Question related to ppm (parts per million)....?
919420
3.2 ppm is 3.2 mg per liter.  Multiply by 0.47 liters (470ml) to get 1.5 mg or 0.0015 g.
458856
I don't quite understand the question. I am a project engineer for a chcolate manufacturer.. $3m will not get you far if you need to build a factory
1041556
The answer to your question is indeterminate, for several reasons\n1) This is a trade secret that competitors would like to know the answer to.  Hire a trade spy.\n2) What are the L x W x H dimensions of the product?. Costs will be determined by the size of the "bath" vessel required.\n3) The surface area-to-volume ratio will determine the melt-spread time, product cooling time required, etc.\n4) Is the production order for 1 ton per day? 2 tons? 10 tons?\n5) WHAT Clarity (letter) per WHAT Rating per WHAT GPS Standard (#) ?
950446
Cost to produce/market the product.\nGoing rate for similar products.
1039016
if 1 ppm = 1 g/m3, 1 ppb would be 1 mg/m3, because 1 g/m3 = 1000000 ug/m3.\nppm = parts per million, billion\ng = grams, mg = milligrams (10^-3), ug = micrograms (10^-6)\nm3 = cubic meters
144611
You might find this paper useful\nhttp://arxiv.org/abs/math-ph/0106027
757978
Like the person said above.  0.038% of the atmosphere is made up of CO2.  That's 3.8 parts per million.  So that's why it won't take very much for people to double the amount of CO2 that is naturally occuring.
670957
parts per million are 10 000 times smaller than parts per hundred (percent).  divide parts per hundred by 10 000 to get parts per hundred.\n\nsee http://en.wikipedia.org/wiki/Parts_per_notation
847153
PV = nRT  .... n/R = P/RT\n\n800 torr x (1.3158×10^−3 atm/1 torr) = 1.053 atm\n\nn/R = (1.053 atm)/(8.206 x 10^-2 L atm/Kmol  x 298 K)\nn/R = 0.043 mol/L\n\nNH3 mass is (14 + 1x3)=17 g/mol\n\ndensity = n/R but we need it in g/mL\nn/R = 0.043 mol/L x (17 g/mol NH3) x (1L/1000mL)\nn/R = 7.31 x 10^-4 g/mL
944167
PV=nRT\nV = nRT/P\nV = (114g/44g/mol)(0.08206L-atm/mol-K)(28+273K)/0.987atm\nV = 64.84L or 0.0648 m^3
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%
1171109
Title : Biodegradable Plastics\n\nSummary : Plastics have become an important part of modern life and are used in different sectors of applications like packaging, building materials, consumer products and much more. Each year about 100 million tons of plastics are produced worldwide.\n\nDemand for plastics in India reached about 4.3 million tons in the year 2001-02 and would increase to about 8 million tons in the year 2006-07. Currently, however, the per capita consumption of plastics in India is only about 3 kg compared to 30-40 kg in the developed countries. The present market in India is of about Rs. 25,000 crore.\n\nHazards of Plastics: Most of today's plastics and synthetic polymers are produced from petrochemicals. As conventional plastics are persistent in the environment, improperly disposed plastic materials are a significant source of environmental pollution, potentially harming life. The plastic sheets or bags do not allow water and air to go into earth which causes reduction in fertility status of soil, preventing degradation of other normal substances, depletion of underground water source and danger to animal life. In the seas too, plastic rubbish - from ropes and nets to the plastic bands from beer packs -choke and entangle the marine mammals.\n\nBio plastics are biodegradable plastics, whose components are derived from renewable raw materials. These plastics can be made from abundant agricultural/animal resources like cellulose, starch, collagen, casein, soy protein polyesters are triglycerides. Large-scale use of these would help in preserving non-renewable resources like petroleum, natural gas and coal and contribute to the problems of waste management. Biodegradable plastics degrade over a period of time if exposed to sun and air.\n\nThough the demand for biodegradable plastics is increasing, acceptance of biodegradable polymers is likely to depend on factors like\n\n Customer response to costs; \n Possible legislation by governments; and \n The achievement of total biodegradability. \n\nImmediate application areas identified in India for biodegradable plastics are Agricultural Mulch, Surgical Implants, Industrial Packaging, Wrapping, Milk Sachets, Foodservice, Personal Care, Pharmaceuticals, Medical Devices, Recreational, etc.\n\nTIFAC report on biodegradable plastics highlights the Indian efforts in this direction, as well as activities at some of the major centres of development in USA/Canada, Germany, Scandinavian countries and Japan. This report also gives brief details of biodegradable plastic technology developed by University of Nebraska, Chronopol, Inc., EPI Environmental Products, Inc., Isolyser Company, Inc., Solpax Limited, Novamont, Federal Cartidge Company and a few other companies. In addition, the reader would find lists of worldwide manufacturers of polymers and additives, international organizations and standards for testing biodegradability.
130925
if you get the million first, the second part is easier. \n\nyeah you don't ask for much ...LOL
672926
ask the suppliers to give you this information if they have not done so already.\nThe manufacturers should/must have and be able to provide this information.
501615
a 12.5% solution is 12.5 parts per 100.  To convert to parts per 1,000,000 you multiply by 10,000.\n\nSo 12.5% solution is 125,000 parts per million.\nTo dilute to 250 parts per million, you need a factor of \n125,000/250 dilution or 500.\n\nSo, depending on how much of the final solution you need, do a 500x dilution.  For example, I would use a volumetric pipet to transfer 1.00 mL of the 12.5% solution into a 500 mL volumetric flask, then dilute to the mark.
644701
Summary: convert your concentration to a mole, mass, or volume fraction and multiply by 1 million.\n\nDiscussion:\nParts per million must be measured in consistent units:\n\nmass per 1 million mass units\nvolume per 1 million volume units\nmoles per 1 million mole units\n\nFor example, if you add 1 milligram of substance to 1 kilogram of solution, the concentration is:\n\n1 mg/1 kg = 0.001 g / 1000g = 1e-6\nMultiply the resulting mass ratio by 1 million and you convert this to ppm\n1e-6*1million = 1 ppm\n\nI don't think it is correct to directly use molarity or molality times 1 million, since both of these are ammount/volume: the units are not consistent.  You have to convert the "volume" in the concentration to a "mass" or "molar" amount consistent with the concentration measurement.\n\nIf you have a concentration in terms of a mole fraction, it is just the mole fraction times 1 million.  A mole fraction of unity (one) is the same as 1 part per part (pure!).\n\nIf you have a concentration in terms of a volume fraction or mass fraction, you just multiply by a million.  It is important to note that whether you do parts by mole, volume, or mass you will get different numbers for a given system.  It is important to be clear which one you are talking about.  You can read more on wikipedia, look up ppm.
363722
Please see this website: http://www.ussteel.com/corp/sheet/hr/pmhsmill.htm