HI THERE,I need urgent help with this problem,thanx in advance,\n\nComposition of a tablet is 325mg acetylsalicyclic acid(ASA),1000 mg citric acid, and 1916 mg sodium bicarbonate. ASA is monoprotic(molar mass=180 gram/mol), Citric acid is triprotic(molar mass=192 g/mol), sodium bicarbonate is a monofunctional base(molar mass= 84 g/mol).When tablet is dropped in water the components dissolve and react.\nCalculate the number of moles of H+ due to acetylsalicyclic acid and citric acid.Calculate the number of moles of base due to sodium bicarbonate.\nWrite the net ionic equations for the reactions of sodium bicarbonate with ASA and citric acid.\n\nThanx again
123573
What are the pka's of all the different species? It sounds like the end product is a mess of conjugate bases, so the pka's are needed to figure out which disassociation reaction will predominate.
363967
_NaOH(aq) + _Fe(ClO4)3(aq) => _Fe(OH)3(s) + _NaClO4(aq)\n\n3  NaOH(aq) + 1  Fe(ClO4)3(aq) => 1  Fe(OH)3(s) + 3  NaClO4(aq)\nThe coefficients of one could be excluded but I put one which is also equivalent.\n\n_((NH4)2)SO4(aq) + _Pb(NO3)2(aq) => _PbSO4(s) + _NH4NO3(aq)\n\n1  ((NH4)2)SO4(aq) + 1  Pb(NO3)2(aq) => 1  PbSO4(s) + 2  NH4NO3(aq)\n\nThese are only the balanced equations.\n\nThe following is the ionic equations:\n(Na)+ (aq) + (OH)- (aq) + (Fe)3+ (aq) + (ClO4)- (aq) => Fe(OH)3(s) + (Na)+ (aq) + (ClO4)- (aq)\nand\n(NH4)+ (aq) + (SO4)2- (aq) + (Pb)2+ (aq) + (NO3)- (aq) => PbSO4(s) + (NH4)+ (aq) + (NO3)- (aq)\n\nThe ionic equations are already balanced.  It is simply the breaking down of the equations into ions.  The solid remains intact while the compounds in aqueous solution become ions.\n\nThe + and - signs are the charge of the ions.  I can't find any other way to write them so I separated the compound with parentheses.  Sorry if it's confusing.\n\nThe net ionic equations are "simplified" ionic equations.  Basically, the things that are repeated on both sides get crossed off.  What is remaining is simply the net ionic equation.\n\n(OH)- (aq) + (Fe)3+ (aq) => Fe(OH)3(s)\nand\n(SO4)2- (aq) + (Pb)2+ (aq) => PbSO4(s)
411718
shawty's NOT good at stoichiometry...has made a fundamental error in the final line so end result is rubbish...\nhow can you end up with MORE Cu than you started with when it's 2 moles : 2moles? \n\n(8/159)x 127 = 6.4 g
525814
11/27:Please post the answer so I can find the error in my work or determine if the answer key is wrong.  Lets see if we can get to the bottom of this one!\n\n\nFirst find the masses of each acid:\n\nHCl:  48.8 ml x 1.19 g/ml x 0.38 = 22.1 g\n\nHNO3:  17.5ml x 1.42 g/ml x 0.70 = 17.4 g\n\nSecond, convert grams to moles:\n\nMass HCl = 35.45 + 1.01 = 36.46 = 36.5 g/mol\nMass HNO3 = 16.0 x3 + 14.0 + 1.0 = 63.0 g/mol\n\nmoles HCl = 22.1 g/36.5 g/mol = 0.605 mol HCl\nmoles HNO3=17.4g/63.0 g/mol = 0.276 mol HNO3\n\nSince you're diluting to one liter, these are also the molarities.\n\n[HCl] = 0.605 mol/L ; [HNO3] = 0.276 mol/L\n\nEach of these acids are strong acids and can be assumed to dissociate completely.  If they were weak acids we would have to account for the acid dissociation equilibria here.  \n\nEach of these acids contribute only 1 equivalent of hydrogen ion.  Thus [H+] is the sum of the two concentrations.\n\n[H+] = [HCl] + [HNO3] = 0.605 + 0.276 = 0.881 mol/L = 0.88 M\n\nReport two sig figs since your mass % vals only have two sig figs.\n\nThe equilibrium equation for water, H+ and OH- is:\n\nH2O <--->  H+  + OH-   K = 1.00 x 10^-14\n\nH2O is a pure liquid and by definition has an activity of unity, so the equilibrium expression is:\n\n[H+][OH-] = 1.00 x 10^-14\n\nNow we can solve for [OH-] :\n\n0.881 X [OH-] = 1.00 x 10^-14\n\n[OH-] = 1.14 x 10^-14 mol/l = 1.1 x 10^-14 M\n\nAgain, two sig figs in final answer.\n\nFinally, pH = - log [H+]\n\nso pH = - log 0.881 = 0.055
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
266815
