what is the oxdn no of phosphorus in pyrophosphoric acid??
603812
The H's are +1, and the O's are -2, and the whole thing is 0, so...\n4(+1) + 2x + 7(-2) = 0\nx=5, so the oxidation number of phosphorus is +5.
157808
It's a molecule of PO4.  These can link to an -OH moiety on a sugar or nucleotide molecule, or they can link to another phosphate group.  When they link to another phostphate group, the bond is a high-energy bond.  Hence AMP, ADP, and ATP are all adensine with either 1, 2, or 3 linked phosphate groups.  ATP has the most energy, and the act of removing either one or two phosphates from the ATP can be coupled with some other process taht isn't energetically favorable, in order to power important changes.  That's why ATP is used as the primary power source of the cell.\n\nThe shape of the phosphate ion is a phosphorus atom surrounded by 4 oxygen atoms.  One oxygen is double-bonded, and the other four oxygens have a negative charge (ie, H3PO4 usually loses its hydrogen in solution, and hence phosphoric acid is highly acidic).\n\nPhosphate in solution is usually denoted as Pi (P with an I subscript) meaning "inorganic phosphate."  There is a lot of this floating around in biological systems.  You'll usually see something like\n\nATP --> ADP + Pi\n\nto denote the loss of a phosphate group by adensine.
905059
It seems that the tripolyphosphate hydrolises to the orthophosphate\n\nThis is the only way I could see it balance BUT I am guessing and balancing ...\n\n3Na5P3O10 + 6H20 → 5Na3PO4 + 4H3PO4
808352
The equilibrium equation is K = (pNO)^2/(pN2*pO2) = 5.0 e-5\nThe conditions for the reaction are:\ninitial: pN2 = 0.79 atm, pO2 = 0.24 atm, pNO2 = 0 atm\nchange: pN2 = -x, pO2 = -x, pNO2 = +2x\nequilibrium: pN2 = 0.79 - x, pO2 = 0.24 -x, pNO2 = 2x\n\nSo, using the equilibrium conditions in the equilibrium equation, you get:\n\nK = (2x)^2 / ((.79-x)*(.24-x))\nK = 4*x^2 / (0.1892 - 1.03x + x^2) = 5.0 E-5\nsolving for x (I'll admit, I used the solver on my calculator)\nx = 0.0015314 or -.0015443\n\nSince the pressure of NO can't be negative, x = 0.0015314 atm\n\nThe final pressures are:\nN2: 0.788 atm\nO2: 0.239 atm\nNO: 0.003 atm
264737
Dissolve the following in 800ml distilled H2O.\n8g of NaCl\n0.2g of KCl\n1.44g of Na2HPO4\n0.24g of KH2PO4\nAdjust pH to 7.4 or 7.6 with HCl.\nAdjust volume to 1L with additional distilled H2O.
328683
http://www.indiana.edu/~bradwood/eagles/phosphate.htm\n\nhttp://pronet-internet.merck.de/Attachment/17795.ProNet.pdf?file
74307
The permanganate is reduced to Mn2+ and the oxalate is oxidized to CO2.\n\n2 MnO4- +  5 C2O42-  +  16 H+ =  2 Mn2+  +  10 CO2  +  8 H2O\n\nYou obviously know enough to ask the question so I'm assuming you can maneuver the numeric placements appropriately.  This forum does not allow me to sub or super set characters.
69413
56.4% of 220 g/mol is ___ grams of P /mol of oxide 1.\n\n43.6% of 284 g/mol is ___ grams of P /mol of oxide 2.\n\nAnalogous for the oxygen content.\n\nMultiply and round to the next whole number.\n\nOk. Oxygen has the molar mass 16. Seeing the difference between the two oxides is 64, oxide 2 holds 4 oxygen atoms more than oxide 1.\n\nLacking a convenient calculator nearby, I'll guess that 43.6%*220 g/mol yields 96 g/mol, or 6 oxygen atoms per mol of oxide 1. That leaves 124 g/mol phosphorous, with a molar mass of 31 this yields 4 atoms of P. P*4 + O*6, or P4O6, three-valent P.\n\nOxide 2 has 4 oxygen atoms more: P4O10, 5-valent P.
532375
Phosphoric Acid, H3PO4.
