What volume (mL) of a 15% (m/v) NaOH solution contains 120 g NaOH?
927975
15%(m/v) means 15grams of NaOH in 100ml of solution.\nHence 120*100/15= 800ml will contain 120 g of NaOH
737901
The question is not worded correctly.  Rewrite the question and I'll be glad to help you.  \n\nYou are asking for the Concentration of NaOH, but it is already given as .10 M as written in your question above.\n\nTo solve a normal titration problem you need 3 of the 4 variables below:\n\nconcentration of base\nvolume of base\nconcentration of acid\nvolume of acid\n\nLet me know.
683950
M1V1=M2V2\nie... 0.24*40.11=M2*0.55\nM2=17.50\nMOLARITY=                                                (MOLECULAR WT/GIVEN WT)*(1000/VOLUMWEOF THE SOLUTION IN MILILITERS)\nIF U CAN SUSBSTITUTE THE VOLUME OF THE SOLUTION THEN U CAN FIND OUT THE MOLECULAR WT\nhope u know how to find out the molar mass fr molecular wt.
320061
Someone else has asked this same question before, and it really isn't that hard . . .\n\n60 ml of 0.250M HCl plus 60 ml of 0.125M NaOH will result in 120 ml of solution total.  All of the NaOH will have been neutralized by the HCl, leaving an excess of HCl, so the solution will be acidic -- the pH will be much less than 7.\n\n60 ml * (1.0 liter/1000 ml) * (0.250 mole/liter HCl) = .015 mole HCl\n\n60 ml * (1.0 liter/1000 ml) * (0.125 mole/liter NaOH) = .0075 moles NaOH\n\nThat means you will have 0.015 - 0.075 = 0.075 moles of HCl left over after the neutralization.\n\nI leave it for you to calculate the pH of a solution of 0.075 moles of HCl in 120 ml of solution.  Remember that pH = -log(base 10)[H+]
511319
NaOH is in excess\n\nSubtract moles of H+ from OH- (.0345 mol OH- - .025 mol H+ = .0095 mol OH-)\n\nDivide by total volume (.0095 mol OH-/.0595 L = 0.1596 M)\n\nFind pOH = -log [OH-] = -log [.1596] = 0.7967\n\nFind pH = 14-.7967 = 13.20
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
472386
(250 X 0.35) + (1350X0.75) + (500X0.15) / (250+1350+500) = \n87.5+1012.5+75/(2100)\n1175 m mol/2100\n1.175 MOLES / 2100ml = 115g of sulfuric acid\n\n1.175moles /2.1 L = 0.58 M SOLUTION
820326
If this is in Normality;\n\n0.1 Normal solution of  NaOH\n2 grams of NaOH  in a liter = 0.05 N\n\nN = weight in grams divided by MW (molecular weight ) = 40 / valence =1 / 1 liter. = 0.05\n\nMW = (Sodium) 23 + 16 (Oxygen) + 1 (hydrogen) = 40\n\nTherefore: Using the formula N1V1 = N2V2\n(0.1)( X) = (0.05) ( 1 liter)\n\nBy cross multiplication:\n\nX = (.0.05) ( 1) divided by ( 0.1) \nX =  0.5 liters or 500 milliliters\n\nbecause  1,000 ml = 1 liter\n\nYou will get the same answer for Molarity because the valence is one.\n\nso the answer is 500 milliliters.
1062341
You are given the volume in ml and the density per ml so,in order to get the answer just multiply 220 ml X 7.800 gm/ml.  This is equal to 1,716 grams or 1.716 kilograms.
954213
KOH  Formula weight is 56.11  following that,\n\n56.11 gm of KOH in 1000 mL gives 1 Molar solution\n\nThen 23 gm in how many mL (?) gives 0.157 Molar solution\n\n\n? = (23 gm x 1000 mL x 1 M) / (56.11 gm x 0.157 M) = 2611 mL
813561
The molecular mass of NaOH is 40.00 gm/mol,  so 8 gm of NaOH contains 8 gm/40.00 gm.mol = 0.20 moles of NaOH.\n\nThe molecular mass of HCl is 36.46 gm/mol, so 8 gm of HCl contains 8gm/36.46 gm/mol = 0.22 moles of HCl.\n\nIf we react these amounts of NaOH and HCl, the NaOH will be the limiting reactant, and we will produce 0.20 moles each of NaCl and H2O.  This reaction will liberate \n\n0.20 mol * 56 kJ/mol = 11.2 kJ of energy as heat\n\nThe heat capacity of the final solution is given by the mass of the final solution (which, by conservation of mass, is equal to the mass of the initial solutions) multiplied by the specific heat capacity of the final solution.  With the assumptions given in the question, the heat capacity of the final solution is:\n\n(58 gm + 258 gm) * 4.184 J/(gm*degC) = 1322 J/degC.\n\n(Note that we don't need to use the density of the final solution.)\n\nThe temperature change of the solution is given by the energy released divided by the heat capacity of the system:\n\n(11.2*10^3 J)/(1322 J/degC) = 8.47 deg C\n\nIf the initial solutions had a temperature of 25 C, the final solution has a temperature of 25 + 8.47 C = 33.47 C.
567447
The volume of the tank is 40 x 30 x 15 = 18000 cubic centimeters, which is the same as 18000 milliliters (mL).  12 Liters = 12000 milliliters, so it will take 18000/12000 = 1.5 minutes.  I am a high school math teacher.
713107
range/amount\n\nrange = 120 mV (90mv - (-30mV))\namount = 2^16 = 64*1024\n\nplug-n-chug
868152
1 mol NaOH and 40 g NaOH are equivalent.
948588
There are 1000 micrograms in a milligram, so 150 mcg = 0.150mg
732310
5.00/0.5 = 1 to 10 dilution\n\n600/10= 60 mL of stock needed\n\nthe rest (540mL) is water
903371
I have a serious feeling that you haven't given us all the info.\nIf you know the density of HCl you can get the first one.\nBut we don't have enough info to even begin to tell you the relevance of the NaOH.   More info, or concentrate on your own homework...