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1

This is not entirely true. Wikipedia lists several possible cathode materials and they are oxides. They may be "metallic" is the sense of being metallic conductors (some oxides are so), but they are ceramic compounds. The anode is typically Li-intercalated carbon (which is also a good conductor). Lithium intercalates readily into carbon, allowing a safer ...


4

While I can't find specific justification for the surface, the first competition involved both gold and silver "racetracks." Drivers gear up for world’s first nanocar race How to build and race a fast nanocar The competition involves propulsion and imaging using STM so the substrate must be conductive. Gold (and silver to a lesser degree) is particularly ...


-2

Contrary to Andselisk's opinion, mercury nitrate Hg(NO3)2 is not decomposed into HgO in water. Mercury nitrate is hydrolyzed into a basic salt, whose formula is Hg(OH)NO3, and the equation is : Hg(NO3)2 + H2O --> HNO3 + Hg(OH)NO3 To get pure mercury from mercuric nitrate, Andselisk recommends heating their substance to 400°C. This technique will ...


4

The process you described would be more appropriately called "reduction of mercury(II) to elemental mercury". Unfortunately, the trick with iron likely won't work (something more inert like copper would be a better choice though). Mercury(II) oxide is weakly basic, so mercury salts in general would easily undergo hydrolysis and form basic oxosalts in ...


0

All ready answered, but I will give the distilled version: Zn and Cd are most commonly found in the +2 OX State (d10) and can not metal-metal bond in this state. +1 OX state is possible for these elements but is rare due to there low ionization energies, but they do form metal-metal bonds in the +1 OX state. Hg has relatively high ionization energy due to ...


-5

It is seen that Au(III) and Au(I) complexes are more stable than Au(II).I have an explanation but that is my observation- Let's consider their electronic configuration. Au → 5d10 6s1 Au+ → 5d10 Au2+ → 5d9 Au3+ → 5d8 So, if we split d-orbital into t2g and eg, we get For Au → For Au+ → For Au2+ → For Au3+ → So, As you can see, Au (I) - ...


-3

I don't think it makes sense to compare Gold with Silver and Copper. In the 6th line of the periodic table, relativistic effects become important. And one of the relativistic effect is a contraction of the lengths. This effect is proportional to the sum of the two first quantum numbers. Gold outer orbitals are 4f 5d and 6s. Its configuration is [Xe] 4f14 ...


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