 Whatever at least you're not using some other types of characters or something, okay? So let's try another one of these average kinetic energies. Let's do Krypton at 28 degrees, okay? Krypton at 28 degrees. What's the average kinetic energy? so KT and this K here is a constant So it's always going to be the same constant and that K is R divided by NA And both of these are going to be given to you And A is Apagodra's number, so 6.0 22 times 10 to the 23rd and in this case we can think of it as what atoms Krypton wanted And R, in this case you're going to be using 10.314 joules per mole And the way to remember to use that R That's because this is the energy R, okay? So if we're looking for energy, it helps you out Okay, so let's try this 1-4 joules per Kelvin and then 1 over NA 2 to 10 to the 23rd of Krypton for one mole So now we can cancel out moles and Kelvin there Okay, and this gives us the units that we've been looking for Joules per atom, okay? So we expect that to be very small though, okay? Because every atom is not going to have very much energy associated with it So let's go ahead and multiply this out Make sense, this answers 6.321 joules per atom So that's how much energy is associated with each atom at 27 degrees Celsius Okay, and like we said, we would expect it to be a very small amount and the interesting thing about this is it doesn't matter Which particle you're talking about like you said earlier, right? Because there's nothing That's associated with Krypton itself within this formula. There's no molar mass or anything like that So What we're really saying is that this was argon helium or anything nitrogen in to or carbon dioxide At 28 degrees Celsius, they would all have per particle the same amount of energy Sorry, sorry, I got so tongue-tied on that one for the last couple times