i have isolated organism that degrade pcp in addin this pcp only as a sole carbon source now i want to estimate the chlorine released by that organisms so im trying with silver nitrate method but, when it is titrated it suddenly changes is colour on addition of pottasium chromate so please give me the correct procedure for estimation of released chlorine send as fast as possible
126252
add 2 parts HCl 1 part H20 to dilute properly
347740
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
550844
ChemDraw from CambridgeSoft Corporation may be help you. Look at www.cambridgesoft.com
188106
Its easy...\nHeat lost by meat = heat gain by contents in calorimeter(ie water)\n\nm X s X delta(t) = M X SX Delta(T)\nm-mass of lead                  M-mass of water\ns-specific heat of lead         S-Specific heat of Water\nt=change in temp=(373-x)              t-Change in Temp = (303-x) because you dont know final temerature\ns of lead= 0.032cal/g (i think)  S of water= 1cal/g\n\nsubstitute these values .. you will  get 53 degrees C as final answer\n\nI hope you got the point.\nGood luck\n\nNote: there may be some calculation mistakes, but the theory is the same
486296
First, find out the number of moles of Ag in the AgNO3:\n.07090 L * 0.2010 mol / L = .0142509 mol AgNO3 = 0.01425 mol Ag+\nThere is a 1:1 ratio of Ag+ to Cl-, so there is 0.01425 mol of Cl- in the unknown compound.  The atomic weight of chlorine is 35.45 g/mol, so there are 0.01425 mol * 35.45 g/mol = 0.5052 g of chlorine in the compound.\n\n0.5052g / 3.000g = 0.1684 = 16.84% Chlorine
324994
Ag+ will precipitate in HCL solution as AgCl, but as for Ag(NO3), i am not so sure.
383614
Well, it depends on whether the substance can dissolve in water. Generally, for ionic compounds, here are a few rules:\nAll nitrates are soluble.\nAll chlorides are soluble except for AgCl and lead (II) chloride (which is only soluble in hot water).\nAll sulphates are soluble except barium sulphate, lead (II) sulphate and calcium sulphate (slightly soluble).\nAll carbonates are insoluble EXCEPT sodium carbonate, potassium carbonate and ammonium carbonate.\nAll compounds of sodium, potassium and ammonium are soluble.\n\nThus barium carbonate is insoluble, (s). Silver nitrate, AgNO3, is soluble, (aq).\nSilver nitrate solution, acidified with sulphuric acid, is usually used in lab to test for presence of chloride ions. Formation of a white ppt of silver chloride indicates presence of chloride ions.\nHope this helps. :)
665249
sounds like an AP question.  So you are trying to calculate Ksp for silver acetate. \n\nAgC2H3O2 <--> Ag+  + C2H3O2-\n\nSolubility will be darn low so from there set up your ice chart and go to town.
230790
That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
423162
is this your school homework,  are there suppossed to be other questions before these because my computer shows them starting from No. 4.\nnway,\n4.  filter the solution while still hot because as the soltion cools crystallization occurs, and you will nloose your product on the fiolter paper.\n5. crystallization can be induced by reduction in temperature i.e put in an ice bath or you can use a technique called seeding where you put a few crystals of the expected substance in the solution, the other method is to increase the concentration of the solution by evaporating it, to saturation point, ensure you do not evaporate it dry.\n6.i do not know what you started with so i can not tell.\n7.slow cooling will definately give bigger crystalls.\n8.sodium chloride reacts with silver nitrate to form a white precipitate, actually silver nitrate solution is used to analyse NaCl in industrial setting.\n\nope you will find these information sufficient.
107966
The aqueous solubility of AgNO3 at 20C is 67.8 (solubility is 100*m2/(m1+m2) where m2 = mass of solute and m1 = mass of solvent).\n\nRearraging the equation gives us m2 = (m1*67.8/100)/(1-67.8/100).  or\nm2 = (m1*.678)/(1-.678) = 210.56g of AgNO3 @ 20C\n\nAt 50C the aqueous solubility of AgNO3 is 79.2 so\nm2 = (m1*.792)/(1-.792) = 380.77g of AgNO3 @ 50C\n\nTo find the amount needed to make the second solution saturated we must subtract answer 1 from answer 2.\n\nWe find that 170.21g of AgNO3 must be added to make the solution saturated at 50C.
