Chemistry question - please help!!?
964894
IF U ARRANGE THE IONS IN ORDER OF increasing oxidation potentials then it's first Mg+2 , Zn+2, Cu+2\nso in a soln of these 3 ions Cu ions have highest reduction potential i.e.\nhigh tendency to accept e~ (to get reduced)\nso Cu will get first discharged & not Mg\nif Cu ions get exhausted then Zn will get deposited then Mg
944583
a) [O 2-]\nb) [ I - ]\nc) [Na+]\nd) [Al 3+]\ne) [Ni 2+]\nf) [Mg 2+]
775000
2 MnO4-(aq ) + 5 H2C2O4(aq ) + 6 H3O+(aq ) --> 2Mn2+(aq ) + 10 CO2(aq ) + 14 H2O
539827
(a) 2 NaClO3 -> 2 NaCl + 3 O2\n\n(b) 3 MnCl2 + 2 Al -> 3 Mn + 2 AlCl3\n\n(c) C4H10 + 13/2 O2 -> 5 H20 + 4 CO2 \n                  or\n(c) 2 C4H10 + 13 O2 -> 10 H20 + 8 CO2\n\nBalancing is quite simple for these simple equations. You just have to see the left hand side pick up an element, see how many atoms of that element are there in it and multiply the R.H.S compound containing that element with a number which will be equal to no. of atoms in the l.h.s. Then again repeat the procedure by picking another element.....see it's quite difficult to explain you with writing aneways i hope it helps u :)
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.
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
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.
711408
I am not a chemistry student but I make this research for you and maybe it helps you http://www.chem.latech.edu/~deddy/chem104/104Antacid.htm\nhttp://www.chemcool.com/regents/acidsbasessalts/aim3.htm\n\njust copy those links and you should be on your way to solve ur problem.
215160
http://www-ec.njit.edu/~skawinsk/chem243/chap11/chap11.htm\nthis may help–near the bottom
612107
a. 2k+ = .20moles , SO42-=.10moles\n\nb.Fe3+=.10moles,3NO3-= .30moles\n\nc.2Al3+=.20moles,3SO42-=.30moles\n\nd. Ni2+=.10moles,SO42-=.10moles
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
782516
a) 3SiO2 +2 Al --->3 Si + 2Al2O3\n\nb)AgNO3 +HCl---> AgCl +HNO3\n\nc) Mg + CuSO4 ---> MgSO4 + Cu\n\nd) Pb(NO3)4 +4 NaI --->4 NaNO3 + PbI4 (s)
1171989
http://en.wikipedia.org/wiki/Chemistry\n \nthis nust be helpful for you
835111
4 NH3  +  5O2  --->  4 NO  +  6H2O\n2 NO  +  O2 ---> 2 NO2  +  N2O4\n3 NO2  +  H2O ---> 2 HNO3  +  NO
866145
HCl + AgNO3-----> AgCl +HNO3\nPb(OOC-CH3)2  + 2LiCl -------> PbCl2  + 2Li(OOC-CH3)\nNa3PO4 + CrCl3 -------> Cr(PO4) + 3NaCl\nH2SO4 + BaCl2 ----------> BaSO4 + 2HCl\nAl2(SO4)3  + 6NH4OH ------> 2Al(OH)3  + 3(NH4)2SO4\n2AgNO3 + H2S -------> Ag2S  + 2HNO3\nCaCl2   + Pb(NO3)2 -------> Ca(NO3)2  + PbCl2\nCa(NO3)2 + Na2CO3 ------>CaCO3  + 2 NaNO3
976540
a) +1\nb) 0   (in molecule state)\nc) 0\nd) -3\ne) 0\nf) +4 of Carbon\ng) +2 of Mg\nh) +1 of Chlorine (interesting!)\ni) +6 of Sulphur
412530
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)