STRUCTURAL FORMULA OF BUPRENORPHINE:\nBuprenorphine hydrochloride is a white powder, weakly acidic with limited solubility in water (17mg/mL). Chemically, buprenorphine is 17-(cyclopropylmethyl)-? -(1,1-dimethylethyl)-4, 5-epoxy-18, 19-dihydro-3-hydroxy-6-methoxy- α -methyl-6, 14-ethenomorphinan-7-methanol, hydrochloride [5α, 7α(S)]-. Buprenorphine hydrochloride has the molecular formula C29 H41 NO4 HCl and the molecular weight is 504.10.\n\nSTRUCTURAL FORMULA OF NALOXONE:\nNaloxone hydrochloride is a white to slightly off-white powder and is soluble in water, in dilute acids and in strong alkali. Chemically, naloxone is 17-Allyl-4,5 α -epoxy-3, 14-dihydroxymorphinan-6-one hydrochloride. Naloxone hydrochloride has the molecular formula C19 H21 NO4 HCl .2H2 O and the molecular weight is 399.87.\n\nBUPRENORPHINE HCI and NALOXONE HCL DIHYDRATE ARE AT A RATIO OF 4:1 WITH 8mg BUPRENORPHINE AND 2mg NALOXONE
218592
The two should be separable by HPLC
393267
Actually it is not a solubility problem, but a buffer.\n\nLet's use the Henderson-Hasselbalch equation for simplicity\n\npH=pKa+log[CH3COONa]/[CH3COOH]\n\nIf V is the volume of the solution in L then\n[CH3COOH]= mole/V= 0.010/V\n[CH3COONa]= mole/V =x/V\n\nso substitute in the equation and you get\n\n5.37=4.74+ log((x/V)/(0.01/V))=>\n0.63=log(x/0.01) =>\nx/0.01= 10^0.63 =>\nx =0.01*10^0.63= 0.0427 mole =42.7 mmole or since MW=82\nm= 0.0427*82= 3.50 g
103184
Assume we have 100g hydroxylamine hydrochloride. Then we would have 5.80 g H, 20.16 g N, 23.02 g O and 51.02 g Cl.\n\nFind the moles of each element in the compound: \n\n5.80 g H / 1.008 g/mol H = 5.75 mol H\n\n20.16 g N / 14.01 g/mol N = 1.44 mol N\n\n23.02 g O / 16.00 g/mol O = 1.44 mol O\n\n51.02 g Cl / 35.45 g/mol Cl = 1.44 mol Cl\n\nSo the formula is: N1.44 H5.75 O1.44 Cl1.44\n\nNow divide by the smallest number: N1.44/1.44 H5.75/1.44 O1.44/1.44 Cl1.44/1.44\n\nSo the empirical formula is: NH4OCl
455273
Pv = nRT\nn = m/M (M Molecular weight)\nPV = (m/M)RT\nP = (m/V)RT/M\nPM/RT = m/v = Density.\nPlug the numbers, watch the units and get the density.\n\nOh a numerical answer, yes, ...42.
173640
I thought there was a site on neural chemistry for kids, but the site I found was about neuroscience but did not talk about neural chemistry. \n\nfound this on the site listed below, they talk about how neural chemistry works  but it a commerical site, so they are trying to sell you something.\n\nEndorphin is the brain's painkiller, and it is 3 times more potent than morphine. Scientific research over the last several decades has led to the revolutionary discovery of opiate-like chemicals in the body that associate with opiate specific receptors in the brain and spinal cord, including Serotonin, a hormone manufactured by your brain.
323164
Osmotic pressure = MRT\nM = 7.55/(0.8206 * 298) = 0.309 M\nmoles = molarity * volume = 0.309 * 0.012 = 0.003708.\nMolar mass = mass/moles = 0.60/0.003708 = 161.8 g/mol
422055
Most of the info provided is superfluous.\nSimply put, you burned something and got 4.161g H2O (MW=18.016), all H coming from the sample.\nSo you formed 4.161/18.016 = 0.23096 mols H2O, and since each mol H2O contains 2 mols H, your sample contained 0.4619 mols of H.\n\nBTW, it is easy to deduce the empirical formula of the compound:\nNo hydrocarbon can have a MW of 60.095, and since only CO2 and H2O were formed, the compound  must contain O...this works out to be C3H8O, which means it is probably propanol (n or iso) or methyl ethyl ether.
