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3 years ago

What mixture can be separated by using a filter?

Chemistry

1 answer:

nexus9112 [7]3 years ago

4 0

<h2>Answer: </h2>

Option 2nd - a mixture of solid particles in a liquid.

<h2>Explanation: </h2>

For a Mixture of two immiscible liquids (for example water and oil) it is not possible to filter them with a filter as both of the liquid will pass through it.
For a mixture of a solid and a liquid (for example sand and water) it can be easily separated through a filter as on filtration solid will be left on the filter and the liquid will pass through it.
For a mixture of two miscible liquids (for example milk and water) both of them will form a single identity, and on filtration both of them will pass through the filter. Hence it is not possible to separate them through filter.
For a mixture of liquid and gas (for example drinking soda or Pepsi) filter is not needed for their separation as it will be separated when exposed in air.

Result: Option 2 is the correct answer.

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Acetaminophen is the generic name of the pain reliever in Tylenol and some other headache remedies- The compoundhas the molecula

slega [8]

<u>Answer:</u> The mass percentage of carbon element is 63.57 %, hydrogen element is 5.96 %, nitrogen element is 9.28 % and oxygen element is 21.19 %

<u>Explanation:</u>

We are given:

A chemical compound having chemical formula of $C_8H_9NO_2$ . This compound is made by the combination of carbon, hydrogen, nitrogen and oxygen elements.

1 mole of acetaminophen contains 8 moles of carbon atom, 9 moles of hydrogen atom, 1 mole of nitrogen atom and 2 moles of oxygen atom

Mass of carbon element = $(8\times 12)=96g/mol$

Mass of hydrogen element = $(9\times 1)=9g/mol$

Mass of nitrogen element = $(1\times 14)=14g/mol$

Mass of oxygen element = $(2\times 16)=32g/mol$

Mass of the compound = 151 g/mol

To calculate the percentage composition of an element in a compound, we use the equation:

$\%\text{ composition of an element}=\frac{\text{Mass of element}}{\text{Mass of compound}}\times 100$ ........(1)

<u>For carbon element:</u>

Putting values in equation 1, we get:

$\% \text{ composition of carbon element}=\frac{96}{151}\times 100\\\\\% \text{ composition of carbon element}=63.57\%$

<u>For hydrogen element:</u>

Putting values in equation 1, we get:

$\% \text{ composition of hydrogen element}=\frac{9}{151}\times 100\\\\\% \text{ composition of hydrogen element}=5.96\%$

<u>For nitrogen element:</u>

Putting values in equation 1, we get:

$\% \text{ composition of nitrogen element}=\frac{14}{151}\times 100\\\\\% \text{ composition of nitrogen element}=9.28\%$

<u>For oxygen element:</u>

Putting values in equation 1, we get:

$\% \text{ composition of oxygen element}=\frac{32}{151}\times 100\\\\\% \text{ composition of oxygen element}=21.19\%$

Hence, the mass percentage of carbon element is 63.57 %, hydrogen element is 5.96 %, nitrogen element is 9.28 % and oxygen element is 21.19 %

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3 years ago

A 1.0 M H2S solution has a pH = 3.75 at equilibrium. What is the value of Ka?

MrRa [10]

The answer is 3.16*10-8

7 0

3 years ago

Read 2 more answers

At a certain temperature the vapor pressure of pure benzene is measured to be . Suppose a solution is prepared by mixing of benz

Marianna [84]

Answer:

P(C₆H₆) = 0.2961 atm

Explanation:

I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:

<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>

<em />

Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:

P = Xₐ * P° (1)

Where:

P: Partial pressure

Xₐ: molar fraction

P°: Vapour pressure

We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.

To get the moles: n = m / MM

To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:

MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol

MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol

Let's determine the moles of each compound:

moles (C₆H₆) = 79.2 / 78 = 1.02 moles

moles (C₇H₁₆) = 115 / 100 = 1.15 moles

moles in solution = 1.02 + 1.15 = 2.17 moles

To get the molar fractions, we use the following expression:

Xₐ = moles(C₆H₆) / moles in solution

Xₐ = 1.02 / 2.17 = 0.47

Finally, the partial pressure is:

P(C₆H₆) = 0.47 * 0.63

Hope this helps

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