PLZ help.. What does the integral trace on an NMR spectrum indicate?
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Answer C. The relative number of protons in the magnetic environment.
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i made a little comment for you on your integral answer:)
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By using a spectrometer you're measureing the amount of light that the solution in the cuvette absorbs at a certain wavelength, which in turn tells you how much of the solute is dissolved in the solution.  If you get fingerprints on the cuvette, this could intefere with the light absorbtion, and thus your readings would be off.
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In each technique you are looking at different spin 1/2 nuclei.\n\nIn general, 1H NMR spectra are much more sensitive, 1H NMR spectra can be aquired much more quickly than 13C, and overall 1H NMR spectra are much more useful for structural determination than 13C. \n\nAn important fact about 1H spectra is that you will see spin coupling between protons three bonds away due to hyperfine interactions. This is what actually provides information about how the atoms in a molecule are connected. Further, 1H spectra are always acquired in a way that the area under a peak is proportional to the number of protons in that particular chemical environment. This can also be very helpful in structure determination.\n\n13C NMR is less useful for several reasons. First, it's not very sensitive. While you can easily acquire a 1H spectrum in 2-3 minutes, 13C spectra routinely require several hours. \n\nTo partially overcome this, a NOE transfer from 1H - 13C is routinely used. This greatly increases the sensitivity of 13C nuclei near 1H nuclei, but doesn't change the\nsensitivity of 13C nuclei far from 1H nuclei. This causes two problems.\n\nThe first problem is that you can no longer integrate the areas under the peaks to find out the relative number of 13C in that position. Although there are ways of avoiding this problem, those techniques usually require days to acquire a spectrum instead of a few hours. \n\nThe second problem is that the farther a 13C nucleus is from a 1H nucleus, the weaker its signal will be. In some cases, if a 13C nucleus is a long ways away from a 1H and has a long relaxation time, it can essentially disappear from the spectrum. This is particularly true for carbonyl groups, and it can be very annoying when analyzing a 13C spectrum.\n\nAnother major problem with 13C is that the natural abundance of 13C is so low that essentially all 13C atoms are surrounded by 12C atoms. The means that you don't see the same spin coupling with neighboring nuclei that you can see with a 1H spectrum. As a result, you really don't have any information at all about how your 13C atoms are bonded together.\n\nIn practice, chemists use 1H NMR all the time. 13C NMR is sometimes used, but due to the drawbacks discussed about it generally isn't very helpful and is only used for journal which require 13C spectra of new compounds.
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I'm going off memory here so be sure to check this.  I think the formula is the same, but the structure may be different.  Just trying to point you in the right direction.
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This looks close enough to the techniques used to show that the roots are close to the critical line. The actual estimates are difficult, though, and nobody has been able to actually give a proof using these techniques (obviously). Go ahead and try it. Perhaps you can see through to a proof. If so, you will be famous. If not, you will have learned somthing.
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if we assume that the following reaction occured\n\nA(l)----->A(g)\n\n[that means no A is not a polymer in liquid phase, the only thing occured  is the evaporation],\n\ntry plot ln(P) against 1/T, most likely you will have a straight line and the slope is - delta H/R, please notice the sign.
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You're right, spectroscopy is the proper method...Below you'll find a link to a quick and simple explanation.
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I'm sorry but did you see this in My_Computer properties?\n\n\n\n--- ► λ ◄ \n\nN.B. I'm still a learner.
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i'm not completely sure, but i think that it has something to do with negative spins and all that. \n\nthe best thing to do would be to ask your teacher.  he's the one that would now the best...
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oneguyin2005 has the right idea, but the wrong terminology. The purpose of such labeling is to use a radioactive isotope of the given element in the biological system. (or else the other atom would have differnet biological activity.) You can trace where the radioactive atom is, relative to the others, and it can help you elucidate structure
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Its either d or e. It should be a deshielding effect by the pi electrons. There is some gobbledy-gook showing up in my browser so I can't tell d or e is the pi symbol.
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Typically this refers to magetic fields.  If you align a large magnet (like a NMR - nuclear magnetic resonance instrument) protons or electrons (ESR - electron spin resonance) are aligned to a certain angle and the like.  Once the magnet is removed, the electrons and protons can start "precessing" and this is recorded via the sound it emits and it is then Fourier transformed into a typical spectra that can be interpreted. \n\nAtoms in degenerate orbitals will align under a big enough magnetic field.
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Here is what i found.  I'm not sure how right this is, but you can check it out.  Good Luck with your testing.
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your formula refers to several different structures which of course would give different NMR spectra. It could be 2-bromo-X-methylpyridine where X is the 3,4,5 or 6 position or Bromoaniline, with the various positional isomers. If you don't know the name go to the sigma website and find the structure/name (it may even have the NMR).
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ask your chem prof or tutor ..it's hard to explain over the internet!
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emm...is this your homework??well..it seems that the question is not from you (individually). Hmm..for more understanding, check out this website:\n\nhttp://faculty.clintoncc.suny.edu/faculty/Michael.Gregory/files/Bio%20101/Bio%20101%20Lectures/Biochemistry/biochemi.htm\n\nI hope it solves your dillema, good luck :)