O3(g) -> O2(g) + O(g)\nO3(g) + O(g) -> 2O2(g)\n\nif (triangle H) is negative for the reaction in part b, which is more stable form of oxygen, O3 or O2?
130961
O2 is always more stable
428081
It is simply sum of products minus sum of reactants\n\n[H2O + N2] - [O2 + NH3]\n\nSince this is a 1 to 1 ratio reaction, the coefficients are one. (if there was a coefficient for NH3, you must multiply the enthalpy value of NH3 by that # because that is how many moles you have, and the value given is for only one mole.)
532296
The reaction\n\nF2 + O2 ===> 2F2O\n\ndH = 2 dH F2O - (dH F2 + dH O2) wher d is the delta symbol and using the heats of formation\n\nSince F2 and O2 are in the elemental state, the heat of formation is 0 (by definition)\n\ndH = 2 x 24.5 = 49 kJ/mol\n\nDelta H for the above is 49 kJ/mol\n\nI got the heat of formation for F2O (difluorine monoxide) from Wikipedia.
167487
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
448805
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
366895
Br+H20--->HBr+O\n\nNi(II)O+H20--->Ni+H202\n\nKO2+H2O--->KOH + H2\n\nHope that helps.
433923
First write the two half reactions:\n\n3 e(-) + Bi(OH)3 ---> Bi\nSnO2(=) ---> snO3(=) + 2 e(-)\n\nMultiply the first equation by 2 and the second by 3 and sum to get:\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi\n\n12 O in the left, 9 on the right, so add 3H2O\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi + 3H2O\n\nCheck that H is OK.  Yes. Done\n\nSo the sum is 13. (d) is the answer.
428206
C6 H12 O6 (arrow) 3CO2 + 6H2O
547086
The total amounts of oxygen atoms equal to six.\n\nHowever, 3 O2 means there are three molecules of diatomic oxygen atoms, and 2 O3 means that there are two molecules of triatomic oxygen atoms.\n\n3 O2 ( "-" shows bonds)\n\nO-O  O-O  O-O = 6 total\n\nO-O-O  O-O-O = 6 total
505326
its actually pretty simple. U multiply by -1 when u flip over an equation...so the reactants become the products and the products the reactants. that way instead of having a negative number u will have a positive number for the kJ. as in this problem. U have to flip #1 so u can cancel a few things with #2. so it becomes..\n1)3 H2O(l) + 2 CO2(g) AH---->CH3CH2OH(l) + 3 O2(g) (multiply by -1 gives u a postive +1367kJ. part two stays the same.\n2) C2H4(g) + 3 O2 ---> 2 H2O(l) + 2 CO2 (g) AH= -1411 kJ\n\nNow the 3 O2 from #1 and #2 cancel since they are on opposite sides. Now the 2 H20 cancel from #2 and you are left with only one in #1 since they are also on opposite sides...which after canceling..leaves u with your original equation.\n\nC2H4(g) + H2O(l) ------> CH3CH2OH(l)\n\nnow since u flipped over #1.....and u have a positive +1367kJ and\n#2 is left the same...its still at negative -1411kJ. u add them together and u get an enthalpy change of -44kJ. sorry its pretty confusing to explain..if anything message me and i willl try to help.
547667
BALANCED . . . \n\n2Sr + O2 ---> 2SrO\n\n. . . and solved . . .\n\nGet an "A".
399229
You need to write first the half reactions:\n\nAl ---> Al(+3)\nO2 ---> 2O(=)\n\nAdd the electrons (this is the answer):\n\nAl ---> Al(+3) + 3e(-)\n4e(-) + O2 ---> 2O(=) \n\n{let's check}\nMultiply the first equation by 4 and the second by 3 and sum to eliminate the electrons:\n\n4Al + 3O2 ---> 4Al(+3) + 6O(=)\n\nRecombining the Al(+3) and the O(=) you get:\n\n4Al + 3O2 ---> 2Al2O3.
363608
1) this one is already balanced for you!!\n\n2)Count the number of Mg and O on the right side, and count the number of Mg and O on the right side.  On the left, you have 2 O and 1 Mg, and on the right you have 1 Mg and 1 O.  To make the oxygen on both sides equal, you need to double the MgO.  Now on the left, you have 2 O and 1 Mg, and on the right, you have 2Mg and 2 O.  Now you need to multiply the Mg by two so the Mg on both sides is equal.  and you're done!!!! \n2Mg + 02 --> 2MgO\n\n3) On the left, you have 1 Zn, 1H, and 1 Cl.  on the right you have 1 Zn, 2Cl and 2H.  the Zn are balanced and you just need to multply the HCl on the left to balance the whole equ.\nZn + 2HCl --->ZnCl2 + H2\n\n4) On the left, you have 1 P and 2O, and on the left you have 4 P and 10 O. If you multiply the P on the right by 4, and the O on the right by 10, you have 4P and 10 O on both sides\n4P + 5O2 ---->P4O10\n\n5)  On the left, you have 1 Fe 2 Cl, and pn the left, you have 1 Fe and 3 Cl.  The easiest way to balance this is multiply the Cl on the right by the amount of the left one and vice versa. when you do that, you have 1 Fe and 6Cl on the right, and you have 2Fe and 6Cl on the left. To finish the equation, multiply the Fe by two.\n2Fe + 3Cl2---->2FeCl3\n\nVoila!  You have balanced 5 equations!\n\ni hope this was helpful!
227207
You need to take more care in writing equations when you post your questions. \n\nPart a) is difficult to decipher - you will need to re post it.\n\nPart b) is presumably:\n\nCH3COOH(l) + 2O2(g) --> 2CO2(g) + 2H2O(l)\n\nPart c) is presumably:\n\nFe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)\n\nI would next use heats of formation to calculate the enthalpy change in the above reactions using the formula:\n\ndelta Ho = sum(heats of formation of products) - sum (heats of formation of reactants).\n\nIf you have a data book or data sheet, these values can be found there (Note: heats of formation of elements are zero.) \n\nIf you want web sites that might help you, try googling "enthalpy change problems" +tutorial.\n\nI hope this helps you.
481545
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
449384
Li2O + H2O --> 2 LiOH\n\nIf you include the H2 it will never ballance.  You will not end up with H2 as a product anyway since OH will react with Li and the remaining H will react with the LiO hence 2 of LiOH.  Now on both sides of the equation you have 2 - Li, 2-O and 2-H.  If you did want to include a water molecule on the produst side then the equation would read as follcws:\n\nLi2O + 2H2O --> 2LiOH + H2O\n\nthis equation ballances with 2-Li, 3-O and 4-H on each side\n\nThe first equation shaow the ideal combination of the two compuonds (what most teachers are looking for) and the seccond equation shows a instance where there is not enough Li2OH to completely react with the all of the water
531671
C3H8 + 5O2 --> 3CO2 + 4H2O