Molar mass of volatile liquid experiment question?
816061
It is tough to answer this question without knowing what the experimental procedure was or what the aluminum foil was used for.  More detail please.
365444
0.500L x 0.250moles/L x 58.44g/mol = 7.31g\n(a)
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
761497
1) The first one is wrong on the last person who answered, O2 weighs 32g/mol because there are 2 O's in it, 16 each.  So in 1.33x10^4 (13,300mol) you have 4.256x10^5g of O2 (425.6kg).\n\n2) Recognize that hydrogen, oxygen, and chlorine are diatomic gasses, so they always occur in pairs (H2, O2, Cl2).  Others like this are nitrogen (N2), fluorine (F2), bromine (liquid @ room temp. Br2) and Iodine (solid at room temp, I2).  So your equations would be:\n\n2H2+O2-->2H2O\nH2+Cl2-->2HCl\n\n3) PV=nRT and solve for n, number of moles.  Put everything in correct units, liters, atmospheres, moles, and kelvins, and R is a constant R=0.0821 L*atm/mol K.  Kelvin temp=Celcius temp +273\n\nSolving for n:\nn=PV/RT=(1.09atm)(0.942L)/(0.0821L*atm/mol*K)(287K)\nn=0.0436 mol\n\nSo, when we usually want to know a number of moles given a mass and having a molar mass we can look up, we say,\nn=mass/molar mass\n\nin this case, we need to find molar mass, so:\nmass/n=molar mass\n3.17g/0.0436mol=72.71g/mol
880546
you should learn how to do your own homework hun you'll never be a chemist with someone else doing your work.\n\n\nuse the molar mass from the periodic table (units = g/1mol)\n\nCl = 35.5\nH = 1.00\n\nMolar mass, formula weight, molecular weight, formula mass = 36.5 grams HCl/1 mol HCl\n\n\nThis is a simple problem is you can't do this problem maybe you should try changing majors.  I hear history or english is nice.
445032
Click on "Add details," post the problem, and, as SMeghead said, someone might be able to help.\n\n(EDIT: You didn't provide too much information. From what you gave, about all we can say is that 12.5% of the solute - 1% from 8% - evaporated. If you need to know how many grams, milliliters, cubic centimeters, etc. evaporated, we'll need more information.)
436874
The molecular mass of C12 H22 O11 is 342 (sum of individual atoms).  There is 100g of it, so by criss-cross\n\n        If  342 g of C12 H22 O11 = 1 mole of  C12 H22 O11 \n        Then  100g of  C12 H22 O11 = ? mole of  C12 H22 O11\n\n              ?= 0.2923 moles of  C12 H22 O11 \n\nMolarity (M)= number of moles of  C12 H22 O11/volume of solution (liter) \n\n           M= 0.2923 moles/0.2 liters = 1.46 M(olarity) of\n           C12 H22 011\n\nHope this helps
951445
... what's the specific gravity of the mixture?\nSince the total volume is the sum of the two ingredients, there was no volume loss due to mixing. (Sometimes there is; water and ethanol are an example.) Then the total mass is 6*0.8 + 5 * 1 = 9.8 kg and the sp. gr. = 9.8/11 = 0.8909 kg/L.
804090
ooo this will be fun!\n\nok..\nA)\n2.207g of H2O MW. 18.0\nmol = weight of sample (g) / molar weight (g/mol) (n=m/M)\nn = 2.207 / 18\nn = 0.1226 moles of H2O\n\nnow, since there is 2 hydrogens in H2O, then when the molecule is broken down, there is twice as many moles of hydrogen then the overall molecule,\ntherefore,\n\n2x 0.1226 = 0.2452 moles of H.\n\ndetermining number of moles of CO2 is also helpful,\n4.313g, n = 4.313 / (12.01 + 16 +16) = 0.0980moles\napprox, 5(moles H2O) :  4(moles CO2)\n\nB) \nworking the number of moles of this compound will be useful:\nn = 1.816 / 74.122 = 0.0245 moles, which is 1/10 of the moles of H produced. (or 1:5 of the moles produced of water)\n\nif you think about it, the reaction is something like this:\nCH2 + O2 --> CO2 + H2O\nclearly the above doesn't make sense, because 1) it isn't balanced, and 2) the reactant needs a higher molecular wt.. now thinking about it, the following can be deteremined:\nand from the ratio of moles found above:\n       1     : (excess):  4        :    5\nC4H10 + 13/2O2 --> 4CO2 + 5H2O\n\nnow, I amoung most allow fractions in equations, but if you don't, multiply through by 2.\n2C4H10 + 13O2 --> 8CO2 + 10H2O\n\nnow, you want the empirical forumla,\ntherefore,\ndivide by highest common factor,\nC2H5 \n\nthe empircal fomula is just a ratio and doesn't make huge sense, as the formula for an alkane is CnH2n+2\nhope that makes sense :)
649234
hmm no one answered this and ill give it a shot. \ntook AP chem last year but lets see what i can remember...\nsome possibilities:\nsplattering\nerror in massing\nOOOi think this is it: naphthaline is volatile [according to wikipedea], which means that it evaporates really easily. \nthats probably the answer u are loooking for.\nif u dont know what volatile means, look it up, ask ur teacher, or go on wikipedia.\nso basically the error would be that the mass of the naphthalene would be less after a period of time becuase some of it had evaporated.\n\n=P read more and figured out a nother problem:\nNaphthalene easily sublimes at room temperature without vacuum.\nsubliming = cahnge from solid to gas wiht no intermediate liquid phase.\nso i guess mroe of the mass is loss that way.\nseems like naphthalene like being a gas, huh? =P\n\nhope that helped
48128
d = density\nM = molar mass or mass per mol\nV = volume\nn = number of mols\nR = gas constant\nT = temperature\nP = pressure\n\nd=nM/V \n\nV=nRT/P\n\nd=nMP/nRT\n\nd= MP/RT \n= (28x10^-3 kg/mol x 300x10^3 N/m^2)/(8.314 J/(mol K)x450K) \n\n= 2.25 kg/m^3
732197
The density can be assumed to be 1.0 g/ml.  \n\nSo you have 25 ml x 1.0 g/ml = 25 g H2O\n\nThe molar mass of water is 18 g/mol, so you have:\n\n25 g H2O/18 g H2O/mol H2O = 1.4 mol H2O
482555
AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit
313383
Molarity= moles/liters
882624
Update from my other answer:\n\nThe reaction is  ethanol + K2Cr2O7 -> products. This means that 1 mol ethanol will react with 1 mol K2Cr2O7. 1 mol K2Cr2O7 weighs 2*39.1 + 2*52 + 7*16 = 294.2 grams. 1.1 mg was consumed which is 3.74 * 10^-6 mol. Therefore the testsample had also 3.74 * 10^-6 mol of ethanol. Ethanol is C2H5OH, and therefore 1 mol weighs 2*12 +6 + 16 = 46 gram. \nTherefore there was 46 * 3.74 * 10^-6 = 1.72 x 10^-4 gram = 0.172 mg of ethanol.
770958
1/1000 x molar mass (grams/litre) and solute in 1 litre
979624
Molar mass of ( CaCl2x2H2O)= mass of (Ca) + 2mass of (Cl)  + 4mass of (H) + 2mass of (O)\n\n\nMolar mass of ( CaCl2x2H2O)= (40g) + 2x(35.5g) + 4x(1g) + 2x(16g) \n answer \nb) 147 g/mole