in chem i am working on a data sheet for statistics of popcorn, i need to find the deviation?
149503
Right, so feed the numbers into your calculator and press the SD button.\n\nOr enter it into Excel and use teh standard deviation function.
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Try looking at a substances MSDS.  That is Material Data Safety Sheet.  If it is a pure substance that will be helpful. Just put the name of the substance and MSDS into yahoo and you'll find a sheet
804468
You need to look up the entropy for all compounds, multiply by the entropy by the stochiometric coefficients (e.g. 4 for NH3, 3 for O2). Then subtract the left side from the right side.\n\nVery simple actually, once you find the data.
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I am pondering this question a bit. I might have a better answer for you, but here is a start:\n- - - - - - - - - - - -  - \n\nYour data is not exactly amenable to standard quantitative analysis. You can't really form an average, for example. \n\nYou can find the mode. The most common answer is, diet. So the mode of your data is "diet". \n\nThe data categories, as they stand, are not represented by ordinal numbers. So it is a bit hard to even describe the distribution, let alone the dispersion of the distribution. What you could do, is say that 35% preferred diet, and 65% did not. Then you could try to estimate the error bounds on these figures, given some assumptions. Or do a robust estimate using a jackknife or bootstrap.\n\n- - - - - -  - - \n\nHere is another idea. You might want to know the average number of people that are in a given category. That is, you might want to know how many people like a given type of soda. You have 5 categories, so the average is \n\n(20)/5=4\n\nThere are an average of 4 people in each category.\n\nNow you can find the standard deviation.\n\nFirst find an estimate of the variance V using the standard formula:\n\nV=(1/4)*[(1-4)^2 + (3-4)^2 +(3-4)^2 +(7-4)^2 +(6-4)^2 ]\nV=(1/4)*[9 + 1 + 1 + 9 +4 ]\nV=(1/4)*[24]\nV=6\n\nThen take the square root\n\nS=sqrt(V)= sqrt(6)\nS=sqrt(6) = 2.44948974\n\nThe average soft drink is liked by 4 people, but there is a standard deviation of about 2.4.\n\nSo one can say a soft drink is liked by 4 += 2.4 people (if we set the error bounds to be 1 standard deviation, but some people set it at 2).\n\nYou could even convert this to percentages. Then you could say that a soft drink is liked by 20% of the people, +- 12%.\n\nThe standard deviation assumes that the distribution is normal, that is a bell-shaped curve (aka Gaussian). However, it probably is not. A robust estimate of the dispersion of the dispersion can be formed just resampling methods, like the bootstrap or the jackknife. I have placed a link to these if you are interested, but I think it might be too advanced for your needs.\n\nAnother thing to do is to look at the median, and the median average deviation, or the median average deviation away from the median. A reference discusses these and other measures of dispersion if you are interested.\n\nI still feel that one might be able to do more with this. I will think some more and if I have any good ideas I will post them. I am not a statistician, so it is likely that one of them might have a much better suggestion right away.
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Look at the number of hydrogen carbonate molcules in the sample and divide the mass (Of the sample) by the mass of the hydrogen carbonate.
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Enter the number into MS Excel, then use the standard deviation formula from the "function" option. Your answer will depend on whether those numbers represent the population or just a sample from a population.
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You might try DAP. You can find it here along with a lot of other free statistical software.\nhttp://members.aol.com/johnp71/javasta2.html
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Use the gas law\nPV=nRT
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http://www.edu.pe.ca/montaguehigh/grass/chem/notes/ptable.html\nhit print screen.
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If a random sample size of 25 corresponds to a standard deviation of 30, then you can expect a random sample of size 25N to roughly correspond to a standard deviation of 30 / sqrt(N).\n\nIf 99% of the area under the bell curve is between -Z and +Z standard deviations, then you want 30 Z / sqrt(N) to be less than or equal to 0.5.\n\nMaybe a little more because of the variation of the estimated standard deviation from the true standard deviation.\n\nAlso check that the theory and the measurements both suggest it is appropriate to use the normal or Gaussian distribution.
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If you mean the mean square deviation (which im guessng you do as the mean is just reffered to as the mean...not mean deviation) the it is quite similar to the standard deviation just it requires less work.\nAll you need to do is find the some of the squares for your data set. This means for your first bit of data take the mean from it and square the result then do the same to the second, third and so on. Now add all the results up. If you have a scientific calculator find the instruction as the calculator can do this for you which is very hand and much quicker!!\nOk now thats done all you need is to divide the answer by the amount of data you have (so for 1,2,5,4 you would divide by 4)\nHope that helped!
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Look up the site chemfinder.com.  its a good website if you have not heard of it already.
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Its' probably a rounding error. It would depend on which masses for the elements you use, how precise you want to be.
495873
Standard plot of what?  What is the analysis parameter you are interested in measuring?
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deviation= (refractive index -1)*angle of prism\n\nif prism angle is small...\nand\ndependence of rf with wavelength is given by cauchy's formula\n\nrf= A/wavelength....  A is constant\n\nso dependence (deviation) proportional to [(1/wavelength) -1)]
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There is ChemDraw, that you can buy\n\nChemSketch is offered online as freeware or shareware (I forget)\n\nIf you want a program that does computational chemistry calculations (DF Theory or Ad Hoc, etc) you might look at buying Spartan.
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X = sample mean, U = theoretical mean, S = standard deviation, and N = number of data...\n\ni assume u know how to calculate X and S from above. \n\n.95 = P(1.96< (X-U)/(S/sqrt N) < 1.96)...solve for U..U will be in between to values, which means that you are 95% confident that U will lie in that range.