I need help Balancing & writting these equations...?
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)
744922
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
290031
Balancing chemical equations is easy, but it's a bit of a trial and error process. Basically, you've got the the arrow in the middle, so the number of each element on one side of the arrow has to equal the number of that particluar element on the other side of the arrow. So start off by balancing any unique elements that only appear once on a particular side. Then balance your nitrogens, oxygens and carbons in that order. The last thing that should be balanced is your hydrogens. I can go into further detail if you need me to, I just don't know how much you know. \n\nFor your other question and yield is how much of something you get from a reaction. For instance, if I mix 100 moles of one substance and 100 moles of another substance, I can calculate a theoretical and experimental yield. For instance, I can say that I expect all 200 moles to react and form a single product. Since i expect it all to react, my THEORETICAL yield is 200 moles or 100%. Say I actually run the experiment and I discover that only 100 moles of my intended product formed, then my EXPERIMENTAL yield is 100 moles or 50%.
594770
Well we are not going to do your homework for you. But we will help you along. why not start out by writing down everything you know. they writing down what you need to find. Then look at your equations and find one that uses what you know to find one of the values you need to find.\n\nSo write it up and put it on here and we will help you along.
814943
There is no balanced equation for that question.\n\nYou need to show your products going to reactants for the world to be able to balance the equation.\n\nGood Luck!
236391
1S = 1T + 1B\n2S +1T = 8B\n\nrewrite 1st equation as 1S-1T=1B and add 2nd equation. get\n3S=9B, so S=3B\n\n3 buttons balance 1 spool
119628
So, usually the best way to do these is to just fiddle with them until you get them right, but for the sake of it, let's go with a systematic approach using algebra.  First we assign a variable in front of each of the terms on the left and right like:\n\nx Pb  +  y H2O  +  z O2  -->  a Pb(OH)2\n\nSo now we'll write down equations to make sure that we have the same numbers of both sides of each *element*.  I'll do this by writing down the variable in front of each compound in your equation (so x, y z and a) and then multiplying those by the number of times an element exists in that compound.  (So when I'm writing the equation for hydrogen, the H20 term will be 2*y, because y is the coefficient in front of H20 and H occurs twice in that compound).\n\nRemember, the number of any element on the reactants (left) side equals the number of that element on the products (right) side.\n\nFor Pb: x = a, because Pb shows up only once on each side with a subscript of 1, those two compounds will have to be present in the same molar amounts.\n\nFor Hydrogen: 2y = 2a\n\nFor Oxygen: y + 2*z = 2 * a\n\nSo now we have three equations:\n\nx = y\ny = a\ny + 2z = 2a\n\nBut four unknowns, which is fine because we'll just solve for them in terms of one of the variables and then simply choose the last variable to make all the numbers nice.  So solving the equations in terms of a:\n\nx = a\ny = a\nz = (2a - y)/2 = (2a - a)/2 = a/2\n\nSo now we have x, y and z in terms of a free parameter, a.  We'll just choose a = 2, for example, to make x, y and z all turn out as integers.  So now we've chosen:\n\nx = 2\ny = 2\nz = 1\na = 2\n\nSo the balanced equation is:\n\n2 Pb + 2 H2O + O2 --> 2 Pb(OH)2\n\nMake sense?
533429
well..write them down ...try to derive them (if their derivation is simple enough or else skip it )......\n\ntry to connect each formula with other one with a hint ....a small hint will help u to recollect other one ....\n\ntry to do at least one problem based on each formula
713860
1.Xc=1/2pifc\nc=1/2pifXc\n\n2.ib=ibq+rt(2is)\nib(1-q)=rt(2is)\nsquaring\n[ib(1-q)]^2=2is\ns=[ib(1-q)]^2/2i\n\n3.Q=Xl/R\nXl=Q/R\n\n4.Lt=L1+L2+2Lm\n2Lm=Lt-L1-L2\nLm=[Lt-L1-L2]/2\n\n5.Es=E1+E2+E3\nE2=[Es-E1-E3]\n\n6.R1=i2R^2/i2\ni2R^2=i2R1/i2\nR=rt[i2R1/i2]\n\n7.R1/Rx=R2/R3\nRx=R1R3/R2\n\n8.y=mx+b\nb=y-mx\n\n9.i2=i1R1/i1
210386
I think the point may be to use synthetic division.  Look that up in your text.
844394
If you have Microsoft Office Discs, then load the Equation Editor. It will allow you to write all of these.
976256
BALANCED . . . \n\n2Sr + O2 ---> 2SrO\n\n. . . and solved . . .\n\nGet an "A".
887756
Basic balancing of chemical reactions. You can always add CO2 O2 and H2O to either side. Start with the Carbon. balance the Oxygen last since it is contained in CO2 and H2O.\n\n1) 2 C6H6 + 15 O2 ==> 12 CO2 + 6 H2O\n2) C12H22011 + 12 O2 ==> 12 CO2 + 11 H20\n3) C3H6O + 4 O2 ==> 3 CO2 + 3 H2O\n4) C4H10O + 6 O2 ==> 4 CO2 + 5 H2O\n5) C4H10O + 6 O2 ==> 4CO2 + 5 H2O
771697
Try Google or Yahoo.  You'll find worked examples there.  I think it needs diagrams to understand; they and the accompanying explanations are too complex to attempt to reproduce here.
577084
Acid + base = salt + water.\n\nUnbalanced eqn is: Al(OH)3 + H2CO3 --> Al2(CO3)3 + H2O
851400
the first equ is balanced, b/c on the left side u have Pb-1, C-1 O-3 on the right side u have Pb-1 C-1 O-3 so since both side are equal it is balanced\n\nans for question 2 BeCl2 + 3O2------Be(ClO3)2\nBe-1                                  Be-1\nCl-2                                    Cl-2\nO-was 2 now 6                                     O-6
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