What does the equation look like when liquid pentane combusts?
965162
Pentane is a Highly Flammable Liquid which may form Explosive Mixtures with air.  It chemically combines with the oxygen in the atmosphere to give carbon dioxide and water:\n\nC5H12 (l) + 8O2 (g) --> 5CO2 (g) + 6H2O (g)
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
570402
Nothing...unless you light a match, then...BOOM!  In the case of Oxygen and Hydrogen, you'll get pure water afterwards...that should put out the fire you just started :)\n(Not really)
351293
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!
565832
Condesates formed under high temp. and pressure can reach flash points suddenly. They usually combust with a flame whose characteristics depends on the material.\n\nIt can be used as reaction indicators or flashes.
714714
A balanced equation for this combustion is:\n\nC3H8 + 5 O2 ----> 3 CO2 + 4 H2O\n\nSo, every mole of C3H8 requires 5 moles of O2 to burn.\n\n# of moles of C3H8 = 138 / (12*3+8) =3.136 mole\n\n# of moles of O2 required = 3.136*5 = 15.682 moles\n\nUsing ideal gas law, PV = nRT, we get\n\nV = nRT / P = 15.682*8.314*(25+273) / (100*1000)\nV = 0.389 m^3
228649
CO2 + H2O + light = C6H12O6 + O2
52130
E=mc∆T is the equation you want to use for this situation, where E is energy, m is mass, c is the specific heat, and ∆T is the change in temperature.  Now, the heat lost by the alloy is equal to the heat gained by the water, right?  So look what we can do:\nmc∆T=mc∆T\nwhere the left side of the equation is talking about the metal, and the right side of the equation is talking about the water.  Let's start pluggin in values.  We'll leave the left c value as a variable because that's what we're solving for.  \n36.0 g *c*(178 oC -30.6 oC)=185.0 g*(4.184 J/g*K)*(30.6 oC-25.0 oC)\nThe trickiest thing about that step is the temperatures: the metal started at 178 and goes down to 30.6; the water starts at 25 and goes up to 30.6.  \nNow evaluate that equation and solve for c. \nc=.817 J/g*K
847829
C(3)H(8)+5O(2)=>3CO(2)+4H(2)O\n\nfuel+oxygen=>carbon dioxide +water\n\n(combustion (the rapid oxidation of a substance giving out energy in the forms of light and heat) is burning, which needs O2!. The fuel reacts with oxygen )\n\nbrackets-subscript
744648
Use Gibbs free energy equation\n\nDeltaG=DeltaH - TDeltaS\n\nYou want Delta G to be negative so if Delta H positive then you need a greater TDeltaS \nDelta S is -25 so in order for the reaction to be spontaneous the T has to be less than zero
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.
605005
NaHCO3(s) + CH3COOH(aq) → CH3COONa(aq) + CO2(g) + H2O(l) \nThis is the real chemical reaction.
221143
They're gases (although butane forms a liquid under a little pressure) and the rest are liquids, at room temperature. Pentane is highly volatile, though.
199900
The easiest way is to simply find literature listing delta H for the reaction.  Otherwise you have to do one of 2 things.  \n\n1. Find the enthalpy of formation for the product of the reaction which should be what you're looking for.\n\n2. Use the equation G=H-TS, plugging in the Gibbs free energy, temperature, and entropy values for the reaction,then solving for H, the enthalpy
1118780
Your question is very open.\n\nheat equations very general can be:\napplied heat = specific heat x mass x increase of temperature\napplied heat = heat capacity x increase of temperature\napplied heat = mass x specific melt heat\napplied heat = mass x specific evaporation heat (at boiling point)\nHeat from a supply = power x time.\nCombinations and others are possible.\n\nwave equation can be x = A sin( 2pi t/T)\nThere are more possibilities.
975732
Substance: Water\nLatent Heat Melting J/g:  335\nMelting Temp°C: 0\nLatent Heat Boiling J/g:  2272\nBoiling Temp °C: 100 \nHeat of vaporization (kJ/mol): 40.65
927436
P1T1/V1 = P2T2/V2\nIn your case there is (apparently) no change in pressure, so the above equation will simplify to:\n\nT1/V1 = T2/V2\n\nV2 = T2V1/T1.\nPlug the numbers you have and find V2.  Remember to convert temperatures to Kelvin.