545262
these are called the classical "work horses" of plant and aniamal cells respectivly.  Regardless chlorplast is reposnsible the energy conversionin photsysnthesis and mitochondia is used in cell metabolism to provide energy through ATP-ADP energy relationships. Any basic science book would do this and so would a Wikepedia or googlesearch. I woudl suggest you go there to get spsecifics, or try the library in your neighbourhood
353846
The reaction in question is:\n\nPCl5(g) <-> PCl3(g) + Cl2(g)\n\nfor which we can write an equilibrium constant, the value of which we are told is 11.5 (this really should have units of atm):\n\nK = 11.5 atm = (p_PCl3)*(p_Cl2)/(p_PCl5)\n\nwhere p_xxx stand for the partial pressure of the species "xxx" at equilibrium.\n\nThe molecular mass of PCl5 is 208.24 gm/mol, so 3.243 gm of PCl5 contains 3.243/208.24 = 1.557*10^-2 moles of this compound.\n\nWe will assume all the gases in this problem obey the ideal gas law.  Using this law, we can calculate what the pressure of 1.577*10^-2 moles of PCl5 would be if it were sealed in a  438 ml (= 0.438 liter) container and heated to 600K.\n\nWe have that pV = nRT\n\nwhere p is the pressure\nV is the volume\nn is the number of moles\nT is the absolute temperature\nR is the universal gas constant = 8.2057*10^-2 liter*atm/(mol*K)\n\np = nRT/V\n\nPlugging in the values appropriate to this problem, we have:\n\np = [1.577*10^-2 mol * 8.2057*10^-2 liter*atm/(mol*K) * 600 K]/(0.438 liter)\n\np = 1.75 atm (only report 3 significant figures because the volume was only given with this many).  This is the answer for part (a).\n\nFor the remaining parts of this problem, let's define "x" as the degree of dissociation of PCl5 at equilibrium.  x can range from 0 (no dissociation) to 1 (completely dissociated).\n\nx = 1 - (p_PCl5 at equilibrium)/(p_PCl5 before dissociation)\n\n(p_PCl5 at equilibrium) = (1-x)*(p_PCl5 before dissociation)\n\n\nWe know from the stoichiometry of the reaction, that each mole of PCl5 that dissociates produces one mole each of Cl2 and PCl3 gas.  Because we're assuming all the gases behave ideally, this means that the partial pressures of Cl2 and PCl3 will each increase by the amount that the partial pressure of PCl5 decreases.  We also know that the reaction produces Cl2 and PCl3 in a 1:1 molar ratio that their partial pressure must be equal at equilibrium.  This means that:\n\n(p_PCl5 before dissociation) = (p_PCl5 at equilibrium) + \n(p_PCl3 at equilibrium)\n\nand\n\n(p_PCl5 before dissociation) = (p_PCl5 at equilibrium) + \n(p_Cl2 at equilibrium)\n\nThese can be rewritten in terms of the degree of dissociation of PCl5 as:\n\n(p_PCl3 at equilibrium) = x*(p_PCl5 before dissociation)\nand\n(p_Cl2 at equilibrium) = x*(p_PCl5 before dissociation)\n\nFrom above, we also have that:\n\n(p_PCl5 at equilibrium) = (1-x)*(p_PCl5 before dissociation)\n\n\nGoing back to the equilibrium constant, we have that:\n\n11.5 atm = (p_PCl3)*(p_Cl2)/(p_PCl5), \n\nwhere all the pressures refer to the equilibrium partial pressures.  Plugging in the three expressions above for the partial pressures expressed in terms of the degree of dissociation and the initial pressure of PCl5, we have that:\n\n11.5 atm = x*(p_PCl5 before dissociation)* x*(p_PCl5 before dissociation)/(1-x)*(p_PCl5 before dissociation)\n\n11.5 atm = x^2 * (p_PCl5 before dissociation)/(1-x)\n\nWe already calculated that pPCl5 before dissociation would be 1.75 atm, so:\n\n\n11.5 = x^2 * 1.75/(1-x)\n\n11.5 - 11.5x = 1.75*x^2\n\n1.75*x^2 + 11.5*x - 11.5 = 0\n\nUsing the quadratic formula to solve for x, we find that:\n\nx = 0.882 or x = -7.453.  Only the positive root has a physical meaning, so x = 0.882 (i.e., the PCl5 is 88.2% dissociated.)  This is the answer to part (d).\n\nWe can now plug this value of x back into the three equations above to calculate the partial pressures of all the gas species:\n\n(p_PCl3 at equilibrium) = 0.882*1.75 atm\n(p_PCl3 at equilibrium) = 1.54 atm\n\n(p_Cl2 at equilibrium) = 0.882*1.75 atm\n(p_Cl2 at equilibrium) = 1.54 atm\n\n(p_PCl5 at equilibrium) = (1-0.882)*1.75 atm\n(p_PCl5 at equilibrium) = 0.207 atm\n\n\nThe total pressure is given by the sum of the partial pressures:\n\np_total = 1.54 atm + 1.54 atm + 0.207 atm = 3.29 atm.
