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Bavarian

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Posted (edited) · Report post

Another way to visualize it

Glass A | Glass B

start 10cl 100% A | 10cl 100% B

1st pour 7cl 100%A | 10cl 100%B + 3cl 100%A = (13cl 23%A 77%B)

2nd pour 7cl 100%A + 3cl (23%A 77%B) | 10cl (23%A 77%B)

7cl A + 0.7cl A + 2.3cl B

7.7cl A + 2.3cl B

10cl (77% A 23%B) same percentages but opposite liquids

It doesn't matter what the amount is you pour as long as each pour is the same. Keep in mind this doesn't make the glasses contain the same solution it only give equal parts of the opposite liquid in each one.

Edited by Branamal
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Posted (edited) · Report post

being a chemist when you say how much I think concentration, Assuming even concentrations, the math does work out

you wind up with 0.2307692 the original concentration of tonic in the fernet with the fernet maintaining 0.7692307 times its original concentration. the exact reverse is true for the tonic. if you want to check it for yourself the only equation you need is: concentration 1 * volume 1 = concentration 2 * volume 2 where concentration 1 is your initial concentration and volume 1 is the volume of the amount you pour out then concentration 2 is your unknown and volume 2 is the total volume in the final glass. give it a try.

Edited by NeilsBohr
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Posted · Report post

:( boohoo poor me from down under didnt even understand the question ...........what is cl anyways????
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Posted · Report post

:( boohoo poor me from down under didnt even understand the question ...........what is cl anyways????

cl = centilitre

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Posted · Report post

First to anyone applying the C1V1 = C2V2 formula, it doesn't work - you're not only changing V2, but also C2 (i.e. I just converted everything to Molarity because I like working with moles :) - 10 moles/1L = x moles/1.3L is the natural route to go, BUT if you solve for unknown concentration "X", you're neglecting the amount of tonic poured into the fernet).

It's been too long since General Chemistry for me (ok, so maybe the truth is I've just killed too many brain cells), so I can't work this out using solubility rules. HOWEVER, using logic you pour tonic into fernet. You then pour the same volume of fernet solution back into the tonic, but this solution also contains tonic. You are therefore removing some of the original tonic poured into the fernet (while at same time pouring same amount of fernet into tonic as tonic originally poured into fernet).

That being case, there is more fernet in tonic :)

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Posted · Report post

Oops - that's supposed to be 10 moles * 1 L = x moles * 1.3 L :)

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Posted · Report post

Once again I crash headlong into a problem with math without taking a step back to see what logic is at the root of the puzzle.

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Posted · Report post

Bravo largeneal! I was thinking the same thing but kept second guessing due to all of the complicated mathematical solutions. :)

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Posted · Report post

First to anyone applying the C1V1 = C2V2 formula, it doesn't work - you're not only changing V2, but also C2 (i.e. I just converted everything to Molarity because I like working with moles :) - 10 moles/1L = x moles/1.3L is the natural route to go, BUT if you solve for unknown concentration "X", you're neglecting the amount of tonic poured into the fernet).

It's been too long since General Chemistry for me (ok, so maybe the truth is I've just killed too many brain cells), so I can't work this out using solubility rules. HOWEVER, using logic you pour tonic into fernet. You then pour the same volume of fernet solution back into the tonic, but this solution also contains tonic. You are therefore removing some of the original tonic poured into the fernet (while at same time pouring same amount of fernet into tonic as tonic originally poured into fernet).

That being case, there is more fernet in tonic :)

Are you saying there is more fernet in tonic than tonic in fernet?

If so....

Try this.

Try removing an amount of tonic from the glass of tonic, then try to remove and amount of fernet from the glass of fernet, such that: when you pour the removed tonic into the original fernet and the removed fernet into the original tonic, each glass contains 10cl and the original amount of removed tonic does not equal the original amount of removed fernet.

It cannot be done because the total amount of fernet in the problem must be equal to the total amount of tonic at all times.

Therefore, if there is more fernet in tonic than tonic in fernet; than the total amount of fernet would somehow be greater than the total amount of tonic. This is not possible, so the ratio must remain the same.

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Posted · Report post

a sponge or a towel

Actually I found Fernet to be more.

If you take 3 cl of tonic and put it into Fernet, spread out 3cl evenly in the fernet. Meaning distribute 3cl to every 1.

1-1-1-1-1-1-1-1-1-1

,3-.3-.3-.3-.3-.3-.3-.3-.3-.3

Ratio is 1:0.3

Now take 3cl out, thats ( After a lot of math) 2.25 Fernet and .75 (,728934) And mix it back in.

Therfore you have 2.25 Fernet into Tonic

Add up the remaining dispersed tonic thats in Fernet and its 2.1

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Posted · Report post

The answer to this puzzle is wrong. You can try it at home by simply dying water and going from there.

First off, you need to understand that in mixtures, particles disperse evenly, but it does not mean that they become 5/5 or te liesk unless even amounts of both substances are mixed.

At first the glasses are both 100 percent After the 3cl of tonic is poured into the fernet we have in Tonic1 a ratio of 7 parts tonic out of 7 parts total. In the second glass we get a ratio of 3 parts tonic of 13 particles altogether and 10 parts fernet out of 13 particles altogether. That is a ratio of 3 parts tonic to 10 parts fernet. Meaning that every molecule is now 3 atoms of tonic and 10 atoms of fernet. The ratios are NOT equal by any means. It also means that the 3cl of your newly formed mixture that you pour back into the glass of 7 parts tonic to 7 parts total is ALSO a ratio of 3 parts tonic to 10 parts fernet.

You add the 3/10 ratio of tonic/fernet to the 7/0 ratio of tonic/fernet and you a ration of 10 to 10 for Tonic1.

But in your second glass you still have a ratio of 3/10 making them uneven.

If you have any feedback I would greatly appreciate it.

Thank you.

Jenna

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Posted · Report post

The answer to this puzzle is wrong.

Nope.

Meaning that every molecule is now 3 atoms of tonic and 10 atoms of fernet.

No, you can't make new molecules that way.

But in your second glass you still have a ratio of 3/10 making them uneven.

If you have any feedback I would greatly appreciate it.

Thank you.

Jenna

Let's do this step by step.

10 cl of tonic in one glass (Glass 1) and 10 cl of fernet in the other (Glass 2).

Pour 3 cl of tonic to the glass with fernet and mix thoroughly. Glass 1 now has 7 cl of tonic and Glass 2 has 10 cl of fernet and 3 cl of tonic. Glass 2 has 3.33... times more fernet than tonic.

If we remove 3 cl of the mixture from Glass 2, about 2.3 cl of what we removed will be fernet and about .7 cl tonic. That means that Glass 2 now contains about 7.7 cl of fernet and 2.3 cl of tonic.

Glass 1 has 7 cl of tonic. If we add that 2.3 cl of fernet and .7 cl of tonic in to Glass 1, Glass 1 will now contain 2.3 cl of fernet and 7.7 cl of tonic.

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Posted · Report post

The math is interesting; when you place the the 3 cl of tonic into the glass of fernet the glass now contains 3/13 parts tonic. Assuming that the tonic is mixed evenly throughout, when you take the 3cl the second time, 10/13 of it is fernent, which is approximatly 2.3 cl. The remaining tonic in the fernent is 10/13 of the 10 cl, which is approximatly 2.3 cl as well.

10/13 of 10 cl is 2.3 cl?

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Posted · Report post

10/13 of 10 cl is 2.3 cl?

"when you take the 3cl the second time, 10/13 of it is fernent, which is approximately 2.3 cl"

10/13 of 3 is about 2.3.

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