Wait a second.. You mean to tell me you all haven't heard of the FTV premier edition?? They reprinted the Beta P9, 1996 World Champion, proposal, and a couple of ante cards I forgot. All in FTV foil, without a marker.
Wait a second.. You mean to tell me you all haven't heard of the FTV premier edition?? They reprinted the Beta P9, 1996 World Champion, proposal, and a couple of ante cards I forgot. All in FTV foil, without a marker.
/s
They did not. The one your talking about came with
Wonder how much a foil lotus would even go for if popper sold the only one in the world
6.02X10^23
Nice touch, avagadro...lol
I'd like to pay in cash (For those of you who wonder: it's the amount of C-14 (carbon) molecules in 14 grams of the stuff.)
It's not just restricted to carbon. Avogadro's number is the basis for the mole, which is a standard unit of measurement used for stoichiometry in chemistry. It's hard to explain how big a deal it is without doing a full-blown chemistry lesson, though. Suffice to say, you can find out the volume of a gaseous product a reaction will produce provided, say, the mass of a reactant you have, a balanced chemical equation, and various conversions to moles.
Quote from: Brawler_1337 on September 23, 2015, 01:44:06 PM
Quote from: Kaylesh on September 23, 2015, 01:26:44 PM
Quote from: Double-O-Scotch on September 23, 2015, 12:34:44 PM
Quote from: Splicer on September 22, 2015, 08:37:29 PM
Quote from: General Kiwi on September 22, 2015, 07:26:46 PM
Wonder how much a foil lotus would even go for if popper sold the only one in the world
6.02X10^23
Nice touch, avagadro...lol
I'd like to pay in cash (For those of you who wonder: it's the amount of C-14 (carbon) molecules in 14 grams of the stuff.)
It's not just restricted to carbon. Avogadro's number is the basis for the mole, which is a standard unit of measurement used for stoichiometry in chemistry. It's hard to explain how big a deal it is without doing a full-blown chemistry lesson, though. Suffice to say, you can find out the volume of a gaseous product a reaction will produce provided, say, the mass of a reactant you have, a balanced chemical equation, and various conversions to moles.
True. I admit to being a tad short in my explanation there, giving only the definition.