I need to find out how Acrylonitrile is drawn as a 4-unit long "head to tail" polymer and a 4-unit long "head to head, tail to tail" polymer.\n\nThanks
209601
A polymer is a chain of the same repeating molecule(s).  Suppose the molecule is represented as A - B - C.  Let A = the head and C = the tail.\n\nA 'head to tail' polymer  would look like:\n\nA - B - C - A - B - C - A - B - C \n\nA 'head to head polymer would look like: \n\nC - B - A - A - B - C\n\nAnd a 'tail to tail' would look like: \n\nA - B - C - C - B - A\n\nNow apply these concepts to the actual structure of Acrylonitrile.\n\nBest wishes.
221760
try the site http://vivisimo.com it's a meta search engine\ntype in "energetic polymers" and you will get your results!
387042
1. The ratio of h to r is not dependent on the can being made of aluminium. The surface area is just a concept - it cannot be aluminium.\n\n2. Is a cube (assuming that no aluminium is used in forming the edges) \n\n3. You need to do this bit.\n\nFurther Investigation  - you need to do this bit , it looks as though if you limit your investigation to regular prisms that will be O.K., investigating cones would be impressive. \n\nIn so far as the wastage is concerned. If you find out what the area+wastage is for the optimally sized circular cylinders and compare with optimally sized hexagonal prisms then this may be sufficient?
670172
I don't understand 100% your question. But one of the best sources of information for any kind of chemicals is NIST.\nBelow you have some data on polyethylene.\n\nhttp://webbook.nist.gov/cgi/cbook.cgi?ID=C9002884&Units=SI&Mask=4#Thermo-Phase\nhttp://webbook.nist.gov/cgi/cbook.cgi?ID=C9002884&Units=SI&Mask=2#Thermo-Condensed\nhttp://webbook.nist.gov/cgi/cbook.cgi?ID=C9002884&Units=SI&Type=SS\n\nGood luck!
644359
Some plastics (called acrylics) have various acrylate compounds in them.\n\nMost common plastics have neither thioacrylate nor thiopropionate in them.
537157
http://www.mcponline.org/cgi/content/abstract/3/4/327
395738
Starting material:\nCH3CH2CHBrCH2Br ?\n\nProduct wanted:\nCH3CH2CH2CH=CHCH3 ?  Cis or Trans?\n\nIf the starting material is 1,2 Dibromobutane, then treating it with base (eg. NaNH2) will give:\n\n CH3CH2CHBr–CH2Br   +   base   ——>  CH3CH2CH=CHBr   +   base   ——>  CH3CH2C≡CH\n\nIf this is done with excess NaNH2 in liquid ammonia or dry diethyl ether, then the sodium salt will be produced:\n\n(CH3CH2C≡C)- Na+\n\n(CH3CH2C≡C)- Na+ + BrCH2CH3  ——>  CH3CH2C≡CCH2CH3 \n\nCH3CH2C≡CCH2CH3 when heated gives the more stable compound CH3CH2CH2C≡CCH3  (this is 2-hexyne).  The path will be different from here depending on if you want a cis or trans product.  A controlled catalytic addition (Lindlar catalyst) of H2 will give the Cis form.  Na in liquid ammonia will give the Trans.\n\nThere might be an easier way, but this path is a classic.
