can i use the I.E. of Hydrogen from the previous answer to determine wavelength? 1312kj/mol?
924166
yes u can. use the equation E = hf where E is the energy and h is the planck constant. once u find f which is the frequecy...insert into equation velocity = wavelength x frequency...assuming u noe the velocity of the hydrogen!! :)\nOH but by the way since ur answer is in kj/mol u need to convert it to kj with the avogadro number which i forget...i'm sure u can find out!
264542
We can use the formula c=v(lambda) where c=speed of light (=3.0x10^8 m/s), v is the frequency, and lambda is the wavelength\n\nSo we get:\n3x10^8 m/s = (3.22x10^16 s^-1)(lambda)\nlamda = 9.32x10^-9 m\nLambda = 932 nm
840473
Need more specifics as to your problem.\n\nHowever, I assume you are trying to do a quantitation on a UV or vis absorbing substance.\n\nMake up  a set of standards of known concentration of your analyte in the solvent you will be analyzing in.  Then do some scans over the whole wavelength range. \n\nUsually pick the wavelength at the top of the largest absorbance peak, unless that is a wavelength which will have interferences from other substances present.  If so, then pick a wavelength with good absorbance that is not subject to such interference.
283860
Frequency = speed of light/wavelength.\n\nMake sure you keep the units straight when performing the calculations or it may not make sense.\n\nSo for your problem its (3 x 10^17 )/200 = 1.5 x 10^15 Hz
787541
Keq = [HF]^2/[H2][F2] = (0.4)^2/0.5)(0.1) = 0.16/0.05 = 3.2\nNow add 0.150 mole F2.
350595
C whether this helps You,\nhttp://web.umr.edu/~kpeaslee/classes/Met221/KeyHW2.pdf#search='measuring%20SDAS%20in%20Aluminum%20alloy%20casting'\n\nhttp://www.aluplanet.com/eng/info_eco_doc_PO.asp?Mnu=141&Itm=163&Grp=0&Tgl=162&Doc=3804
464906
did you also prepare a standard solution? because the way i do it, is that i have a standard so i can compare the 2 absorbtions (same wavelength)...You know the equation..
804090
ooo this will be fun!\n\nok..\nA)\n2.207g of H2O MW. 18.0\nmol = weight of sample (g) / molar weight (g/mol) (n=m/M)\nn = 2.207 / 18\nn = 0.1226 moles of H2O\n\nnow, since there is 2 hydrogens in H2O, then when the molecule is broken down, there is twice as many moles of hydrogen then the overall molecule,\ntherefore,\n\n2x 0.1226 = 0.2452 moles of H.\n\ndetermining number of moles of CO2 is also helpful,\n4.313g, n = 4.313 / (12.01 + 16 +16) = 0.0980moles\napprox, 5(moles H2O) :  4(moles CO2)\n\nB) \nworking the number of moles of this compound will be useful:\nn = 1.816 / 74.122 = 0.0245 moles, which is 1/10 of the moles of H produced. (or 1:5 of the moles produced of water)\n\nif you think about it, the reaction is something like this:\nCH2 + O2 --> CO2 + H2O\nclearly the above doesn't make sense, because 1) it isn't balanced, and 2) the reactant needs a higher molecular wt.. now thinking about it, the following can be deteremined:\nand from the ratio of moles found above:\n       1     : (excess):  4        :    5\nC4H10 + 13/2O2 --> 4CO2 + 5H2O\n\nnow, I amoung most allow fractions in equations, but if you don't, multiply through by 2.\n2C4H10 + 13O2 --> 8CO2 + 10H2O\n\nnow, you want the empirical forumla,\ntherefore,\ndivide by highest common factor,\nC2H5 \n\nthe empircal fomula is just a ratio and doesn't make huge sense, as the formula for an alkane is CnH2n+2\nhope that makes sense :)
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
991807
uh a pH meter will help with the hydrogen iodide because it's a strong acid.  As for the I2 I have no idea...
245193
Use your buffering equation and plug in the numbers you know to get your answer.  You might have to do reverse calulations to find your answer.
827543
The equations involved are:\na) freq*wavelength = velocity of wave\nb) energy of radiation E = h*freq, where h is Plank's constant.\nYou can work out from these that the choice should be a.
753451
A bit vague but a you could work out the rate from a titration, also the molar value from the same experiment.
452634
N2 +3H2 ---> 2NH3\n\nso, H2 Mwt is 2g/mol and NH3 Mwt is 34g/mol\nas per above eqn, 6g of H2 will give u 34g of NH3\nSo, 9.988g of H2 will give 56.6g of NH3\n\nYou got it?
125296
use spectrophotometric method(uv-vis spectrometry).Few things u shud remember.....\n1) Use Labert-Beer's law\n2) u shud know the molar extingtion coefficient value of that particular species\n3) concentration of the species\n4) Calculate lambdamax. using LB law\nIt is an indirect method.\n\nUV-VIS. Spectroscopy will give u the lambdamax value directly...alongwith absorbance
517715
Yes, if L is wavelength, then it is L/2.
796834
try asking this in the chemistry section next time....