what is E/AMU, and how to calculate ionization density?
464790
looks like E/AMU is the charge/mass ratio of a proton.  e is usually used to represent the elementary charge (of either proton or electron) and an amu is defined as the mass of a proton.\n\nI'm not sure about the ionization density.  You'd probably want to measure the charge of the material.  Then divide the charge by the volume to get the density.
822591
kinetic energy of a moving object = .5(mass)(velocity^2).\n\nfirst you would want to find the mass of a hydrogen atom. which is 1 amu. which is 1.66053886 × 10-27 kilograms. plug that into the equation and you get (1.66053886 × 10-27)(.5)(3055^2).
750464
My chemistry is pretty rusty, but let's see.\n\nI'll assume that you have an infinite amt of O2, and that you have 68.12 grams of NH3.\n\nThe equation isn't balanced, so you need to do that first (you have 3 hydrogens on one side, but two on the other).  Balance to get:\n\n2(NH3) + (2.5)(O2) --> 2NO + 3(H20)\n\n4(NH3) + 5(O2) --> 4NO + 6(H20)\n\nNow, given that you have 68.12 grams of NH3, you need to convert this into moles of NH3, as Stoichiometry deals with molar ratios.  To get the conversion ratio, look at the periodic table.  Under each element, there should be an atomic mass.  For oxygen is 16.something, Hydrogen 1.008 ( i think ).  Find it for N and H.  (this is where I get hazy on the math).  I believe: take 3 * atomic mass of hydrogen + atomic mass of nitrogen, let's call it M  (it should be in grams/mole)  So to get moles, multiply 68.12 * (1/M).  \n\nNow, our equation above shows that for every 4 moles of NH3, we get 6 of H20.  # moles NH3 * (6/4) = # moles of H20.  Now, do the reverse operation of H20 to get how many grams is produced.  H20 has an atomic mass of something like 18.016.  So, # moles H20 * 18.016 = # grams H20\n\n\nI hope that's somewhat correct, and if not, leads you in the right direction at least
359119
look at column 7.  all of the densities appear there in grams/cm3.\n\nhttp://www.science.co.il/PTelements.asp
587954
Measurements (In distance I'm assuming) would give you volume.\n\nMass / Volume = Density\n\nIf you have volume & density, you can rewrite the equation:\n\nMass = Volume * Density
365764
Your second statement is correct, i.e proton mass+electron mass = constant mass and energy.\nok, This is with reference with another question you asked about mass defect(and chose my answer as the best, thanks). The sum of individual masses is greater than the actual mass of the atom. I am sure of this because this is what explains the stability of an atom(less energy, more stable).\nI'll give you an example which I took from my chem book\n\nMe = 0.0005486 amu(atomic mass unit)\nMp = 1.007277 amu\nMn = 1.008665 amu\nMe,Mp,Mn are masses of electron, proton and neutron respectively and 1 amu = 1.6605 x 10^-27 Kg\n\nOne He atom contains two electrons, two protons and two neutrons.\nSumming up the values of the constituents you get 4.032981amu as the expected mass of the atom.\nHowever the actual mass of the atom is 4.002603amu\nDifference = 0.030378 amu\n\nThe formula to find the binding evergy(energy required to bring the constituents together) is: mass defect x 931.48MeV\n\nHere that would be 0.030378 x 931.48 = 28.296 MeV\ni.e that is the equivalent of the mass defect in energy.\nThe '931.48' part is calculated using Einstein's formula E=mc^2.
911771
Which energy?\n\nCheck out the Periodic Table below.  For Hydrogen, the Ionization Potential is listed as 13.598 volts.  The electron configuration (energy level of the outer most electron) is 1s^1 (one electron in the 1s orbital).\n\nThe rows are the Levels, the columns are the individual orbitals:\nI & II are the "s" orbitals (He is "out of place").\nIIIB to IIB (under Sc to Zn) are the "d" orbitals.\nIII to 0 (under B to Ne) are the "p" orbotals.\nThe Lanthanides & Actinides are the "f" orbitals.
