HELP!!!!!! chemistry geniuses out there!! please!?
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.
17774
It doesn't give me room to list them all and explain them on here so let me direct you to a website I found that does a pretty good job:\nhttp://vergil.chemistry.gatech.edu/notes/quantrev/node20.html
834501
First let's find how many moles of each one we have\n\nmole= M*V, so for \nFeCl3 =0.234*25*10^-3= 5.85*10^-3\nNaOH =0.453*42.5*10^-3 =19.25*10^-3\n\nFeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl\n\nso\n\n1mole FeCl3 reacts with 3mole NaOH\n5.85*10^-3 mole react with x\n\nThus x =3*5.85*10^-3=17.55*10^-3 mole NaOH but we have 19.25*10^-3 mole NaOH, therefore NaOH is in excess\nTherefore the amount of Fe(OH)3 will be calculated according to the limiting reageant FeCl3.\n\nso from the reaction\n\n1 mole FeCl3 gives 1 mole Fe(OH)3\nor better yet, since the molecular weight of Fe(OH)3 is 106.85 \n1 mole FeCl3 gives 106.85 g Fe(OH)3\n5.85*10^-3 mole give x g\nx= 106.85*5.85*10^-3= 0.625 g\n\nWe calculated that 17.55*10^-3 mole NaOH reacted.\nSo we have remaining (19.25-17.55) *10^-3= 1.7*10^-3 mole NaOH\n\nThese are in a total volume of 25.0+42.5=67.5 ml= 67.5*10^-3L.\n\nSo the concentration of the excess NaOH is\nC=mole/V= 1.7*10^-3/67.5*10^-3= 0.025 M
149472
Hey I'm studying physical and chemical equilibrium too, currently.The answer is that it will definitely have units when the concentrations have different powers.Just try and work it out with the units and it will be very clear.
782516
a) 3SiO2 +2 Al --->3 Si + 2Al2O3\n\nb)AgNO3 +HCl---> AgCl +HNO3\n\nc) Mg + CuSO4 ---> MgSO4 + Cu\n\nd) Pb(NO3)4 +4 NaI --->4 NaNO3 + PbI4 (s)
475338
1) FIND THE REDOX-REACTION\n\nChromate (+VI) is reduced to 2 Cr(III) meaning this that 6 electrons are needed.\n\nThe corresponding oxidation on the alpha-C-atom in Ethanol goes from oxidation-level (-I) to oxidation-level (+III) in acetic acid, meaning it delivers 4 electrons per molecule.\n\nSo we find a stochiometric electron equilibrium if 12 electrons are exchanged:\n\n2 Cr2O7(2-) + 3 CH3CH2OH  =  4 Cr(3+) + 3 CH3COOH\n\n2) THE PROCESS MUST BE ELECTRICALLY NEUTRAL\n\nWe add the K-Ions having the left side of the equation neutral and the right side with 16 loads positive:\n\n2 K2Cr2O7 + 3 CH3CH2OH  = 4 Cr(3+) + 4 K(+) + 3 CH3COOH\n\nIn order to have the left side also with 16 loads positive, we add 16 protons, meaning this, that the reaction needs acid. Since we use sulfuric acid we also add 8 sulfat-ions on both sides and hereby can neutralize the product side:\n\n2 K2Cr2O7 + 8 H2SO4 + 3 CH3CH2OH  =\n                         2 Cr2(SO4)3 + 2 K2SO4 + 3 CH3COOH\n\n3) STOCHIOMETRIC BALANCE OF THE ELEMENTS H and O\n\nTaking into consideration that unused protons finally end up as water, we add on the right side 11 H2O and hereby is also reached the balance of oxigen:\n\n2 K2Cr2O7 + 8 H2SO4 + 3 CH3CH2OH  =\n                  2 Cr2(SO4)3 + 2 K2SO4 + 3 CH3COOH + 11 H2O\n\nSo now good luck for counting the atoms!
367815
Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
72603
Wow, I hope these are right.  I wish i had a solubility chart or something (Which you probably have)  \n\nCuSO4(aq) + Na2S(aq) ---> CuS (s) + NaSO4 (aq)\nNet ionic - Cu2+ (aq) + S2- (aq)  ---> CuS (s)\n\nNiCl2 + 2KNO3 --> Ni(NO3)2 (aq) + 2KCl (aq)\nnet ionic - none (if NiCl2 isn't aq, then this is wrong)\n\nCa(OH)2 + 2HC2H3O2 ---> Ca(C2H3O2)2 (aq) + 2H2O (l) \nnet ionic - H+ (aq) + OH-(aq) ---> H2O (l)\nactually, since the acetic acid is a weak acid, this might be a little different.  If you learned it this way, then your net ionic is:\nHC2H3O2 (aq) + OH- (aq) ---> H2O (l) + C2H3O2- (aq)\n\nHope this helped
750179
NaCl2 + NaCO3 ==> NaCO3 + NaCl2 NOTE: these two reactants will not react because the reactants and the products are identical\n\nCu + O2 ==> CuO2   This is already balenced\nCuSO3 + Fe ==> FeSO3 + Cu   The charge is not given so procuts could be different\n\n2HCl + 2Zn ==> 2ZnCl + H2\nMnO2 + H2O2 ==> No reaction: similar to NaCl2+NaCo3\n\nGood luck
880509
Hey, Blond fuss?  Good luck if U can't find help @ this late hour. Keep working on it.  Something from what U learned so far must b in your head - find it. No one else working on this problem awake?\n  I'm tired (12:35 a.m. here @ UMAss Amherst).  Had 1st-yr Chem 100 in 1978, not Chem 1111 !! Barely remember orbitals.  U do have books there, don't U?   Library maybe not open now as ours is.    \n Aces 2 computer. So U have awhile 2 figure it out.  Good luck.
734957
Na+ Cl-\nC +4 O-2 each\nK+ Br-\nN-3 H+1 each\nMg+2 O-2
730330
Different chemical composition, one has a reaction the other doesn't..☺
433211
This is simple-\n3Mg(S) + N2(G) (2 is subscript) Yields (arrow) Mg3N2(S)
436191
Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.
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.
835717
Hmmm.. I'm not trying to be stingy, but I still prefer using Yahoo! Answer since that there is a bunch of good chemist in this community, to asking for a help from there.\n\n\npeace\nvixklen