what does the capital "N" mean? is it the same thing as "M" which stands for molarity??\nexample 0.0108N Sodium Thiosulfate is in my sample....
124400
the N means newtons i kno that
512472
molarity  = moles / liters\n\nmoles = given mass / molecular mass\n\n10 g H2SO4 / 98 g per mole = about  0.1 moles\n\n0.1 moles / 1 L = 0.1 M solution
350524
ANSWER: 0.482 M\n\n\nEXPLANATION:\n\nM = mol solute / L soln\n\nTherefore, you need to find the mol of solute and L of soln to find M.  \n\nTo find mol solute:\n10800mg Na(1g Na/1000mg Na)(1mol Na/22.99g Na) = 0.4698mol Na\n\nTo find L of soln:\n1kg soln(1000g soln/1kg soln)(1ml soln/1.025g soln)(1L soln/1000ml soln) = 0.9756L\n\nUsing the two values found plug in to M = mol solute / L soln\n\n0.4698mol Na / 0.9756L soln = 0.482 M
418373
"1g/18g of H2O=.0556g/mol\n.57g/(63.55*32.07*16*4) -->> .57/ 130435.104= 4.3699x10^-6" \n\nemm, I dont really understand what u'r trying to do here, molar mass of CuSO4 should be calculate this way:\n\nM[CuSO4] = M[Cu] + M [S] + 4*M[O], not M[Cu]* M [S]*4*M[O].\n\nANY molar mass has the unit of g mol^-1, not g. If you divide 1g by 18g, u will have a UNITLESS number.\n\n"So now we divide both of them by teh lowest mole which for this equation is .0556g, right? But is the CuSO4 right? (The Mole)" \n \nit should be 0.0556 mol, not 0556g. No, it's the mole of water lost.\n\nThis is what u should do. \n\nfind out the MOLAR MASS [unit g/mol] of CuSO4 and H2O [CuSO4 should be around 160 g/mol, H2O should be around 18 g/mol]\n\nthen, find out the MOLE of CuSO4 and H2O in the compound by dividing the mass of CuSO4/ H2O by their molar mass.\n\nThen divide the mole of H2O in the compound by the mole of CuSO4 in the compound, find out ur ratio.\n\nthe ratio turns out to be 15.6, a very strange number in my point of view [not an integer, and it's a little bit large for this compound]
422184
Yes, you are correct.\nEveryone else who has posted so far is mistaken.\n\nGiven:\nM = mol of solute / liters of solution.\n\nFirst Step:\n\nSolving for x (liters of solution), you get:\n.40 / .096 = x\nx = 4.167 L\n\nSecond step.\nL => mL\n4.167 / 1000 = 4,167 ml or app. 4000 mL\n\nYour answer is approx. correct because you were asked for mL of water, which you found, not L of solution, which everyone else seemed to think.  You may lose points for not solving to the correct amount of significant figures though.
516097
to what?  Are you looking for the molecular weight of the monoprotic acid?  If so, remember M (molarity) is in moles/lter) and your colume of NaOH is in mL so you'ww want to convert that to liters.\n\n\n.01524 liters x .1578 moles/liter = 0.002405 moles\n\nso therefore you have 3.567  g / 0.002405 moles = 1483 gram/mole\n\nuse dimensional analysis and keep track pf the units to find the answer
99988
Answer was found in the Handbook of Chemistry and Physics\n\nKMn04 is only slightly solube and as you have aready found the\nsolubility constant.\n\nNaMnO4 is list in the hand book as Very Soluble both in cold and hot water, and did not give a constant. That means that a lot of NaMnO4 can be dissolve into water, somewhat like table salt, NaCl\n\nNaMnO4 is often found in liquid form at about 40% in water, and it is a dark purple liquid. you can get this data sheet \nby just type in Yahoo search "sodium permanganate solubility"\n\ngood luck\n\nDavid
345637
FW of NaCl is 58.44  then\n\n   58.44 gm in 1000 mL = 1 Molar\n\n   4.78  gm in 200 mL = ?\n\ndo cross multiplication\n\n(4.78 gm x 1000 mL x 1 M) / (58.44 gm x 200 mL) = 0.4089 M\n\n(gm/gm and mL/mL cacels out so you get answer in M)\n\n\nNow you are taking 28 mL of 0.4089 M solution and diluting it to make final volume of 100.  Use M1V1 = M2V2\n\n0.4089 M x 28 mL = ? x 100 mL\n\n\nThen  ? = (0.4089 M x 28 mL) / 100 mL    { mL/mL cancels out and you get answer in Molarity)\n\n\n? = 0.1145 Molar is your answer
213284
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.
