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laila [671]
3 years ago
10

After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of

38 percent. The correct value for the percentage of water in the hydrate is 51 percent. Which of the following is the most likely explanation for this difference.A. Strong initial heating caused some of the hydrate sample to spatter out of the crucibleB. The dehydrated sample absorbed moisture after heatingC. The amount of the hydrate sample used was too smallD. The crucible was not heated to constant mass before useE. Excess heating caused the dehydration sample to decompose
Chemistry
1 answer:
zzz [600]3 years ago
4 0

Answer:

A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible.

Explanation:

Hi

The percentage of water in the sample is lower than expected.

A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible:

If part of the sample is splashed from the crucible the mass of water detected will be less.

B. The dehydrated sample absorbed moisture after heating:

If the sample absorbs water after heating the percentage of water would be higher than expected.

C. The amount of the hydrate sample used was too small:

Depending on the sample size, different procedures can be chosen for analysis.

D. The crucible was not heated to constant mass before use:

In many occasions the crucible is heated next to the sample and not in previous form.

E. Excess heating caused the dehydration sample to decompose:

If the sample decomposes during heating, the analysis should be discarded.

success with your homework

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drek231 [11]

Answer: La piedra tarda 2.47 segundos en caer al suelo, y cae a 29.64 metros de la base del edificio.

Explanation:

Ok, tenemos 2 sistemas de ecuaciones para este problema.

Primero, el vertical:

La aceleración es la aceleración gravitatoria, entonces:

a = -9.8m/s^2

para la velocidad podemos integrar sobre el tiempo y obtenemos

v = (-9.8m/s^2)*t + v0

donde v0 es la velocidad inicial, pero la velocidad inicial es solo horizontal, entonces v0 = 0.

Para la posición integramos de vuelta:

p = (1/2)*(-9.8m/s^2)*t^2 + p0

donde p0 es la posición inicial, en este caso, 30m

p = (-4.9m/s^2)*t^2 + 30m

la piedra va a llegar al piso cuando la posición vertical sea igual a 0m

p = 0m =  (-4.9m/s^2)*t^2 + 30m

de aca podemos despejar el tiempo que la piedra tarda en llegar al suelo.

t = √(30/4.9) segundos = 2.47 s.

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Aceleración nula, pues no hay ninguna fuerza actuando en la dirección horizontal.

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v = 0*t + v0 = v0

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v = 12m/s

para la posición integramos de vuelta:

p = 12m/s*t + p0

en este caso podemos asumir que estamos inicialmente en el punto x = x0, asi que la posición inicial es p0 = x0.

p = 12m/s*t + x0

entonces, si queremos calcular la distancia entre la base del edificio y el punto donde cae la piedra, tenemos que calcular:

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D = 12m/s*2.47s + x0 - (12m/s*0s + x0)

D = 29.64m

La piedra tarda 2.47 segundos en caer al suelo, y cae a 29.64 metros de la base del edificio.

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