amphibole

On the electrical resistance of liquid helium M.Wolfke W.H.Keesom Show more https://doi.org/10.1016/S0031-8914(36)80356-5Get rights and content Summary The specific electrical resistance of liquid helium was found from 1.28°K to 4.22°K to be at least 1015 Ω cm. _________________________ From a paper on the electrical conductivity (or lack thereof) of Helium. I expect Argon will behave in much the same way.

What were the tools made of? Were they an alloy of copper and iron? If not, it is highly unlikely (read: impossible) that you would get copper oxide if there was no copper in the tool to begin with. If there was copper present in the mix somewhere, the red color might come from copper (I) oxide, which has a red color. If iron was present in the sample, the orange-red color is almost certainly due to the presence of ferric acetate.

At this point, I already have everything I need. For future batches, I'm just going to stick with heating up and decomposing the copper (II) carbonate. It's a quick and easy process that leaves me with a pretty pure product.

Valid question! I am not implying that I made an oil, but the substance did exhibit some seemingly strange properties in that it did not readily dissolve in the acetone and instead rolled around at the bottom of the beaker in what I thought would best be described as oily-looking beads. Additionally, it did not exhibit any strange behavior until I poured in acetone, not water. It behaved as expected when it was in the aqueous solution. I say that I am certain that I made indium (III) chloride because of a few reasons. 1) I made a solution of copper (II) chloride and displaced the copper from it with indium metal. Simple single displacement, very similar to the classic "copper displaces silver from AgNO3" demonstration. 2 In + 3 CuCl2 -> 3 Cu + 2 InCl3 I was left with a clear liquid and removed the copper metal flakes by decanting the liquid into another clean beaker. This was a pretty quick process, and I was around to watch the whole reaction, so I have no reason to believe that the anion somehow changed. Thus, I have no reason to believe that the anion is not Cl. 2) When I eventually got the solute to crystallize, I performed a flame test on it and the flame was dark blue. Also, I purchased the starting material (indium metal) from a reputable metal dealer. So, I have no reason to believe that the cation is not In. I am doubtful that it is the indium (I) cation, because the resultant crystals are colorless, which is not the case with indium (I) chloride. I am just curious as to why I observed an "oil-like" behavior in this substance when I poured the acetone over it.

Title says it all. I was trying to crystallize some InCl3 and came across something I had never seen before - an inorganic substance 'oiling out'. It was just rolling around in dense oil beads in the acetone at the bottom of my beaker. I evaporated the solvent and washed it with acetone a few more times, and it became more clumpy and granular with each washing. Upon drying, a flame test proved that it had the indium cation in it, so I have little reason to believe that it isn't indium (III) chloride (unless it somehow reacted with the acetone, but I am skeptical of this). Does anyone know what happened to make it behave as an oil? I've never seen that kind of behavior in an inorganic metal salt - plenty of times with organic molecules, but never with a simple salt like InCl3.

Yeah, I actually purchased some from a ceramics supplier, which prompted me to ask this question in the first place. It's beautiful to look at, but I sure wish I could figure out just what's going on!

Yeah, I went to the wiki page for it and that was the part I was confused about. I believe that a phosphor is a term used to describe any material that emits light without heat

Holmium (III) Oxide exhibits some pretty dramatic color changes (see this attached photo), but I am uncertain about the mechanism that causes this change to occur. The literature on it is slim, and what exists on the subject is dense and jargon-y, as well as being behind a paywall that my school doesn't subscribe to. Can anyone here explain this phenomenon to me?

While I'm unsure about the nature of the redox reaction, the balanced stoichiometric equation between Mn and HNO3 should be something like this: 3Mn + 8HNO3 → 3Mn(NO3)2 + 2NO + 4H2O Hopefully that helps.

I think the error was a combination of "not leaching the powder" and "I didn't heat it at a high enough temperature". I ran two more trials today, one following the same steps as before, except I heated it at a greater temperature (approx. 260C), as well as a second run where I repeated the first trial (same temperature as the first go - around 200C), but I leached it with the bicarbonate solution before heating. Both trials resulted in a darker product. Thanks.

I am attempting to prepare some CuO for various experiments and am running into a bit of a sore problem. I have, on hand, an excess of Copper (II) Sulfate pentahydrate and baking soda, and decided to employ them as precursors to CuO. To my best knowledge, the reaction should resemble this: 2(CuSO4.5H2O) + 4NaHCO3 -> Cu2(OH)2CO3 + 2Na2SO4 + 3CO2 + 11H2O After the reaction was complete (i.e. no more fizzing) , I was left with the precipitate of the basic copper carbonate (Cu2(OH)2CO3). I then washed this precipitate several times and then heated it up to decompose it to CuO, H2O and carbon dioxide. I believe the reaction should go: Cu2(OH)2CO3 + heat -> 2CuO + H2O + CO2 However, upon heating, I was left with a powder visually similar to the one in the attached picture. It is a sickly sort of yellowy/grey-ish color instead of the presupposed black. What kind of errors or impurities might I be encountering here? Did some of the Na2SO4 contaminate it? Or am I missing something? Thanks.