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3 years ago

Which method would be effective for separating salt and water in a mixture?

1 answer:

6 0

A seperation method that would be most successful woth the separation of a salt and water mixture would be evaporation for these reasons, (1) There are no magnetic elements involved with salt nor with water, (2) Observing colour differences would not have much affect on separating the two, and (3) Chromatography is used to separate two liquids, such as speparating two dyes, not salt and water.

Evaporation:
The water will evaporate, leaving you with salt as a solid.

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3 years ago

If you have 2 Moles of C3H6 and 10 Moles of O2, which reactant is limiting the following reaction: 2C3H6 + 9O2 → 6CO2 + 6H2O?

Hatshy [7]

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The limiting reactant is propene, $C_3H_6$ .

Explanation:

$2C_3H_6 + 9O_2 \rightarrow 6CO_2 + 6H_2O$

Moles of nitrogen propene = 2 mol

Moles of oxygen = 10 mol

According to reaction, 2 moles of propene reacts with 9 moles of oxygen gas, then 2 moles of propene will react with:

$=\frac{9}{2}\times 2mol=9\text{mol of oxygen gas}$

According to the question, we have 10 moles of oxygen gas, which is more than 9 moles of oxygen gas. This indicates that propene is present in a limiting amount hence, it is a limiting reactant.

The limiting reactant is propene, hence the correct answer is the $C_3H_6$ .

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3 years ago

A gas that exerts a pressure of 537 torr in a container with a volume of 5.30 L will exert a pressure of 255 torr when transferr

ArbitrLikvidat [17]

Answer:- The gas needs to be transferred to a container with a volume of 11.2 L.

Solution:- From Boyle's law. "At constant temperature, Volume is inversely proportional to the pressure."

It means, the volume is decreased if the pressure is increased and vice versa.

Here, the Pressure is decreasing from 537 torr to 255 torr. So, the volume must increase and calculated by using the equation:

$P_1V_1=P_2V_2$

Where, $P_1$ is initial pressure and $P_2$ is final pressure. Similarly, $V_1$ is initial volume and $V_2$ is final volume.

Let's plug in the values in the equation:

(537 torr)(5.30 L) = (255 torr)( $V_2$ )

$V_2=(\frac{537 torr*5.30 L}{255 torr})$

$V_2$ = 11.2 L

So, the new volume of the container needs to be 11.2 L.

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3 years ago