153981
I think you are referring to ABS Acrylonitrile Butadiene Styrene, not polystyrene.\n\nand its cost is a function of purchase volume, EAU, stocking, lead time, colorants, and other additives
428609
57) P1V1/T1 = P2V2/T2\n(temp in kelvins)\n140 x 8.00 / 293 = 1 x V2 / 273\nV2= 1043.55 L but sig figs is 3 so 1040 L\n\n93) uhh I'm just learning this myself.. very confused.. something about london dispersion, hydrogen bonding, dipole-dipole forces? perhaps someone else can take this part... and give them the 10 points I don't really care
345087
CH2OHCl: 66,45g/mol\nDo you mean 4dm^3 or a cube with 4dm length  edges?\nIf you mean 4dm^3 then\n4dm^3= 4*10^3 cm^3 = 4*10^3 mL= 4L\n2M= ?/4L\n?= 8 moles of CH2OHCl \nif you meant 4dm edges then\n4dm= 4*10^1cm\n(4*10^1cm)^3= 6*10^4cm^3 (actually it is 6,4 but because of the lack of figures you must report 6 instead of 6,4)\n6*10^4cm^3= 6*10^4 mL = 6*10^1 L\nTherefore\n2M= ?/6*10^1L\n?= 1,2*10^2 moles of CH2OHCl \n\nPart II\nCH4 + 2O2 → CO2 + 2H2O\n7dm of O (again: 7dm edges or 7dm^3???)\nfor 7dm sides\n(7*10^1cm)^3= 3*10^2 L \nO density = 1,43 g/L\nTherefore: 5*10^2g or 0,5 kg of Oxygen\n4g CH4*(1mol CH4/16g CH4)*(1mol CO2/1mol CH4)= 1*10^1g of CO2\nIf you compare the amount of Oxygen and methane and the molar ratios there is no need to do the same for O. Obviously CH4 is the limiting reagent\nCO2 density: 1,98*10^-3 g/mL\n1,98kg/m^3\nIf 5dm (again the same problem) were produced, then\nfor 5dm edges: 2*10^2g of CO2 (obviously not this one)\nfor 5dm^3: 5*10^3 cm^3 of CO2: 1*10^1g of CO2 this is it!)\nNow the yield\nIf you would have added more figures we would have had\n(9,9g/11g)*100= 90% yield\nBut because of the lack of figures we end up with\n(1*10^1g /1*10^1g)*100= 1*10^2 % (you got the maximum yield)
128486
I'll go out on a limb and suppose that you need some very simple projects that can demonstrate polymer properties to the layperson.\n\nPhysics:  \n\nIf you take a rubber band, an elastomer, suspend a weight from it, and gently heat the rubber band with something like a hair drier, you can demonstrate that polymer chains prefer to adopt a random coil configuration rather than a straight, linear configuration. (I'll leave out the basic thermodynamic details so you can delve deeper into the subject.)\n\n\nChemistry:\n\nTo demonstrate the differences in size between polymers and small molecules, look up the size of a single water molecule and then get a piece of foam rubber.  The water molecule's size is on the order of Angstroms, while the foam rubber is essentially one gigantic molecule.\n\nIf you have access to sebacoyl chloride and hexamethylene diamine, you can make Nylon-6 by using equal parts of both reactants.  The polymer forms at the interface of the two liquids, and you can use tweezers to pull the polymer out of the reaction vessel.  NOTE:  Proper safety precautions MUST be followed as both chemicals are irritants. \n\nI've left out sources and a lot of other details so you can research the physics/chemistry of the examples given here.
425580
I taught and explained the process to my children using four different colors of Lego blocks in two different heights.  I used shorter blocks to represent the C and G, and taller ones for the A and T.  If you glue them in the correct matching pairs to two pieces of ribbon, you can match up sides and part them again.  I undid the bottom half of the ribbon chain so they could see how free neucleotides were able to come in and attach.  You have to be abit careful since the Legos tend to come apart if handled too roughly, but at least they can visualize the differences and how the four types combine.   Since the legos were glued to the ribbon, I could also twist it to recreate the double helix.
525167
this looks like the closest thing that I could find that looks like how my old technical manual shows on drawing threads:\n\nhttp://honolulu.hawaii.edu/distance/aec80/threads/20c-ctl.htm\n\nI'll look for more tommorrow\n\naddon\nif you want a 3d thread look around here:\nhttp://discussion.autodesk.com/search.jspa?numResults=25&source=forumlist%7C39&q=3d+thread+spring&inputEntered=true&objID=c39\n\n3d springs are easy to do in 2007 there is a helix command that lets you make a 3d helix that you can use for the path of an extrusion
443242
Here is a good guide\n\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Limiting-Reagent.html
248472
Try searching on http://Americanplasticscouncil.org\nOr on http://www.HDPE.com
121209
In addition polymerization, many small units of the same type called monomers combine to form a long chain of a polymer, with simple covalent bonds and no loss of a molecule. For this polymerization to occur, the monomer must have a C=C double bond, because polymerization takes place across the double bond E.g. formation of polyethene from ethene; polypropene from propene etc.\n\nIn condensation polymerization, two same monomers or two different monomers combine to form a polymer, with the elimination of a small molecule like H20. Here, a double bond need not be present. E.g. formation of proteins from aminoacids etc.
36974
I use a 1 KB DNA ladder for electrophoresis and it comes with a paper to actually tell you which band represents which size fragment. Maybe you could look on the tech company's website. If you do not need to know the actual fragment size of each band, maybe you could just compare to make sure the band you need is there in the first place.