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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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Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon
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<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: Cu_2O\text{ and }CuO

  • <u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = \frac{2.26}{2.26}=1

For Oxygen  = \frac{4.55}{2.26}=2.01\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is CO_2

  • <u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = \frac{3.57}{3.57}=1

For Oxygen  = \frac{3.57}{3.57}=1

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is CO

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = \frac{1}{2}\times 1=\frac{1}{2}

Now, taking the ratio of carbon atoms in both the samples, we get:

C_1:C_2=\frac{1}{2}:1=1:2

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0
3 years ago
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