Molarity, Molality, Normality?
921065
you know you have got 152g total. So 1.8g/ml means you have got 152/1.8 for volume.Now just use MW KOH. Molarity will be moles KOH per litre, molality will be moles KOH/kg of solvent and Normality in this case will be the same as molarity but written  with symbol N.\n\nAs an aside the density of this solution would not be anywhere near 1.8g/ml, are you sure you have this right?
867444
do you know the molarity of vinegar?\nUsed M1 x V1 = M2 x V2\n(1) is vinegar and (2) is NaOH
166678
in giving appropriate amount of drugs you have to know a little about concentration of solution or colloid (syrup) molality, molarity, normality, percentage by mass, by volume, by mass-volume,etc; thermochemistry as well. in physics, knowledge in heat and thermodynamics, properties of matter, hydrodynamics (flow of blood), gas laws, etc are some of the basics that you are going to use.
1134846
stoichiometry is easy, there are many websites that can tell you exactly what do to for different types of problems\nhttp://www.chemtutor.com/\nhere is the specific website for help on mols and stoichiometry\nhope that helps!\nhttp://www.chemtutor.com/mols.htm
761477
Actually you have an electrolyte, so the drop in the freezing point  is\n\nΔT=i*Kf*m\n\nwhere i is the van't Hoff coefficient and (for simplicity) for strong electrolytes it is equal to the number of ions that are produced by its dissociation, according to the molecular formula.\nNaCl dissociates completely to two ions (Na+ and Cl-) thus i=2\n\nfor water Kf=1.86\n\nm is the molaLity and not molaRity. \nYour solution is 2 M = 2mole solute/L of solution \nmolality is mole solute/kg of solvent. You need additional info in order to convert molarity to molality\n\nIf you had 2 molaL and not molaR solution then m=2, or if we assume due to lack of more info that in this case molaRity=molaLity...\n\nΔT=i*Kf*m =2*1.86*2 = 7.44\n For pure water Tfwater=0 C, so for the solution \nTf= Twater-ΔT= 0-7.44= -7.44 C
741136
(a): Si+, P2+\n\n(b) Ar+, S-\n\n(c) Ca+, Cl2-\n\nBut the other answers that you gave are also correct.
938873
concentration\nattention\ndetection\ndetention\nattraction\nintention\nseduction
969572
M (molarity) is defined as "mole per L (mole/L)" of solution\n\nglucose = C6H12O6, its molecular weight is 180.16g/mole (say about 180g/mole)\n\n1000mM = 1000mmole / L\n--> (1000mmole / L ) * (180g / mole)\n--> cancel out the "mole"\n--> 180000 mg/L\n\nsame logic for 970mM = 970mmole / L\n--> (970 mmole / L) * (180g / mole) = 174600 mg / L\n\nnote: you can leave the "m (milli)" there in your calculations, because, remember, "m" simply means 10^(-3), which is 0.001
298095
If the freezing point of the solution had been incorrectly read .3̊ lower than the true freezing point, then the observed ΔTf would have been greater than it should have been. This would have caused the calculated molality to be too high. If the calculated molality were too high, then the number of moles present as calculated would be too high. An\nexaggerated number of moles would cause the calculated molar mass to be too low, as the same weight of the substance would be distributed against more moles of that substance.\nThe substance would seem lighter than it is.\n\nI'm sure this will help,\nUltimeaciax
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).
639379
The boiling point elevation constant of water is required\n\nThe formula is T = Kfm where m is molality and T is rise in temp.
876892
Pressure, volume/specific volume, temperature, entropy, enthalpy, specific heat, atomic weight, and thats all I can think of off the top of my head
524657
molarity = moles/volume\n\n125 g C2H5OH / 46 g/mol = 2.717 moles\n\nmolarity = 2.717/0.250 = 10.87 M
973272
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).
836092
is it General Chemitry 2 ?  If it is gen chem 2, I swear I hate this course. I am taking it again in the spring semester. Molarity, Molality...all non sense!!
903092
1) b)\n\n2) [H2O]^2 / 2[H2] [O2]\n  \n   [NO]^2 / [N2] [O2]\n\n3) b)\n\n4) b)
780378
vapor\n\ngas\n\nliquid\n\nvapor\n\nliquid\n\n\nnot 100% sure but i'm pretty good at science