How many joules of heat energy were added?(Specific heat of Al =0.90 Jg^-1K^-1)\n\n2) Using the equation below:\n\nC(s) + O2(g)->CO2(g)     Enthalpy Heat = -390 kJ\nMn(s)+ O2(g)->MnO2(s)    Enthalpy Heat = -520 kJ\nwhat is Enthalpy Heat (in kJ) for the following reaction?\n\nMnO2(s)+ C(s)->Mn(s)+Co2(g)
122259
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
327828
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.
447667
The answer is B.  The enthalpy ( heat content ) of the barium hydroxide and water is more than the enthalpy of the solution.  The reason is because heat is lost in the exothermic reaction\nTrust me I did a chemistry degree!\n\nGuy below me has the right idea, but has used the wrong equation hence the answer is B and NOT A\n\nEnthalpy is a measure of energy content really.  How can the final solution have more energy content if heat has been lost during the reaction?\n\nThe equation below is wrong it should be:\n\nEnthalpy products - Enthalpy reactants = delta H\n\nSee 3rd line of http://www.ausetute.com.au/enthchan.html \n\nThis is why the guy below has the opposite answer.\n\n\nThe simple way of looking at it is \n\nReactants have energy, they lose some during the reaction, this means the products will have even less energy content hence B!!
421768
Your reaction evolved 500 calories (you raised 100 grams of water five degrees) , or 500*4.184 J = 2092 Joules.\n= 2.092 kJ\n\n\n2.092 kJ / 0.05225 mol = 40.0 kJ/mol
250069
Formula:  J = K*m*(T2-T1)\n\nSince both K & m are 1 for your problem, J = 125 Joules\n\nK is the specific heat defined as 1 for water.  Varies with other materials.  Also this formula is not strictly true for 1 liter unless you assume the 1 liter weighs 1 kg.  Liters are volume and the formula is for mass.
521824
(300 K) 80.4 W·m−1·K−1=Iron\n(300 K) 401 W·m−1·K−1=Copper\nI found these answers on Wikipedia (online dictionary).\nI would go with copper. I am wondering since there is 300k if this is a trick question and the answer is "they both heat up at the same rate."
174199
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
488379
Remember your equation for specific heat capacity::\n\nHeat content = mass x spec heat capacity x delta T\n\nq = m*C*dT\n\n225 J = (15 g) * (0.80 J/gram-deg C)* dT \n\nTherefore\n\ndT (in C) = 225 J  /  (15 g)(0.8 J/g-C) = 18.75 deg C\n\nMind your units and make sure they cancel.  I did.
119746
The reaction should be written as follows,\n2 KClO3(s) --> 2 KCl(s) + 3 O2(g) H = -89.4 kJ\n\nYou either did not include the arrow or it is not displaying properly.\n\nPotassium Chlorate decomposes with heat to form Potassium Chloride and Oxygen gas, no where in this reaction is elemental Potassium involved.\n\nThe standard enthalpy of formation of any pure element in its natural state is zero. Oxygen is a diatomic molecule and thus, Oxygen gas (O2) has a enthalpy of formation of 0 kJ in this reaction, no matter how much of it you have, it is still zero.\n\nSince the Potassium Chloride is the only product left that has any non-zero enthalpy of formation, we know that the over all change in enthalpy is equal to the enthalpy of formation of KCl * the number of moles produced.\nIf the over all change in enthalpy is -89.4 kJ and 2 moles of KCl are produced, then the enthapy of formation of the KCl must be 44.7 kJ/mole less than the enthalpy of formation of the Potassium Chlorate.
241242
This sounds like homework, so I'll give you the ideas but not the data.\n\nThe enthalpy of the mixture at standard conditions is the sum of standard heats of formation of the components of the mix. It will be given in kJ/mol (Ar is 0 kJ/mol being an element, the other two you can look up in tables or on internet). \n\nThen you need to adjust for the sensible energy difference compared to standard conditions : you will also need to look up the specific heat of each component (usually expressed in kJ / mol ºK) and multiply the specific heat of each component by the amount and by the difference in temperature. Make sure you have everything in the same units.\n\nRemember if you are working in moles its easier because to a first approximation, one mole of any gas occupies the same volume (if you undrestand how to allow for deviations from the ideal gas laws, you wouldn't be asking this question) and I assume that the percentages are volume percent, not by weight\n\nEnjoy
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
505326
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.
28470
-278.0 + 2*(-321.0) + 2*(-784.0) = -2488 kJ\n\nI am assuming that the enthalpy was actually given in kJ/mol.
136872
You havent really provided many details as to how you want to set it up.\n\nQ = m*c*(delta T)\nQ = heat energy\nm = mass\nc = specific heat\ndelta t = change in temperature (= final - initial temp)\n\nRemember, in a closed system, heat gained by one thing equals the heat lost by another.\n\nHeat is a form of energy, meausred in Joules (sometimes calories, but the official SI unit is Joules),\nDepending on the value you use for the specific heat, it is common to use grams or kilograms for the mass measurement.\nAgain, depending on the value of the specific heat, most commonly, you'll use temperature in terms of degrees C (or perhaps K, a change of 1 degree C = change of 1 K).
47929
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 .
465989
To solve this question we need to know the specific heat of the resulting KI solution.  To make things easier, we can probably assume that the specific heat of the solution is the same as the specific heat of water (c = 4.184 J/g degree C).\n\nThe energy (Q) released when a given mass, m, cools a given number of degrees C, (delta T), with a known specific heat, c, is:\nQ = mc(delta T)\n\nWe know the mass of the water, we know its specific heat, we know how many degrees its temperature changed, now just plug in and get an answer,\nQ = (60 g)(4.184 J/g degree C)(14 degree C)\nQ = 3514.56 Joules\nQ = 3.51 kJ\n\nBut this amount of heat energy was due to 28.7 grams of KI dissolving, and we want to know the heat given off by 1 mole, so we need to convert 28.7 grams of KI into moles of KI using the molar mass (166 g/mol).\n28.7 g KI / 166 g/mol KI = .173 moles KI\n\nSo if 3.51 kJ of energy was absorbed when .173 moles of KI dissolved, if one mole of KI dissolves, 20.3 kJ would be absorbed....the answer which you were looking for.\n\nBy the way....\nthe "(triangle) H" you are referring to is the change in the enthalpy of the reaction.  The "triangle" is the Greek letter "Delta".  It is more common to see it written out as "delta" or "change in" if one cannot use the actual Greek alphabet in typing, as opposed to "triangle".  Δ
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