PLEASE READ NOW/STOICHIOMETREY! I beg you :-(?
830588
If I understand your question, you start out with 2.15 g of AgNO3 and you isolate 1.16 g of AgCL.\n\n(It saves time if you include the M.Wt. of the materials.)\n\nAg = 107.87g/mole\nNO3- has 14.0067 + 3*15.9994 = 62.005 g/mole\nCL = 35.453 g/mole\nAgNO3 = 169.875 g/mole\nAgCL = 143.323 g/mole\n\nIn theory, for every mole of AgNO3 you should get 1 mole of AgCl\n\n(2.15 g AgNO3)*(1 mole AgNO3/169.875g) = 0.0126 mole AgNO3, and also the same number of moles of AgCL\n\n0.0126 mole AgCL * (143.323 g AgCl/mole AgCl) = 1.81 g AgCl in theory (rounded)\n\nYou isolated only 1.16 g, therefore\n\n  (1.16/1.81)*100% = 64 % yield.\n\nYou say you want % error, I don' t know if that is the correct term but here is what the error term is:\n\n [abs (1.16 - 1.81)/1.81]*100% = 35.9%\n\nplease check for errors
861843
Start with the Ca then balance the number of PO4 ions, then balance the Oxygen and the hydrogens should balance themselves out.  Its just trial and error. Monkey with it until it works.
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
980488
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD
215160
http://www-ec.njit.edu/~skawinsk/chem243/chap11/chap11.htm\nthis may help–near the bottom
613443
Niuchemist is mostly right and deserves the vote, but there is one correction I would make-heavier molecules don't necessarily have less "kinetic energy", they move more slowly-it is the speed of movement that is the critical thing, not the kinetic energy, which is a product of speed *and* mass.
1057769
I'm a teacher and i always give the students the "Rope example"\n\nimagine equilibrium as a rope pulled with the same force both ways! \n\nnow if you look at the your chemical reaction energy (heath) is release from the reaction -----> so if you take the heath out (so you cool the system) the reaction ------> will be more power full so equilibrium will be moved in C and D direction.\n\nIf you heath the system (you'll have a lot of heath that needs to go somehow --- and the only way it can go is by the <--------- reaction -- which is a reaction that use heath ).\n\nThe same with pressure (but you need to have different amounts of molecules -- in your reaction you have the same -- 5 in the reaction ----> and 5 in the reaction <----)\n\nif you speak about concentration use the rope example. You put more of A(increase the concentration of A) , that A need to be consumed in order to keep the equilibrium.\n\nif you take A out of system (so reduce the concentration of A) you'll encourage the <---- reaction (reaction that release more A)\n\nI don't know if you'll understand my spelling and grammar (I'm not the aching chemistry in English). \nI'm just curious what grade are you on? because  I teach this subject to 10th grade students (16 years old )\n\nIf you want me to try to rephrase this answer just let me know \njustplaincornnel@yahoo.com
147341
emm...is this your homework??well..it seems that the question is not from you (individually). Hmm..for more understanding, check out this website:\n\nhttp://faculty.clintoncc.suny.edu/faculty/Michael.Gregory/files/Bio%20101/Bio%20101%20Lectures/Biochemistry/biochemi.htm\n\nI hope it solves your dillema, good luck :)
207715
Hello,\n\nThe idea behind Job's method is to prepare a series of reaction mixtures (for example solutions) in which the total number of moles of the reactants is fixed but with different ratio. One then plots the number of moles of product vs the ratio. The maximum of the curve will occur at the correct stoichiometric ratio.\n\nIn your case, one could prepare solutions containing a total number of moles (i.e. moles of Ni + moles EDA) equal to 1 with the following composition:\n\nmoles Ni     moles EDA Ratio Ni/EDA\n0.9                0.1           9        \n0.8                0.2           4\n0.6                0.4          1.5\n0.4                0.6          0.666\n0.2                0.8          0.25\n0.0                1.0          0.0\n\nOne then needs to measure the amount of Ni-EDA complex and plot that amount agains the ratio. The maximum will tell you which is the right stoichiometric ratio. Hope this helps. See the web site below for a more complete explanation.\n\nOne then measures the
675016
This is a calorimetry problem! There is an equation for it:       \nq = mc(change in temperature)\n\nq = the heat required \nm= mass\nc= specific heat \n\nAlso, write out the reaction and products.
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
93462
I hope this helps:\n\nhttp://chemistryformulas.com/
795154
2CO(g) + O2(g) --> 2CO2(g)
668373
Ok abaddon...you sound baddly cute.Your coconut seems hard to CRACk.Pipe down a bit to here would ya?LetS operate in the same state.Is the Last Kingdom Political then?Or what is mathematics doing to solve the issue?I didnt do calculus in school but you are talking differential calculus and quantum stuff(joke).So pipe low...be forthright a little
149472
Hey I'm studying physical and chemical equilibrium too, currently.The answer is that it will definitely have units when the concentrations have different powers.Just try and work it out with the units and it will be very clear.
924669
We need the source of phosphorous for this equation.  Without it, we can't help you.
687306
Check the Michaelis-Menten equation.It's too long to write it all down for you ;)