982228
Ionization potential is defined as the energy require to turn a neutral atom into an ion. Usually, this refers to the first ionization energy, i.e., the energy to remove the 1st electron in an atom. Using the same idea, the 2nd ionization energy, for example, refers to the energy that is required to remove the 2nd electon after the removal of the first electron.\n\nIn this case, you are likely to refer to the first ionization potential of group 1 metal, so let's just focus on this. Group 1 metals (Li, Na, etc.) all have 1 valence electron. To turn to the most stable form, these metals have to lose that only one electron to form a 1+ ion, which has the same electron configuration as a noble gas. As you go down the periodic table, more electron shells are added to the elements, i.e., the atomic radius increases. As you may know, the negative electrons are attracted to the positive protons in the core. As the size of the atom gets bigger, it is harder for the core to experience the outermost electron (this is due to shielding effect by the inner shells), and thus the attraction between the inner core and the outermost electrons are weaker for bigger elements. As a result, it is easier to lose the valence electrons for the bigger elements than the smaller elements (provided that they all have the same valence electrons). Therefore, you should expect K to be more reactive than Li, for example.
401395
Way to go Arbitrage!  Oh to funny guy- yeah we get that E=MC2  IS a formula- but if you really want to "get it" Check out Maxwells equations. It's a great time to be a science nerd with all this info pouring in now! Figure it out & go up one on Albert!
497711
M=Ka/x^2\n =1.8e-5/(0.01)^2\n =0.18 M\n\n(x=fraction of ionized acetic acid)
921703
Just let the units tell you what to do.  H=mass* T*sp heat.  with T in K
421381
Several Units actually and they can be a bit confusing\n\nRad, Rem, Seiverts, grey, etc etc\n\n\nhere is the break down\n\nKerma = Kinetic Energy transferred to charged particles in Rads / Grays.\nRad = radiation absorbed dose\n= 100 ergs of energy / gm of tissue\n= 0.01 joule of energy / kg of tissue\n1 Gray = 100 Rads = 1 joule of energy / kg of tissue\n1 Rad = 0.01 Gray\n1cGy (centi Gray) = 1 Rad\n1 mGy (milli Gray ) = 0.1 Rad\nRem = radiation equivalent man\n= rad x RBE (QF) / (relative biological effectiveness or Quality factor)\nSeivert = Gray x QF\n1 Seivert = 100 rems \n(1000 milliSv or mSv)\n1 mSv = 0.1 rem ( for Xrays which have a qualifying factor of 1) = 1 mGy\n\n\n\nconfusing huh?  took 2 college terms on it myself
393625
mass # =#protons + # neutrons\nand # protons = # electrons in an electronically neutral atom.\nso you can therefore find the # nuetrons\n#neutrons=mass#-# protons
561926
XeF  + e-  ============> Xe  + F-      E°/V = 3.4 with reference to the standard hydrogen electrode
1045656
An electron passing through a potential difference of 150 kV picks up 150keV of energy.  So:\n150keV=h*c/lambda\nlambda =h*c/150keV.\nh=6.626*10^-34 J*sec\nc=3*10^8m/sec\n1eV=1.6*10^-19 J\n\nSo, lambda=((6.626*10^-34)(3*10^8))/((150000)(1.6*10^-19))\n\nlambda=8.28*10^-12 m, or 8.28 nm
669901
Break this up into three questions to Yahoo Answers:\n\n1. What is the charge e of an electron?\n\n2. What is the mass m of an electron?\n\nand after those are answered,\n\n3. What is [...] divided by [...]?\n\nYou can "spend" more points that way. :-) :-)\n\np.s. And the answer is: The e/m ratio of an electron is a constant of nature the numerical value of which depends on the units used.
58191
Density - 1.151 \nNot easy to find