BenSon

That is one awsome collection how big are those platinum group samples? they must have cost an arm and a leg... ~Scott

I'd think that the carbon did come from the air as CO2 and heres my guess of the equation.... 2Cu + 3H2O + CO2 = Cu2CO3(OH)2 + 2H2 ~Scott

Its also fun to ignite the hydrogen gas with oxygen to make water, don't ignite too much at once before you've done it with small amounts to get the idea of how much it burns. ~Scott

This is probably a stupid question but... I thought there was a certain amount of ionized water in all water OH- and H+ ions, something like 10^-7(?) conc in neutral water. So does de-ionised water have no ions at all or is it just talking about removing all ions that haven't come from the water? ~Scott

Welcome I'm new here myself.... NaCl is hard to electrolyse straight because its so hot when it melts to sodium will evaporate instead of being a liquid. I was thinking of doing this myself with some K/Na salts. Their hydroxides have much lower melting point 380C and 322C respectively and you wont have to deal with Cl2 gas. Your metals will also react readily with the atmosphere more so at those temperatures so an inert atmosphere is a good idea it was sugested to me to use Argon don't know if thats easy for you to get? Anyway i hope that helps good luck to ya ~Scott

I haven't done this for a while but...ravio suggested this already but a titration may be a good idea. I'm not sure if you have the gear for that though anyway here is a pretty through site on it http://www.dartmouth.edu/~chemlab/techniques/titration.html ~Scott

lol...I found their website if you haven't seen the show before then this will give you an idea of the kind of stuff they do. http://www.tvtome.com/MythBusters/guide.html ~Scott

Google mythbusters maybe they'll have a movie of it. Yeah maybe it would be better to use something not full of sugar, btu either may it does work. ~Scott

Your right, dehydrated borax is a chalkey white substance. If you measure how much borax you are going to react before you hydrate it then you won't have to factor the water in. However if you plan to hydrate it then measure it out, then you will have to add the water into the molecular mass. Also make sure not to add more water then enough to hydrate the borax or that could effect the amounts. ~Scott

I think my best option would be to just melt down some hydroxide and run a charge through it.... see what happens. If it works, then worry about gathering it up in oil. ~Scott

You had the way to find the molecular mass right but you said it was borax decahydrate. That is Na2B4O7.10H2O you need to add the weight of 10 H2O's as well then you'll have it right. As far as the Propyl formate thing goes i've been trying to reseach it but i havn't come up with anything yet i could have a stab at the equation but i have a feeling that it would be wrong... H2CO2 + C3H8 = C4H8O2 + H2 however in your secong post you said propanol not propane so that would be H2CO2 + C3H7OH = C4H8O2 + H2O But like i said before these equations could be bogus i can't find any reference to them online so yeah... Anyhow you said you got amber liquid on top of and oily layer well i dont know whats up with that i could find about as much info of the colour of C4H8O2 as i could about making it. sorry i couldn't be of more help... ~Scott

This is going to sound weird, but on Mythbusters they tested a leading brand of chrome cleaner against coke. Coke absolutely killed the chrome cleaner. What they did was pour coke all over a chrome bumper bar and scrub it with Aluminium foil... it worked really well!!! don't believe me? give it a try... ~Scott

I got 382 for the molecular mass?... But everything else you said was right i'll have to think about that separation. ~Scott

Here's the first one you wanted 2Na2B4O7 + 2H2SO4 + 10H2O = 8H3BO3 + 2Na2SO4 I think thats right but ive never done the reaction before. About the ratios you can work that out with this formula... Mass(g) = Number of moles X Molecular mass I'm not sure how much chem you've done before but just write back if theres a problem. As for propyl formate i'm too tired to do both 2nite... ~Scott

Well the NaOH definately sounds like a good option i was originaly going for K but its hydroxide is only 380C as well so either one i go for i'll use the hydroxide. I also wont have to worry about the Cl2 gas now. I'll look into those solvents, and the argon may be a problem but this is sounding less impossible. Thanks for the help.

http://cwx.prenhall.com/bookbind/pubbooks/blb/chapter23/medialib/blb2304.html Man that looks hard to build... ~Scott

That would be sooo fun... but sounds very hard. However... The hardest part to me sounds like keeping it under the argon atmosphere (how expensive is that stuff anyway?) I guess i could just keep a jet of it shooting at the sodium electrode. Then when im done just quickley throw the whole electrode into the oil...Man that sounds dodgy... Have you ever electroliesd alakali salts? ~Scott

I dont see any way i can do this reaction at home withour spending too much money ~Scott

Oh ok also how would you get the NH3 as a liquid at high temps?... Anyway i've done thermite before but i'd think that the high temp would probably cause the K to react violently with the atmosphere even if no water was present. ~Scott

Fair enough...

where do the other metals come into play? sorry. ~Scott

I've used ammonia before but im not sure if the conc was 3% or not. I was thinking maybe collect some gas in a syringe and pressurise it, to react it with the KNO3. Then id only have to use very small amounts cutting down on the amount i breathed in but thats probably to hard... ~Scott

NH3(aq) is a weak base in equilibrium. NH3(aq) + H2O(l) <=> NH4+ + OH- So wouldnt it be fair to say that the reactants are still NH4+ and OH-, even if its not in the form of NH4OH? just curious. NH3(aq) +HCL(g) + H2O = NH4CL + H2O Primarygun- The ionic equation for the reaction between NaOH and HCL is Na+ + OH- + H+ + Cl- = H2O + NaCl Half equations NaOH = Na+ + OH- HCl = H+ + Cl- ~Scott

Ah now that one is easy... HCl(aq) + NaCO3(s) = HCO3(aq) + NaCl(aq) HCl = H+ + Cl- NaCO3= Na+ + CO3- 2HCO3(aq) quickly decomposes into H2O(l) and CO2(g) So it is more like HCl + NaCO3 = H2O + CO2 + NaCl (<= I know its not balanced) ~Scott

I was just thinking that if the HCL is gas, and the NH4OH is aqueous, then wouldn't the HCL have to become aqueous in order to react with the NH4OH? I think the ionic equation would be the same either way unless theres some rule im not remembering. ~Scott