which is the balanced equation for the fission of  U-235?
409660
235 Uranium + 1 neutron --> 140 Cesium + 93 Rubidium + 3 neutrons
435218
Iso- as a prefix means "the same". Thus isobar (same pressure), isosceles triangle (two sides the same length) and isotope two or more versions of the same element which all have the same number of protons (or it would be a different element!) but varying numbers of neutrons in the nucleus.\n\nAs protons are positively charged and neutrons have no charge (electrons have a negative charge) neutrons are needed as "packing materials" to help keep the positive charges fom repelling one another and making the nucleus unstable.\n\nWith elements of low atomic number, the number of neutrons will be the same or just greater than the number of protons. But with elements of high atomic number, more packing materials are needed, and the ratio of neutrons to protons begins to climb towards 1.5:1\n\nThere is only a narrow range of options for that ratio for isotpes that are stable. Those with too few or too many neutrons, iutside that range, are unstable and decay.\n\nCompare Carbon 12 and Carbon 14 with Uranium 235 and Uranium 238\n\nCarbon 12 has 6 protons and 6 neutrons and is stable\n\nCarbon 14 has 6 protons and 8 neutrons and is unstable, it decays\n\nU235 has 92 protons, 143 neutrons and is unstable.\n\nU238 has 92 protons, 146 neutrons and is unstable\n\nAs regards silver (Ag) oy has two stable isotopes Ag 107 (47p, 60n) and Ag 109 (47p, 62n) But very short-lived isotopes varying in atomic mass from the neutron-deficient Ag 93 (47p, 46 n) to Ag 123 (47p 76n) also exist.
835018
about 3000 kilo J\nit depends on the atom
798483
Here it is:\n\n(98,42)Mo + (1,0)n ---> (99,42)Mo
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
908504
Take 100 g of natural Cu, then if x is the mass of the isotope with atomic mass 62.930, the mass of the isotope with atomic mass 64.928 must be (100 - x) g. So we have the equation:\n\n62.930*x/100 + 64.928*(100 - x)/100 = 63.540\n\nSolving for x:\n\nx = 69.45% of the isotope with atomic mass = 62.930, and:\n\n100 - x = 100 - 69.45 = 30.55% of the isotope with atomic mass = 64.928
496427
Not sure of the question but... uranium ore is mostly composed of the isotope U-238 which does not support self-sustaining fission reaction very well.  Enrichment is the process of extracting and purifying U-235 which is much more potent for use in weapons.\n\nWell, rockeytj - it's a physics question.  It's not like anyone's going to do it tonight with a blender and some toothpicks.  It takes zillions dollars worth of equipment, not to mention tons and tons of uranium ore.
386839
Only U-235 and Pu-239 and slow neutrons. fast neutrons just pass through. Uranium has some spontaneous fission. \nCollisions with moderate size atoms will leave the new atom spinning and it can then fission. \nAlphas lose a lot of energy just overcoming the coumb barrier then the protons/neutrons are absorbed more than splitting.
257364
When an atomic bomb detonates, a nuclear fission reaction occurs which actually transforms a very small amount of mass into energy.\nUsing the well-known formula,\nE=mc^2,\nwe can calculate the amount of energy associated with that mass that is "lost".\n\nOnly a very very tiny amount of mass is actually trasformed into energy in the reaction, but since that mass is multiplied by such a huge number (the speed of light squared), the energy released is gigantic.\n\nUsually, A-bomb yeilds are given in terms of [kilo / mega] tons of TNT to measure the energy given off.\nThe yeild on the atomic bomb detonated over Hiroshima Japan in 1945 ('Little Boy') was about 16 kilotons of TNT [high end estimate].  1 ton of TNT releases about 4.25 E9 Joules of energy.  Using this conversion, we can calculate that the Hiroshima bomb released about 6.8 E13 Joules.  By comparison, the entire average enegry consumption of the United States per year is on the order of 10 to the 20 Joules, so 1 A-bomb packs a pretty good punch.\n\nSince 1945, atomic bombs have gotton much MUCH more powerful in addition to the Hydrogen bomb which is exceedingly more powerful than the atomic bomb.
385240
The amount of energy generated by nuclear fission depends on the particular element involved and the preciseness of the split. For example, the splitting of one Uranium-235 atom can produce 200 MeV of energy (equivalent to 3.2 x 10E-11 joules) - while the splitting of a Plutonium-239 atom can create 210 MeV of energy.\n\nA good explanation of nuclear fission can be found in the article I've linked below:
956102
Here is the equation\n\n(246,92)U ---> (4,2)He + (242,90)Th + 2Gamma
101213
You may already know the answer, but I'll give you credit for wording it precisely.\n\nThe answer is more than one (on average), because neutron losses mean it would be impossible to sustain a critical nuclear reaction (that is, a constant rate of nuclear reactions) if you had one or less than one (on average) per reaction, since it takes at least one neutron to cause a nuclear reaction.\n\nThe exact reaction that occurs and the byproducts produced will determine exactly how many you get per individual reaction, but *on average* you need to net >1.0 neutrons per reaction.  The more losses you have (due to poor shielding, absorbant material, etc.), the higher the number needs to be, but something like 1.2-1.5 should be fine.