387181
N=aM where a is the valency (depending on the type of chemical and reaction, e.g. number of H+/OH- for acid/bases per molecule, number of electrons gained/lost in redox reactions).\n\nFor AgNO3 a=1 (it is Ag+ so for redox or precipitation reactions it gives one equivalent group per molecule)\n\nSo M= N/a = N/1 =N\n\n1 M is 169.87 g/L =16.987g/100mL =16.987%\n\nUse M1*V1=M2*V2\n16.987*V1=1*V2\nV1 is the volume in mL of the stock solution that you will use to prepare V2 mL of 1%\nYou didn't specify how much you want to prepare...\n\nAlternatively you could have converted 1% to mole/L and do the math with molarities.
497557
20 ml of gastric juice  of (........) molarity <-- acid \n29.8ml of NaOH of (.0210M) <-- base   \n\nMOLE of acid = MOLE of Base  \n\n.0298L  X .0210 M = 6.258 X10^ -4 mole of base.. \nnow,  u know that mole base = mole of acid only at equivalence point... \n\n6.258X10^ -4 moles of acid / .025 L ( this is what was used in begining)  = .025032 Molarity..  \n\nacid base problem .. memorize the units..   V = Liters, moles = moles, Molarity = moles/Liters ...    this problem they give u Liters, u need to know how much moles in beginning... to get Molarity
431629
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.
612775
A cubic centimeter (cc) is always equal to a millileter, nomater what material you are measuring. But a cc or mL or pure water at STP (standard temperature and pressure) is equal to one gram, which is what I think Mac C was hinting at. It was all devised this way when they made the metric system.
479981
I'll help you..this is weird because I just asked this same question and did this for my chemistry homework online also...I wonder if we're in the same class.\n\n3 862.8 nanometers- thats the answer, but you're going to write\n3.9 x 10^3 nm in the box.\n\nI actually think that we go to the same school...csu stan? probably..well good luck.
141472
Molarity (M) is moles of substance/liters solution\nyou need to know how many mols of FeCl3, or the molar mass and how many grams of FeCl3, to get the molarity\n\nmolar mass of FeCl3 is (55.845 + 3*35.453) or 162.204 g/mol\n\nassuming that you had 10g of FeCl3\n10g * 1/162.204 g/mol = .06165 Mols\n\nthen Molarity is just .06165mols/.02L = 3.08 M\n\nBUT you have to be able to find the number of mols to get the molarity\n\nSame idea for KI, except the molar mass for that is 39.098+126.904 or 166.002g/mol
65549
molality = moles of solute / kg of solvent.\n\n200 grams of water  = .2 kg of solvent.\n\n.3 molality = x moles / .2 kg\nsolve for the number of moles (x).\n\nThen multiply the number of moles by the Molar mess of Silver Nitrate and that is your answer.
537161
ok, here's what I get:\n\nKa = 1.8 x 10^-5 = [H+][C2H3O2-]/[HC2H3O2]\n\nKa = 1.8 x 10^-5 = x(0.40+x)/(0.80-x)\n\nbut we'll assume the dilution is negligible...\n\nKa = 1.8 x 10^-5 = x(0.40)/(0.80)\n\nx = 1.8 x 10^-5 M /((0.40)/(0.80)) = [H+] = 3.6 x 10^-5\n\npH = -log(3.6 x 10^-5) = 4.44\n\nSee if that agrees with some of your numbers