what's the formula for magensium carbide?
144641
MgC(2)... the 2 is supposed to be subscript.
778176
Use the periodic table of the elements.  The "weights" are in grams per mol.  Just use algebra and solve for the number of mols.  Then you know how many atoms there are by the number of mols which is a definite number.
844942
Phase - gas \nDensity  (0 °C, 101.325 kPa)\n9.73 g/L \nMelting point  202 K\n(-71 °C, -96 °F) \nBoiling point 211.3 K\n(-61.7 °C, -79.1 °F) \nCritical point 377 K, 6.28 MPa \nHeat of fusion 3.247 kJ·mol−1 \nHeat of vaporization 18.10 kJ·mol−1 \nHeat capacity (25 °C) 20.786 J·mol−1·K−1 \nVapor pressure P/Pa 1 10 100 1 k 10 k 100 k \nat T/K 110 121 134 152 176 211 \n \nAtomic properties \nCrystal structure cubic face centered \nOxidation states 0 \nElectronegativity no data (Pauling scale) \nIonization energies 1st: 1037 kJ/mol \nAtomic radius (calc.) 120 pm \nCovalent radius 145 pm \nMiscellaneous \nMagnetic ordering non-magnetic \nThermal conductivity (300 K) 3.61 mW·m−1·K−1 \nCAS registry number 10043-92-2 \nSelected isotopes \nIsotopes of radon iso NA half-life DM DE (MeV) DP \n211Rn syn 14.6 h Epsilon 2.892 211At \nAlpha 5.965 207Po \n222Rn 100% 3.824 d Alpha 5.590 218Po \n \n\n\nthis is what wikipedia had to say. hope it helps!
770269
For aA+bB -> cC+dD\n\nE=E0-(RT/nF)* ln([C]^c*[D]^d / ([A]^a*[B]^b))\n\nwhere n=number of electrons taking part in the reaction\n\nSolids are not considered in the equation, so in your case\n\nE=E0 -(RT/nF)*ln([Fe2+]/[Cd+2])\n\nYou can find the standard reduction potentials for the Fe+2 and Cd+2 at \nhttp://www.northland.cc.mn.us/Chemistry/standard_reduction_potentials.htm\nThen E0= E0(Cd2+)-E0(Fe2+)\nR=8.314\nn=2\nF=96500\n\nYou have to specify T ...
663777
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
650254
it is C as in use the formula for CI for one sample proprtion from your book(i better get the best answer choice cause this is the fifth one i am answering)\nand the above guy is a moron
553400
take the mole fraction and multiply by the atomic weight of each which you can find on the periodic table. For carbon, it would be 12 x .04 = .48 Once you have all three, sum them and divide each by this sum. So, say you find Ch and I for the other two. Then the percentage by weight of Carbon would be the number you got for Carbon (.48) divided by (.48 + Ch + I).\n\nTo double check, your weight percentages should all add up to 100%.
881597
Cr2(SO4)3 + 6KOH  --> 2Cr(OH)3 + 3K2SO4
440961
Zr 4\nhttp://en.wikipedia.org/wiki/Zirconium\n\nW 6\nhttp://en.wikipedia.org/wiki/Tungsten\n\nMo 6\nhttp://en.wikipedia.org/wiki/Molybdenum
498666
Aaaaaaaaalright... Formulas depend on what you're solving for but since they all involve using the ratio of moles it shouldn't be too different from equation to equation.\n\nLet's go over the mole first to help you better understand what's going on in stoichiometry.  The mole is the amount of a substance that contains as many particles as there are atoms in exactly 12g of carbon-12.\n\nWHOA there... hold on for a second.  The mole is simply a counting unit, just like a dozen is.  We don't usually buy 12 or 24 doughnuts; we order half dozen, one dozen, or two dozen.  Similarly, a chemist may want 1 mol of carbon, or 2 mol of iron, or 2.567 mol of calcium.  \n\nThe number of particles in a mole is known as Avogadro's number, named after an Italian chemist Avogadro. (Go figure)\n\nAvogadro's number is 6.022 x 10^23... which is the number of particles in exactly one mole of a pure substance.  One mol of a substance is going to equal its AMU on the periodic table, or, the mass. \n\nFor example, one mol of Helium is 4g.  If you wanted to find out how many grams are in 3 mols of Helium, you would do a conversion like this:\n\n3 mols He x 4g He/1 mol He\n\nIn words that would be... 3 mols of Helium multiplied by however many grams per mol Helium is, which is 4 grams per mole.  