How many milliliters of 0.200 M HNO3 are needed to completely neutralize 60.0 ml of 0.100 M Mg(OH)2 solution?
964051
First you must know the equation of the reaction, so you can establish the mole connections between the involved substances\n\n2HNO3(aq) + Mg(OH)2(aq) → Mg(NO3)2(aq) + 2H2O(l)\n\nNow, you have 60,0 mL of 0,100 M Mg(OH)2(aq)\nTherefore:\n0,100M*(60,0 mL)(1L/1000 mL)= 6,00*10^-3 moles of pure Mg(OH)2\nNow, 2 moles of HNO3(aq) react with 1 mol of Mg(OH)2(aq)\n\n6,00*10^-3 moles Mg(OH)2*(2 moles HNO3/1mol Mg(OH)2 )= 0,0120 moles of HNO3(aq)\nMolarity = moles solute/volume of solution\n0,200M= 0,0120 moles /?\n?= 2,40*10^-3 L of solution= \n2,40 mL of 0.200M HNO3(aq)  are needed to completely neutralize 60.0 ml of 0.100 M Mg(OH)2 solution
924787
First of all\n3 moles of NO2 give you 2 moles of HNO3, while 1 mol of water gives you the same amount of product. We must find out which reactant gives you the least product, and that will be your maximum amount of product.\n36,0g H2O*(1mol H2O/18g H2O)*(2moles HNO3/1mol H2O)*(63g HNO3/1mol HNO3)= 252 g of HNO3\n\nNow with NO2\n92,0g NO2*(1mol NO2/46g NO2)*(2 moles HNO3/ 3moles NO2)*(63g HNO3/1mol HNO3)= 84,0g HNO3\nTherefore, the biggest amount of HNO3 you can expect to get is 84,0g.
773826
the answer that i got is 3.10 M. \n\n3.25 g NaOH/40.00 g NaOH = .0813 mol NaOH\n\n26.2 mL / 1000mL = .0262 L\n\nM= .0813 mol / .0262 L = 3.10 M
572470
a.  4H20 + 4NO2 + 02--------->4HNO3 + 2 H2O\n      Al(OH)3+ 3HCl------------> AlCl3+ 3H2O\nb. not sufficient as Al(OH)3 is 12/78 moles which is able to react with .462 moles of HCl and here we have .9444 moles of HCl\n\n9. it is done by equilliberium reaction\n and u written Na reaction with H2 to form hydride
959929
A.\nmole of HCl availible = 1.0 * 100 = 100 mmole\nmolarity of diluted HCl = 100 / 1000 = 0.10 M\n\nB.\nmole of HCl availible = 1.0 * 10 = 10 mmole\nmolarity of diluted HCl = 10 / 500 = 0.02 M\n\nC.\nmole of HCl availible = 1.0 * 50 = 50 mmole\nmolarity of diluted HCl = 50 / 100 = 0.50 M\n\npeace\nvixklen
9807
Well lots of chemistry questions lately....finals coming up? :)  Good Luck!!!  \n\nTo calculate this type of question first of all calculate the #of moles of HCl used.  So \nyou used 2100 mL = 2.1 L HCl \nThis HCl is 0.3 mol/L so you used:\n\n2.1 L HCl* (0.3 mol HCl/L) = 0.63 mol HCl\n\nnow use the ratio of Cl/HCl  (which is 1 to 1)\n\n0.63 mol HCL *(1mol Cl/1mol HCl)= 0.63 mol Cl ion\n\nSo far so good I hope.  \n\nNow do your solution diltion to calculate volume,\n2.1L + .2385L = 2.3385 L of mixed solution\n\nNow simply divide \n0.63 mol Cl/2.3385 L solution = 0.269 mol Cl ion in final solution.\n\nAgain these numbers are quick and dirty but the method of solving this type of problem is the same.
944660
You are using the wrong number of moles...\nHNO3 mol = (40.5ml/1000 ml/L)(0.620 mol/L) = 0.02511\n\nfrom chemical eqn...\n\n(3 mol H2/6 mol HNO3) 0.02511 mol HNO3 = 0.012555 mol H2\n\nNow...\n\nPV = nRT\n\nT = 27.5 + 273.15 = 300.65K\nP = 0.755 atm\nn = 0.012555 mol \nR = 0.08206 L atm/mol K\n\nPV = nRT\nV = nRT/P = (0.012555 x 0.08206 x 300.65)/0.755 atm = 0.41 L
242296
22.49 L = 2 moles of HNO3 will react with one  mole of K2HPO4\n\nyou only have .0014 mol  of K2HPO4 available\n\nSO, YOU WILL NEED 31.45 mls of HNO3 to react with .0014mol of K2HPO4
887934
40 ml 3% H2O2 contains 3% H2O2 which is 1.2ml\n.\n20 ml 3% H2O2 means it has 20*3%= 0.6ml H2O2\nso 20ml 3%H2O2 + 20ml distilled water contains 0.6ml H2O2 in 40ml. So, 0.6/40=1.5%H2O2\n\nsimilarly, 10ml 3%H2O2 has 0.3ml H2O2.\nSo 10ml 3% H2O2 +30ml dist Water makes a 40ml solution with 0.3 ml H2O2 making it 0.3/40=0.75%H2O2 solution.\n\nSame steps for the 5ml case.
