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dsp73
3 years ago
9

Calcula el %m/v de alcohol en una mezcla utilizada para la desinfección de manos formada por: 15 ml de agua (densidad=1g/ml), 10

5 g de etanol (densidad: 0,798 g/ml) y 4,5 gramos de jabón líquido (densidad= 1,5 g/ml)
Chemistry
1 answer:
mr Goodwill [35]3 years ago
3 0

Answer:

%m/v =70%

Explanation:

El %m/v es una unidad de concentración que se define como cien veces la división entre la masa de una sustancia (En gramos) y el volumen total en el que esta sustancia se encuentra (en mL).

En el problema, debemos hallar la masa de etanol (Alcohol) y el volumen total de la solución.

<em>Masa alcohol: </em>

Ya te la dan en el problema: 105g

<em>Volumen solución:</em>

Volumen agua: 15mL

Volumen etanol: 105g × (1mL / 0.798g) = 131mL

Volumen Jabón líquido: 4.5g × (1mL / 1.5g) = 3mL

Volumen: 15mL + 131mL + 3mL

149mL

Así, el %m/v de alcohol en la solución es:

%m/v = (105g / 149mL) × 100

<h3>%m/v =70%</h3>

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The question is incomplete. Her eis the complete question.

Steam reforming methane  (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

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K_{c} = 2.10^{-2}

The concentration equilibrium constant for the process is K_{c} = 2.10^{-2}.

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