. Exactly what weight of dry ice is needed to react with 2 mmol of phenyl magnesium bromide?
541101
First you need the equation for the reaction of CO2 with phenyl magnesium bromide to get the mole ratio. (i get one to one.)\n\nSecond, calculate the number of millimoles of CO2 required to react.\n\nLastly, multiply the number of millimoles of CO2 by its molar mass. I get 88 mg.
436191
Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.
255396
FeCl3 is the limited reactant and the mass of the excess \nreactant present in the vessel is .55g Mg
106494
First, distill off the alcohol. With ethanol, the best you can get is 95% purity, the remaining 5% is water. There are cheap zeolites and hygroscopic salts such as CaCl2 or NaOH available which can further reduce the amount of water. If you are still not happy, you can get absolute alcohols by using magnesium. And magnesium ribbons or turnings are relatively cheap, too. Place (50-100 g) magnesium in a flask, put some amount of alcohol just to cover the magnesium. Leave for a few hours until you see white magnesium alkoxide being formed on the surface of the metal. Caution: exothermic reaction. Use the tallest distillation column you can find. Do not stir the mixture. Just leave it there. When you have now a lot of magnesium alkoxide you can add more alcohol (up to 2 liters). The alkoxide will destroy any water in the alcohol and you now can distill off your absolute alcohols. For methanol and ethanol, formation of the alkoxide takes only a few hours. For other alcohols, you might need to activate the magnesium by adding some solid iodine and heating to activate the magnesium. Don't distill to dryness. The remaining alkoxide reacts violently with water. Use alcohol-water mixture when disposing them. Sounds fun, doesn't it? =)
318519
One mole of NaOH weighs 40gms. 200mmol is 0.2 of that and 100ml is 0.1 of a litre. \nSo you would need 40 X 0.2 X 0.1gms of NaOH.=0.8gms\n\nIn reality you would likely dissolve 8gms of NaOH into a 1L volumetric flask and fill it to the mark. You would then measure out 100 mls of this solution. Weighing out 0.8 gms of NaOH is inaccurate because it reacts with CO2 in the air and absorbs water vapour.
980695
It depends on how much dry ice you have and what is the temp of the dry ice.
954876
NiCl2. mass percentage would be same as relative molecular mass percentage.\n\nCl2=71g/mol\nNiCl2=130g/mol\n\n71/130 x 100% = ??
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
122744
Mg + 2 CH3COOH = Mg(CH3COO)2 + H2
739325
According to Sigma-Aldrich the solubility of MgCl2.6H2O is 1M at 20C (or 20.3030 g/100ml solution), so it shouldn't be possible to prepare 4.5 M MgCl2 in just water. If there are more solutes involved which might increase the solubility of MgCl2 try dissolving those first and then the MgCl2.\n\nYet I saw some papers using 4.5 or 2.5 M...\n\nI don't know why there is this discrepancy between sigma-aldrich and the solubility provided in the link in the above answer which would correspond to 8 M if you consider 100ml of solution and not water. In reality you would have a significant change in volume when you add so much salt and the solubility would be much less but I don't know exactly how much. If you consider 167g of MgCl2.6H2O then you are adding 88.8 g of water which means that the total volume is at least 189 ml and thus the maximum concentration is 4.3 M.\n\nWikipedia gives a different number; 54.2g/100ml at 20C, without specifying if it is for the anhydrous or hexahydrate... so it would be 5.6 or 2.6M respectively... \n\nWhat a mess !!!\n\nMaybe you should also post the question in biology, since someone who has been doing such purifications can clarify the issue.\n\nAnother idea would be to prepare a 100ml saturated solution according to the lowest reported solubility (that of sigma-aldrich) using a 250 ml volumetric cylinder, and then from a pre-weighed amount of MgCl2.6H2O keep adding powder slowly until it doesn't dissolve any more. Weigh the powder that is left in order to find how much you dissolved, measure the final volume and calculate the molarity. If you achieve higher than 4.5 M, dilute.\n\nOR\n\nprepare a 4 M if the amount of powder that you couldn't dissolve is very small.\n\nor \n\nprepare 450ml of 1M and evaporate to 100ml of solution while praying that there will be no precipitation...\nThat's very desperate but I am running out of ideas...\n\nI must say the whole story is very fishy (finding very different solubility values)\n\nI'd go mad too...
