How can we titrate nitric acid with hyrdofluoric acid mixing ratio 4 nitric acid and 1 hydrofluoric acid?
462489
put one of the acids in the titer, the other in the flask underneath.  Use a magnetic stirrer in the flask to ensure good mixing and ensure that you take required precautions (like goggles gloves lab coat ...possibly in a fume cupboard).\n\nThe two acids don't react with each other but is this 4:1 mix not acqua regia?
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
929685
When you have weak electrolytes, (equillibria in general) you should make the following table.\n\n1. write the reaction\n\nHNO2 <=> H+ + NO2-\n\n2.Put the initial amounts (expressed as concentration) of each molecule. In this case we have only the concentration of HNO2, so\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\n\n3.Assume that an unknown amount "x" of the reactants react.\nThen write also how much of each product you expect to be produced according to the stoichiometry. In this case 1 mole of HNO2 gives one mole of H+  and NO2-, so x mole/L HNO2 will  give x mole/L of each and our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\n\n4. Ka is an expression for the concentrations at equilibrium. So let's see what are those. For the reactants you will have initial-what reacted. We have only HNO2 with 0.5-x. For the products it will be the initial+what was produced. We didn't have any intial so it is only what was produced (x for both of the products). Thus the table in its final form is \n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. .. .. x .. .. .x\n\nKa= [H+][NO2-]/[HNO2] = x^2/(0.5-x)\nNow you either solve the quadratic equation and find x, or we assume that 0.5 >>x and practically 0.5-x=0.5.\nThen Ka=x^2/0.5=4*10^-14 =>\nx=squareroot(2*10^-14)= 1.4*10^-7. It is much smaller than 0.5 so our assumption is valid (otherwise we would have to solve the quadratic and get an exact solution).\n\nFor a simple level of chemistry we would now say\npH=-log[H+]= -logx =-log(1.4*10^-7) = 6.85\n\nHowever for more advanced and accurate expectations we see that the H+ coming from HNO2 is found to be in the same order of magnitude as that coming from the self dissociation of water (10^-7 for distilled water). Therefore in this case we cannot ignore the self dissociation of water (as we usually do) for calculating the pH. \nThere are 2 approaches for this case, here is one of them.\nRemember that we didn't put an initial amount for H+ since in most cases we ignore the self dissociation of water. Now we will consider that we had pure water, so the initial concentration of H+ is 10^-7 and we want to see how much the HNO2 will dissociate under these conditions so our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5 .. .. .. 10^-7\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. 10^-7+x. .x\n\nand Ka=(10^-7+x)x/(0.5-x)\nNow we have to solve the quadratic so we can't do any simplification\n\n0.5Ka-Kax=10^-7x +x^2 =>\nx^2+ (10^-7+Ka)x-0.5Ka=0\n\nThe only acceptable solution is the positive one\nx=0.5(- (10^-7+Ka)+ squareroot (D))\nwhere D=(10^-7+Ka)^2-4*(-0.5Ka)=\n=(10^-7+4*10^-14)^2-4*(-0.5*4*10^-14)\n=9*10^-14\n\nand x=0.5*(-((10^-7))+4*10^-14) +3*10^-7) =10^-7\n\nso now pH=-log(10^-7+x) =-log(2*10^-7) = 6.70
766259
4 NH3 + 3 O2 ---> 2 N2 + 6 H20
1061977
(H+) + NH3 -> NH4+\n It makes an ammonium ion, if you are talking about a hydrogen ion and not H2.
371296
well seeing that the molarity of the two solution is a direct ratio of how many molecules there are in the solution, i think that the ration would be 0.15/0.20\n\nso assuming that the reation is 1 to 1 (i cant remember so if it isnt 1 to 1, then i'm wrong) then you would need 18.75 mL of the 0.200 M HCl
491215
-Fluorinate toluene with F2 (highly controlled) or FCl or F5Cl, or DAST or one of these other nice fluorinators - gives the trifluoride on the methyl carbon\n-Nitrate it with nitric acid/sulfuric acid - CF3 is withdrawing so nitro goes on the 3-position\n-Reduce the nitro with Sn/HCl or H2/Pd\n-Nitrate again - the amino protonates and directs to the 5-position\n-Titrate the water to basic pH and extract with solvent like ether.  Concentrate and purify - probably crystallizes nicely.
