What is 250.0000001 divided by the fine-structure constant, approximately?
353253
Fine-structure constant, alfa = 1/137,03599911\n\nSo, 250.0000001/alfa =\n= 250.0000001 * 137,03599911 =\n= 34258,9997912
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
753355
125.6\n\nhttp://www.webelements.com/webelements/elements/text/Ag/ionz.html
665207
12.8 & -0.15\n\nEach multiplied by (1/41) gives\n0.312195122 & -0.003658536
86803
0.77(with a little horizontal line right above the last 7)
919497
Simply.  The answer is\n\nU89sjn33hHYkPM0iL
802629
Let's assume that the energy of the Cl - Cl bond is 328 kJ/mol.\nSo that's how much energyis needed to break it.\n\n328 kJ/mol = 5.49 x 10^ -19 J/molecule\n\nUse E = hf  (E = energy, h = Planck's constant) \nto find the frequency (f) of the light with that much energy.\n\nThen, use the universal wave equation v= f(lambda) where v is the speed of light, and lamba is the wavelength, in metres.\n\nGood luck !
727822
It is 63 x 0.66 + 65 x 0.34.  Do you see why?  The actual atomic weight found in the periodic table will differ from this by a tad because of nuclear binding energy.
915896
if ur question is 4*565*(56c55)\nthen it is\np=4*565*1   (because 56c55=56!/55!1!)\np=2260\nelse \nmadness
819160
em energy radiated is proportional to T^4, so it would be : ((273.15+9627)/ (273.15+5227))^4 =10.5
752020
A: 53\nB: Approx. 127 AMU\nC:  Atomic Number\n\nThe Element is Iodine by the way.
913686
This is a simple ratio problem. The equation would be Dgalaxy/Dearth=Dearth/x so x=(Dearth)^2/Dgalaxy. Asssuming a parsec is 3.6 light years, and a light year is 5.87 trillion miles and the diameter of the Earth is 8,000 miles, then the Earth would be 0.001 millimeter or one micron in size, which is the size of an average bacterium, maybe? Very cool. Oops! I re-did my calculation and I see that I made a factor of ten error. The correct answer is about 0.1 microns, or the size of some viruses. But tell me how you would make such a model of the Earth?
821366
2.7*10^-5\n7.041*10^5
7018
58.93\nJust go to your Yahoo search page and type in 'Periodic Table'
744231
Well, mv = h/λ\nm = h/(λv)\n= 6.626E-34 / (3.31E-15 x 0.01 x 3E8)\n= 6.673E-26\nConvert this to Da\n= 6.673E-26 / 1.66E-27\n= 40.2\nYou know, this is Ca. (look for it in the periodic table)
1176062
10^100 - 10^-100... That's pretty close.. hehe
502482
1000000000000000000000000000 = 1 octillion = 10^27