I'm having problems solving this problem: What is the energy change (in kJ) for the combustion of 1.145 grams of ethane using 12.5 grams of oxygen?
120659
first figure out the number of moles of each substance, in this case: ethane = 0.038 and oxygen = 0.39, so oxygen is in excess and will not limit the reaction.\n\nThe standard enthalpy of combustion of ethane = -1423KJ/mol\n\nas you have 0.038mols of ethane the energy change is 0.038x-1423 = -54.07JK/mol
512036
(Triangle) or (Delta) H for this process will equal:\n\n60g * Cp of water (4.1855 J/(C*g) * (Delta T)\n\n60*4.1855*14* = 3515.82 J\n\nThe change in enthalpy or (triangle H) is 3515.82J
194507
* 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.
534765
initial energy : (mv^2)/2 = (0.047*56^2)/2 = 73.696 J\nfinal energy : mgh +KE= 0.047*9.81*17 + KE = 73.696\n\nKE = 73.696 - 7.83819 = 65.85781 = (mv^2)/2\nv = squareroot(KE*2/m)\nv= 52.94 m/s
151380
it is based on arrhenius equation\nK=A e ^ (-Ea/RT)\nA- frequency factor\nEa- activation energy (proportional to Delta H)\nR -Gas constant\nT -Temperature\nSince you dont know A- fequncy fator, yu need to do a lengthy calculation\n\nThe modified arrhenium equation will be ( by considering two situations T1, T2 there will be two rates k1, k2 but A-frequency factor is smae for one reaction so it's a constant..)\n\nEa= (R X T1 X T2 X Ln (k1/k2) ) / (T1-T2)\n\nEa= delta H = 17,390 J\nR = 8.314 J/K\nK1 = 1.6 X 10E-4\nK2= ?\nT1 = 298\nT2= 273+97= 370K\n  After substituing all these values, you will get K2= 6.27 E-4\nwhich is in agreement with theory (At high Temperatures, rate should be high..!)\n\nIs this info will help you?\nGood luck!
433019
Conservation of energy:  KE=PE\n1/2*Mb*Vb^2= (Mb+Mw)*g*h\nVb^2=2*(Mb+Mw)*g*h/Mb\nVb^2=2*(0.01+2)*10*0.1/0.01\nVb^2=402\nVb=20.05 m/s
471505
Heat released = 480*2.34 J\n\nSo 0.02 mol releases this much \n\n1 mol will release 480*2.34/0.02 = 56.16 kJ
40487
First use the ideal gas law to find out how much oxygen gas reacts with the butane.\n\nPV=nRT where P is pressure in Pa, V is volume in meters cubed (m^3), n is amount in moles, T is temperature in Kelvin (K), and R is the ideal gas constant 8.314 J/mol K.\n\nn=PV/RT=[(101325 Pa)(0.015 m^3)]/[(8.314 J/mol K)(298 K)]\nn=0.613 moles O2\n\nNow write the balanced equation for butane combustion.\n\n2 C4H10 + 13 O2 --> 8 CO2 + 10 H2O\n\nNow use stoichiometry to find the moles of butane.\n\n(0.613 mol O2)(2 mol C4H10/13 mol O2)= 0.094 moles C4H10.\n\nNow convert moles C4H10 to grams by multiplying by molar mass. The molar mass for butane is 58 g/mol.\n\nMass butane = (0.094 moles)(58 g/mol) = 5.47 grams butane
472247
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!
386187
just look at your units..that normally hleps\n\nhf ...kj / mole\n3.22 kJ / 6.02 Kj/mol =    x moles\nxmoles * 18 gram / mol.....  Thats your answer
371445
CH4 + 2O2-> CO2 + 2H2O
29609
I'll answer it, but you'd best pick me for the best answer if I'm doing all this work. ;)\n\nConvert C to K:\n\n178.0C = 451.15K\n25C = 298.15K\n30.6 = 303.75K\n\nTotal energy of water initially: Ei\n\nEi = 185.0g * 298.15K * 4.184J/g*K = 2.308E5 J\n\nTotal energy of water, final: Ef\n\nEf = 185.0g * 303.75g * 4.184J/g*K = 2.315E5 J\n\nEf - Ei = 700J\n\nLet Q be the specific heat of the alloy\n\nQ * 36.0g * 451.15K = Q * 36.0g * 303.75K + 700J\n\nQ*{(36.0g * 451.15K)-(36.0g * 303.75K)} = 700J\nQ = 700J/{(36.0g * 451.15K)-(36.0g * 303.75K)} \nQ = 700J/(16241g*K - 10935g*K)\nQ = 700J/3506g*K\nQ = 0.200 J/g*K
367871
I don't know the conversion, but I know how they are related:\n\nA Joule, J, is the SI unit of energy and is defined as one kg.m2/s2. The prefix "kilo" means 1,000, so one kJ = 1,000 J. As the energies associated with a single molecule or atom are quite small, we often find it easier to discuss the energy found in one mole of the substance, hence "per mole". To get the energy for one molecule, divide kJ/mol by Avogadro's number, 6.022 x 1023.
52397
if I understand the question:\nto go from cal/gK to cal/gC you'd convert the temp from K to C. that's simply subtract 273. degrees C = degrees K - 273.\nif you're trying to go from j/kg to j/mol you need to know the identity of the substance you're talking about. specifically you must know the molecular weight of the substance. first convert kg to grams(multiply by 1000) then divide by the M Wt of the compound.
78615
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
208982
E = Fd = (2E-9 N) (200E-12 m) = x J\nx J (1 kJ / 1000 J) = xE-3 kJ\n\nI think this is already per mol...
385265
omg i just had this test today too!!!!!\nu use hess's law to find the change--->write the equation, balance it \nand then use the moles (coefficients) to multiply the change is s for each element.\nhess's law says that the change in s (or h, or G, its usable for any of the thermo concepts) is equal to the sum of the entropy of the products minus the sum of the entropy of the reactants. its pretty easy after that--just plug in the values and there you go! \ngood luck! this was a hard test for me..so confusing.