how many grams of NaOH are needed to prepare 250 mL of 0.205 M NaOH?
829104
Well, it's 0.205 moles/liter and you're making 1/4 liter, so you need 0.205/4 or 0.05125 moles.  The molar mass of NaOH is equal to the sum of the molar masses of Na, O, and H, or 22.990 + 15.999 + 1.007 = 39.996 g/mole.  The mass on NaOH needed is equal to 0.05125mole*39.996 g/mole = 2.050 grams.
116087
keeping with the same unit fractor harry is correct.\n \n .2 ml
683322
Well, I suppose 1.25 g is 1250 mg, so seeing there's 1250 mg in each ml of solution, taking out 0.2 ml of the solution will net you 250 mg of the substance.
165248
(.002mol /20ml) (1000ml / 1L ) = .10 M NaOH\n\n\n pOH =  -log 0.10 =  1\n\n pH = 14.00 - pOH   = 14 - 1  = 13 = D
258981
The convertion is with reference to a certain medium...\n\nIf the medium is ordinary fresh WATER ...\n\n1 gram (gm.) = 1 millilitre (ml.)\n\n1,000 milligram (mg.) = 1 gm.\n\nTherefore:\n\nIt will take 1,000 mg. to produce 1 ml. of water.\nIt will take 1,000 gm. to produce 1 liter of water.
900861
As usual, we first check for a balanced equation.  It isn't, so we fix it: 2 LiOH + CO2 -> Li2CO3 + H2O.  Also as usual, get the molecular weights: LiOH = 24, CO2 = 44, and the others don't matter.  There are 4 moles of CO2, so we need 8 moles of LiOH, and that is 192 grams.  Extra credit: find the error in the next response.
592167
We assume NaOH is a strong base and therefore dissociates completely at 60 degrees C. a concentration of .002M solution of NaOH dissociating completely contributes .002 mols/liter of OH ions to the solution. Since pOH = -log[concentration of OH-], pOH = -log.002 = 2.699 (with correct logarithmic sig figs). pH is 14 - pOH, so the pH of the solution will be 14-2.699 = 11.301.
470567
To do this problem we need to do some convertions.  Convertion factors are simply ratios that allow us to change the units of a certial quantity.  For example: the molar concentration (molarity) is a convertion factor that is a ratio of the moles of dissolved substance to the volume of solution (moles/L). Using the concentration  we can convbert moles to liters or liters to moles depending on what we have to start with.  Another convertion factor that we will need is the molar mass of KOH.  This is the ratio of the mass of KOH to the moles of KOH (this is usually measeured in grams per mole (g/mol)).  We can find this number by looking up the molar masses of K, O, and H on a periodic table and adding them together. K =39.0983g/mol, O = 15.9994g/mol, and H = 1.00794g/mol. \nAdding these together we have a molar mass of 56.10564g/mol for KOH.\nSchematicaly this is the road map we will follow to find our answer:\n\n-------(moles/L)-------(g/mol)\nvolume <------> mole <-------> mass\n\nWhere volume, moles, and mass are actuall ammounts and the convertion factors between them are on the arrows.\n\nNow for your specific problem we will start with the given volume 750mL.  First we will convert it from mL to L then convert it to moles of KOH then finally to grams of KOH.\n\n750ml(L/1000mL)(15.0mol KOH/L)(56.1056g KOH/mol KOH)= 631.188g KOH\n\nTo three significant figures our answer is 631g KOH.  This is a big number but a 15.0 M solution is very concentrated so it is not surprizing.
889019
mass of 1 mol NaOCl = 22.98 + 16.00+35.45 = 74.43 g/mol\n\n18.6/74.43 =0.25 mol\n\n0.25/0.800 = 0.3125 M\n\n0.3125M
798345
Approximately a 1:1 ratio - 225 grams = 225 ml.
354060
0.2 gal. or 25.6 fluid oz.
775430
25% citric acid will have a concentration of 250 g/l.\n\nDividing by 192.1 g/mol we will get:\n\n(250 g/l) / (192.1 g/mol) = 1.3 mol/l = 1.3 M.
905904
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
685262
the answer depends on how many H+ need neutralising by the NaOH.\nassuming it is only 1 per molecule of the acid then you can use the following equations:\n\nno.of moles = (vol/1000)*concn\nno. of moles = (40.11/1000)*0.240 = 0.0096264\n\nno. of moles = mass(g) / RMM\ntherefore\n0.0096264 = 0.55 / RMM\ntherefore RMM = 0.55 / 0.0096264\nRMM = 57.134\n\nif you have 2H+ that need neutralizing then the molecular weight = 2*57.134 = 114.268 etc etc
202477
I'm not sure if I understand your question.\n\nIf you are wanting to make a 0.92 M NaCl solution out of 0.39 M NaCl, well you are going to have to work hard.  Basically, you would be boiling the water and having to measure the mass of the flask.  This is a very tedious process.  Bottom line, you cannot use the dilution formula to make it happen.\n\nNow, if you are wanting to make a 0.39 M NaCl solution from a 0.92 M NaCl solution, then you use:\n\nM1V1 = M2V2\n\n0.92 * V1 = 0.39 * (.1)\n\nremember .1 L is 100 ml.\n\nSolve for V1\n\nV1 = 0.42 L\nV1 = 42 mL\n\nSo, take 42 ml of the 0.92 M NaCl and dilute it up to 100 ml.\n\nHope it helps!
888417
Well 50 mL is a volume. And grams is a mass. So therefore, they are completely different ways to measure things. It's like me asking you to convert 140 pounds into inches. It can't be done! \n\n\nBut, If someone asks you to convert 50 mL of 2-propanol (example) to grams of 2-propanol, that is a completely different question. You have 50 mL of 2-propanol and you look up the density. It is .785 g/mL. so you multiply 50 mL x .785 g/mL. mL cancel, and you are left with the grams.
833268
I'll tell you how to figure this one out...\n\nStart by always converting to moles...\nConvert 25g of LiOH to moles using its molecular weight.\n\nOnce you have that, you set up dimensional analysis to make your units cancel...\n\n0.139M is the same as 0.139 mol/L \n\nso... your answer above in mol * (L/0.139 mol) * (1000mL/L) = your final answer... the units all cancel and you are left with mL