621533
NaOH ----------------> Na+   +   OH-\nFind the number of moles of NaOH\nMoles of NaOH = 0.10M * 0.150L = 0.015 moles\nSince the coefficients in fornt of NaOH, Na+ and OH- are all 1, therefore if 0.015 moles of NaOH disssociates, 0.015 moles of Na+ and 0.015 moles of OH- are produced.\n\nSimilarly we will work with HCl\nHCl --------------------> H+      +      Cl-\nMoles of HCL = 0.30M * 0.05L = 0.015 moles\nThis means the 0.015 moles of H+ will be produced\n\nSince H+ and OH- have equal amiunts, they will completely neutralize each other leaving the solution at neutral pH. Since the neutral pH is 7, therefore, the pH of this solution will be 7.
1125846
NaVO3, MW = 122 g/mol\n\n100 mM = 0.1 M, so 0.1 moles in 1 L.  Dissolve 12.2g in water, dilute to 1L.
754271
2H2 + O2 --> 2H2O\nPV = nRT\nP = 2 + 3 = 5\nT = 25 + 273 = 298\n5.00 atm x 55.0 L = n x .08206 x 298\nn = 11.25 mol H2 and O2 mixture\n11.25 mol x (2 mol H2O / 3 mol) = 7.50 mol H2O\n\nP x 55.0 = 7.50 x .08206 x 298\nP = 3.33 atm
877279
First you need to determine your limiting reagent in the first step.\n\nMolecular mass of NH3 = 17 g/mol, O2 = 32 g/mol\nMol of NH3 = 5,700,000 g/17 g/mol\nMol of NH3 = 3.35E5 mol\n\nMol of O2 = 57,000,000 g/32 g/mol\nMol of O2 = 1.78E6 mol\n\nSo based on the first equation (1), NH3 is the limiting reagent.\nFor every 4 mol of NH3, one gets 4 mol of NO.  So the first reaction produced 3.35E5 mol of NO.\n\nThe second reaction produces 2 mol of NO2 for every 2 mol of NO used, so the second reaction produced 3.35E5 mol of NO2.\n\nThe third reaction produces 2 mol of HNO3 for every 3 mol of NO2 used, so this final reaction will produce 2/3 x 3.35E5 mol of HNO3 = 2.23E5 mol.\n\nThen use molecular mass of HNO3 to calculate how much of it (mass) that you made.\nHNO3 molecular mass = 63 g/mol\nmass = mol x molar mass\nmass = 2.23E5 x 63 g/mol\nmass = 1.41E7 g of HNO3\n\nOf course this assumes 100% yields in all the above noted reactions.
741771
0.5M = 500mM\nv x c=v'xc'\nv x 500 = 50 x 1\n0.1mL of the 0.5M solution diluted to 50mL will be 1mM
459258
start with computation of molarity of solution:\nM_BaOH = (51.24g/ MW BAOH)/ 1.2 L\n\nthen \nusing M1V1 = M2V2\nM_BaOH (x) = (0.1) (1L)\nsolve for x
820519
First write the balanced chemical equation:\n\nPb(NO3)2 + 2NaBr --->  PbBr2  +  2NaNO3 \n\nThen, find the total volume of the solution: 200mL + 200mL = 400mL \n\nFrom the density, the mass is:   density = mass/volume   mass = density * volume = 1.0 g/mL * 400 mL = 400 g\n\nQ = mC(Tf - Ti) ; where Q = heat released, m = mass of solution,, C = specific heat capacity, and (Tf - Ti) = change in temperature\n\nSo,  Q = 400g * 4.2 J/gK * (2.44 + 273)  (Since C is J/gKelvin, the temperature rise also has to be in Kelvin)\n\nQ = 462.5 KJoules\n\nHope this helps...
773833
2NaOH + H2SO4  = Na2SO4 +2H2O                                      Sodium Hydroxide is 0.5moles.Two Moles neutralise One mole of Acid.   0.5Moles neutralise of Alkali neutralises 0.25Moles of Acid                                                                                           50mls of 1M SULPHURIC ACID = .05M THEREFORE MOLAR CONCENTRATION IS 5