375237
You added 0.04 liter * 0.07891 mol/liter = 3.156*10^-3 moles of Ag as AgNO3 to the dissolved unknown sample.  \n\nHowever, there was excess Ag added.  The "extra" Ag amounted to:\n\n0.01127 liter * 0.1 mol/liter = 1.127*10^-3 moles Ag\n\n\nSo the net amount of Ag that reacted with the AsO4(3-) ion was \n\n3.156*10^-3 - 1.127*10^-3 = 2.029*10^-3 moles Ag.\n\nIt takes three moles of Ag for every one mole of As to make Ag3AsO4, so the precipitate consisted of \n\n2.029*10^-3 moles/3 = 6.765*10^-4 moles Ag3AsO4, which, of course, contains 6.765*10^-4 moles of As.\n\nThe molecular mass of As2O3 is 197.8414 gm/mol, but each formula unit of this compound contains 2 As atoms, while the Ag3AsO4 precipitate only contains 1 As atom per formula unit.  If we rewrite the formula of As2O3 as AsO1.5, it would have a molecular mass of 197.8414/2 gm/mol = 98.921 gm/mol.\n\nThe precipitate contains 6.765*10^-4 moles of As, and this amount of As as an oxide would have a mass of:\n\n6.765*10^-4 moles * 98.921 gm/mol =  gm As2O3\n\nIf the initial sample had a mass of 1.223 grams, then this mass of As2O3 constituted a mass fraction of:\n\n6.692*10^-2/1.223 = 5.472*10^-2 = 5.472 wt %
411153
Chlorine is a highly reactive element, and undergoes reaction with a wide variety of other elements and compounds.\n\nChlorine is a good bleaching agent, due to its oxidising properties.\n\nChlorine is soluble in water (which solution is called Chlorine Water) and this loses its yellow colour on standing in sunlight, due to the formation of a mixture of Hypochlorous Acid and Hydrochloric Acid.\n\nCl2 + H2O ==> HOCl + Hcl \nWhen Chlorine Water (i.e. a solution of chlorine gas in water) in a flask inverted in a basin of the same liquid is exposed to bright sunlight, the Chlorine is decomposed and a solution of Hydrochloric Acid remains.\nH2O + Cl2 ==> HCl + HClO\nThe Hydrochloric Acid, HClO, is not very stable and the solution readily decomcomposes, especially when exposed to sunlight, yielding Oxygen.\n2 HClO ==> 2 HCl + O2 \n\nhope this helps u to find out the answer......have fun...enjoy.......
449561
Yes, you're somewhat right.. It's just a replacement of the 6 water molecules with 6 ammonia molecules.\n\nBut for each 'step' in a synthesis you must start with the product of the PREVIOUS step. So rewrite your equations.\n\nBy the way you're missing the chlorine in your equations as well.
173200
http://www.chembio.uoguelph.ca/educmat/Chm258/enzplots.htm\n\nsee if that site helps any.  If not.  I can give you a very confused book answer from my Biochemical Calculations text.  It might be hard to explain online though, sorry.  The text I use (as a graduate student--biotechnology) is Biochemical Calculation 2nd Ed. by Irwin H. Segel, it's a great book.