question on homework chemistry?
112894
Haven't had chem in a while but maybe it would help to write out all the individual symbols.  Your second example would look like\n\nHHHPOOOO + MgOOHH  = MgMgMgPPOOOOOOOO + HHO\n\nthen you can see that the left side needs three Mg so you'd need 3 of the Mg(OH)2 to balance.  The left side also needs two P so there needs to be two H3PO4 there.  This makes a total of 12 H on the left, so to balance that you need 6 H2O on the right.  This also gives 14 O on the right to balance the 14 O you'd now have on the left.  So\n2 H3PO4 + 3 Mg(OH)2 = Mg3(PO4)2 + 6 H2O
664298
molecular: \nHCl + K2CrO4 + SnCl2 --> CrCl3 + SnCl4 + KCl + H2O\n\nnet ionic: H+ + CrO42- + Sn2+ --> Cr3+ + H2O + Sn4+\n\nreduction half reaction: \n3 e- + 3 H+ + 5 HCl + K2CrO4 --> CrCl3 + 2 KCl + 4 H2O\noxidation half reaction:\n2 HCl + SnCl2 --> SnCl4 + 2 H+ + 2 e-\n\nBalancing them:\n6 e- + 6 H+ + 10 HCl + 2 K2CrO4 --> 2 CrCl3 + 4 KCl + 8 H2O\n6 HCl + 3 SnCl2 --> 3 SnCl4 + 6 H+ + 6 e-\n-----------------------------------------------------------\n6 H+ + 16 HCl + 2 K2CrO4 + 3 SnCl2 --> 2 CrCl3 + 4 KCl + 8 H2O + 3 SnCl4
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
490327
solution A has a hydrogen ion concentration of 10^-5\nsolution B has a hydrogen ion concentration of 10^-3\nAll you have to do is 10^-ph\n\nIs this homework?
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
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)
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 :)
58231
http://www.studentquestions.com/homework_ans/4244.html
315267
hey kid do your own homework
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
827688
a) 2C2H6 + 7O2  --> 4CO2 + 6H2O\n\nb) C6H12 + 9O2  --> 6CO2 + 6H2O\n\nc) C5H12 + 8O2  --> 5CO2 + 6H2O\n\nYou must apply Markovnikov rule for the next addition reactions:\n\na) CH2=CHCH2CH3 + HCl  --> CH3CH(Cl)CH2CH3\n\nb) CH2=C(CH3)CH3 + HBr  --> CH3-CBr(CH3)CH3
770691
CoCl2(aq) + FeSO4(aq) = CoSO4 (aq) + FeCl2(aq)
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
799504
FeS + 2HBr --> FeBr2 + H2S\nFe2+ + S2- + 2H+ + 2Br- --> Fe2+ + 2Br- + H2S\n2H+ + S2- --> H2S\n\nK2CO3 + CuCl2 --> 2KCl + CuCO3\n2K+ + CO32- + Cu2+ + 2Cl- --> 2K+ + 2Cl- + CuCO3\nCu2+ + CO32- --> CuCO3\n\npeace\nvixklen
740547
you've posted this question about six times! Are you not intending to do your own homework at any point of the day or do you insist on never learning how to do it by getting the rest of the world to do it?!\n\nDo your own homework and learn something!
701443
1 - the reaction is reversed and multiplied by 4, so the enthalpy change is 11508 kJ\n\n2 - The reaction is a combination of the other three (I'll label them A, B and C, in the order you have them listed, for simplicity)\nThe original reaction = 2A - 2B -2C (the first one is just doubled, the next two are doubles and reversed), so 2*321.8 - 2*571.7 - 2*288.6 = -2542.2 kJ\n\n3 - This one is done similarly to question 2.  A - 0.5B + 1.5C\n180.8 - 0.5*(-906.2) + 1.5*(-483.6) = -91.5 kJ\n\nHope this helps
777814
Q1)\nCa3 (PO4)2  (MW)molecular weight = 3*40 + 2 * (95) = 310 the oxygen weight is 16*4 = 64 .so % of O is (64/310) *100 = 20.6%\n\nQ2)\nSO4 (NH4)2 ---> MW = 32+4*16+ 2*18 =132 Mass of sulfur 32 \nhence  % S = 100*32/132 =24.2\n\nQ3) \n\nFe2 (SO4)3 ----> MW = 56*2+96*3=400 \n\n% of iron 100*112/400 = 28%