Using the 12M sodium hydroxide solution, prepare 350 ml of a 2.5 M sodium hydroxide solution.
131890
M(conc.base) x vol.(conc.base) = M(dil. base) x vol.(dil. base)\n12 x vol. = 2.5 x 350\n vol = 72.9 ml \n\nIn a graduated cylinder, take 72.9 ml of concentrated NaOH. And in a standard flask, take 350 - 72.9 = 277.1 ml of water. To this flask, add the concentrated NaOH (72.9ml) from the graduated cylinder and shake the flask for uniform concentration. Now you've prepared 350 ml of diluted NaOH solution.
397107
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
659833
I tried to write you, but it said your email was unconfirmed?  So I changed the answer.  If you need help in chemistry feel free to write me and I'll try to help.  If I can write you back.  Otherwise keep asking questions.\nI think you'll do fine in Chem.  You've already got half the battle done, you're doing homework.\nDid you convert everything?  That's where people usually go wrong.\n\nThe formula-you'll just be plugging things into this:\nosmotic pressure=cRT\n\nThese are the terms in the formula: \nc=the concentration of ions in mol/L=\n(73 g/L of NaCl/58.43 g/mol of NaCl)--->1.25 mol/1l*2(there are 2 ions in the salt, and it dissociates in water, so 1.25 mol of Na and 1.25 mol of Cl. Which makes 2.5 mol of ion in solution\n\nR=the constant that makes all of the units work out=0.082 Latm/Kmol\n\nT=temperature in K=298K\n\nPlug all your numbers in:\nosmotic pressure=1.25mol/L*2*.082Latm/Kmol*298K=61atm\n\n\nIf you want, you can then convert atm to some other unit, if the answer is in torr or something.  I like to make sure that all of the units in the question cancel to give me the one I want in the end,it's a good, quick answer check.  I didn't use significant digits, you'll have to change that.
512759
Do you mean M instead of N?\nIf you mean M, then\n1,00M/L*0,500 L= 0,500M\n0,500 moles NaOH*(40g NaOH/1 mol NaOH)= 20,0g of NaOH\nIt is c).
550260
2H3PO4+6NaOH-------->6H2O+2Na3PO4\n\nThere has to be two moles of phosphoric acid and six moles of sodium hydroxide for it to balance out
450181
! litre of solution contains 6.0 moles\nTherefore\n125 mL contains 6.0/1000 * 125 moles\nie 0.75 moles ← Option a
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Since Na+ and OH- ratio is 1 to 1 in the solution, the concentration of OH- would be 0.1M as well. \n\nC = n x V\n\nNumber of Na+ ions is equal to the number of OH- ions. So n is both same. If the volume is fixed, the concentration should be 0.1M.
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The answer will depend on th strength of the solution. At an equal concentration,  sodium hydroxide will have a much higher pH than an solution of sodium carbonate.\n\nA 10% solution of sodium carbonate will have a much higher pH than 0.00001 M solution of sodium hydroxide.
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b a solution, the salt will dissolve
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2. 2 Na + 2 H2O --> H2 + 2 NaOH\n3. 2 HNO3 + Ca(OH)2 --> Ca(NO3)2 + H2O\n4. C5H12 +  8 O2 --> 6 H2O + 5 CO2
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This time i ans ur question but next time plz do ur homewrk ur self.\n\n1.\nWeight of NaOH = (4*.05) = .2 grams\nMoles of HCl =  Moles of NaOH = (.2/40) = .005\nVol. of HCl = .005*22.4 = .112 l = 112 ml\n\n2.\nWell let the grams be x\n1000x = 74.5*2.25*(75+x)\n(1000-167.25)x = 167.25*75\nx = 14.98 grams
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I think you should keep it simple yet smart. First get a piece of pure elemental Sodium and then let it drop on warm water. Afterwards explain how the chemical reaction released Sodium Oxide and Hydrogen Gas.
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As this is clearly homework I will only do one as an example.\n\nNa+(aq) S=59.0\nCl-(aq) S=56.5\nNaCl(s) S=72.13\n\nSo S for dissolving solid NaCl = 56.5+72.13-59.0=69.63 J K-1 mol-1\n\nDissolving is mainly entropy driven - the process is actually slightly endothermic.
139178
The amount is diluted by a factor of 25.  So the resulting solution would have a concentration of 11.6M divided by 25 or \nor 0.464M.
368078
Fill the 5 litre jar with water and pour it into the 3 litre jar. what is left over is 2litre exactly. do this twice and you have 4litres.
24810
use sodium hydrolic solution
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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.