354191
First off, you need the concentration of the NaOH solution.  From that you can determine the number of moles of NaOH that were used.  This is equal to volume (V) time the concentration, in mol/L (C).\nn = V * C = 0.02260 L * ? mol/L = ? mol\nn = 0.02200 L * ? mol/L = ? mol\n\nThe reaction of KHP (C8H5O4K) and NaOH is this:\nHOOC-C6H4-COOK + NaOH -> NaOOC-C6H4-COOK + H2O\nSo, 1 mol of NaOH will react with 1 mol of KHP.\nThat means that the number of moles of NaOH you calculated above is also equal to the number of moles of KHP.\nThe molecular weight of KHP is 12*8 + 1*5 + 16*4 + 39 = 204 g/mol\nMultiply that by the number of moles: 204 g/mol * ? mol = g actual KHP\n\nYou can divide that number by the 0.911 and 0.950 g and then multiply by 100% to find the purity.\n\nHere’s an example if your NaOH solution was 0.1 M (M means ‘molar,’ or mol/L):\nn = 0.02260 L * 0.1M = 0.002260 mol of NaOH = 0.002260 mol of KHP\n204 g/mol * 0.002260 mol of KHP = 0.461 g pure KHP\n0.461 g pure KHP / 0.911 g impure KHP * 100% = 50.6 % KHP\n\nHope this helps!
248180
(delta)H(rxn) = sum of heat of formation of products - sum of heat of formation of reactants.\n\na.  Hf Br2(g) = 30.91 kJ/mol, Hf Br2(l) = 0, so (delta)H = -30.91 kJ/mol\n\nb.  Hf CaCO3(s) = -1207 kJ/mol, Hf CaO(s) = -635.5 kJ/mol, Hf CO2(g) = -393.5 kJ/mol, so (delta)H = -635.5 - 393.5 + 1207 = +178 kJ/mol\n\nc.  Hf NO(g) = 90.25 kJ/mol, Hf O2(g) = 0, Hf NO2(g) = 33.2 kJ/mol, so (delta)H = 2 * 33.2 - 2 * 90.25 = -114.1 kJ/mol
260846
It is Oxycodone. You can see a pic of it here.\nhttp://search.drugs.com/images/pills/custom/pill5068-1.jpg
216467
Naltrexone is the generic name for Revia or Naltrexone Hydrochloride.
490949
Please see the ALL CAPS part of the medication warning below for:\n\nSUBOXONE:\n\nDrug interactions:\n\nDeaths have occurred when buprenorphine/naloxone has been misused, especially when used in combination with benzodiazepines (e.g., diazepam, lorazepam) or other depressants such as alcohol or additional narcotics.Also report the use of other drugs that cause drowsiness such as: medicine for sleep (e.g., sedatives), tranquilizers, anti-anxiety drugs (e.g., temazepam), narcotic pain relievers (e.g., codeine), psychiatric medicines (e.g., phenothiazines such as chlorpromazine, or tricyclics such as amitriptyline), anti-seizure drugs (e.g., carbamazepine), MUSCLE RELAXANTS, antihistamines that cause drowsiness (e.g., diphenhydramine).Do not start or stop any medicine without doctor or pharmacist approval.
296014
The Chemical formula for Polyvinyl Chloride is ~(CH2-CHCl)~\na better picture can be seen here at http://en.wikipedia.org/wiki/Polyvinyl_chloride#Preparation
404184
1) I know that cyclopropane is C3H6\n\n50c=323K\n22.414 l/mole at STP\n22.414*323/273/.984=26.95 l\n1l 1/26.95 moles\n1.56g 1/26.95 moles\n1.56*26.95=42 g/mole\n\nC:H mass ratio =.857+.143=6:1\nmole ratio C 6/12=.5\nH: 1/1=1\nmole ratio is 1 C to 2 H\nor CH2\nc=12, H2=2 12+2=14\n42/14=3\nso formula is C3H6
657470
methyl3-benzoyloxy-8-methyl-8-azabicyclo [3.2.1]octane-4-carboxylate  (thats the systematic name)\n\nFormula: C17H21NO4
503167
STP is 1 atm, 273 K.\n\nmoles of HCl = PV/RT = 1 atm * 1.1 L/ (0.08206 * 273K) = 0.049 moles\n\nSince HCl and NaOH is a 1:1 reaction, 1 mole HCl gives 1 mole NaCl, so you get 0.049 moles NaCl (MW 58.5), or 2.87g.
507465
1) NH(4)Cl - Ammonium chloride\n\n2) HClO(2) - Chlorous acid\n\n3) Ca(BrO(3))2 - Calcium Bromate\n\n4) BeSO(4) - Beryllium sulphate\n\n5) (NH(4)) (3)N - Ammonium nitride\n\n6) RbI - Rubidium iodide