how do i calculate the second ionization energy of He    in Kj/mol?
338598
i don't know how to calculate it, but i get 5250 Kj/mol from
956102
Here is the equation\n\n(246,92)U ---> (4,2)He + (242,90)Th + 2Gamma
881354
The first ionization energy of Mg is 737.7 kJ/mol which is 176.2 kCal/mol.
919041
We're basically looking for a ratio of ionization energy E(n), where n =4 and n = 1 (ground state). Electron energy levels of a hydrogen atom is given by:\n\nE(n) = 13.6/n^2 eV\n\nTherefore,\n\nE(4)/E(1) =  1/16\n\nIt takes only 1/16 of 13.6 eV , or 0.85 eV, to ionize the ground state electron from a hydrogen atom on this very hot planet.
759238
just look at your units..that normally hleps\n\nhf ...kj / mole\n3.22 kJ / 6.02 Kj/mol =    x moles\nxmoles * 18 gram / mol.....  Thats your answer
703814
211_____207_____4\n      Bi  -->__Tl +____a    (a helium nucleus) \n83_______81_____2
212309
The half-lives will vary depending on the isotope.  Go to the library and look them up in "The Handbook of Chemistry and Physics" in the reference section.
848448
I think this is what you are looking for:\n\n(1)\nLet E be for energy (to avoid confussion with k's).\nLet α be a proprotional sign.\n E α M*v²    (from information given).\n E = k M*v² \n 30 = k [(3x10^-3)*(3²)]\n 30 = k [72x10^-3]\n 30/ 72x10^-3 = k\n k = 416∙666....\nNow use the constant value of k to calculate your answer.\n E = (416∙666....)(4x10^-3)(6²)\n E = (416∙666....)(4x10^-3)(36)\n E = 60 Joules of energy.\n\n(2)\n Z α x²(1/y^3)\n Z = k [x²(1/y^3)]\n 2/9 = k [2²(1/3^3)]\n 2/9 = k [4(1/27)]\n k = 2/9*27/4\n k = 54/18\n k = 3\nNow use the constant value of k to calculate your answer.\n Z = k [x²(1/y^3)]\n Z = 3 [1²(1/4^3)]\n Z = 3(1/64)\n Z = 3/64\n\n(3)\n R α L*1/D²\n R = k(L*1/D²)\n 3∙5 = k(50*1/(0∙38x10^-2)²\n 3∙5 = k(50*1/0∙00001444...)\n 3∙5 = k(50*69252∙07756)\n 3∙5 = k(3462603∙878)\n 3∙5 / (3462603∙878) = k\n k = 0∙000 001 01\nNow use the constant value of k to calculate your answer.\n R = ( 0∙000 001 01..)(75*1/(0∙625x10^-2)²)\n R = ( 0∙000 001 01..)(75*1/0∙000 039 062)\n R = ( 0∙000 001 01..)(1920 000)\n R = 1∙940736 Ω\n R ≈ 1∙94 Ω
701742
I do not have a reference, but can tell you that Ethanote (Acetate) ion has the negative charge delocalised over 2 oxygen atoms.  This means that the bond energy is between that of a carbon-oxygen single bond and that of a carbon-oxygen double bond.\n\nThe bond energy for the carbon-oxygen (double) bond in CO2 is 799 kJ/mol while a single bond C-O is about 358 kJ/mol, so the answer is between these.  It is probably in the ball park of 580 kJ/mol.\n\nThe true or more accurate answer is a very complexed Physical Chemistry calculation.
