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docker41 [41]
3 years ago
11

A homogeneous portion of a mixture that is characterized by uniform properties and capable of being separated by mechanical mean

s is called a _____.
product
phase
reactant
solvent
Chemistry
2 answers:
guapka [62]3 years ago
8 0
The answer that would best complete the given statement above would be the term PHASE. Here is the complete statement. <span>A homogeneous portion of a mixture that is characterized by uniform properties and capable of being separated by mechanical means is called a PHASE. Hope this helps.</span>
kozerog [31]3 years ago
8 0

Answer: The correct answer is phase.

Explanation:

Phase is defined as region of space in which properties of matter are distributed uniformly with distinct boundary due which they are mechanically separable.

In a chemical reaction , when to substances reacts together they are called reactants and new substance formed from the reactions is called products.

Solvent is defined those substance which capable of dissolving other substance in them.

Hence, the correct answer is phase.

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A methanol-water mixture is to be flash distilled at 1 atm. If the feed is 25 mole %methanol, what are the liquid and vapor comp
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Answer:

Explanation:

Given that:

The distillation is carried out at a pressure of 1 atm

The feed harbors 25% mole of methanol (z)

The total moles of feed is usually 100 moles

In the system, we have both methanol and water

Using the total mole balance for the distillation column.

Fz = Lx + Vy

where;

F = amount of feed

z = mole fraction of ethanol (in feed)

L = amount of liquid product out of the column

V = amount of vapor product out of the column

x = mole fraction of methanol out of the liquid

y  = mole fraction of methanol out of the vapor.

SO;

(a)

If all the feed is vaporized, then the vapor will likely have the same composition as the feed.

(b)

If no vaporization of the feed takes place, then the bottoms moving out of the column contains the same composition as the feed.

(c)

If 1/3 of the feed is vaporized; then 2/3 of the feed is at the bottom.

The balance equation would be:

Fz = (\dfrac{2}{3}F) x + (\dfrac{1}{3}F)y \\ \\ z = \dfrac{2}{3}x+\dfrac{1}{3}y

Replacing z = 0.25; we have:

0.25 = \dfrac{2}{3}x+\dfrac{1}{3}y

0.75 = 2x + y

(d)

If 2/3 of the feed is vaporized;

Then:

Fz = (\dfrac{1}{3}F) x + (\dfrac{2}{3}F)y \\ \\ z = \dfrac{1}{3}x+\dfrac{2}{3}y

replacing z = 0.25

0.25 = \dfrac{1}{3}x+\dfrac{2}{3}y

0.75 = x + 2y

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Answer:

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Explanation:

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