thermochemical equation question (enthalpy)?
919460
its actually pretty simple. U multiply by -1 when u flip over an equation...so the reactants become the products and the products the reactants. that way instead of having a negative number u will have a positive number for the kJ. as in this problem. U have to flip #1 so u can cancel a few things with #2. so it becomes..\n1)3 H2O(l) + 2 CO2(g) AH---->CH3CH2OH(l) + 3 O2(g) (multiply by -1 gives u a postive +1367kJ. part two stays the same.\n2) C2H4(g) + 3 O2 ---> 2 H2O(l) + 2 CO2 (g) AH= -1411 kJ\n\nNow the 3 O2 from #1 and #2 cancel since they are on opposite sides. Now the 2 H20 cancel from #2 and you are left with only one in #1 since they are also on opposite sides...which after canceling..leaves u with your original equation.\n\nC2H4(g) + H2O(l) ------> CH3CH2OH(l)\n\nnow since u flipped over #1.....and u have a positive +1367kJ and\n#2 is left the same...its still at negative -1411kJ. u add them together and u get an enthalpy change of -44kJ. sorry its pretty confusing to explain..if anything message me and i willl try to help.
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!
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
533621
(delta)H(rxn) = sum of heat of formation of products - sum of heat of formation of reactants.\n\na.  Hf Br2(g) = 30.91 kJ/mol, Hf Br2(l) = 0, so (delta)H = -30.91 kJ/mol\n\nb.  Hf CaCO3(s) = -1207 kJ/mol, Hf CaO(s) = -635.5 kJ/mol, Hf CO2(g) = -393.5 kJ/mol, so (delta)H = -635.5 - 393.5 + 1207 = +178 kJ/mol\n\nc.  Hf NO(g) = 90.25 kJ/mol, Hf O2(g) = 0, Hf NO2(g) = 33.2 kJ/mol, so (delta)H = 2 * 33.2 - 2 * 90.25 = -114.1 kJ/mol
116913
I think it is this:\n\n            \nC12H22O11(s) --- HNO3---->  12C(s) = 11H2O(g) + heat
395237
you need to have more information that that to calculate a lattice enthalpy. eg Electron affinity and ionisation energy, chlorine homolytic dissociation etc\n\nThe final answer is -2223KJ/mol
464223
'c' is the right answer.\nuse I law of thermo...\nU=Q-W\nU=Q-Q/4\nU=3Q/4\nBut U=n*C*T where n is moles,C is molar heat cap.for const volume.\nLet us assume that the heat absorbed during the process as\nQ=n* A*T\nwhere A can be defined as molar heat capacity for the process.\nso , n*C*T=3*n*A*T/4\nA=4C/3\nFor a diatomic gas C=5R/2(C=no.of degrees of freedom*R/2)\nA=4*5R/3*2\nA=10R/3
401299
i already answered parts a and b in the last question like this.\n\nc) enthalpy = mass of water x specific heat of water x change in temp\n\nE=0.1*4.18*3.5\nE=1.463KJ\n\nmultiply by two to get answer per mole = 2.926KJ/mol\n\nd) main source of error is the glass beaker since it is leak out the heat produced by the reaction, the reaction would be better carried out in a calorimeter or in a polystyrene beaker with a lid.\n\ne)  the temperature change would be the same as 1mol of KOH reacts with 1mol of HCl therefore with the extra HCl the KOH will act as a limiting reagent and u will get the same reaction as before.
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
410173
This formula might help\n\nchange in S=(m1)(c1)(ln(Tf/Tc))+(m2)(c2)(ln(Tf/Tc))\nm1 & m2=substance mass in this case oxygyn gass (O2)=15.999(2)\nc1 & c2 are spcific heats\nTc =cold temp\nTh = Hot Temp\nTf=((m1)(c1)(Tc)+(m2)(c2)(Th))/((m1)(c1)+(m2)(c2))\n\nI'm assuming m^3/kg is the pressure (Volume/Mass) ???\n\nhope this helps!!
824412
The temperature difference is 500 K.  If we add 26/deg for NH3, and subtract 0.5 x 27/deg for N2 and 1.5 x 29/deg for H2, we will net out at -0.5 -4.5 = -5.0 J/K per mole of product, so (if i've got the signs right) we will be at -46,000 -2,500 joules/mole, or -48,500 joules per mole.  It is more straightforward to compute the enthalpy changes separately and add them up, but by saving the multiplication till later I can do the numbers in my head.
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 .
72400
-278.0 + 2*(-321.0) + 2*(-784.0) = -2488 kJ\n\nI am assuming that the enthalpy was actually given in kJ/mol.
360824
Since you have 3 points plot those(number of carbon in each compound vs molar enthalpy)\n\nSO ethanol would be 2, -0.215\n\nThen you could try drawing a line of best fit(assuming the relationship is linear) between the three and extending the line till it reaches 5 carbon
496291
You need to take more care in writing equations when you post your questions. \n\nPart a) is difficult to decipher - you will need to re post it.\n\nPart b) is presumably:\n\nCH3COOH(l) + 2O2(g) --> 2CO2(g) + 2H2O(l)\n\nPart c) is presumably:\n\nFe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)\n\nI would next use heats of formation to calculate the enthalpy change in the above reactions using the formula:\n\ndelta Ho = sum(heats of formation of products) - sum (heats of formation of reactants).\n\nIf you have a data book or data sheet, these values can be found there (Note: heats of formation of elements are zero.) \n\nIf you want web sites that might help you, try googling "enthalpy change problems" +tutorial.\n\nI hope this helps you.
1064976
Q (heat energy) = m(mass in gram)*c(specific heat in J/gm/K)*t (change in temp)\nQ = 25 * 1 * (36-24)\nQ = 300 J
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