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3 years ago

What can you use to filter out dish soap from water

Chemistry

1 answer:

Alinara [238K]3 years ago

7 0

Answer:

One way to remove soap from water is to have it react with other substances. When the reactions occur, a solid called a precipitate is sometimes formed. A precipitate can be filtered out of the water.

Explanation:

1 cup sugar; Final soap suds height; 9.1 cm

2 cup Table Salt; Final soap suds height; 1.2 cm

3 cup Epsom salt; Final soap suds height; 0.2 cm

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How much of a sample remains after three half-lives have occurred? 1/16 of the original sample 1/9 of the original sample 1/8 of

Drupady [299]

For this question, assume that you have 1 compound. This compound is divided in half once, so you are left with 0.5. That 0.5 that remains is divided in half again, this is the second half-life, and you are left with 0.25. The final half life involves dividing 0.25 in half, which means you are left with 0.125. For the answer to make sense, you need to know your conversions between decimals and fractions. To make it simple, if you have 0.125 and you times it by 8, you are left with your initial value of 1. Therefore, after three half-lives, you are left with 1/8th of the compound.

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3 years ago

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Which statement BEST describes the policies of the Freedmen's Bureau? Question 1 options: The Freedmen's Bureau wanted to encour

olga2289 [7]

<h2>ANSWER:</h2>

The Freedman's Bureau wanted to help newly freed slaves participate in American society.

<h2>EXPLANATION:</h2>

The Freedmen's Bureau, formally known as the Bureau of Refugees, Freedmen and Abandoned Lands, was set up in 1865 by Congress to help a large number of previous dark slaves and poor whites in the South in the outcome of the Civil War. The Freedmen's Bureau given nourishment, lodging, and therapeutic guide built up schools and offered lawful help. It additionally endeavored to settle previous slaves ashore appropriated or surrendered amid the war. Be that as it may, the department was kept from completely doing its projects because of a lack of assets and faculty, alongside the legislative issues of race and Reconstruction.

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3 years ago

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Please help I don’t have a lot of time left

Harrizon [31]

Answer:

So sorry if I was wrong but I think it's B. Because from the source states,

https://socratic.org/questions/which-group-on-the-periodic-table-is-the-least-reactive-why

"The least reactive elements are those who have a full outermost valence shell ie they have 8 electrons in the outer shell so elements such as helium, neon, radon or the transition elements."

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3 years ago

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A sample of radium-226 contains 1.0×108 atoms of radium-226. How many atoms of radium-226 will remain in the sample after three

cricket20 [7]

After 3 half lives it will be (1/2)^3 or 1/8 of original

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4 0

3 years ago

compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of

arlik [135]

<u>Answer:</u> The empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

<u>Explanation:</u>

We are given:

Mass of $CO_2=3.926g$

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

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

<u>For Carbon:</u>

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

<u>For Fluorine:</u>

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is $CH_2F$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

3 0

3 years ago