Well, let's see here.  You have a "2-liter" bottle that is really 2.10 liters, with 250 ml of water in it.  You want to add enough Alka Seltzer tablets to generate 4.50 atm at 28°C.  I presume you're thinking of making some kind of soda bottle rocket . . .\n\nThis involves using the Ideal Gas Law:  PV = nRT, and what you need to solve for first is the number of moles (n) of CO2 that you need to generate to make the pressure in the bottle 4.5 atm.\n\nWe'll assume that the bottle is rigid -- the volume won't change -- and that the temperature is constant at 28°C.  So there's three of the four variables in the equation:\n\nn = PV/RT\n\nP = 4.50 atm\nV = 2.1 liters - 0.25 liters (the volume of the water), so \nV = 1.85 liters\nT = 28°C = 291K\nR = 8.20568x10^-2 liter-atm/K-mole\n\nso n = (4.50 atm)*(1.85 liter)/(291K)*(8.20568x10^-2)\nn = 0.35 moles of CO2 are required\n\nNow, you have Alka Seltzer tablets that are 35% sodium bicarbonate -- that is where the CO2 will come from.  Assuming all of your sodium bicarbonate will react (100% return), according to the reaction:\n\nHCO3- + H+  =====> H2O + CO2\n\nyou would need 0.35 moles of HCO3- to generate 0.35 moles of CO2.  All of that would come from NaHCO3:\n\n0.35 moles of NaHCO3 = 0.35 * 84.007 grams/mole = 29.4 grams of sodium bicarbonate\n\nIf one Alka Seltzer tablet has a mass of 3.065 grams, and 35% of it is bicarbonate, then you have 1.07 grams of bicarbonate per tablet.  \n\nSo you need *at least*\n\n29.4 grams/(1.07 grams per tablet) = 27.48 tablets\n\nto generate enough CO2 to get a pressure of 4.5 atm.  Make it 27.5 just to be sure.\n\nThat's a lot of Alka Seltzer!  Stand back, 'thar she blows!"
526510
You need to make sure that there is the same number and kind of atoms on each side of the arrow.\n\nCa(OH)2 + MgCl2 ---> Mg(OH)2 + CaCl2, both Mg and Ca stay+2\n\nBaO2(s) + H2SO4 ---> BaSO4(s) +H2O2 \n\nAg2S --->2 Ag+  +  S-2\n\nHC2H3O2 + MgCO3 (s) ---> H2O + CO2 + Mg+C2H302-\n\nI may not have written net ionic form, but I did get all the atoms on each side I am unsure of the reaction producing hydrogen peroxide, but you do have to account for all atoms.
606097
mix concetrated sulphuric acid to the sodium acetate, the reaction will produce acetic acid and sodium sulphate.\n\nif you take all the water out of the sodium acetate, you will end up with 100% pure acetic acid.\n\nH2SO4 + CH2COONa => CH3COOH + Na2SO4\n(unbalanced equation)
219453
AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit
63081
[Ca(NO3)2]=n/V  ---> n=[]*V= 0.390M * 0.044L=0.01716 mol\nn=m/M ---> m=n*M= 0.01716mol * 164 g/mol=2.81424 g
529024
I'm not going to solve these for you.  My homework days are far behind me.  I will try to teach you how to solve them, though.\n\n1.  Keep your eye on the moles and don't get confused by the changing volumes.  The final solution is 0.00436 M  and you know the final volume so calculate how much KNO3 is there (=0.00436 M x Vf, in liters).  That will tell you how much was in 25 mL of the second solution.\nSo how much was in the total second solution?  Multiply the moles above by five (25 mL goes into 125 mL five times.)  This is also how many moles was in original 20 mL solution.\nSo, to find the molarity of the original solution, divide the number of moles by the volume, in liters (20 mL = 0.02 L).\n\n2.  Same thing goes for this one.  Multiply 1.66 M by .037 L (remember that M is in terms of liters, not millilieters!) to get the moles in the first solution.  Do the same for the second.  Add the two volumes together to get the resulting volume (in liters!!!)  Divide the sum of the moles in both solutions by the total combined volume and you have your answer.\n\n3.  Same would go for this one if you hadn't left out what volume of solution the Pb++ was in.  Anyway, the way to do this one is to notice that the reaction of Pb+2 ion and SO4-2 ion is a one-to-one proposition.  So, if you figure out how many moles of Na2SO4 you have in 0.009 grams of the stuff, you will also calculate how many moles of Pb(NO3)2 you have.  Divide that by the volume in liters and you have your answer.  [Nice touch by Teach to throw in univalent species to cross you all up.  These are irrelevant to the reaction you're considering.  Still, I like that.  Makes you think.]\n\nGood luck.
