the products in a reaction have a total heat content of 458 kJ/mol and the reactants have a total heat content
363124
detlaH=H(products) - H(Reactants)\n\n200KJ/mol
482555
AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit
881597
Cr2(SO4)3 + 6KOH  --> 2Cr(OH)3 + 3K2SO4
775809
I believe the units of your supposed answer are wrong.  They should be kJ/mol and not kJ.\n\nHeat = m * Cp * DeltaT = 60 * 4.186 * (13.2 - 27.2) = - 3516 J\n\nThis heat is absorbed by 28.7 g of KI.  Per mole (MW(KI) = 166 g/mol), the heat will then be:\n\n28.7 g ===> 3516 J\n166 g ===> x\n\nx = 3516 * 166 / 28.7 = 20,340 J/mol = 20.3 kJ/mol
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
777003
CH3COOH + NaOH ---> NaCH3COO + H2O
813561
The molecular mass of NaOH is 40.00 gm/mol,  so 8 gm of NaOH contains 8 gm/40.00 gm.mol = 0.20 moles of NaOH.\n\nThe molecular mass of HCl is 36.46 gm/mol, so 8 gm of HCl contains 8gm/36.46 gm/mol = 0.22 moles of HCl.\n\nIf we react these amounts of NaOH and HCl, the NaOH will be the limiting reactant, and we will produce 0.20 moles each of NaCl and H2O.  This reaction will liberate \n\n0.20 mol * 56 kJ/mol = 11.2 kJ of energy as heat\n\nThe heat capacity of the final solution is given by the mass of the final solution (which, by conservation of mass, is equal to the mass of the initial solutions) multiplied by the specific heat capacity of the final solution.  With the assumptions given in the question, the heat capacity of the final solution is:\n\n(58 gm + 258 gm) * 4.184 J/(gm*degC) = 1322 J/degC.\n\n(Note that we don't need to use the density of the final solution.)\n\nThe temperature change of the solution is given by the energy released divided by the heat capacity of the system:\n\n(11.2*10^3 J)/(1322 J/degC) = 8.47 deg C\n\nIf the initial solutions had a temperature of 25 C, the final solution has a temperature of 25 + 8.47 C = 33.47 C.
984673
molar mass Fe=55.85\nmoles iron=490/55.85\nmoles iron=8.774\n4 moles are required for the reaction\n8.774/4 provides the number of times the reaction can be completed\nthis equals 2.19\nmultiply this by -1.65103\nand you get -3.62 kJ.  This being negative tells you the reaction is exothermic and that the positive 3.62kJ is the amount given off
309732
i.e. it is easier to go backwards than forwards, so the product has more energy ( i.e. higher up the energy scale / nearer the top/ easier to get to the transition state ).\n\nHeat of reaction = heats of formation of products - reactants = 85 - 80 = +5kJ mol.... endothermic ( makes sense if products contain more energy than reactants )
372026
CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)
1022281
d)  both a and c\n\nbecasue in a chemical reaction   everythin on the product and teh reactant must be the same\n\nthe mass and atoms must be same
97497
There are temperature changes is most reactions becuase there is heat generated, known as the heat of reaction.
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
390022
the reaction is exothermic because the resultant solution or products is warmer than the reactants, assuming that it didn't warm up in the enviroment. In addition all neutralisation reactions are exothermic.\n\nThe solutions were mixed rapidly so that the concentration gradiant increases as quickly as possible preventing the reactants absorbing the heat from the reaction before the reaction is finished.  ie the initial rate of reaction being as high as possible.
863456
I assume the reaction in question is:\n\nA(g) <-> 2B(g)\n\nThe equilibrium constant, Keq, for this reaction is given by:\n\nKeq = ([B]^2)/[A]\n\nwhere the quantities in square brackets mean "the concentration of" the relevant species at equilibrium (or, in the case of gases, one commonly uses equilibrium partial pressures instead of concentrations).\n\nPlugging in the values we are given for the equilibrium concentrations, we have that:\n\nKeq = (0.154^2)/0.381\n\nKeq = 0.0622
809514
Here you are:\n\nFirst write the combustion equation and balance it:\n\nH3NNH2(l) + O2(g) ---> N2(g) + 2H2O(g).\nThe heat will be the diffrence of enthalpies of formation between the products and the reactants. Do not forget to multiply by the stoichiometric coefficient.\n\nRemember that the enthalpies of formation of O2 and N2 are 0.\n\nYou will also need to find the enthalpy of formation of H2O(g) in the table of your book.  It is -241.8 kJ/mol\n\nHf(H2NNH2) = [2*(-241.8) + 1*0] - [1*50.6 + 1*0]\n\nHf(H2NNH2) = -483.6 - 50.6 = - 534.2 kJ/mol