740045
http://www.wellesley.edu/Chemistry/chem120/titrate1a.xls\n\n\nstep by step
149153
This is a limiting reagent problem. You must calculate the moles of NH3 produces from the given amount of grams of N2 and H2.\n\n3.00g N2 x 1mol/28.02g x 2 mol NH3/1 mol N2 = .214 mol NH3\n\n.54g H2 x 1 mol/2.02g H2 x 2 mol NH3/3 mol H2 = .178 mol NH3\n\nFrom this calculation you now know that the H2 is the limiting reagent. The next part to calculate the grams of NH3 goes like this:\n\n.178 mol NH3 x 17.04g/1 mol = 3.03 g NH3
948758
1)write out the equation\n2)find the no of mol of both reactants\n3)compare the stoichiometry ratio of reactant from 1) to find the limiting reagent\n4)compare the mol ratio of the limiting reagent with NH3 to find the no of mol of NH3 produced\n5)theoretical yield=17(no of mol of NH3)
253380
P3(NO3)3 \n\nNO3 = nitrate NO4 does not
283215
[Cu(NH3)4]SO4*H2O (s) + 2HCl(aq) --->\n\nCu(NH3)4 +2 (aq) + H2SO4(aq) + H2O(l) + 2Cl-(aq)
788534
N2 + 3H2 -> 2NH3\n\nN2 has a gfm of 28g/mol and H2 has a gfm of 2g/mol.  (you do the math to find the moles I will make a rough guess) you've got about .5 mols of N2 and about 1 mole of H2, so H2 is your limiting reagent.  your ratio of H2 to NH3 is 2molNH3/3moleH2 and you multiply that by the one mole of H2.\n\nNH3's gfm is 17g/mol so .667molNH3 x 17g/molNH3 ~ 12g\n\n*edit* make sure you do the math over because I was guessing and estimating where you should not *edit*
866145
HCl + AgNO3-----> AgCl +HNO3\nPb(OOC-CH3)2  + 2LiCl -------> PbCl2  + 2Li(OOC-CH3)\nNa3PO4 + CrCl3 -------> Cr(PO4) + 3NaCl\nH2SO4 + BaCl2 ----------> BaSO4 + 2HCl\nAl2(SO4)3  + 6NH4OH ------> 2Al(OH)3  + 3(NH4)2SO4\n2AgNO3 + H2S -------> Ag2S  + 2HNO3\nCaCl2   + Pb(NO3)2 -------> Ca(NO3)2  + PbCl2\nCa(NO3)2 + Na2CO3 ------>CaCO3  + 2 NaNO3
805524
Hydrogen and nitrogen react to form ammonia in what is called the Haber process.\n\nThe reaction is as follows:\nN2 + 3 H2 <--> 2 NH3\n\nThis reaction will give about a 85% yield at 200°C (473 °K) and 200 atm of pressure.  This is typically what is done in industry.\n\nWith special considerations to safety, you can get about 100% yield at 200°C and 750 atm of pressure.\n\nThe only remaining question is, how much do you want to make?\nI will assume 100 kg per batch.  You can recalculate for different amounts.\n\nThe molecular weight of ammonia is 17 g/mol, so 100 kg will have 100 kg / 0.017 kg/mol = 5882 mol\n\nTo produce 5882 mol of NH3, you need 2941 mol of N2 and 17646 mol of H2\nSo, the maximum number of moles of gas will be at the beginning of the reaction: 2941 + 17646 mol = 20587 mol\n\nUsing the ideal gas law:\nPV = nRT, where R is the gas constant and n is the number of moles\nV = nRT/P = 20587 mol * .082057 L*atm / mol*K * 473°K / 750 atm = 1065 L = 1.065 m3
139469
16.45 mL HNO3 solution * [1 liter/1000 mL] * [0.302 moles HNO3/1 liter HNO3 solution] * [1 mol NaOH/1 mol HNO3] * [1 liter NaOH solution/0.250 mol NaOH] * [1000 mL/1 liter]\n\n=67.0 mL NaOH solution\n\nEDIT: [Sorry, the answer above was wrong; I don't know how I got that. The answer is 19.87 mL]\nEND EDIT\n\nBasically, you just use a whole bunch of ratios relating all the units and measurements together.\nIf you write all that out, all the units should cancel except the "mL NaOH solution"\nMight want to check my answer though...\n\n\nNext question:\n[16.45 ml HNO3 solution] * [1 liter/1000 ml] * [0.302 mol HNO3/1 liter HNO3 solution] * [1000 millimoles solution/1 mol solution]\n\n=4.97 millimoles HNO3 solution
754704
You should write NH4(+) otherwise you will not get the oxidation number right.  H is 1 so H4 is 4.\n\nIn order to be left with 1(+) you need 4 - 3 = 1.  So N will be -3.\n\nCheck -3 + 4*1 = +1 OK.