476263
you should learn how to do your own homework hun you'll never be a chemist with someone else doing your work.\n\n\nuse the molar mass from the periodic table (units = g/1mol)\n\nCl = 35.5\nH = 1.00\n\nMolar mass, formula weight, molecular weight, formula mass = 36.5 grams HCl/1 mol HCl\n\n\nThis is a simple problem is you can't do this problem maybe you should try changing majors.  I hear history or english is nice.
265564
Ba(NO3)2 + Na2CO3-->BaCO3 +2NaNO3\n\nto do a double displacement follow this formula\n\nAB + CD --> AD + CB
481031
Okay first, you have to deduce how many moles of acid react with the base.\n\n1 mol HCl ----------> 1 Mol NaOH\n\n1 H+ --------> 1 OH-\n\nTherefore:\n\nMolarity Acid X Volume Acid = Molarity Base X Volume Base\n\nMaVa = MbVb\nMa(20cm3) = 1.2 X 25 cm3\nMa =  30/20\nMa = 1.5 moles per dm-3\n\nThat gives the MOLARITY of the acid. The concentration is in grams per dm-3. \n\n1.5 moles of HCl = (Molar Mass) X 1.5\n                            = 36.46 X 1.5\n                            = 54.69g\n\nSo effectively;\n\nThe Molarity = 1.5 moles dm-3\nThe Concentration = 54.69 grams per dm-3\n\nWhich ever you need!
68864
To determine pH for this mixture, one must know the molarity for the HCl, the molarity for the acetic acid, and the Ka for the acetic acid.  The goal is to find the concentration of the hydronium ion in solution.  Thus, for a strong acid, since it completely dissociates, the molarity equals the concentration of the hydronium ion.  But, it is more complicated for the acetic acid, which is weak.  We need to solve for the hydronium concentration, so we set up a Ka equilibrium equation, which with all the values added looks like:  Ka= (x^2)/ (Molarity of Acid).  Solve for x.  This gives you the hydronium ion concentration for the weak acid.  Add the two concentrations together and take the negative log of that number.\nRemember to use correct significant figures.  Also, it is not necessary to subtract x from the molarity of the acid in the equation, because since the Ka is small enough, when the math is done the percent ionization is less than 5%, so the x can be ignored without incurring significant errors.
68419
When a salt derived from a strong acid and a weak base dissolves in water, the solution becomes acidic. For example, consider the process\n \nNH4Cl(s) ----H2O ---> NH4+(aq) + Cl-(aq)\n\nThe Cl- ion has no affinity for H+. The ammonium ion NH4+ is  a weak conjugate acid of the base NH4+(aq) and ionizes as follows \n\nNH4+(aq) + H2O(l) <=> NH3(aq) + H3O+(aq)\n\nor simply  \n\nNH4+(aq) <=> NH3(aq)  + H+(aq)\n\nThe equilibrium constant ( or ionization constant) for this process is given by \n\n\n\nKa = Kw / kb  =   1.0*10^-14 / 1.8*10^-5 = 5.6*10^-10\n\n\n      x^2 / .10 =  5.6*10^-10\n\n           x^2  =   5.6*10^-11\n\n             x  =    7.5*10^-6 M  =  [H+]\n\npH  =  -log [H+]\n    \n    =  5.12\n\n     my answer is  C , NH4Cl
232765
Write out the balanced reaction which occurs.\n\nSince both NaOH and H2SO4 are strong electrolytes (strong base, strong acid), we can assume both dissassociate completely and the reaction produced Sodium Sulfate and water,\n\n2NaOH + H2SO4 --> 2H2O + Na2SO4\n\nFind limiting reactant in order to find which  will be left over (either NaOH or H2SO4).\nIn order to do this, you need to know both the concentrations of the NaOH and the H2SO4 in order to find the initial number of moles of each.\n\nOnce you know how many moles of either OH- or H+ are left over, you can then find the pH, or pOH like this,\n\npH = -log ([H+])\npOH = -log ([OH-])\n\npH + pOH = 14\n\n*Remember to take into account that the final volume of solution is the sum of the intial volumes of each othe reactant solutions when calculating pH.\n\nEDIT:\nYou know, it just occured to me, I dont know why I even brought up pH, that was not what the question was asking.  but everything I said still stands in order to calculate the molarity...you just dont have to go the one extra step and calculate the pH by taking the negative log of the molarity concentration of H+ ions.\n\nAlso, for all of you out there who are just given answers with not other explanation, it would be nice if you stated what the concentration was of....salt? NaOH?  H2SO4?\nBecause of this ommision, I would highly inclined to disreguard their answers as numerically incorrect as well.