Does anybody know a lot about Calculating energy changes in chemistry?
475093
see dear ,\nwhile calculating the energy changes we need to know the hidden heat { latent heat } that is absorbed during the change in state . for eg when we are calculating the heat absorbed or given out during a change in state , like ice to water , just first calculate the latent heat ie \nq=mL{ heat=mass *latent heat } now add this value to \nq=mct{heat=mass*specific heat*change in temp} keep in mind that specific heat is a constant value \nwater's specific heat is 4180j/kg/degree Celsius.
861726
You will have to first find the standard enthalpy values of the reactants and products. They're given in the appendices of any standard GenChem or PhysChem books.\n\nThen, you subtract the sum of enthalpies of reactants from the sum of enthalpies of the product to get the enthalpy of the reaction.
770269
For aA+bB -> cC+dD\n\nE=E0-(RT/nF)* ln([C]^c*[D]^d / ([A]^a*[B]^b))\n\nwhere n=number of electrons taking part in the reaction\n\nSolids are not considered in the equation, so in your case\n\nE=E0 -(RT/nF)*ln([Fe2+]/[Cd+2])\n\nYou can find the standard reduction potentials for the Fe+2 and Cd+2 at \nhttp://www.northland.cc.mn.us/Chemistry/standard_reduction_potentials.htm\nThen E0= E0(Cd2+)-E0(Fe2+)\nR=8.314\nn=2\nF=96500\n\nYou have to specify T ...
860351
The right chemical reaction is as follow:\n      6CO2+12H2O+energy=C6H12O6+6O2+6H2O\nPlz do not get confused with this reaction\n     6CO2 + 6H2O + light (energy) --> C6H12O6 + 6O2\nBecause 1st one is latest one and you can't get in some conventional books.Later one is wrong.
132159
Open up your thermodynamics book and do your own homework.\nHint.. add the energy required to raise the temp from 100C to 112C plus the energy used to change state, plus the energy used to go from 76C to 100C.  You should have all the info you need in your problem, and the energy required to do the change is in your book.
158900
Yup. They are not simple...\nChemistry...\nhttp://www.chm.davidson.edu/ChemistryApplets/equilibria/EquilibriumConstant.html\nhttp://en.wikipedia.org/wiki/Chemical_equilibrium\nThey use the equilibrium constant expression (Kc) here...\nhttp://www.cofc.edu/genchemlab/equilibrium.htm
840804
Others ways to change an element (nuclear chemistry)\n\nProton emission \nK^40 ------> Ar^40 + e\n\nElectron Capture.\nHg^197 + e-  ------->  Au^197\n\nAlpha Radiation 4/2He+2\nU^238 ----->  (4/2He2+) +  Th^234\n\n\nBeta Radiation which causes a neutron in the nucleus to spontanously decay into a proton pluse an electron which is then ejected\n\nI^131 ------> Xe^131 +  e-
320537
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.
888187
something about chemistry
605311
chemistry, quantum mechanics...
753535
Just use the formula\n\nE = m*Cp*(Delta T)\n\nE=energy\nm=mass\nCp= Specific Heat\nDelta T = change in temperature\n\nCp= 0.51998 J/(C*g)\n\nGood Luck with the test!
185616
The guy who answered before was right but i wanna explain it. An exothermice reaction is one in which heat energy is lost to the surroundings, thus at the end of the reaction, some heat would have been lost. For heat to have been lost, the product has to have lesser heat than the reactant.Take not that heat here is the same as Enthalphy.\nSince the Enthalphy heat value = heat of product-heat of reation = a negative number because heat of product is lesser.\nthus the answer is i nd ii\n\n2. To calculate the heat absorbed or evolved in a system, the formulais Q(quantity of heat) = m*c*T, where m is the mass, c is the specific heat capacity, and T is the change in temperature, thuis this three are neede. Density is not important in this calculation\nso the answwers are i and iii
753892
Calorimeter Info center\n\nhttp://www.scipeeps.com/Sci-Chemistry_Topics_C_-_Cl/Calorimeter.html
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
824613
Rate of chemical reactions depend on - temperature, pressure, concentration and the use of catalyst. The rxn constant (generally used in equilibriums) is often calculated to able to see the rate of the reaction. \n\nX + Y <----> 2V + Z\n\neq constant =  ([V^2]x[Z])/[X]x[Z]) \n\nThe higher the constant, the higher the rate. Equilibrium constant only changes if the there's a temperature change. Concentration changes doesn't effect the constant however it effects the rate. Although pressure changes lead to temperature changes, it's assumed that pressure changes don't effect the constant. If the constant is greater than 10^6, rxn will take place at reasonable rate.
891159
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
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!