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Finger [1]
2 years ago
14

In an extraction procedure, it is advisable to: save all layers until the experiment is complete. throw away all layers as soon

as you have extracted them. put the aqueous layer down the drain. put the organic layer in the aqueous waste.
Chemistry
1 answer:
LenKa [72]2 years ago
4 0

All the layers should be kept until the experiment is complete.

<h3>What is extraction?</h3>

In chemistry, solvent extraction is accomplished by adding a sample containing the substance to be separated into a system of two solvents, an aqueous layer and an organic layer.

It is important to note that all the layers should be kept until the experiment is complete. No layer ought to be discarded before the work is completed.

Learn more about solvent extraction:brainly.com/question/11041092

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If you have 10.0 grams of citric acid with enough baking soda (nahco3 how many moles of carbon dioxide can you produce?
kap26 [50]
Easy stoichiometry conversion :)

So, for stoichiometry, we always start with our "given". In this case, it would be the 10.0 grams of NaHCO3. This unit always goes over 1.

So, our first step would look like this:

10.0
------
  1

Next, we need to cancel out grams to get to moles. To do this, we will do grams of citric acid on the BOTTOM of the next step, so it cancels out. This unit in grams will be the mass of NaHCO3, which is 84.007. Then, we will do our unit of moles on top. Since this is unknown, it will be 1.

So, our 2nd step would look like this:

1 mole CO2
-----------------
84.007g NaHCO3

When we put it together: our complete stoichiometry problem would look like this:

10.0g NaHCO3     1mol CO2
---------------------- x -------------------------
            1                  84.007g NaHCO3

Now to find our answer, all we need to do is:
Multiply the two top numbers together (which is 10.0)
Multiply the two bottom numbers together (Which is 84.007)

And then....

Divide the top answer by the bottom answer.

10.0/84.007 is 0.119

So, from 10.0 grams of citric acid, we have 0.119 moles of CO2.

Hope I could help!
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Answer:

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