Chemistry homework help?
105926
Initial equation:\nAl(OH)3 + H2C03 --> Al2(CO3)3 + H2O\n\nOn the right side, there are 3 CO3 ions, so we must put a "3" in front of H2C03 on the left hand side of the equation. Similarly, there are 2 Al ions on the right hand side of the equation, so we must write "2" in front of Al(OH)3 on the left hand side of the equation.\n\n2Al(OH)3 + 3H2C03 --> Al2(CO3)3 + H20\n\nNow, we see that the number of "O"s from OH ions and the number of "H"s from the OH ions (in Al(OH)3) and H ions (in H2CO3) is NOT EQUAL to the H2O on the right.\n\nOK so number of "O"s from OH ions = 2 X 3 = 6\nNumber of "H"s from OH ions and from H2C03 = (2 X 3) + (3 X 2) = 12\n\nOne H2O molecule has 2 Hs and 1 O, so 6H20 molecules will have 12 Hs and 6 Os.\n\nThus your final equation will give\n\n2Al(OH)3 + 3H2CO3 --> Al2(C03)3 + 6H20\n\nI hope I made this clear enough. I suggest that you read the chapter again or ask your teacher about this.
317042
www.chemfinder.com might help
574078
This might help :)\nhttp://www.chemicalforums.com/index.php?board=8;action=display;threadid=3263
27684
Protons, neutrons, and electrons.
835111
4 NH3  +  5O2  --->  4 NO  +  6H2O\n2 NO  +  O2 ---> 2 NO2  +  N2O4\n3 NO2  +  H2O ---> 2 HNO3  +  NO
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
791178
I HAVE A CHEMISTRY EXAM tomorrow. Actually a small test.lol\nwant answers go to the site below!
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
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 :)
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
780875
a)4 Fe + 3O2--->2 Fe2O3\n\nCH4 + 2O2 ---> CO2 +2 H2O
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
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)
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
942137
I have two problems with this: First, this is not an avenue for high school kids to get their homework done for them and second, the thought that someone would take yahoo answers 'points' as payment for doing a highschool kids homwork makes me feel ill. \n\n(i)All that aside, the reducing agent is the compound/element that looses electrons. \n\n(ii) First balance the equation, then find then number of moles of potassium permagenate used in the titration, then convert to the equivalent number of mols of the Be complex. \n\n(iii) multiply answer for (ii) by 50\n\n(iv) use the difference between your asnwer for (iii) and the given sample mass to find mass percent.