463421
KE = mv\nWrong: energy <> momentum\n\n2y = (2 m)cos(kx)\nRight: a lenght =  lenghth x a number between -1 and +1\n\nvf = vi + ax\nWrong: x should be t\n\ndensity = volume/mass\nWrong: fraction should be inverse
805169
alright...\nconsider the R.H.S.\nh2o has two o-h bonds which account for 2x467 and since thre are 2 mol. u get 2x2x467--------(1)\nco2 has two c=o bonds which account for 2x799 and since u hav 4 mol. u get\n4x2x799-------------(2)\naddin 1 and 2 u get 1868+6392=8260\nconsider the L.H.S.\nu hav in c2h2, two c-h bonds which account for 2x413--------(3)\nand one triple bond accounting for 1x839---------(4)\nnow u hav 2x(3+4)=2x(826+839)=3330------(5)\nand u hav 5x(o=o) which acc for 5x(495)=2475----(6)\naddin 5 and 6 u get 5805\ntherefore ur ans is 5805-8260=-2455\nand since u want for one mol of acetylene u get -1227.5 kj/mole
184012
Determining the number of valance electrons\nValance electrons \n \nHelium atom model\nThis helium (He) model displays two valence electrons\nlocated in its outermost energy level.\nHelium is a member of the noble gases and contains\ntwo protons, neutrons, and electrons. \nTo determine the quantity of valance electrons an element has, one must look at the family (vertical column) in which the element is categorized. With the exception of groups 3–12 (transition metals), the number within the unit's place identifies how many valance electrons are contained within the elements listed under that particular column.\n\nGroup Quantity of valance electrons \nGroup one (1) (alkali metals) One \nGroup two (2) (alkaline earth metals) Two \nThe (transition metals) (3-12) One to two \nGroup three (13) (boron group) Three \nGroup four (14) (carbon group) Four \nGroup five (15) (nitrogen group) Five \nGroup six (16) (oxygen group) Six \nGroup seven (17) (halogens) Seven \nGroup eight / zero (18) (noble gases) *Eight \n\nThe asterisk (*) is used to indicate that all of the noble gases contain eight valance electrons with the exception of helium (He), which only contains two.\n\n\nValance electrons in chemical reactions\nThe chemical behavior of atoms is largely due to interactions between electrons. Electrons of an atom remain within certain, predictable electron configurations. Electrons fall into shells based on their relative energy level which is usually visualized as their mean distance from the nucleus. The valance electrons have the greatest influence on chemical behavior. Core electrons (those not in the outer shell) play a role, but it is usually in terms of a secondary effect due to screening of the positive charge in the atomic nucleus.\n\n \nThe atomic orbital wavefunctions of hydrogen. The principal quantum number is at the right of each row and the azimuthal quantum number is denoted by letter at top of each column.Each shell, numbered from the one closest to the nucleus (lowest in energy), can hold up to a specific number of electrons due to its differing sublevel and orbital capacity:\n\nShell number Electron capacity Sublevels  Orbitals \nOne 2 S  1 \nTwo 8 S, P  4 \nThree 18 S, P, D  9 \nFour 32 S, P, D, F  16 \n\nTo determine the electron capacity of a shell, the formula 2n² is used, where n is the shell number or principal quantum number. Electrons fill orbitals and shells from the inside out, beginning with shell one. Whichever occupied shell is currently most outward is the valence shell, even if it only has one electron.\n\nThe reason why shells fill up in order is that the energy levels of electrons in the innermost shells are significantly lower than the energy levels of electrons in outer shells. So if the inner shells were not completely full, the electron in an outer shell would quickly "fall" into the inner shell (with the emission of a photon that would carry away the difference in the energy).\n\nThe number of electrons in an atom's outermost valence shell governs its bonding behavior. Therefore, elements with the same number of valence electrons are grouped together in the periodic table of the elements. As a general rule, the fewer electrons in an atom's valence shell, the more reactive it is. Group 1 metals are therefore very reactive, with caesium, rubidium, and francium being the most reactive of all metals.\n\nEvery atom is much more stable, or less reactive, with a full valence shell. This can be achieved one of two ways: an atom can either share electrons with neighboring atoms, a covalent bond, or it can remove electrons from other atoms, an ionic bond. Another form of ionic bonding involves an atom giving some of its electrons to another atom; this also works because it can end up with a full valence by giving up its entire outer shell. By moving electrons, the two atoms become linked. This is known as chemical bonding and serves to build atoms into molecules or ionic compound
352406
The predicted adiabatic electron affinities, obtained at the DZP++ BHLYP level of theory, are 0.91 eV (AsH), 1.09 eV (AsH2), -1.03 eV (AsH3), 0.91 eV (AsH4) and -0.033 eV (AsH5). The adiabatic electron affinity at this level of theory for AsH5 is much smaller (0.37 eV) than the one obtained using other functionals. BHLYP vertical detachment energy for AsH2 (1.08 eV) is in good agreement with experiment, while other values range from -1.03 eV (AsH3) to 1.30 eV (AsH4).
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
205383
Divide each mass by their molar weight to get the number of moles of each gas. 0.856g He/(4.003 g He/mole) = 0.214 mole of He. Do that for all the masses and add them up. I got a total of 1.034 moles. \nFrom that you could find the total pressure using PV=nRT.\nNext, find the mole fraction (I think this is the same as molar concentration) which is the # of moles of each gas/total number of moles.\nThen multiply each mole fraction by the total pressure to get each partial pressure. \nHope this helped. Let me know if you want me to go through each step.
817827
energy levels/orbitals
525644
Alpha particle is a helium nucleus, so He-4.\n\nAl-27 + He-4 ---->  H-1 + Si-30\n\nBalance the atomic number and the atomic mass on both sides and you should find that you've generated an atom of silicon-30.