The ionization energy for a single atom of hydrogen? i will take either formula or answer?
962029
1312 kJ is for ONE MOLE of hydrogen atoms to be ionized.  You'd have to divide by Avogadro's number to find kJ per atom.\n\nIf you want an explanation instead of just an answer, here it is:\n\nThe energy of a Bohr orbit is given by the equation:\n\nE = (-hcRH)/n^2, where h=Planck's constant, c = speed of light, RH = Rydberg constant for H.  The term "-hcRH" is -2.18E-21 kJ.\n\nFor an atom of hydrogen to be ionized, its electron has to go from n=1 (the lowest energy orbit) to n=infinity (the boundary of the atom).  Plugging in n=1, the energy becomes -2.18E-21 kJ.  Plugging in n=infinity, the energy goes to zero.  So the energy needed for an electron to travel from n=1 to n=infinity is E(final) - E(initial), which is 0 - (-2.18E-21 kJ) = 2.18E-21 kJ.  This is the ionization energy for one atom.
369573
I think if you put that sort of question you are familliar with SCHRODINGER ecuation. It is a complex ecuation simply described by\n\nNabla (a small triangle with the basis on the upper side) of H (hamiltonian - a complex math analysis operator) equals to zero.\n\nThis equation was solved only for the Hydrogen atom (in my physics books from college the solution occupy 8 A4 pages ...) and the results were extrapolated for others atoms (helium, lithium, berylium, etc.) but is just an extrapolation. \n\nFor other atom than Hydrogen the Schrodinger equation can not be solved with the mathematical apparatus we dispose today ...\n\nSo, the Schrodinger equation describes the POSSIBILITY to find an electron in a region of space at one moment. The shape of this spatial arrangement is called orbital (wave function) and we have S orbitals (spherical shape - maximum 2 electrons), P orbitals (spatial simetry over Ox, Oy and Oz with a shape similar to mathematical sign for infinitum on every axis - maximum 6 alectrons in 3 pairs), D orbitals (maximum 10 electrons in 5 pairs) and F orbitals (maximum 14 electrons in 7 pairs)\n\nSo the Sulphur electronical configuration is:\n1S(2)2S(2)2P(6)3S(2)3P(4)\n\nFrom the last shell is probably the easier to extract an electron. The distance from the nucleus dictate the strenght of atraction ... So, an 3S electron is harder to remove than an 3P electon but easier to remove than a 2P electron\n\nDONE!
922532
E = 423.6 kJ / mol ; the amount of energy needed, but this is per mole, you want it per atom, so you use Avagadro's number:\nE = 423,600 J/ mol / (6.02E23 atoms/mol) = 7.036 E-19 J/atom\n\nh is planck's constant; 6.626068 × 10-34 J * s\nc is the velocity of light in a vacuum 3.00 E8 m / s\n\nThe E__ is a shorthand way of writing * 10^__, so E8 means * 100,000,000 and E-34 means * 0.0000 0000 0000 0000 0000 0000 0000 0000 1 (which makes it a very small number)\n\nYou can rearrange the equation like this:\nE = hc/λ\nλ * E = h * c\nλ = h*c / E\n\nλ = 6.626068 E-34 J * s * 3.00 E8 m/s / (7.036 E-19 J/atom * 1 atom)\n\nλ = 2.825 E-7 m = 28.25 E-6 m = 28.25 μm
575905
It is  1.008 g/mol. for hydrogen atoms\n\nOne hydrogen atom weights on the average 1.67323 * 10^-24 grams.\n\nHydrogen gas contains 2 atoms so it weights 3.4646 *10^-24 grams.\n\nAverage mass is average abundunce of all the isotopes of hydrogen
589038
The answer is c) +2. because if the atom has 12 protons, it is going to have 12 electrons to make it neutral. by taking away 2 electrons, it is giving the atom a positive charge, with 12 protons (positive charge) being more than 10 electrons (negative charge) giving an overall charge of +2.\n* protons - electrons = charge. =    +12-10=+2     =   c
816137
a) v^2-u^2=2as\n    v^2-52.4^2=2*(-9.8)*28.5\nv^2= 2745.8-558.6\n     =2187.2\nv=46.76m/s\n\nKinetic Energy=1/2mv^2\n                     =1/2*46.9*46.76^2\n                      =51,290 Joules\n\nb)height=28.5-5.06=23.44m\n\nv^2-u^2=2as\n\nv^2-52.4^2=2*(-9.8)*23.44\nv^2=2745.8-459.42\n     =2286.4\nv=47.8m/s
1022778
Mg2+ + 2HCl --> MgCl2 + H2 (gas)\n\nOkay, hydrogen.  As you can see in the above equation, for one atom of magnesium you'll produce two atoms of hydrogen.   \n\nLet's work out how much hydrogen you'll get in grams.\nAnd let's say that you have 1 gram of Mg.  First, what you need to do is work out how many atoms are in 1 gram of Mg.  \n\nThis is easy.  Simply:\n\n1 / atomic mass x Avagadro's constant = number of atoms.\n\nFor Mg, this equals:\n\n1 / 24.31 x 6.0221415×10^23 = 0.2477228 x 10^23 atoms.\n\nWe know that for every atom of Mg you'll produce 2 atoms of hydrogen.  So, that means you'll produce:\n\n2 x 0.2477228 x 10^23 = 0.4954456 x 10^23 atoms of hydrogen.\n\nTo work out how much 0.4954456 x 10^23 atoms of hydrogen weighs you just calculate, like this:\n\n0.4954456 / 6.0221415 x 1.008 = 0.0829 grams.\n\nAnd that's it.  1 gram of Mg will produce 0.0829 grams of hydrogen.