929172
3Mg(NO3)2 + 2Na3PO4 => Mg3(PO4)2 (prec.) + 6NaNO3\n\nAqueous solution is assumed.  The magnesium phosphate is insoluble and precipitates out.  This is normally indicated with a downward arrow right after its formula, but no such symbol exists on this keyboard.
67952
Dictionary of chemical formulas \nThis is a list of chemical compounds with chemical formulas and CAS numbers, indexed by formula. This complements alternative listings to be found at list of inorganic compounds, list of organic compounds and inorganic compounds by element. If you are adding to this list please consult the rules by which the list is organised.\n\n\n\n\nAlPO4    Aluminium phosphate
439611
Yes!\n\nComparison of the Ammonium Oxalate Extraction of Fertilizer for K2O, Method 983.02, with the Ammonium Citrate/EDTA Extraction, Method 993.31 \nNatalie F. Newlon \n\nThe direct available method for phosphorus in fertilizers, Method 960.03, is an Official Method for extraction of both P2O5 and K2O. Because K2O is much more soluble and easier to extract than P2O5, the conditions of the extraction procedure are set for efficient extraction of P2O5, and the extraction of K2O is, in effect, a bonus. In 1993, a modified extraction procedure, using an extractant of ammonium citrate/ethylenediaminetetraacetic acid (EDTA) solution, was evaluated for P2O5 and then collaboratively studied. This now-modified extraction procedure for phosphorus is Official Method 993.31, which is used extensively as an alternative to Method 960.03 for phosphorus because it is less labor-intensive. Method 993.31 was also evaluated in our laboratory for K2O extraction, but it was never formally collaboratively studied for K2O. To demonstrate the suitability of Method 993.31 as an extraction procedure for K2O, as well as P2O5 in fertilizers, Method 993.31 extraction for K2O was compared with Method 983.02, the traditional ammonium oxalate extraction procedure for K2O. A wide range of fertilizer samples was extracted by using both procedures. The t-test for samples containing <30% K2O had a t-value of 1.23 and a probability of 0.22, which indicates there was no significant difference in the results obtained by the 2 extraction methods. The t-test for samples containing >30% K2O had a t-value of 0.30 and a probability of 0.77, also indicating no significant difference. It is recommended that Method 983.02 be modified to include the ammonium citrate/EDTA extraction procedure as an additional option for extraction of K2O from commercial fertilizers.
766811
d. \n\nIt's a double replacement (acid/base) reaction.  The Sr and the H are both positive and they switch.  Since Sr has a 2+ charge and the PO4 has a 3- charge, you use a common multiple of 6.  The other product would be H(OH) or water.
700768
Count the hydrogen and oxygen atoms produced by the reaction.  In 2 mol H3PO4, you have 6 H's.  In 6 mol NaOH, you have 6 H's and 6 O's (you actually have six OH(-) molecules, but I don't want to get too deep here).  Twelve H's and six O's make six H2O's.\n\n(c) 6.0 mol
995333
just use it frequently, and you will get familiar with those name\nBut a i have a little trick\nin the most of negative ion with -ate ending, the non-oxygen atom usually has the oxidation number with the same number with its group number\ncarbonate (CO3)2-, C's ox number=+4,\nsulphate (SO4)2-, S's ox number =+6\ntry nitrate,phosphate,etc.\nbut oxohalogen anion is exception\nperhalat (XO4)-(e.g perchlorate=(ClO4)-), halat (XO3)- (e.g chlorate (ClO3)-, halit (XO2)-, hipohalit (XO)-\nThe non-oxygen atom in negative ion with -ite ending are usually has ox number of 2 number lower than that of -ate ions
1022781
adenosine tri-phosphate\n\nSignificance: http://en.wikipedia.org/wiki/Adenosine_triphosphate
773855
Well this is not a typical acid-base reaction in aqueous solution.\n\nThis is cooking the hell out of the calcium carbonate.  I guess it is more thermodynamically stable as calcium oxide.  The carbonate moiety is not stable in that heat.  An oxyen stays with the calcium and the carbon and oxygen reassemble as CO2 as they leave the compound.\n\nedit\n\nThink you would get same pattern with other alkaline-earths/polyatomic ion compounds like BaCO3 or even alkali metals and transition metal compounds, although with different molar quantities like Li2(CO3).  Sulfates, chlorates would probably do the same thing.