i am doinng my coursework and i really want to know the exact molecular formula of nitrophenol phosphate to find the concentration ...
218020
I assume you are talking about the formula and not the structure.\n\nThe molecular formula is C6H5NO5P\n\nStructure (if in case you want it too) is as follows:\n\nDraw a benzene ring (phenyl group) minus 1 hydrogen (C6H5). To the Carbon without the hydrogen, attach a Hydroxyl group (-OH). On the other Carbon positions, whether it is the ortho positions, meta positions, or para position, erase the hydrogen and attach a nitrate group (-NO2). On the Oxygen with the double bond to Nitrogen on the Nitrate group, erase one of the bonds and attach this bond to the phosphate group (-PO3).\n\nI'm not sure if you are talking about a nitro-phospho bond or a hydroxy-phosphate bond, but the difference is only 1 hydrogen.
537598
treat it as a quadratic equation and use the quadratic formula:\n\nfirst subtract A from both sides:\n2pr^2 +2prh - A = 0\n\nthen plug the coefficients: 2p, 2ph, -A into the\nquadratic formula for a, b, c, respectively and simplify:\n\nr = [- b +/- sqrt(b^2 - 4ac)]/(2a)\n\ni'll leave it to you to do the tedious steps to get\nr = [ph +/- sqrt(p^2h^2 + 2pA)]/(2p)
383299
2P + 5S ==> P2S5\nBalanced equation\n3.2g of P = 3.2/30.97molar mass= 0.103 mol\n4.1g of S = 4.1/32.065molar mass= 0.128 mol\nand in your balanced equation you have a 2:5 ratio where 2 moles of phorphorus react with 5 moles of Sulfur.\nTherefore you have excess of phorphorus and a limit of Sulfur since you have 0.103 mol of P and its a 2:5 ratio with the Sulfur which only has 0.128 mol
530173
A=E*d*C\n\n(d=length of the path of the light through the solution)\n\nIf you have only one value of absorbance then obviously there is no way of determining the concentration without the molar extiction coefficient (E).\n\nIn practice even if you know E, you don't use it to find the unknown concentration. Instead you measure the A of standard solutions and do a standard curve. From the equation of that curve (which you find using linear regression) and the value of A of your sample you calculate the unknown amount. \nThe standard curve doesn't need to be A vs mole/lt. It can be A vs anything e.g. g/ml, mg/ml, ppm or even just g, mole, etc. the only thing that changes is what the slope represents.\nFor example for A vs mole/lt the slope corresponds to E*d, for A vs g/lt the slope corresponds to E*d/MW, etc.\n\nIf you  always have the same volume of sample you can do the curve A vs grams directly. Then if V is the volume of the sample, the slope corresponds to E*d*V/MW\n\nEven if your standard curve is expressed in moles/lt you can convert it to grams. You just need the MW of the compound and the volume, not E...
550260
2H3PO4+6NaOH-------->6H2O+2Na3PO4\n\nThere has to be two moles of phosphoric acid and six moles of sodium hydroxide for it to balance out
343336
1) Balance the reaction equation:\nPb(NO3)2  + 2 LiBr -> PbBr2 + 2LiNO3\n\n2) Calculate the amounts we start with\nPb(NO3)2 :  2.05 L * 0.0550 mol/L = 0.1128 mol\nLiBr : 1.52 L * 0.0618 mol/L = 0.0939 mol\nAs we need 2 moles of the latter for 1 mole of the former, the reaction is limited by the amount of LiBr\n\n3) Calculate the amount of PbBr2 produced\n0.0939 mol of LiBr will react with 0.470 mol of Pb(NO3)2 to produce 0.0470 mol of PbBr2. This is 0.0470 * (207.2 + 2* 79.9) =17.249 gram\n\n4) Calculate the yield\nThe yield is 15.784 / 17.249 = 91.51%
534322
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
536609
We will need to find if there is a limiting material.\n\nCO + 2H2 ------> CH3OH\n\n(1.61 g H2)*(1 mole H2/2.016 g) = 0.7986 mole H2\n\n(0.7986 mole H2)*(0.5 mole CH3OH/mole H2) = 0.3993 mole CH3OH at most\n\n(15.95 g CO)*(1 mole CO/28g) = 0.5696 mole CO\n\n(0.5696 mole CO)*(1 mole CH3OH/1mole CO) = 0.5696 mole CH3OH at most\n\nH2 is the limiting reagent, therefore:\n0.3993 mole CH3OH * (32.032 g/mole CH3OH) = 12.79 g\n\nPlease check for errors, I went by memory on what the atomic and molecular weights were, you may want to check this.
