gold content in plated wire
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You need to know the thickness of the wire. From that you can calculate the surface area and then just multiply it with the thickness of the plating and density of gold and you will have the amount of gold.
Göran
Göran
[stt]12,138g[/stt]
Wrong! See my post below
Wrong! See my post below
Barren Realms 007
In Remembrance
Sounds like the perfect reason to build a large sulphuric cell out of plate steel.
Or better yet cut the material up and sell some of it on ebay in lot's.
I've got a set of cutters that would make short work out of that lot in a short amount of time. 8)
Or better yet cut the material up and sell some of it on ebay in lot's.
I've got a set of cutters that would make short work out of that lot in a short amount of time. 8)
A
Anonymous
Guest
It would be realistically more profitable to send it to someone to leech- even if you paid 25% of the gold value you would be better off overall when your time and materials are taken into account.
I think all these answers are wrong, including mine. I think the correct answer is 12.33 grams.
I converted everything to inches for my covenience:
1mm = 1/25.4 = .03937" dia. of wire
Imagine you are able to peel all the gold off in one piece. The dimensions of this sheet would be:
Width = .03937 X pi = .12368"
Thickness = .2um = .000040 X .2 = .000008"
Length = 1000 X 39.37 = 39,370"
Therefore, total cubic inches of Au = .12368 X .000008 X 39370 = .038954 in3
There are 10.18 tr.oz. of Au in one cublc inch and 31.1g per tr.oz.
Therefore, one cubic inch of gold = 10.18 tr.oz. X 31.1 = 316 6g
Therefore, the total Au = .038954 X 316.6 = 12.33 g
I converted everything to inches for my covenience:
1mm = 1/25.4 = .03937" dia. of wire
Imagine you are able to peel all the gold off in one piece. The dimensions of this sheet would be:
Width = .03937 X pi = .12368"
Thickness = .2um = .000040 X .2 = .000008"
Length = 1000 X 39.37 = 39,370"
Therefore, total cubic inches of Au = .12368 X .000008 X 39370 = .038954 in3
There are 10.18 tr.oz. of Au in one cublc inch and 31.1g per tr.oz.
Therefore, one cubic inch of gold = 10.18 tr.oz. X 31.1 = 316 6g
Therefore, the total Au = .038954 X 316.6 = 12.33 g
jason_recliner
Well-known member
All the different answers just depend on the precision of the input numbers. I can see one answer where pi = 3.1 and when I was working in magnetic electronics, pi was 3!
Using 4 digit precision of pi = 3.141 and density of gold = 19.32 is probably already more precise than the material will be accurate.
The microns and metres cancel each other out. 0.2 x 3.14 x 19.32 = 12.136824. Or if you prefer, 12.
Using 4 digit precision of pi = 3.141 and density of gold = 19.32 is probably already more precise than the material will be accurate.
The microns and metres cancel each other out. 0.2 x 3.14 x 19.32 = 12.136824. Or if you prefer, 12.
jason_recliner
Well-known member
Eamonn, just that the units are 1000 metres length, millimetres diameter and microns thickness.
1000 x 1000 / 1000000. That's how it works in my head. Everyone has their own easiest methods. High school was too long ago to remember what 'cancels out' is really called.
1000 x 1000 / 1000000. That's how it works in my head. Everyone has their own easiest methods. High school was too long ago to remember what 'cancels out' is really called.
Using metric, I would first convert all to cm
Dia = 1mm = .1cm; circumference = pi X Dia = .31416cm
1000m = 100,000cm
.2um = .0000002m = .00002cm
Therefore, there are .31416 x 100000 x .00002 = .628cm3 of gold
The density of Au is 19.3 g/cm3
Therefore, the weight of Au is .628 X 19.3 = 12.1g
_____________________________________________
Dia = 1mm = .1cm; circumference = pi X Dia = .31416cm
1000m = 100,000cm
.2um = .0000002m = .00002cm
Therefore, there are .31416 x 100000 x .00002 = .628cm3 of gold
The density of Au is 19.3 g/cm3
Therefore, the weight of Au is .628 X 19.3 = 12.1g
_____________________________________________
That's what I said in the beginning. :mrgreen:
Göran
Göran
Significant Figuresgoldsilverpro said:
If you really want to be technically "correct" in your answer, you must know the number of "significant figures" in the numbers used to compute this answer. If you've ever had a college physics or chemistry lab, this was most probably taught in the first session. If you know the whole truth about this, I have included links for a couple of very good articles on this subject.
In general, there are 2 types of significant figures, those in the numbers resulting from a count and those resulting from a measurement. In essence, the only ones that are really significant in your answer are those derived from a measurement. Every measurement contains error and that must be reflected in answer. If you have a count of 113 chickens, you don't have 112.7 or 113.1 chickens. Your answer can have no more significant figures than the number with the least significant figures used to compute your answer.
For example, let's say I want to know the circumference of a steel rod, where the circumference is defined as the diameter multiplied by pi. On my calculator, pi is built in to 10 digits - 3.141592654. I measure the diameter of the rod with a digital micrometer that is, say, accurate to the nearest 1/1000". Let's say the rod measures 1.051". I enter pi on my calculator and multiply it times 1.051. The answer given is 3.301813879. Is this correct? The answer is no. The correct answer is 3.302. This has 4 significant figures (count them), the same number of significant figures that is in the number in the calculation with the lowest amount of significant figures - 1.051. Note that the number 3.302 is 3.3018 rounded off. The number written as 3.302 always means a number somewhere between 3.3016 and 3.3024.
In the wire problem, all 3 factors have different significant figures, mainly due to the tolerances inherent in their manufacture. The one with the biggest tolerance, percentage wise, is the plating thickness. It's not easy to hold plating to an exact thickness. Therefore, the best answer is probably 12g and not 12.3 or 12.1. It might even be 10g (with the 2 rounded off), if the plating thickness in reality has only 1 significant figure.
http://www.chemteam.info/SigFigs/SigFigRules.html
https://www.hccfl.edu/media/43516/sigfigs.pdf
A
Anonymous
Guest
g_axelsson said:
Hey even a broken clock is right twice a day mate. 8) 8)
can't you just drill a hole in some tool steel a hair smaller than the gold and pull all the wire through it, peeling the gold and some of the substrate off. Then simmer that lot of material in nitric and melt it?
Almost like a copper wire stripper, except we're stripping some gold and a little in copper.
Lou
Almost like a copper wire stripper, except we're stripping some gold and a little in copper.
Lou
Good thought experiment but it sure doesn't sound feasible to me. Might polish up the gold though. When they run bonding wire through a series of progressively smaller dies immersed in soapy water, none of the gold comes off, but the wire gets stretched. A troy ounce becomes 2-4 miles long, depending on the final diameter.Lou said:
Lou that's how gold filled is made or as we call in the UK rolled gold, it will not strip the metals only reduce the diameter, a little of the Au may be left behind but only a very very small amount.
I think this part of the science is physics which leaves me totally baffled, hell I struggle with chemistry :shock:
I think this part of the science is physics which leaves me totally baffled, hell I struggle with chemistry :shock:
