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d1i1m1o1n [39]
3 years ago
14

"Para enfriar algo rápidamente se hace una mezcla de hielo con sal o, si tiene precaución, alcohol. El punto de congelación baja

rá y el hielo se derretirá rápidamente. Pese a aparentar haberse perdido el frío, la mezcla formada estará en realidad a unos cuantos grados bajo cero y será mucho más efectiva para enfriar que los cubos de hielo sólidos". El fenómeno relatado corresponde a una consecuencia de:
Chemistry
1 answer:
Marizza181 [45]3 years ago
6 0

La pregunta está incompleta, la pregunta completa es;

"Para enfriar algo rápidamente, haga una mezcla de hielo cd sal o, si

tenga cuidado, alcohol. El punto de congelación bajará y el hielo

derretir rápidamente. A pesar de parecer haber perdido el frío, la mezcla

formado será en realidad unos pocos grados bajo cero y será

mucho más eficaz para enfriar que los cubitos de hielo sólido. "La

fenómeno reportado corresponde a una consecuencia de: *

una. Aumento del punto de ebullición

B. disminución de la presión de vapor

C. aumento de la presión de vapor

D. disminución del punto de congelación

Answer:

D

Explanation:

Esta pregunta recuerda las propiedades coligativas. Las propiedades coligativas son propiedades de sustancias que dependen de la cantidad de sustancia presente.

El punto de congelación es una propiedad coligativa. Se sabe que la presencia de impurezas reduce el punto de fusión y congelación. Esto se llama depresión del punto de congelación.

Por lo tanto, la observación registrada en la pregunta es el resultado de la disminución del punto de congelación.

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Answer:

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Explanation:

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

Sample 1: The 1.15 M sample  has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

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<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1:  1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

<u>Step 3:</u> Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams  ( in 55mL)

<u>Step 4</u>: Calculate moles of AgBr

AgNO3 reacts with KBr  

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

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