Moles of water in hydrated salt?
918628
find the grams of water=5.3-2.89=2.41\nmolecular weight of water=18\nno.of moles=2.41/18\n=0.134\nhope that helps!!!\nbest of luck \nmay u get a gud score on the lab
385501
mostly mineral, you'll probably find sodium chloride (salt) in it, but it depends on which water body is evaporated.
898790
ok...\nweigh a sample of hydrated gypsum,\nput it in an oven set at 110 degrees C for a day or so. will evaporate the H2O\nre-weigh.\nfrom the difference, the weight of h2o is found, and then number of moles can be found.\nmoles = mass / MW.\n\nthen determine number of moles in gypsum sample left.\nthen you will have a ratio.\nand that ratio should be 1:2\nie calcium sulfate dihydrate, CaSO4·2H2O.\n\ncheers
240496
Okay, so molar heat capacity differs from normal heat capacity in the simple fact we're talking about a mole.  So water has a usual heat capacity of 4.186 J/(g*K).  All you have to do is multiply by the molar mass of water (18.0152.)  That should give you a molar heat capacity of 75.41 J/(mol*K)
807347
Since it takes 1 mole of O2 to react with H4 and get H2O + H2O. I would have to say that .264 moles of H2O were created. water is 18gper mole so .264*18g=4.752g of water.
661119
sodium chloride or salt
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
885670
Get the chemical equation first:\nHCl + NaOH --> NaCl + H2O\n\nMoles of HCl used = (50/1000)dm3 X (1.0mol/dm3) = 0.05 mol\nMoles of NaOH used = (50/1000)dm3 x (1.0mol/dm3) = 0.05 mol\n\nFrom chemical equation, 1 mole of HCl needs 1 mole of NaOH for complete neutralization.\nFrom what you are given, you have the exact amount for complete neutralization (meaning, there is no excess reagents).\nFrom chemical equation again, 1 mole of HCl or NaOH produces 1 mole of water.\nThus you had exactly 0.05 mole of water produced here.\nHope this helps. :)
697284
if we assume that 20 drps is equal to 1 mL then the correct answer is 360 drps equals 1 mole of water.
295517
First you need to know what a mole is. A mole of HCl is it's formula mass, which is 36.5g. So, if you put 36.5g of HCl into one litre of water, you would have 1.0 molar solution. For 0.1 molar you use 3.65g HCl.
39876
To answer this question you must look at your given data. 13% of 1.1 = .143 so you have .143g NaCl for every ml solution. Molarity is moles solute/liters solution. You just calculate how many grams NaCl are in a liter and convert to moles, do the division and your done.
913013
0.9% NaCl means 0.9 g NaCl in 100 mL SOLUTION, so you need 9 g NaCl for 1000 mL (1L)\n\nTake a volumetric cylinder, putsome water in it, dissolve the salt and add water to a final volume of 1L and mix. If you just add 9 g of NaCl in 1L of water the final volume of the solution will be different than 1L and thus your concentration will deviate from 0.9%.
755325
The first thing we want to account for is that there are 6.02 x 10^23 molecules of H2O in one mole of H2O.  This is Avagadro's number, and it applies for the number of molecules in a mole, the number of atoms in a mole, etc.  So we want to see the problem up as follows:\n\nWe will start with our given quantity of molecules (2.48 x 10^26 molecules of H2O) and multiply this number by the ratio 1 mole of H2O/6.02 x 10^23 molecules H2O.  Essentially we are dividing 2.48 x 10^26 by 6.02 x 10^23.  The answer will be left with moles of H2O because the molecules of H2O will cancel each other, and we will be left with an answer.  \n\nFor a frame of referance, this answer is 4.12 x 10^2, or is it's not put into scientific notation, it is 412. molecules of H2O.
225870
Take 33g of the salt and dissolve it in 67 g of water to get 33% salt solution.\n\nIts not the same as 0.3 N solution
215367
1 mole occupies-----22.4lts\n??????   -----------5.6lts\n\n5.6/22.4=0.25
523531
At the heart of the issue is a phenomenon called Boiling Point Elevation. The boiling point of water is elevated when solutes are dissolved in the water. The boiling point also depends on the atmospheric pressure, but for the this question, I'm going to assume that we are at one atmosphere of pressure so that the boiling point of water is 100 degrees centigrade. \n\nSome people put salt in water when they make pasta, believing that the dissolved salt raises the boiling point of the water, and that the water is actually hotter than 100 degrees centigrade when the pasta is added. I say that they believe this because I'm doubt people add enough salt to result in a significant elevation in the boiling point. For water, the boiling point is elevated by 0.513 degrees centigrade for each mole of dissolved solute-ions per kilogram of solution. A mole of salt (NaCl) is 58.44 grams, and since one mole of NaCl disassociates into two moles of ions (Na+ ions and Cl- ions), 58.44 grams of salt (or 2 ounces) would be required to raise the boiling point of one liter (kilogram) of water (at one atmosphere of pressure) to 101.026 degrees. That would be some pretty salty water. \n\nThere is a related phenomenon known as Freezing Point Depression, in which the freezing point of water is lowered by 1.86 degress centigrade for each mole of solute-ions per kilogram of solution. We exploit the phenomenon of Freezing Point Depression by using salt to melt snow and ice on roads in the winter and to make ice-cream. \n\nNow, how does this relate to your question about the boiling point of blood? Well blood is basically a water solution. The dissolved solutes are more complex than a simple NaCl solution, but the principle of boiling point elevation (and freezing point depression) should still apply. The salt (NaCl) concentration of blood is 0.9%. This is approximately 0.154 moles of NaCl per kilogram (liter), or 0.308 moles of dissolved solute per liter. This should result in a boiling point elevation of a whopping 0.158 degrees centigrade. \n\nNow, there are other molecules dissolved in blood besides Na+ and Cl- ions, so the boiling point may be slightly different, but I think that you can see that it isn't going to be noticeably higher than 100 degrees (at one atmosphere of pressure). \n\nSo the answer to your question is basically, at the same temperature that water does. I hope that isn't too disappointing. If you want more information on Boiling Point Elevation or Freezing Point Depression, you should be able to find it in any college-level chemistry text book.
558840
When you dissolve a salt in water, the freezing point of the resulting solution is lower than the freezing point of the pure solvent (water).\nThe amount of the freezing point depressure depends on two factors, the concentration of the solute (salt) and the freezing point depression constant of the solvent.\n\nThe formula to calculate the freezing point depression in therms of the change in the original freezing point (delta T) is,\n(delta T) = k_f * m\nwhere k_f is the freezing point depression constant and m is the concentration of the solute in terms of molarity.\n\nIt is mathmatically possible for the concentration of the solute to be so greate at to create a change in the freezing point of water of more than 32 degrees F.\n\nThe freezing point depression constant of water is -1.86 degrees C per molality.  So mathmatically, if one had a solute concentration of about 9.57 molality or greater, the freezing point of water would be below 0 degree F.