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Kryger [21]
3 years ago
7

During an experiment, it is observed that iron powder sinks in water and sulfur powder floats in water. What will most likely be

observed when sulfur powder and iron powder are mixed together and then added to water?
Sulfur will float and iron will sink because each will retain its properties.
Iron will float and sulfur will sink because each will lose its properties.
Both will float because they are physically mixed.
Both will float because they react chemically.

someone pls help me, i literally don't understand chem
Ill give 50 points
Chemistry
2 answers:
lutik1710 [3]3 years ago
7 0

Sulfur will float, and iron will sink because each will retain its properties.

“Iron will float, and sulfur will sink because each will lose its properties” is <em>incorrect</em>. A substance will lose its properties only if it reacts to form a new substance.

“Both will float because they are physically mixed” is <em>incorrect</em>. The substances in a mixture retain their properties.

“Both will float because they react chemically” is <em>incorrect</em>. Iron and sulfur do not react at room temperature. Even if they did, the iron sulfide would sink.

irga5000 [103]3 years ago
6 0

Answer:

Sulfur will float and iron will sink because each will retain its properties.

Explanation:

This is the correct answer because sulfur powder and iron powder are two different things, with different properties. Even if they are added to water and mixed around, they still retain their properties since no actual chemical activity was performed on them (they were just mixed by physical means). Hence, the sulfur will float back up, as it does initially, and the iron will sink, as it had initially.

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A chemist wants to extract a solute from 100 mL of water using only 300 mL of ether. The partition coefficient between ether and
devlian [24]

Answer:

a. X = 0,909

b. X = 0,965

c X = 0,997

Explanation:

The partition coefficient (k) is defined as:

k = Solute in ether / solute in water

a. 3,34 = \frac{\frac{X}{300mL} }{\frac{1-X}{100mL} }

Where X is the fraction of solute extracted

3,34 = X / 3-3X

10,02-10,02X = X

10,02 = 11,02X

<em>X = 0,909</em>

b. First extraction:

3,34 = \frac{\frac{X}{100mL} }{\frac{1-X}{100mL} }

3,34 = X / 1-X

3,34 - 3,34X = X

3,34 = 4,34X

<em>X = 0,770</em>

That means solute in water will be: 1-0,770 = 0,23

Second extraction:

The second extraction will extract the same fraction of solute, as now you have 0,230 of solute in water you will extract:

0,230×0,770 = <em>0,177</em>

Third extraction:

In the same way, the third extraction will extract:

(0,230-0,177)×0,770 = <em>0,018</em>

Fraction in water×Fraction extracted

That means total solute extracted is:

0,770 + 0,177 + 0,018 = <em>0,965</em>

c. Extracting with 50mL  of ether:

First extraction

3,34 = \frac{\frac{X}{50mL} }{\frac{1-X}{100mL} }

3,34 = 2X / 1-X

3,34 - 3,34X = 2X

3,34 = 5,34X

<em>X = 0,625</em>

Second extraction:

(1-0,625)×0,625= <em>0,234</em>

Third Extraction:

(1-0,625-0,234)×0,625= <em>0,088</em>

Fourth extraction:

(1-0,625-0,234-0,088)×0,625= <em>0,033</em>

Fifth extraction:

(1-0,625-0,234-0,088-0,033)×0,625= <em>0,013</em>

Sixth extraction:

(1-0,625-0,234-0,088-0,033-0,013)×0,625= <em>0,004</em>

Total extractions gives:

0,625+0,234+0,088+0,033+0,013+0,004 = <em>0,997</em>

I hope it helps!

8 0
4 years ago
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