what volume of iron would be needed to balance a 1.13cm3 sample of lead?

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stephen asked the Naked Scientists:
What volume of iron would be needed to balance a 1.13 cm3 sample of lead on a two-pan balance?

What do you think?
« Last Edit: 11/10/2010 13:30:03 by _system »


Offline Bill.D.Katt.

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Find the density of lead, and the density of iron, and compute. Wikipedia is great for this. I feel like this was a homework problem.


Offline imatfaal

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This does read like homework - so here is the method, if not the answer. 

First you need to find the density of both iron and lead.  You can find these on wikipedia (if you are allowed to use this as a reference) otherwise maybe try your library's encyclopedia.  These density figures should have units of grammes per centimetre cubed or g/cm3 (sometimes written as g.cm-3; its the same thing).

m - Mass. V - Volume . ρ (thats a greek letter rho) - Density.  And, of course, Pb is lead and Fe is iron.

Density equals mass divided by volume ρ = m / V
so mass equals volume times density  m = V.ρ

You need your masses of lead and iron to be the same to balance the scales so we can say that

Volume(lead) times Density (lead) equals Volumes (iron) times density (iron)   V(Pb)(Pb) = V(Fe)(Fe)

We can rearrange that to  (V(Pb)(Pb)) / ρ(Fe) = V(Fe).

I hope I haven't made a mistake ( I am sure it will be pointed out very quickly if I have).  Remember to check your units and then re-check your units; ie make sure you are dealing with same grammes and cm3 all the way through!  Good luck. 
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