How many grams of KOH are needed to create 750mL of a 15.0 M solution?
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.
497859
A)\n\nConvert HCl into millimoles. .2M * 1 mL = .2 mmol. Now, you've diluted it to a volume of 5 L. So you convert back, keeping in mind that you need to divide by mL for the units to cancel:\n\n.2mmol/5000 mL = 4*10^-5 H30+\n\nKnowing that Kw = [H30] * [OH-], and that Kw = 1*10^-14, you can figure out OH- and pOH. (pOH is the -log of OH-)\n\nThe pH of the HCl soln is 4.39, based on -log (4*10^-5)\n\nB)\n\nSodium Oxide, like most alkali metal oxides, reacts with water to create sodium hydroxide an \n\nNa2O + H20 --> 2NaOH \n\nSo it's going to be .2 mol of OH- created, for every .1 mol created. So it's .2 mol NaOH/ 1L = .2M NaOH.\n\nThe OH- is .2M\nThe H+ is figured out using Kw = H+ * OH- (see above)\nThe pOH is 0.70\nThe pH, since pH + pOH = 14 at 25 degrees, is 13.3
154637
first you convert to grams = 15000 grams\nthen 44 grams per mole\n22.4 liters per mole\n\nanswer= 15.2 liters CO2\n\n=D\n\nhope that helped
328014
1 ml = 1 cc\n1 ml = 15-16 gtts (drops)
968613
If you want to make 1 Liter of this final solution, you would need to take 721 mL of your 52% formaldehyde solution, add 120 mL of methanol to it and bring the final volume up to 1 Liter with water.  You should be able to scale this up or down to make any particular volume you need.\n\nWhat isn't coming out right in your calculations?
745889
We can solve this using a system of equations.\n\nLet x be the volume of the 25% solution. Let y be the volume of the 60% solution.\n\nThe sums of the volumes must total 140 ml:\n\nx + y = 140.\n\nThe mass of CuS04 in the desired solution must equal the amounts added through the use of both solutions.\n\n30 percent of 140ml gives 42 ml of "pure" copper sulfate solution.\n\nThe 25% solution would contribute .25x of "pure" copper sulfate solution. The 60% solution would contribute .60y of "pure" copper sulfate solution, so:\n\n0.25x + 0.60y = 42.\n\nHere is the system:\n\nx + y = 140\n0.25x + 0.60y = 42\n\nThere are many different ways to solve the system. I'll use substitution here:\n\nx + y = 140, so y = 140 - x\n\n0.25x + 0.6(140 - x) = 42\n0.25x + 84 - 0.6x = 42\n-0.35x = -42\nx = 120\ny = 20\n\nSince x is the amount of 25% solution and y is the amount of 60% solution, we need 120 ml of 25% solution and 20 ml of 60% solution.\n\nThis appears to match your answer!
443106
5-milligrams-per-milliliter, so 15 x 5 = 75. 75 millimeters. Next time pay attention in math.
1145574
0.150 M NaOH\n\n0.150mol / L  X  40 g / mol  X  1 L / 1000 mL  X  200 mL  =  1.2 grams of solute (NaOH)
748321
You're kidding right? \n\n1L (1000ml/ 1L) x (0.5g/1ml) = 500g
1048754
it depends on NaOH consentration\nit must be more than 7\nthis formula may help you\npH = 14 + log [NaOH]\n14 figure come from negative logatherm of Kw, 10^-14\n[NaOH]=NaOH concentration in mole/L
808549
M=molesof solute/Liter of solution\n\nM= molarity
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
896861
Remember your equation for specific heat capacity::\n\nHeat content = mass x spec heat capacity x delta T\n\nq = m*C*dT\n\n225 J = (15 g) * (0.80 J/gram-deg C)* dT \n\nTherefore\n\ndT (in C) = 225 J  /  (15 g)(0.8 J/g-C) = 18.75 deg C\n\nMind your units and make sure they cancel.  I did.
910910
Firstly, please forgive me for not having a periodic table at hand. :)\n\nSecondly, please forgive yourself as even the most advanced scientists in the world sometimes forget how to do this calculation.\n\nThirdly, it is such a simple calculation, that it is no wonder that it is so easily forgotten.\n\n1.5% solution (mass/volume) means 1.5 g of KBr in 100 ml of solution\n\nIt's that simple. :)\n\nEnjoy your science. :)\n\nSincerely,\n\nCurtis
1185900
H2SO4 + 2KOH --> K2SO4 + 2H2O
646163
Presuming that the base is monobasic.\n\na) with 25 ml added, you are at the equivalence point.  The pH is 7.\nb) after adding 25.01 ml, you will have 0.01 ml of base that did not get reacted.  The molarity of the base in solution will then be 0.1 * (0.01/50.01) = 2.0 x 10^-5.  Your pOH is 4.70, pH would be 9.30\nc) Now, it's 0.1 ml of unreacted base.  Molarity is 0.1 * (0.1/50.1) = 2.0 x 10^-4.  Your pOH is 3.70, pH is 10.30.\nd) Now 1 ml of unreacted base.  Molarity is 0.1 * (1/51) = 2.0 x 10^-3.  pOH is 2.70, pH is 11.30.\ne) Now 25 ml of unreacted base.  Molarity is 0.1 * (25/75) = 3.33 x 10^-2.  pOH is 1.48, pH is 12.52.\n\nIn each case, you have to figure out the amount of unreacted titrant and recalculate the molarity as if you were simply diluting it, using McVc = MdVd, where the volume is now the total volume of the solution (the unreacted amount plus the 25 ml of acid and the 25 ml of base that have neutralized each other).
929350
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).