352912
If you have the reaction:\n\naA + bB ↔ cC + dD\n\nThen:\n\nKp = ((PC)^c * (PD)^d) / ((PA)^a * (PB)^b) (PC is the partial pressure of C, PD is the partial pressure of D, etc.)\n\nKp = 109 = ((PNOBr)^2) / ((PNO)^2 * (PBr2)^1)\n\n109 = (0.0717)^2 / ((PNO)^2 * 0.0108)\n\n1.18(PNO)^2 = 0.00514\n\n(PNO)^2 = 0.00436\n\nPNO = 0.0660 atm\n\n*** The partial pressure of NO at equilibrium = 6.60 x 10^(-2) atm
316676
Here's how to solve it (I am convinced that you should do your own homework, but hints are OK as long as you learn something):\n\nStart solving by finding the molecular mass of the compound.\nA. Volume occupied by one mol of gas is 22.4 L at standard temperature and pressure (STP: 760 mmHg and 273 K (0 C))\nTo find the volume of gas that weighs 2.57 g at STP, use formula: P1V1/T1=P2V2/T2, where R is 0.08206, P1=319 mmHg, V1=2 L, T1=298 K; P2=760 mmHg, T2=273 K and V2 is unknown (it will be somewhat less than 1 L).\nAfter you get this volume, see how many mols do you have in it: divide 22.4 by V2 you just got. Then multiply this number by 2.57 and get the molecular mass of your compound (M(X)).\n\n2. Multipy M(X) separately by 0.6481, 0.136 and 0.2159 to get number of atomic weight units that are allocated to C, H, and O respectively. \nThen divide first number by 12, 2nd by 1 and 3rd by 16 to see how many atoms it amounts to. The resulting integer numbers are X, Y, Z.\n\n3. CxHyOz is the empirical formula of your compound. Google it to get the possible molecular ones. This is a trick question, so there will be more than one but less than 8. All of them are alcohols or ethers.
669677
P4 is the pure amorphic form of phosphorous (white) in which 4 atoms of phosphorous are bonded together in a tetrahedion form. in this form all vallience electrons have been used.\n\nin a compound such as sodium phosphate NaPO4 there is only one atom of sodium and phosphorous and there are four atoms of oxygen directly connected to the phosphorous.
500145
That is the molecular formula for nitrous acid, written HNO2.
71101
To do this, you need to employ the ideal of PV=nRT.  You have the values for P, V, R, and T.  Plug those into the equation to find moles.  Then, plug the number of moles in with STP values, and solve for Volume at STP.  That's it.
137554
molecular weight's 305.4188 g/mol
210282
go here and you'll find all your answers and more\n\nhttp://en.wikipedia.org/wiki/Phosphorus
462343
2 HNO3(aq) + Sr(OH)2(aq) --> Sr(NO3)2 (aq) + 2 H2O\n\n2 H3PO4(aq) + 3 Ba(OH)2(aq) --> Ba3(PO4)2 (cr)* + 6 H2O\n\n* Barium Phosphate will precipiate out of an aqueous solution in this reaction
253862
If you know the grams, which you do on one...its easy.  To find the grams from ml, you need to know the density.  Density is in grams/ml, so if the density was 2 g/ml and you had 4.5 ml, then you would have 9 grams.  So you take the grams and you need to find moles to calculate percent yield.  To do this, find the total molecular weight (for instance, water is 18 b/c of two H's which are 1 and one O which is 16).  When you have the grams and molecular weight, divide the grams by the molecular weight.  This will give you the moles, since the molecular weight is grams per mol.  Do this for every reagent or thing you use.  Whicever one has the lowest amount of moles is your limiting reagent because once that is gone, the reaction stops!  Now the theoretical yield is found by doing stoiciometry.  So whatever your limiting reagent is...that number is your theoretical yield for this problem.  Then weight the product when you are done, and find the amount of moles you have by oncve again dividing how many grams you have by the molecular weight.  Then divide that by your Theoretical yield and that is your percent yield!  Looks harder than it is.