The mols would cancel out leaving us with the units in grams.  Do 3x4 and you get 12 grams of Helium in 3 mols of it.\n\nIf you want to find out how many mols of something you have you would take the amount of grams you have of it and set up another conversion like this:\n\nAmount of g x 1 mol/gram in that mol\n\nFor example, if you have 20 grams of Helium and you want to find out how many mols of it you have, the equation would look like this:\n\n20g Helium x 1 mol/4g He per mol\n\nThe grams would cancel out, leaving the answer in mols.  You would then take 20 divided by 4 and get 5 mols of Helium.\n\n(Hoping this is making sense)\n\nNow stoichiometry is just pretty much comparisons of stuff. (Mols)\n\nAlright, when you're asked to calculate the amount in moles of one substance that will react with or be produced from the given amont in moles of another substance, you must use this simple mole conversion problem.\n\namount of given substance (mols) ----> amount of unknown substance (mols)\n\nIn order to solve this type of equation, use this conversion factor.\n\ngiven quantity x mole ratio = unknown quantity\n\n\nLet's try an example now.\n\nCOsub2 + 2LiOH ---> Lisub2COsub3\nsub=subscript\n\nLet's calculate how many moles of Lithium Hydroxide are required to react with 20 mol of COsub2\n\nGiven information = amount of COsub2 = 20 mols\nUnknown = amount of LiOH in moles\n\nSince the equation is balanced, we need to use the ratio that would give us LiOH in the final answer, therefore, meaning LiOH must be on the top of the ratio.\n\nThe correct ratio would be (mols LiOH/mols COsub2)\n\nIn order to calculate this, set up the equation now. (given q x ratio = unknown)\n\nmols COsub2 x (mols LiOH/mols COsub2) = mols LiOH \n\nsince the mols of COsub2 are on the bottom, it cancels out with the mols of COsub2 on the other side of the multiplication side because that's the rule in math, therefore leaving you with the units of LiOH.\n\nsubstitute the values into the equation and calculate the answer.\n\n20 mols COsub2 x (2mols LiOH/1 mol COsub2) = 40 mol LiOH\n\nsince the mols of COsub2 cancel out, you're left with 20 x 2 mols LIOH which equals 40 mols of LiOH.\n\nThe ratio of LiOH/COsub2, 2:1, was made from the balanced equation because for every mol of COsub2 you have you need 2 mols of LiOH to react with it to produce the right amount of products.\n\nHope this helped, good luck.
784503
i m surprised none of these  answers are correct \nthe correct answer is Metalliod
214547
[Cr(en)3]Cl3  =  Tris(ethylenediamine)chromium (III)\ntrichloride\n\n[Fe(CO)3]  =  Iron carbonate or Iron (II) Carbonate\n\n[CoCl(NH3)5]Cl2  =  pentaamminechlorocobalt(III) chloride but it is [Co(NH3)5Cl]Cl2\n\nK3[Al(C2O4)3]  =  potassium triethanedioatealuminate(III)\n\nK3[Co(NO2)6]  =  potassium hexanitritocobaltate(III)\n\n[Fe(H2O)6]++  =  Hexaaquairon(II) chloride\n\nK[Ag(CN)2]  =  Potassium dicyanoargentate(I)
315781
you are in the wrong section!!! this is math section...u need to go chemistry section....\n\nAnyway, the formula for cobalt(III)chloride plus ammonia is........\n\nCoCl3 + NH3 = NCo + 3HCl
853340
KCl + Cu(Cr2O7)\n\nDichromate has 2 Chromium +6s to counter the 7 oxygen -2s.  +2 + -2 = 0\n\nPotassium Chlorite and Copper Dichromate
798483
Here it is:\n\n(98,42)Mo + (1,0)n ---> (99,42)Mo
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
381600
Five (5). \n\nMore details:\n\nName: Boron\nSymbol: B\nAtomic Number: 5\nAtomic Mass: 10.811 amu\nMelting Point: 2300.0 °C (2573.15 K, 4172.0 °F)\nBoiling Point: 2550.0 °C (2823.15 K, 4622.0 °F)\nNumber of Protons/Electrons: 5\nNumber of Neutrons: 6\nClassification: Metalloid\nCrystal Structure: Rhombohedral\nDensity @ 293 K: 2.34 g/cm3\nColor: brownish