889364
i am just starting chemistry, but i think that you multiplied hydrogen (H) by four, when it should be multiplied by 3, like the Oxygen (O).  otherwise it is correct.  you sould round to whatever your teacher tells you or to 6 significant digits, because that is the least amount you have.  the correct answer should be \n\n(considering your atomic weights are correct)\n\n[12.0107 x 3] + [1.00794 x 5] + [15.9994 x 3] + [1.00794 x 3] =  \n\ncalculate it out and it should work fine... (i think)
761199
Although there is a general trend toward an increase in the first ionization energy as we go from left to right across this row, there are two minor inversions in this pattern. The first ionization energy of boron is smaller than beryllium, and the first ionization energy of oxygen is smaller than nitrogen.\n\nTo read about why, follow the link. It's really hard to explain
1031597
That 1kg of hydrogen would convert to about 0.954 kg of helium, the other 0.046 kg is changed to energy.\n\nE=MC^2\nM=0.046 kg\nC=300,000,000 m/s\nE=4.14X10^15 joules\n  = 1,150,000,000 Kilowatt-hours
573821
I think this question is incomplete because the Ka of HF hasn't been provided.  The amount of H+ that is formed from HF can only be calculated from its Ka value.
509278
As mentioned above, two hydrogen atoms are brought together under high temperature and pressure so that they "fuse" into one and create a Helium atom.  What is not mentioned is that the result of this fusing is the creation of a huge amount of energy.  The energy comes from the difference in mass between the two hydrogen atoms and the resulting helium atom which weighs less.  The mass dissappears and is converted into energy in the form of electromagnetic radiation and kenetic energy of various particals.
872530
Magnesium has 2 valence electrons. Hence, the first and second ionization energies are not as much as third ionization energy. Its already difficult to remove its third electron because it would come from an orbital that follows the octet rule (meaning already stable). To visualize:\n\n1st ionization: Mg + energy ---> Mg+  + e-\n2nd ionization: Mg+  + energy ---> Mg2+  + e-\nnote that Mg2+ is the stable form of magnesium ion, it occurs naturally.\n3rd ionization: Mg2+  + energy ---> Mg3+  + e-\nnote that Mg3+ does not occur naturally
800905
One mole of electrons is Na = 6.023x10^23 electrons, so it takes:\n\nE = 208.4x10^3/6.023x10^23 J = 3.46x10^(-19) J to remove 1 electron.\n\nE = h*f, where E the energy of a photon, h = 6.63x10^(-34) J*s the Planck's constant and f the frequency of light. Solving for f:\n\nf = E/h, f = 3.46x10^(-19)/6.63x10^(-34), f = 5.22x10^14 Hz\n\nBut:\n\nc = λ*f, where c = 3x10^8 m/s the speed of light and λ the wavelength. Solving for λ:\n\nλ = c/f, λ = 3x10^8/5.22x10^14, λ = 5.75x10^(-7) m, or\n\nλ = 575 nm
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
1155563
D because\n   when we take the empirical formula of these as\n     %age of Hydrogen devided by its Mr i.e. 25/1=25\n                                                                     \n%age of carbondevided by its Mr i.e. 75/12=6.25\nnow the answers will be devided by the no. which is smaller and as 6.25 is smaller we will use it to devide both the valuesi.e\n                       H :C\n                   25/6.25 : 6.25/6.25\ni.e.               4 :1\nor             1:4\ni.e CH4
246055
51 electrons.\n\nbtw-the atomic number of an element (antimoney is 51) corresponds to the number of electrons in a neutral atom.\n\ni.e.=hydrogen is 1, and has 1 electron in a neutral state.