Does anyone know the equation to work out: organic matter content in soils??!?!?!?!?!?!?
929557
TOTAL SOIL AMOUNT  minus   INORGANIC CONTENT = ORGANIC CONTENT
183731
This doesn't really belong in Earth Sciences.  Biology or Botany are more specific categories for questions about living matter.  However, I'm in a good mood because it's Friday.  Check the link.
841680
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 5.752 g x (12 / 44) = 1.569 g\nhydrogen: 2.354 g x (2 / 18) = 0.262 g\noxygen: 2.356 - 1.569 - .262 = 0.525 g\n\n2) convert to moles.\ncarbon: 1.569  / 12  = 0.131 mol\nhydrogen: 0.262  / 1  = 0.262 mol\noxygen: 0.525 / 16 = 0.0328 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.131  / 0.0328  = 4\nhydrogen: 0.262 l / 0.0328  = 8\noxygen: 0.0328 / 0.0328 = 1\n\nWe have now arrived at the answer: the empirical formula of the substance is C4H8O
642291
http://natres.psu.ac.th/Link/SoilCongress/bdd/symp9/1375-t.pdf
658192
Try these links:\n\nhttp://www.fao.org/ag/agl/agll/wrb/wrbmaps/htm/soilres.htm\n\nhttp://soils.usda.gov/use/worldsoils/mapindex/
675012
Look here:  http://www.analytictech.com/mb021/organic_vs_mechanistic_structure.htm
323054
You sure did pick a really long question for explaining.  And I'll have to argue that all of the pH's do not have correct sig figs but the numbers are right.\n\n(Sorry about the numbers in the tables but this place doesn't seem to allow better tables)\n\nA) 25.0mL of HAc(acetic acid) and 10.0mL of NaOH\n\n*This is a reaction so we have to use moles*\n\nHAc : 25.0mL * (0.100mol/L) * (1L/1000mL)= 0.00250 moles\n\nNaOH : 10.0mL * (0.125mol/L) * (1L/1000mL) = 0.00125 moles\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00125      0.00250        0\nChange:   -0.00125     -0.00125      +0.00125 \nEquilib:      0         0.00125       0.00125\n\nHAc and Ac-(NaAc breaks apart) for an equilibrium\n*Use Molarity when no reaction is taking place (best shown by when no volume is changing)*\n\nBoth HAc and Ac- have same molarity:\n\n(0.00125 moles / 35mL ) * (1000mL/1L) = 0.0357M HAc and Ac-\n\nUsing the equilibrium equation:\n\nEquation:  HAc   <--> Ac-     +  H+         Ka = 1.8 * 10^-5\nInitial:   0.0357M     0.0357M  \nChange:    -x          +x         x\nEquilib:   0.0357-x    0.0357+x    x\n\nKa=1.8*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]\nKa=1.8*10^-5 = [0.00357][x]/[0.0357] usually good approx.\nKa = 1.8 * 10-5 = x = [H+]\n\npH = -log [H+] = -log [1.8*10^-5] = 4.745 (3 sig figs as only after the decimal counts for logs)\n\nB) 25.0mL HAc and 20.0mL NaOH\n\n*See A for HAc moles and double the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00250      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:      0            0          0.00250\n\nCalculate molarity of Ac-:\n\n(0.0250moles /45mL) * (1000mL/1L) = 0.0556M Ac-\n\nUsing the other equilibrium equation (Kb):\n\nEquation:  Ac- + H20   <--> HAc     +  OH-         Initial:   0.0556M   -        0\nChange:    -x        -        +x        + x\nEquilib:   0.0556-x  -       +x         + x\n\nThis gives us the Kb equation\nKb = ([HAc][OH-])/[Ac-] = with approx ([x][x])/[0.0556]\nso Kb = x^2 /0.0556\n\n**but we need to calculate Kb\n**Kw = 1*10^-14 at STP I'll assume this for this problem\n**Kw = KaKb so Kb = Kw/Ka\n**Kb = (1*10^-14)/(1.8*10^-5)= 5.56*10-10 \n\nso x = 5.56 * 10^-6 = [OH-]\n\npOH = -log [OH-] = 5.253\nPH = 14 - pOH (at STP)= 8.747\n\nC)25.0mL HAc and 30.0mL NaOH\n\n*See A for HAc moles and triple the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00375      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:   0.00125            0       0.00250\n\nAfter the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.\n\nCalculate NaOH molarity\n(0.00125moles/55.0mL)*(1000mL/1L)= 0.0227M NaOH\n\nStrong Bases dissociate 100% so [OH-] = 0.0227M\n\npOH= -log[0.0227] = 1.644\npH = 14-pOH = 12.356\n\nThis should show how they calculted the half-neutralization point (A), the neutralization point (B) and after the neutralization point (C).\n\nIf you have questions on this, feel free to send me a message.  I taught this a couple weeks ago so I'm still pretty fresh on the subject.
1154934
Online conversion figures these things out for you!\n\nYou have to choose liquid or dry matter.  But the answer is there.
672778
If it is just about vertical infiltration, you can use Green-Ampht equation or simply Darcy's law if the soil is saturatted.\nThese require, soil properties such as porocity, field capacity, hydraulic conductivity and rainfall intensity.\n\nFor more realistic and regional studies, you can use \nWater balance method, Penman-montith method, \nThe above require all sorts of meteorologic information including solar radiation. The water balance method is much simpler but you need a long-time record. P=M method is very complex so you'll be better off to use HELP 3 model that kind of automates the process. This DOS based model can be downloaded at USGS website.
243163
organic debris in the soil.
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
950743
theoretical amount of product=10g lead acetate/ mw lead acetate x mw lead iodide\n\ntake the result of this formula as the denominator and divide 5.00 grams product/theoretical x 100%
969868
What parameters do you already have?  I guess you have an a/c weight? The CL equation is a standard with CL= L/(1/2 density*V^2*S)  where S is the Area.
973794
moles = concentration* volume\n\nfind the moles of the substance with given concentration and volume\n\n\nwrite and balance the equation. determine the moles equivalence of every reactant then use the equation above
422810
try this: http://www.fao.org/docrep/T0115E/T0115E0j.htm
918628
find the grams of water=5.3-2.89=2.41\nmolecular weight of water=18\nno.of moles=2.41/18\n=0.134\nhope that helps!!!\nbest of luck \nmay u get a gud score on the lab
668245
http://www.google.com.au/search?hl=en&q=soils+india&btnG=Google+Search&meta=