calculate the enthalpy change (in kJ) for the reaction?
970544
Here is what you do: \nDo the same operations as I did on the equations with the H values and you will be OK (Do not forget to use the same multipliers on the H as I did on the reactions).\n\n10PCl3 + 5O2 ---> 10Cl3PO\n10PCl2 + 10Cl2 --- 10PCl5\nSubstract the equations to get (keep coeff positives by moving to the other side):\n\n5O2 + 10PCl5 ---> 10Cl3PO + 10Cl2\nP4 + 5O2 ---> P4O10\nSubstract and get\n10PCl5 + P4O10 ---> P4 + 10Cl3PO + 10Cl2\nP4 + 6Cl2 ---> 4PCl3\n\nSum the two to get (P4 goes):\n\n10PCl5 + P4O10 ---> 10Cl3PO + 4Cl2 + 4PCl3\n\n4 PCl3 + 4Cl2 ---> 4PCl5\n\nSum the two equations\n\n10PCl5 + 6P4O10 ---> 10 Cl3PO.
875216
ok convert 12.6 kcal/mol to J/mol. \nthen get the pressure times delta V from the ideal gas law PV=nRT so you then have delta E = delta H- nRT\nn = # of moles (1)\nR=.0806... J/mol*K\nT= 273.15+25=298.15K\nso delta H and PdeltaV are known so plug in solve for Delta E.\ngood luck!
675016
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.
676662
What did you use to measure the activation energy?\nYou probably used a thermometer which was probably marked to 0.1 degrees. That means that your measurements are probably accurate to 0.05 degrees. Reactant temp may be measured as 25.1 (+/- 0.05) so this means that the temp could be 25.05 or 25.15.\nDo the same for the transition state.\nDraw small vertical bars through each point, e.g for reactants the bar would go from 25.05 to 25.15, with the cross in the middle.\n\nI can't remember the formula, but errors are a lot more pronounced on small measurements. It may be something like the difference between the highest and lowest possible values divided by the recorded value, all multiplied by 100.
434523
* The heat content of a chemical system is called the enthalpy (symbol: H)\n    * The enthalpy change (H) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure.\n    * H = H(products) - H(reactants)\n    * H is specified per mole of substance as in the balanced chemical equation for the reaction\n      The units are usually given as kJ mol-1 (kJ/mol) or sometimes as kcal mol-1 (kcal/mol)\n      1 calorie (1 cal) = 4.184 joules (4.184 J)\n    * Energy changes are measured under standard laboratory conditions\n      25oC (298K) & 101.3kPa (1 atmosphere) \n\nEnthalpy changes are measured in KJ per mole, but experiments on measuring enthalpy changes always measure a temperature change. The temperature change is then converted to enthalpy change using the following formula:\n\nDH    =    m x c x DT\n\nm = mass in grams\n\nc = specific heat capacity in Joules/Kelvin/gram\n\nDT = change in temperature in degree Kelvin\n\nThe experiment  is done by simple calorimeter experiments which involve carrying out the reaction in a heat resistant vessel and measuring the change in temperature. \n\nThis is also a common practical assessment, and you should be aware of the sources of error in your measurements such as heat loss to surroundings, incorrect readings, inaccurate thermometers etc.
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
730160
The change in entropy is + 44.55 J/molK.\nThe enthalpy change is +242 kJ/mol, at standard conditions (SATP).  \nYou can use standard ambient temperature (298 K) for T.\n\n    Gibbs = +242.2 kJ/mol - (13.27 kJ/mol) \n               = +228.9 kJ/mol\n\nIf the temperature increased by more than a factor of about 20,  the second value would increase, and the Gibb's energy would be a negative quantitity.
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)
294164
Thw formular to use here is Q = mcT where Q = quantity of heat, m = mass, c = specific heat capacity, and T is the change in the temperature, thus here Q = 2.0g*(30-25)*0.90 = 9J. Notice that I sed Joules as the unit.\n\n2.For this one, a law which i cant remember the name right now is nedeed. Here we have C(s)+O2(g)=CO2 Enthalphy -390KJ\nand also Mn(s)+O2(g)=MnO2(s) Enthalphy = -520KJ\nFor this law to be appplied, we have to write the second equation the other way to become MnO2= Mn+O2 Enthalphy = 520KJ. Notice that i changed the sign when i reversed the equation. Then O2 could be canceled out on the left side of the first equation and on the right side of the second equation. This was we are adding. Finally adding, we get MnO2(s)+C(s) = Mn + CO2 Enthalphy = -390KJ+520KJ = 130KJ.\nand thats the correct answer
748741
2K + I2 -> 2KI\n\nBa + S -> BaS\n\nMgBr2 + MnI6 -> No Reaction\n\nC7H16 + H2O -> No Reaction\n\n2KI + Br2 -> 2KBr + I2
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!
734516
Easy.  \n\nYou have q = mass * specific heat * change in temp.\n\nFind the specific heat of water and your set.  Q is joules so you need to convert.
1063841
W = PV = RTn, where P = pressure, V = volume, R = gas constant = 8.31 J/mol.K, T = temperature, n = moles of gas involved.\n\n1) Change in P = 4 x 10^5 Pa, then:\n\nChange in W = 4 x 10^5 Pa x 0.250 m^3 = 10^5 Pa.m^3 = 1 x 10^5 Joules.\n\n2) Change in V = 17.00 x 10^(-2), then:\n\nChange in W = 6 x 10^5 Pa x 17.00 x 10^(-2) m^3 = 102.0 x 10^3 Pa.m^3 = 1 x 10^5 Joules (note: there is only one significant number in result)
937733
Enthalpy change of formation of 1 mole of C2H2 is the energy change when 1 mole of it is formed from its elements (2C(s) and H2(g)) in their standard states.\n\n2C(s) + H2(g) ---> C2H2(g)\n\nenthalpy of formation of C2H2= \nenergy absorbed to break H-H bond \n+ \nenergy released to form 1 C=C bond and 4 C-H bonds \n= 610 + 4(-410)\n= -1030 KJ/mol\n(If you are taking A level Chemistry, refer to your data booklet for these values under the section of bond energies)\n\nHope this helps.
404275
divide 200g by the molecular weight of KCL and then divide that by .8
737043
The answer is (e).\n\nTo calculate the enthalpy change, find the product of the specific heat of water, the mass of the water in grams and the temperature change of the water in kelvins. Since the change in degrees celcius equals the change in kelvins, use (-2).\n\nE (enthalpy change) = 4.18*200*(-2) = -1678 J \n\nI hope this helps!
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!