thermochemistry test!! help?
928379
(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
814637
The equation is  2H2 + CO--->CH3OH\nThe first step in a problem like this is to determine if you have enough of each reactant...or is there a limiting reactant.\n\nYou started with...\n1.73g H2 (MW 2.016) = 0.858mols...and\n16.05g CO (MW28.01) =0.573mols\nBut since 0.848mol H2 would react with only 0.429mols CO you have an excess of CO...too much CO, so your limiting reactant is H2.\nYou get one mol of CH3OH (MW 32.04) for each half mol of H2 per equation, so the theoretical yield is...\n0.429 x 32.04 = 13.7g of CH3OH\n\nCheck my figures for accuracy...which will also give you good practice. :-))\n\nEDITED LATER: If you check this work carefully, you will note that the answer is correct, but the statement is not...it should read:\nYou get one-half mol of CH3OH for each mol of H2 per the equation, so the theoretical yield is...\n1/2 x 32.04 x 0.858 = 13.7g CH3OH\n(Did you find this error?)
338000
I only had time to do the first one, btu the answer is \ne-2.  Check your email for the work.
460409
Just guess what the colors would be for different compounds before you actually do the flame test for the hypothesis.  \nOr you could guess whether it is the cation (the positively charged metal ion) or the nonmetal anion that is responsible for the different colors.
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.
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.
701443
1 - the reaction is reversed and multiplied by 4, so the enthalpy change is 11508 kJ\n\n2 - The reaction is a combination of the other three (I'll label them A, B and C, in the order you have them listed, for simplicity)\nThe original reaction = 2A - 2B -2C (the first one is just doubled, the next two are doubles and reversed), so 2*321.8 - 2*571.7 - 2*288.6 = -2542.2 kJ\n\n3 - This one is done similarly to question 2.  A - 0.5B + 1.5C\n180.8 - 0.5*(-906.2) + 1.5*(-483.6) = -91.5 kJ\n\nHope this helps
574078
This might help :)\nhttp://www.chemicalforums.com/index.php?board=8;action=display;threadid=3263
1138290
Look up stuff like Arrhenius equations. I don't have my phys chem text book to hand.\n\nFor part two, I think it's D, as I'm fairly certain it's not any of the others.
49101
Here is one for a software package which can be down loaded for a 30 day test:\n\nhttp://www.chembuddy.com/?left=EBAS&right=balancing-chemical-equations\n\nand another at:\n\nhttp://chemfinder.cambridgesoft.com/
132159
Open up your thermodynamics book and do your own homework.\nHint.. add the energy required to raise the temp from 100C to 112C plus the energy used to change state, plus the energy used to go from 76C to 100C.  You should have all the info you need in your problem, and the energy required to do the change is in your book.
977052
You need a variable in any experiment to compare results. Perhaps you could use cold (ice) water in one container, warm in another and hot or boiling in another. You can then make your prediction on which will dissolve first and have an experiment that will verify whether you have made an accurate prediction.
891159
use the equation\nE(enthalpy change) = m(mass) c(specific heat) DeltaT(change in temperature)\n\ntemperature change is initial temperature minus final\nsince temperature increased then the temp change is negative\n\n-2.00C
955628
For the reaction:\n\nMgO + 2 HCl  -->  MgCl2 + H2O\n\nThe temperature should go up, not down.  Only the initial and final temperature are needed.\n\nTypical values for students are between 110 an 150 kJ/mol MgO.  This is typical, not the best that can be done.
239222
no, reaction times are too slow at that temperature.  (I'm just guessing, but I'm a chemist.  You should be able to look up reaction energies, try Beilstein.)
557678
Ah ha.  Thermite.  Magnesium ribbon would be the usual method to start this reaction.\n\nBe careful with this reaction...
963248
The heat given off by the burning reaction (delta E) is equal to the heat gained by the calorimeter.  (m * change in temp)\n\nYou can calculate the heat released by burning.\n.553 g with a molecular wt. of 88g/mole \n(.553 grams) / (88 (grams / mole)) = 0.00628409091 moles\n(975 (kJ / mole)) * 0.00628409091 moles = 6 126.98864 joules\n\nYou can calculate the heat absorbed by the water (3000 g * .356)\n3 000 * .356 = 1 068 joules\n\nyou can calculate the difference, it must have heated up the apparatus.\n6 127 - 1 068 = 5 059 joules \n\n5059 joules = x grams * .356\n5059 joules / .356 = 14210.67 grams of water\n\nThe metal apparatus was had a heat capacity equivalent to 14210.67 grams of water.