Ecstasy MDMA chemistry, need help with MDP2P?
968631
All information here is to be used at your own risk. The procedures documented in this file, if carried out by unlicensed individuals would violate laws against controlled substances in most countries and could result in criminal charges being filed. If carried out by individuals unskilled at chemistry they could result in serious bodily harm.\n\nhttp://www.erowid.org/chemicals/mdma/mdma_faq.shtml
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
810837
Just remember glucose (C6H12O6). Then using the basics of balancing any chemical equation you can figure out the rest of it.
128803
It depends on what kind. According to the government, ecstacy is MDMA, methylenedioxymethamphetamine. It is a mildly hallucinogenic stimulant. In reality however, most extacy is a small amount of MDMA mixed with methamphetamine, or isn't MDMA at all. It's often caffeine and DXM(dextromethorphan), the hallucinogenic chemical found in cough syrups. Ecstacy is any pill mixing a hallucinogen with a stimulant these days. Unless you know what you're doing and have a test kit, chances are you will not get a pill that's 100% MDMA.\n\nIt's not made in a bathtub, and it's not a date rape drug.\nIt's not cut with ajax, that'd be bad for business.\nIt's not cut with heroine, a dumb ass pays 20 bucks for a pill of ecstacy (me about 6-8 for pure MDMA, less if it's cut with something, proven by a test kit), and an effective dose of heroine plus the other drugs in the pill plus pressing the pill and other labor would cost more than 20 bucks a pill to make, no dealer's going to do that.
967201
Write the dissociation reaction:\n\nAg2Co3(s) <--> 2Ag+1(aq) + CO3-2(aq)\n\nSince this is a solubility equilibrium, the solubility product expression is just the equilibrium expression.\n\nKsp=[Ag2]^2[CO3]\n\nTo find the actual value, you'll need to look it up in a table.\nBTW, it's 8.1*10^-12
701172
1. molar mass of Ar: Mr = 40, so\nm = n x Mr = 8.44 x 40 = 337.6 g\n\n2. molar mass of ZnCl2: Mr = 136.4, so\nn = m/Mr = 288/136.4 = 2.11 mol (approx)\n\n3. The right formula of ethanol is CH3CH2OH and it's molar mass: Mr = 46, so\nm = n x Mr = 6.75 x 46 = 310.5 g\n\n4. Atomic mass of Na: Ar = 23, so we have:\nn = m/Ar = 5.4/23 = 0.235 mol of Na.\nAccording to the chem. equation:\n\n2Na + F2  --> 2NaF\n\n2 mol of Na react with 1 mol of F2, so\n0.235 mol of Na react with 0.235/2 = 0.117 mol of F2\n\nMolar mass of F2: Mr = 19\nm = n x Mr = 0.117 x 19 = 2.23 g of F2
215160
http://www-ec.njit.edu/~skawinsk/chem243/chap11/chap11.htm\nthis may help–near the bottom
759093
It's totally an individual thing-it depends both on the grade of the ecstasy and the chemical make-up of the person taking it. Ya know, the one man's potion is another man's poison...that it is a one time thing is pretty irrelevant. It may pass through or it may fry holes in your brain-never can tell.
198113
This might work:\n\n1. pyrrolidine (to form the enamine)\n2. Chloromethanol*\n3. water\n\n*Don't know what reactants you have, but you could start with ethene:\n\n1. ozone\n2. peroxide in HCOOH\n3. NaBH4\n4. water
133783
Some more linky's for you\n\nhttp://en.wikipedia.org/wiki/Mescaline\nhttp://www.erowid.org/chemicals/mescaline/mescaline.shtml
245982
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
636906
Functionallizing the EDTA to make a substituted EDTA
701882
Ecstasy is awesome.  Ecstasy is NOT addictive, that is a fact.  It is pretty good for sex.  You have to try it to see.  If you pay more than $20 a pill you are getting ripped off.  \n\nAsk a bartender for it, or find a black guy (no offense to blacks) in a club and ask them if they know of where you can get some.  \n\nAlso, a lot of ecstasy is adultrated (has speed or other undesireable drugs in it).  Go to www.pillreports.com, it is the best web site and is very accurate and current.  People report their experiences about different pills and if they have tested them to see if they are pure or not.  I've been on this site like every day because it is hard to find good pills here. Helped me out tremendously.\n\nNo, ecstasy is not $10-$12 a pill in the US (how long ago did you do it)? If you have a friend, maybe you can get it for $15 like I do.  But otherwise, it is $20 a pill!\n\nHow out of touch are you, Angel?  That is a completely valid question.  Web sites like www.erowid.org about drug users' experiences are extremely popular.  Where have you been hiding, under your DARE high school textbook?  :-)
112894
Haven't had chem in a while but maybe it would help to write out all the individual symbols.  Your second example would look like\n\nHHHPOOOO + MgOOHH  = MgMgMgPPOOOOOOOO + HHO\n\nthen you can see that the left side needs three Mg so you'd need 3 of the Mg(OH)2 to balance.  The left side also needs two P so there needs to be two H3PO4 there.  This makes a total of 12 H on the left, so to balance that you need 6 H2O on the right.  This also gives 14 O on the right to balance the 14 O you'd now have on the left.  So\n2 H3PO4 + 3 Mg(OH)2 = Mg3(PO4)2 + 6 H2O
655605
Apply the formula of\n\nM = molar strength wanted ( in M )\nW = molecular weight of the compound in question (in g/mol)\nV = the volume you wish to make (in mL)\nNo. = weight of compund needed (in grams)\n\napply to the formula\n\nNo. = (V * M * W)/1000 \n\nThen dissolve No. into the amoult of V you need (using a gradlated volumetric flask up to the volume you wanted)
968360
Ecstasy is so powerful.  In low doses it can be recreational and okay.  However, it is a powerful hallucinogen, and on the way down from the high, the hallucinating is really bad.  One of my friend said that he thought the trees all around him were going to eat him.  Also, the next day is really great, and you feel great about having done it.  BUT, the day after that is a shithole and you feel really really low and like you want to do it again.  This day makes the experience not worth it.
880891
you can go to wikipedia.com and search for x pills\n\nhttp://en.wikipedia.org/wiki/MDMA\n\nor you can also go to www.drugfree.com\n\nhttp://www.drugfree.org/Portal/drug_guide/Ecstasy