411320
simple is when there is a simple fission, multiple is when there are multiple fissions and binary is when there is just one fission heh...
888620
Nuclear fission—also known as atomic fission—is a process in nuclear physics in which the nucleus of an atom splits into two or more smaller nuclei as fission products, and usually some by-product particles. Hence, fission is a form of elemental transmutation. The by-products include free neutrons, photons usually in the form gamma rays, and other nuclear fragments such as beta particles and alpha particles. Fission of heavy elements is an exothermic reaction and can release substantial amounts of useful energy both as gamma rays and as kinetic energy of the fragments (heating the bulk material where fission takes place).\n\nNuclear fission produces energy for nuclear power and to drive explosion of nuclear weapons. Fission is useful as a power source because some materials, called nuclear fuels, both generate neutrons as part of the fission process and also undergo triggered fission when impacted by a free neutron. Nuclear fuels can be part of a self-sustaining chain reaction that releases energy at a controlled rate in a nuclear reactor or at a very rapid uncontrolled rate in a nuclear weapon.\n\nThe amount of free energy contained in nuclear fuel is millions of times the amount of free energy contained in a similar mass of chemical fuel such as gasoline, making nuclear fission a very tempting source of energy; however, the waste products of nuclear fission are highly radioactive and remain so for millennia, giving rise to a nuclear waste problem. Concerns over nuclear waste accumulation and over the immense destructive potential of nuclear weapons counterbalance the desirable qualities of fission as an energy source, and give rise to intense ongoing political debate over nuclear power.\n\nhere:http://en.wikipedia.org/wiki/Nuclear_fission#Physical_overview
976256
BALANCED . . . \n\n2Sr + O2 ---> 2SrO\n\n. . . and solved . . .\n\nGet an "A".
403672
Maybe you are referring to uranium enrichment?\n\nUranium found in nature consists largely of two isotopes, U-235 and U-238. The production of energy in nuclear reactors is from the `fission' or splitting of the U-235 atoms, a process which releases energy in the form of heat. U-235 is the main fissile isotope of uranium.\n\nNatural uranium contains 0.7% of the U-235 isotope. The remaining 99.3% is mostly the U-238 isotope which does not contribute directly to the fission process (though it does so indirectly by the formation of fissile isotopes of plutonium).\n\nUranium-235 and U-238 are chemically identical, but differ in their physical properties, particularly their mass. The nucleus of the U-235 atom contains 92 protons and 143 neutrons, giving an atomic mass of 235 units. The U-238 nucleus also has 92 protons but has 146 neutrons - three more than U-235, and therefore has a mass of 238 units.\n\nThe difference in mass between U-235 and U-238 allows the isotopes to be separated and makes it possible to increase or "enrich" the percentage of U-235. All present enrichment processes, directly or indirectly, make use of this small mass difference. \n\nA number of enrichment processes have been demonstrated in the laboratory but only two, the gaseous diffusion process and the centrifuge process, are operating on a commercial scale. In both of these, UF6 gas is used as the feed material. Molecules of UF6 with U-235 atoms are about one percent lighter than the rest, and this difference in mass is the basis of both processes.\n\nLarge commercial enrichment plants are in operation in France, Germany, Netherlands, UK, USA, and Russia, with smaller plants elsewhere.
194920
It depends.\n1) In nuclear power plants a moderator is used to reduce the neutrons' velocity and the chain reaction rate is thereby controlled.\n\nA typical 1000 MWe nuclear power station uses only about 24 tonnes per year of uranium fuel.\nwww.aph.gov.au/house/committee/isr/uranium/subs/sub5.pdf\nOR\nhttp://www.uic.com.au/whyu.htm\n\nThe average reaction rate would calculate as follows:\n\n(24 tonnes/year) X (1000 kilograms/tonne) x (year/365 days)=\n                 66 kilograms/day (about 145 pounds a day)\n\n                   OR\n(66 kilograms/day) x day/24 hours x hour/60 minutes x 1000grams/kilogram =\n\n                 46 grams every minute \nThis is the weight of one typical (1.55 ounce) candy bar!\n\n2) In a nuclear bomb the neutrons are not moderated.\nThe reaction time for a bomb is 1/1000 second. (Complicated quote below.)\n\n "An explosion is an interplay between the total amount of energy released, energy density, and the time in which the energy is released. The time factor is perhaps the easiest to define. While there is no exact definition of reaction time for an explosion, we use one millisecond as a reasonable value to distinguish a steady-state regime from an explosive regime."                      \n\nhttp://www.ieer.org/ensec/no-6/fusion.html\n\nThe reaction rate calculation for a very small bomb (10 kilograms):\n\n10 kilograms / 1/1000 seconds = 10,000 kilograms/second !\n                       OR\n                217,391 candy bars per second!\n\nSummary:\n Nuclear Power Plant - Chain Reaction Rate \n 1 candy bar per minute\n\n Nuclear Bomb - Chain Reaction Rate\n 217,391 candy bars per second\n\nI haven't calculated candy bars per minute (Nuclear Bomb) because no one can eat that many in a minute.\n\n- Stan the Answer Man