the molecular weight of NaOH is 40. a buffer contains     200 mmol/1 NaOH. if you were preparing 100 ml of this buffer, what mass of NaOH would you need?
134434
One mole of NaOH weighs 40gms. 200mmol is 0.2 of that and 100ml is 0.1 of a litre. \nSo you would need 40 X 0.2 X 0.1gms of NaOH.=0.8gms\n\nIn reality you would likely dissolve 8gms of NaOH into a 1L volumetric flask and fill it to the mark. You would then measure out 100 mls of this solution. Weighing out 0.8 gms of NaOH is inaccurate because it reacts with CO2 in the air and absorbs water vapour.
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
341313
1) Calculate the number of moles of NaOH in 300 mL\n3.81 g /( 22.99+16+1.008) = 0.0955 mol \n\n2) Calculate the number of moles of NaOH in the 25 mL transfered\n0.0955 mol * 25/300 = 7.96 x 10^-3 mol\n\n3) Calculate the molarity \n7.96 x 10^-3 mol / 100 mL = 0.0796 mol/liter
503167
STP is 1 atm, 273 K.\n\nmoles of HCl = PV/RT = 1 atm * 1.1 L/ (0.08206 * 273K) = 0.049 moles\n\nSince HCl and NaOH is a 1:1 reaction, 1 mole HCl gives 1 mole NaCl, so you get 0.049 moles NaCl (MW 58.5), or 2.87g.
397786
The number of moles of acid used for neutralisation = volume (in Litres) x concentration  =  0.045 x 0.5  =  0.0225 moles.\n\nFrom the equation for the reaction it can be seen that the same number of moles of NaOH are used as the acid.\n\nNaOH  +  HCl  =  NaCl  +  2H2O\n\nSo there must also be 0.0225 moles of NaOH\n\nThe mass of NaOH  =  number of moles  x  formula mass (Na : 23 +  O  : 16  +  H  : 1) = 0.0225 x 40   =  0.9g NaOH\n\nSo the % NaOH in the Lye is 0.9/1.2  x 100   =  75%
479998
Get the chemical equation first:\nHCl + NaOH --> NaCl + H2O\n\nMoles of HCl used = (50/1000)dm3 X (1.0mol/dm3) = 0.05 mol\nMoles of NaOH used = (50/1000)dm3 x (1.0mol/dm3) = 0.05 mol\n\nFrom chemical equation, 1 mole of HCl needs 1 mole of NaOH for complete neutralization.\nFrom what you are given, you have the exact amount for complete neutralization (meaning, there is no excess reagents).\nFrom chemical equation again, 1 mole of HCl or NaOH produces 1 mole of water.\nThus you had exactly 0.05 mole of water produced here.\nHope this helps. :)
536982
0,220M/L*2,20L= 0,484 moles of HCl\n12,00M = 0,484 moles/?\n? = 0,0403 L or 40,3 mL of 12,00 M HCl solution + 2,16 L of pure water, so you can get you 2,20 L of 0,220 M HCl solution
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!
397152
Suspected trick question.  The volumes included are irrelevant.  You know the molecular weight (66000 Da) and you know how many micromoles you need total in each solution.  What does it matter if the one micromole is in 1 ml or 10K gallons?  You know it only contains one micromole.  Divide the molecular weight by 10^6 and you're done.  (or you have not stated the problem correctly)
545418
ok, first go to moles.\n\n24.0g LiOH (1mol/23.948g) = 1.0022mol\n\nnow get liters using molarity\n\n1.0022mol (1L/0.153mol) = 6.55L or 6,550mL \n\nRight, there ya go, remember that you have 3 sig.figs. so you may want to represent significance by using scientific notation. So...\n6.55x10^3mL\n\nIf you're ever stuck, just simply go to moles and that will get you started (and usually gets you some points on tests).
501604
okay...100 g*(mol/12.01g)=8.33 mol C\n\ntake the aforementioned 1mol *(-113kj/2mol)=-56.5 kJ released for 1 mol\n\nthen multiply that -56.5 by 8.33= -470.65 kJ released\nnote it is exothermic and thus negative
633457
To make 100mL: \n\n0.6% from a 10% solution is 6ml of the 10% diluted in 94ml of H2O.  \n\n0.6% from a 25% solution is 2.4ml of the 25% diluted in 97.6ml of H2O.\n\nThe math?\n\n0.6%*100mL / 10% = 6ml\n\n0.6%*100mL / 25% = 2.4ml
431774
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...
270830
The first step, find the moles of both.\nYou should get 25 mmol benzoic acid, and 15.6 mmol NaOH.\n\nThen you must find after the reactions, how many moles remain.  \nYou should get 9.4 mmol benzoic acid, 0 moles NaOH, and 15.4 mmol benzoate.\n\nYou have some benzoic acid and some benzoate ion, a conjugate acid/base pair.  This is a buffer solution, so use the Henderson-Hesselbach equation.  You should get pH = 4.4.
495164
step 1 2NaOH + H2SO4 ---------> Na2SO4 + 2H2O\nstep 2 amount of NaOH = 0.2 x 25/1000= 5.0 x 10^-3 mol\nstep 3 amount of H2SO4 = s.0x10^-3 x 1/2 = 2.5 x 10^-3 mol\nstep 4 mass of H2SO4 = 2.5 x 10^-3 x 98 = 0.245g required\n\nNOTE\n^= power\nin step 4 it is multiplied by 98 because that is the molar mass of H2SO4 ie (2x1)+32+ (16 x 4) =98\n\ni do hope this is correct it has a been about 3 yrs since i did my A Level in chemistry
449732
1) Calculate number of moles of NaOH:\n8.75 g/ (22.99+16.00+1.01) = 4.571 mol\n\n2) Calculate molarity in mol /liter\n4.751 mol / 25.4 mL = 0.187 mol/mL = 1.87 x 10^-4 mol/L
442343
You have 75mL of 0,4000M NaOH solution\n0,4000 M= moles of NaOH/0,075L\nmoles of NaOH = 0,030\n 3NaOH + H3PO4 → 3H2O + Na3PO4\nNaOH reacts with H3PO4 at a ratio of 3 moles : 1 mol\nTherefore: 0,030 m NaOH*(1m H3PO4/ 3m NaOH)= 0,010 moles of H3PO4\nThen\nM = 0,010 moles/0,050L\nM = 0,20\nYou are correct; nevertheless, watch the figures!!!!