How much heat will be released by the combustion of 100g of C?
914456
okay...100 g*(mol/12.01g)=8.33 mol C\n\ntake the aforementioned 1mol *(-113kj/2mol)=-56.5 kJ released for 1 mol\n\nthen multiply that -56.5 by 8.33= -470.65 kJ released\nnote it is exothermic and thus negative
272349
Ca(s) + 2HCl(aq)-->CaCl2 + H2(g)\n\nor...\n\nCalcium Chloride and Hydrogen Gas... you know... the stuff that fuels rockets! :)
872856
It will eventually melt at 2572C, and it will boil at 2850C.\n\nHeat does not cause it to react, typically.
925364
Definitely not water, as those three metals will burn on contact with water (and potassium will explode.) Liquid CO2 might do, as it would cool the burning material extensively, possibly below the ignition point. The sand (SiO2) would be a smothering agent. I'm not sure about the carbon tet.\n\n5 NOV 06, 1522 hrs, GMT.
836375
e. 650cal.\n\nq=mc(change in temp/ cal)\n  = 100 x 1 x 6.5 (since 31.5 - 25.0 = 6.5 deg C)\n  = 650cal.\n\nTake note of the SI unit for 'c'; it determines the necessity to convert temperature from deg C to K, or mass from g to kg.
454506
100 degrees celcius\n212 degrees Fahrenheit
958888
Use the formula:\nQ (heat) = Cp (Heat Capacity) * delta T (Change in Temperature)
418645
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
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.
1032149
Q=MC (delta) T....(5000 G)(.377 J/G degree C)(200 Degree C) which should be.....i get 377 J, which isn't your answer....i just had an exam on this last week, so i thought for sure i am right....the only thing i am not sure about for certain is the specific heat of coal, i got it from a website.....need more info if i am able to help, surely they provided the specific heat of coal....also if latent heat of fusion of vaporization has a part, the answer above is going to be different....sorry dude
121351
Are you tryn to have us do your hw!?!?\n\nThe eqn should be in your book.  I dont feel like digging mine out, but its something like;\n\nCp*g*(T2-T1)=energy\n\nSo what u want to find is (t2-t1) Ima call it x\n\n400*50*(x)=800  do the math.  I think thats it.\n\nfor the 2nd question, this one is easy.  The combustion is energy is for 1 mol (per mol).  If you only have .2 moles, the combustion amount is .2* 1370 .
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
14789
A general form of the equation I beleive you are looking for comes from a basic understanding of thermodynamics.  I am assuming that your ct=(Tfinal-Tinitial)...\n\nSo a general form of the equation would be:\nmCp(Tf-Ti)=Q \n\n"Cp is the specific heat"\n\nWATCH UNITS!!!!!  \n(I am ignoring the conversions in this statement)\n55g*Cp(67degrees UNITS)=343 cal \n\nSolving this equation will give you Cp ... just make sure that you watch units.
436044
Standard enthalpy of formation 227 KJ/mol\nStandard enthalpy of combustion -1300 KJ/mol\nStandard entropy 201 J/K/mol
667010
If there is combustion, there will be SO2.  The temperature will depend on the construction of the system, distance to chimney,etc.  The temperature of the flame will be constant depending on the type of oil.
1164242
C- combustion chamber
709139
Bond enthalpy at 25 C\nC-H = 99 kcal/mole\nC-C = 83 kcal/mole\nC=C = 147 kcal/mole.\n\n(1st C) = 3 * 99 + 83 = 380 kcal/mole\n(2nd C) = 99 + 83 +147 = 329 kcal/mole\n(3rd C) = 2 * 99 + 147 = 345 kcal/mole