[Eclipse]

So I was set this problem for homework, and I thought I might as well share it since it's moderately interesting.

You're tasked with coming up with an experiment to determine the number of atoms in your signature. The catch is that you can only use a top balance with a resolution of two decimal places, and you must assume that the pencil that you sign with writes in pure graphite. Feel free to attempt it.

Spoiler: My Solution

The first thing to note is that the differences in the number of atoms in different signatures, when using a top balance that has a resolution of two decimal places, is negligible. We can assume that all signatures have approximately the same number of atoms. There is a way of making the average more reliable but I'll come to that later.

The task is to find the number of atoms in a signature. We are to assume that the pencil we're signing with writes in pure graphite. Now, our scale only reads to two decimal places. That's obviously not accurate enough to measure the mass of a pencil mark of a single signature. The obvious solution is to do more signatures; say a hundred of them. Then we divide the measurement of the mass by a hundred to get an estimate of the mass of a single signature.

I did mention that there are negligible differences in the signatures of different people. Some might have signatures of slightly smaller or larger masses. The solution to this is to get a hundred different people to sign their name on the paper. This way, you can get the average mass of the signatures of a hundred people. Or you could get fifty people to sign their name twice, which would be easier to do, but a less accurate and varied average. But, the differences would be negligible, and not detectable by the top balance that we'd be using, so it doesn't really matter. I can just sign my own name a hundred times.

First of all you should calibrate the top balance, and then take an accurate measurement of the mass. All masses should be recorded in the same place with the same conditions. Then, after the signatures have been done on the paper, the mass should recorded again. The mass of the graphite on the paper will be the second mass measurement minus the first one. This value would then be divided by a hundred to get the mass of a single signature.

Let's say that the mass of a hundred signatures came out to be 0.02g. The mass of a single signature is then 0.02g / 100 = 2 x 10-4g. There are 6.02214086 x 10^23 atoms in 12g of carbon. To get the number of atoms in 2 x 10-4g of carbon, you'd divide that mass by 12g, and then multiply it by 6.02214086 x 10^23. This gives us an answer of 1.003690143 x 10^19. That's the number of atoms in my signature. That's it.

However, there's another problem we haven't addressed. Percentage inaccuracies. The top balance that we are using only measures to two decimal places. This gives us an error margin of +/- 0.005g. We took two measurements so we do 2 x (0.005g / 0.02g) x 100 = 50% error. That's quite a large error margin, but it's to be expected for a measurement so precise.

This percentage inaccuracy gives us a +/- range of the actual answer. We take our answer and multiply it by 0.5 and 1.5 to get the lower and upper ranges, respectively. This gives us:

5.018450715 x 10^18 to 1.505535215 x 10^19

Again, this range used example data to achieve a numerical value. The actual answer would have to be determined by actually carrying out the above method.

[Rick]

Take two identical pieces of paper.

Get 50 signatures on the pieces of paper.

Inside an airtight system burn the plain paper and the paper with signatures and record the amount of additional gas produced (I.e by using a gas syringe) once the system has cooled.

Take the difference in gas between the two if dm^3 and divide it by 24.

Multiply this by Avogadro's constant and ayy you got da mols.




Obviously I wanted to come up with a different method to you (Yours is clearly the best possible) and I'm not even sure if this one would work, given that O2 gas should be being used up at the same rate as CO2 is produced. But hey.

[meow]

Take two identical pieces of paper.

Get 50 signatures on the pieces of paper.

Inside an airtight system burn the plain paper and the paper with signatures and record the amount of additional gas produced (I.e by using a gas syringe) once the system has cooled.

Take the difference in gas between the two if dm^3 and divide it by 24.

Multiply this by Avogadro's constant and ayy you got da mols.




Obviously I wanted to come up with a different method to you (Yours is clearly the best possible) and I'm not even sure if this one would work, given that O2 gas should be being used up at the same rate as CO2 is produced. But hey.

Are you done grave-digging these fucking retarded threads yet?

[Rick]

Are you done grave-digging these fucking retarded threads yet?

no

barely even grave digging if it's only a couple of threads down in this sub forum because the whole thing is fucking deceased

[Skullmeat]

Grave digging. Please do not reply to threads from users who are banned. Locked.

[Eclipse]

Are you done grave-digging these fucking retarded threads yet?

Grave digging. Please do not reply to threads from users who are banned. Locked.

Jesus christ. It's a nice post, and discussion can carry on without the OP. Sort your shit out.

Also, this.

[meow]

Jesus christ. It's a nice post, and discussion can carry on without the OP. Sort your shit out.

Also, this.

Stop gravedigging this retarded thread

[Eclipse]

Take two identical pieces of paper.

Get 50 signatures on the pieces of paper.

Inside an airtight system burn the plain paper and the paper with signatures and record the amount of additional gas produced (I.e by using a gas syringe) once the system has cooled.

Take the difference in gas between the two if dm^3 and divide it by 24.

Multiply this by Avogadro's constant and ayy you got da mols.




Obviously I wanted to come up with a different method to you (Yours is clearly the best possible) and I'm not even sure if this one would work, given that O2 gas should be being used up at the same rate as CO2 is produced. But hey.

Assuming the paper burns cleanly (You'd need a supply of oxygen), H2O and CO2 would be formed. If there was carbon on the paper, there would be additional CO2 but a very tiny amount. I'd say almost negligible. You'd need a massive amount of signatures to maybe get the result within a few orders of magnitude, and not to mention, you'd need really accurate measuring instruments.

About your last point, since CO2 is heavier than O2, it should sink to the ground, and then a flame test or the lime-water test could be used to differentiate between the CO2 and O2.

Stop gravedigging this retarded thread

I'm interested to hear what classes as a non-retarded thread to you? Calling yourself a shit-poster and telling everyone you're a tranny? Fuck off.

[pvnk]

I like this thread more than I like Skullmeat's thirst for moderation. That is my opinion.

[lux]

some magic shit with electric current