what is the empirical formula for hydrated magnesium chloride?
110757
both MgCl2·6H2O and MgCl2·(H2O)6 are correct formula for hydrated magnesium chloride
358871
Calcium chloride = CaCl2 = Mwt 110.98\n% Ca = (40.08 / 110.98)*100 = 36.11%\n% Cl = (70.90 / 110.98)*100 = 63.89%\n\nCalcium chloride dihydrate = CaCl2.2H2O = Mwt = 147.01\n% Ca = (40.08 / 147.01)*100 = 27.26%\n% Cl = (70.90 / 147.01)*100 = 72.74%\n\nOther hydrates can be calculated in the same way.\n\nHope this helps.
824141
Start off with the original equation\n\nCaCO3(s) + HCl(aq) --> CaCl2(aq) + CO2(g) + H2O\n\n\nStart off by making the Ca equal on both sides, which is fortunately already done\n\nThen make the Cl equal on both sides by putting a 2 in front of HCL\nCaCO3(s) + 2HCl --> CaCl2(aq) + CO2(g) + H2O\n\nThen make the C's on both sides equal, which again, is already done\n\nNext make the O's equal on both sides, which yet again is already done.\n\nFinally you make the H's equal on both sides, which is done also. \n\nSo the answer to this whole question is to put a 2 in front of the HCl. (I'm guessing you meant to put CaCl2 instead of CaC12 in the products)\n\nThat was easy.
87280
MBrO3-9H2O where M can be Na,K,Li, other +1 cations\n\nIf M is other than these you have to adjust subscripts
131743
The formation of PhMgBr is a redox reaction.  The two electrons of magnesium metal are added to the PhBr bond to make Ph-Mg-Br.  Then you added this in solution to the dry ice (solid O=C=O).  The electrons between Ph and Mg add to the carbon of the CO2.  The electrons from one of the pi bonds moves out to the oxygen so that you have PhC(O)O- where the (O) is still doubled bonded to the C.  After the reaction was complete, you added water to protonate the O minus.  This completes the reaction.  Since you are in organic lab, you should still have an organic textbook.  There is likely a mechanism drawn out in any organic textbook you can find.
820243
the chemical formula of a hydrate would be:\n\nthe original anhydrous form x nH2O where n is any whole numbers\n\nfor example: CaCl2 (calcium chloride) may form several different hydrates--\nCaCl2 x H2O\nCaCl2 x 2H2O\nCaCl2 x 6H2O
356281
MgCO3 + 2HCl ---> MgCl2 + CO2 + H2O\n\n2HCl + MgO ---> H2O + MgCl2\n\nBarium hydroxide is a base so can be neutralised with any acid.  An example being HCl.\n\nBa(OH)2 + 2HCl ---> BaCl2 + 2H2O\n\nor H2SO4\n\nBa(OH)2 + H2SO4 ---> BaSO4 + 2H2O
776502
Add the molecular weights of the molecular formula.\n\nCaCl2\n\nCa = 40.08g/mol\nCl (2) = 70.906g/mol\n\nCaCl2 = 110.986g/mol\n\nNow you do the other one.
489210
Potassium Hydroxide and Hydrochloric acid.\n\nKOH + HCl = KCl + H2O
738944
Potassium chloride, Chromium oxide, Barium chlorate, Ammonium chloride, Phosphorus trichloride
883304
Na2EDTA is a complex molecule. Its name stands for ethylenediaminetetraacetic acid – disodium salt. The formula is Na2C10H14N2O8. Even though its name and structure are complex, you only need to know that: EDTA reacts with Ca+2 and Mg+2 in a one to one ratio as follows:\n:Ca+2 + EDTA-2 -----> CaEDTA\n\nMg+2 + EDTA-2 -----> MgEDTA\nThe formula mass of Na2EDTA dihydrate is 372.24 g/mole.Na2EDTA comes as a dihydrate(absorbs water from the atmosphere) which is efflorescent (literally means "bursting into flower" but in this case just means that it forms bubbles), you cannot reliably weigh out the desired number of grams. The salt may be fully, or only partially, hydrated at the time you weigh it. The result is that you cannot be certain that you have prepared the proper strength reagent of EDTA..therefore: you need to standardize the solution  The results of the titration is reported as mg of CaCO3/liter.\nTo convert Calcium plus Magnesium (meq/L) Hardness as CaCO3 (mg/L) 50\nOf course since Fe and Mg and Ca and other metals are also present in water it is better (more accurate) to titrate for each metal species individually and calculate the hardness using the results of the individual results. EDTA hardness is sometimes referred to as "Total hardness"and is reported as " Hardness as mg/l CaCO3."
458687
http://www.altavista.com/web/results?itag=ody&q=structural+formula+of+Dicycloverine+Hydrochloride&kgs=1&kls=0
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
237268
.001565mol / .075L = .021M NaCl
803450
MgCl2(aq) + 2NaOH(aq)  --> Mg(OH)2(s) + 2NaCl(aq)\n\nNet ionic equation:\n\nMg2+(aq) + 2OH-(aq) --> Mg(OH)2(s)
1022778
Mg2+ + 2HCl --> MgCl2 + H2 (gas)\n\nOkay, hydrogen.  As you can see in the above equation, for one atom of magnesium you'll produce two atoms of hydrogen.   \n\nLet's work out how much hydrogen you'll get in grams.\nAnd let's say that you have 1 gram of Mg.  First, what you need to do is work out how many atoms are in 1 gram of Mg.  \n\nThis is easy.  Simply:\n\n1 / atomic mass x Avagadro's constant = number of atoms.\n\nFor Mg, this equals:\n\n1 / 24.31 x 6.0221415×10^23 = 0.2477228 x 10^23 atoms.\n\nWe know that for every atom of Mg you'll produce 2 atoms of hydrogen.  So, that means you'll produce:\n\n2 x 0.2477228 x 10^23 = 0.4954456 x 10^23 atoms of hydrogen.\n\nTo work out how much 0.4954456 x 10^23 atoms of hydrogen weighs you just calculate, like this:\n\n0.4954456 / 6.0221415 x 1.008 = 0.0829 grams.\n\nAnd that's it.  1 gram of Mg will produce 0.0829 grams of hydrogen.
852236
I would say Aluminum Chloride. (AlCl3)