water = 25 g = 0.025 kg

<u>Step 4</u>: Calculate molality

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molality = <u>0.96 molal</u>

Suppose you take a solution and add more solvent, so that the original mass of solvent is doubled.

This means double mass of water = 2*0.025 kg = 0.050 kg

Now molality is 0.024 moles / 0.050 kg = 0.48 molal

When the mass of solvent is doubled, the molality is halved from 0.96 molal to <u>0.48 molal</u>

You take this newsolution and add more solute, so that the original mass of the solute is doubled.

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