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3 years ago

How could you show experimentally that the molecular formula of propene is c3h6, not ch2?

2 answers:

6 0

Answer : By performing the experiment for calculating the elevation in boiling point and finding out the ebuilloscopic constant for the solvent we can find the molecular formula of the unknown solute.

Also by performing the experiment for depression in freezing point one can find the molecular formula for any unknown solute. We can find the cryoscopic constant for the solute and it will help in determining the molecular formula.

katen-ka-za [31]3 years ago

5 0

Answer:

By performing experiments related with the elevation in boiling point and depression in freezing point.

Explanation:

Hello,

In this case, we use the concept of colligative properties of solutions because they depend on the concentration of solute molecules or ions rather than its identity. They are classified as vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure.

Elevation in boiling point and depression in freezing point are suitable for us to determine the molecular formula of propane as shown below:

- Elevation in Boiling Point: by performing an elevation in boiling point experiment, it is possible to quantify ebuilloscopic constant for the solvent, in order words, the molal (referred to molality) boiling point elevation constant of propane, using it as a solute. Then, one applies the formula:

$\Delta T_b=i*Kb*m_{solute}$

To compute the molality and subsequently propane's molecular mass.

- Depression in freezing point: by performing this one, it is possible to determine the molecular mass of the propane. Thus, we find the molal freezing point depression constant of the solvent to subsequently apply the shown below formula:

$\Delta T_f=i*Kf*m_{solute}$

Which is suitable to compute the molality as well as in the previous case.

Best regards.

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Un móvil se mueve con movimiento acelerado. En los segundo 2 y 3 los espacios recorridos son 90 y 120 m, Calcula la velocidad in

faust18 [17]

Answer:

La velocidad inicial es 55 $\frac{m}{s}$ y su aceleración es -10 $\frac{m}{s^{2} }$

Explanation:

Un movimiento es rectilíneo uniformemente variado, cuando la trayectoria del móvil es una línea recta y su velocidad varia la misma cantidad en cada unidad de tiempo . Dicho de otra manera, este movimiento se caracteriza por una trayectoria que es una línea recta y la velocidad cambia su módulo de manera uniforme: aumenta o disminuye en la misma cantidad por cada unidad de tiempo. Y la aceleración es constante y no nula (diferente de cero).

En este caso la posición del objeto esta dada por la expresión:

$x=x0+v0*t+\frac{1}{2} *a*t^{2}$

donde x es la posición del cuerpo en un instante dado, x0 la posición en el instante inicial, v0 la velocidad inicial y a la aceleración.

En este caso, por un lado podes considerar:

x= 90 m
x0= 0 m
v0= ?
t= 2
a= ?

Reemplazando obtenes:

$90=v0*2+\frac{1}{2} *a*2^{2}$

$90=v0*2+\frac{1}{2} *a*4$

$90=v0*2+2*a$

Y por otro lado tenes:

x= 120 m
x0= 0
v0= ?
t= 3
a= ?

Reemplazando obtenes:

$120=v0*3+\frac{1}{2} *a*3^{2}$

$120=v0*3+\frac{1}{2} *a*9$

$120=v0*3+\frac{9}{2} *a$

Por lo que tenes el siguiente sistema de ecuaciones:

$\left \{ {{2*v0+2*a=90} \atop {3*v0+\frac{9}{2} *a=120}} \right.$

Resolviendo por el método de sustitución, que consiste en aislar en una ecuación una de las dos incógnitas para sustituirla en la otra ecuación, obtenes:

Despejando v0 de la primera ecuación:

$v0= \frac{90-2*a}{2}$

Reemplazando en la segunda ecuación:

$120=\frac{90-2*a}{2} *3+\frac{9}{2} *a$

Resolviendo:

$120=(90-2*a)*\frac{3}{2} +\frac{9}{2} *a$

$120=135-3*a +\frac{9}{2} *a$

$120-135=-3*a +\frac{9}{2} *a$

$-15=\frac{3}{2} *a$

$\frac{-15}{\frac{3}{2} } =a$

-10=a

Reemplazando el valor de a en la expresión despejada anteriormente obtenes:

$v0= \frac{90-2*(-10)}{2}$

Resolviendo:

$v0= \frac{90+20}{2}$

$v0= \frac{110}{2}$

v0=55

La velocidad inicial es 55 $\frac{m}{s}$ y su aceleración es -10 $\frac{m}{s^{2} }$

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