What is the enthalpy change, in kJ, for the reaction below?\n\nH-O-O-H(g)->H-O-H(g)+ 1 haft of OXYGEN=O(g)
135472
I think for that you need the bond enthalpy for the O-H bond because that bond is being formed in H-O-H; anyways, first you see the bonds which are being broken in the reactant (in this case it is 0-0 and O-H) and add up the bond enthalpies(it should be positive). Next, see the bonds formed in the products (H-O and O=0) and add their bond enthalpies(it should be negative as energy is released when bonds are formed). Then to calculate the enthalpy change, simply add up the total enthalpies.
389840
In all reactions one uses that 2 OH -> H2O + O, the split of oxygen goes to create the ion. Finally balance the equation:\n \nSO3 + 2 NaOH -> Na2SO4 + H2O\n\nCO2 + 2 KOH -> K2CO3 + H2O \n\nSO2 + Ba(OH)2 -> BaSO3 + H2O
532272
For the reaction:\n\nMgO + 2 HCl  -->  MgCl2 + H2O\n\nThe temperature should go up, not down.  Only the initial and final temperature are needed.\n\nTypical values for students are between 110 an 150 kJ/mol MgO.  This is typical, not the best that can be done.
505986
2C6H14 + 19O2 ----> 12CO2 + 14H2O\n---or--\nC6H14 + (19/2)O2 -----> 6O2 +7H2O
553353
You need to find the heats of formation of each species (will come from a standard table) and use this equation to get the heat of reaction:\n\n(delta)H = sum of heat of formation of products - sum of heat of formation of reactants\n\nEach heat of formation will be multiplied by its stoichiometric ratio, and the physical state will be important (especially that of water, as the heat of formation of steam is higher than that of liquid water).  You will not find a heat of formation for oxygen gas, because by definition there is no heat of formation for an element.
478180
H2+O2-->H2O\nNaOH+HCl-->NaCl+H2O\nKOH-->(on heating,deomposition)K2O+H2O\ni guess this is what u want\nand i f u didn't get it i am sorry......
448805
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
516360
2\nthe reaction is\nCa(OH)2+2HNO3=2H2O+Ca(NO3)2
327828
This is a calorimetry problem! There is an equation for it:       \nq = mc(change in temperature)\n\nq = the heat required \nm= mass\nc= specific heat \n\nAlso, write out the reaction and products.
484139
use the equation\nE(enthalpy change) = m(mass) c(specific heat) DeltaT(change in temperature)\n\ntemperature change is initial temperature minus final\nsince temperature increased then the temp change is negative\n\n-2.00C
157524
detlaH=H(products) - H(Reactants)\n\n200KJ/mol
139991
decomposition reaction\n\n2H2O2 -------- O2   +  2H20
420295
alright...\nconsider the R.H.S.\nh2o has two o-h bonds which account for 2x467 and since thre are 2 mol. u get 2x2x467--------(1)\nco2 has two c=o bonds which account for 2x799 and since u hav 4 mol. u get\n4x2x799-------------(2)\naddin 1 and 2 u get 1868+6392=8260\nconsider the L.H.S.\nu hav in c2h2, two c-h bonds which account for 2x413--------(3)\nand one triple bond accounting for 1x839---------(4)\nnow u hav 2x(3+4)=2x(826+839)=3330------(5)\nand u hav 5x(o=o) which acc for 5x(495)=2475----(6)\naddin 5 and 6 u get 5805\ntherefore ur ans is 5805-8260=-2455\nand since u want for one mol of acetylene u get -1227.5 kj/mole
186303
heat of combustion = -(enthalpy change for reaction )\n          \n = -(heats of formation of products - reactants)\n \n1 mole of ethanol, when burnt completely, produces 2 moles of CO2 and 3H2O i.e delta H = 2(heat of formation of CO2) + 3(heat of formation of H2O ) \n\n= 2(-394 kJ/mol ) + 3(-286 kJ/mol)\n\n= -1646 kJ/mol\n\nheat of formation of ethanol = -277 kJ/mol\n\nso\n\nheat of combustion = - (-1646 - (-277) )\n                   \n                   = - (-1646 + 277 )\n\n                   = 1369 kJ mol
495739
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)
381050
Use the equation:\nDelataH of the reaction = DeltaH(products) - DeltaH(reactants)\n= -242 - (-191)\n= -51kJ/mol
38967
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)