A. 2CO2 + C → CO \n    B. CO2 + C → 2CO \n    C. CO2 + C2 → CO \n    D. CO2 + C → C2O
125914
The answer is B.\n\nIn your beginning reaction CO2 + C --> CO, you have 2 carbons and 2 oxygens on the reactants side, and you have 1 carbon and 1 oxygen on the products side.\n\nIn order to make the equation balanced, you have to make sure there are the same number of each element on both sides of the equation.\n\nSo to make this chemical equation balanced all you have to do is put 2 as the coefficient of the CO on the products side, making the final equation:  CO2 + C --> 2CO\nThis will give you 2 carbons and 2 oxygens on both sides.
515830
Part A:  equilibruim to the right because more co3 added\n\nPart B:  Equilibrum to left becausemore co2 added\n\nPart C:  Equilibrium to right to replace removed h20\n\nEquilibruim dynamic because although appears stable ions constantly moving back and forth just at the same time
337606
1. c\n2. c\n3. c\n4. c\n5. c
167487
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
397846
CH4 + 2 O2 ---> CO2 + 2 H2O
484918
44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
179694
carbon, organic compounds
67349
Actually, that is a somewhat simplified reaction,\nin reality, the water and Carbon Dioxide dont form directly, they form after a secondary reaction (Carbonic acid decomposing).\n\n3Na2CO3 + 2H3PO4 --> 2Na3PO4 + 3H2CO3\nH2CO3 --> H2O + CO2\n\nSo combining them,\n3Na2CO3 + 2H3PO4 --> 2Na3PO4 + 3H2O + 3CO2
524127
a) [O 2-]\nb) [ I - ]\nc) [Na+]\nd) [Al 3+]\ne) [Ni 2+]\nf) [Mg 2+]
213351
reduced: 2H2O + 2e- ---> H2 + 2OH-\n\noxidized: 2I- ---> I2 +2e-
344088
the conversion of water (H2O) and Carbon dioxide (CO2) into Glucose((C6H12O6) and oxygen(O2)\n\n             H2O+CO2 -----> C6H12O6 +O2
448805
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
329410
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
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.
374337
Cl2 + 2 KI → I2 + 2 KCl
414878
1 -> b\n2 -> e\n3 -> c\n4 -> d\n5 -> a\n6 -> f
18255
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2