1016191
technically mgs is a measure of weight while mls is a measure of volume.  So you would need to know the density of the liquid to properly calculate.  Different liquids have different densities.\n\nThat being said, 20 mls of water weighs 20 mgs.  I'm assuming that is a good enough approximation.
795822
Anhydrous magnesium sulfate is used to dry the organic solvent phase isolated after an extraction of your solute (compound) from an aqueous phase.  After isolating the organic phase, there is a certain amount of water present (solubility factor, suspended microdroplets of water, how careful you were at separating the layers).  While it is relatively easy to remove the solvent (ie: ether, ethyl acetate, hexane) by evaporation, it is more difficult to remove traces of water.  Often water becomes hydrogen bonded to your solute resulting in a sticky mess that will affect your weighing measurements.  Water is especially hard to get rid of if your compound is a liquid.  Weighing errors will lead to inaccurate concentration measurements.  Inaccurate concentration measurements lead to less than an "A" grade on your experiment.\n\nThe magnesium sulfate hydrates rapidly (forming both magnesium sulfate monohydrate and magnesium sulfate heptahydrate) and is then easily separated from the organic phase containing your dissolved solute by simple filtration.  Voila, the water is now gone, you have a "dry" organic phase which, when evaporated, will give you a dry compound.
136403
You can calculate the freezing point depression of the solution is you know three things, firstly, the freezing point of the pure solvent, and secondly, the concentration of the solute, and thirdly, the freezing point depression constant of the solvent.\nIn this problem, you know both (One is given, two of them are assumed knowledge).\n\nThe freezing point of pure water is 0 degrees Celcius.\nThe concentration of the solute is 52 percent.  You need to convert this to a molaltiy concentration to use it (or fiind a different freezing point depression constant).\nLook up for your self the k_f value.\n\nthe freezing point depression = k_f * m,\nwhere k_f is the freezing point depression and m is the concentraion of the solute in terms of molality.\n\nmolality = moles of solute / kg is solvent
933723
You don't need a chemical reaction.\nFrozen carbon dioxide, known as dry ice, can do the job.\nIt's in a thermos container, but as soon as it touches the air, it sublimates into gaz, producing the smoke. (To make more, put it in water instead.)  And it's heavier than air, so it stays close to the ground.  And it's not expensive!
680540
Assuming the ice is at 0 degrees, it would take 3.33 kJ (10 grams is 0.556 moles).  If the ice is not at 0 degrees, it will take more energy to raise its temperature.
802874
Magnesium reacts incredibly slowly with cold water, and it takes about a week to collect  even a few cubic centimtres of hydrogen gas. However, if phenolphthalein is added to the water at the end of the experiment, it will turn pink. This confirms that a solution of the not very soluble magnesium hydroxide has been formed.
699816
the first thing you do is figure out how much heat it will take to melt the ice, this is done by multiplying the quantity of ice by the heat of fusion\n\n10g x 335 J/g = 3350J\n\nnow figure out how much the water cools off to give the ice that 3350J that it needs to melt, this is done by multiplying the amount of water by the specific heat and then dividing this into the energy transferred\n\n100g x 4.18J/gC = 418 J/C\n\n3350J / 418 J/C = 8 degrees C\n80-8=72C\n\nso now you have 100g water at 72C and 10g water at 0C, so you can figure out what this mixture will equilibrate to\n\n100 x 72 =7200\n10 x 0 = 0\n7200 / 110 = 65.4C will be the final temp of the mixture\n\nso the answer is D