9. In the following, indicate which SN2 reaction will occur faster. Explain your reasoning.?
535526
a.  1-Bromobutane will undergo a faster reaction than 1-chlorobutane, as bromide is a better leaving group than chloride, due to being a weaker base.\nb.  1-Bromobutane will undergo a faster reaction than 2-bromobutane, as primary halides undergo SN2 reactions faster than secondary halides, due to steric considerations.
803448
This is an SN2 reaction because you have a secondary haloalkane and a polar aprotic solvent so inversion of conformation occurs. The answer is B
619470
o.k., in short, d < a < c < b. This is basic material, so I'll give a generously detailed explanation.\n\na) 2 moles N2 * (28 g N2/ mole N2) = 56g N2\n\nb) 1 mole KCl * [(39.1gK +35.5gCl)/ mole KCl] = 74.6 g KCl\n\nc) 9 x 10^23 molecules CO2 * (1 mole CO2/ 6.02 x 10^23)*([12gC + 16gO + 16gO]/ mole CO2) = 66g CO2\n\nd) 2 moles LiF * [(6.94g Li + 19gF)/ mole LiF] = 51.88 g LiF
610699
Pb(NO3)2(aq) + H2SO4(aq/l)[depending on concentration] --> PbSO4(s) + 2HNO3(aq)\nPb2+(aq) + SO42-(aq) --> PbSO4(s)\n\nI don't see how an redox reaction could occur.
823470
First write the two half reactions:\n\n3 e(-) + Bi(OH)3 ---> Bi\nSnO2(=) ---> snO3(=) + 2 e(-)\n\nMultiply the first equation by 2 and the second by 3 and sum to get:\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi\n\n12 O in the left, 9 on the right, so add 3H2O\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi + 3H2O\n\nCheck that H is OK.  Yes. Done\n\nSo the sum is 13. (d) is the answer.
896646
"The final experiment was to observe the reaction when one small crystal of NH4SCN was added to 0.1 M FeCl3. The color of FeCl3 was yellow, but after NH4SCN was added, the color of contents changed to red.\nIn second experiment, when the Mg was ignited, the resulting flame was hot. Therefore, the release of thermal energy through the interaction of Mg and O was shown.\nFiguring out the type of the last two reactions was difficult, because the chemical equations were not given. However, the possibility that the type of both reactions were replacement reactions was high. The reason being that the chemical reactions were observed as a color change and this means that the composition of the substance is changed.\nThis experiment was very successful, and it was a good opportunity to observe chemical reactions directly."\n\nI would change your paragraph to that.  I would also expand on the other experiments if you know more about them.
1151400
simply put: the higher the concentration the faster the reaction rate. in equilibrium the higher concentration shifts the reaction t o the other side.
1018313
Silver is being oxidized, nitrogen from the nitrate is being reduced.\n\nOxidation half reaction:\nAg --> Ag+  +  e\n\nReduction half reaction:\nNO3- +  4H+ + 3e  -->  NO + 2H2O\n\nNet balanced and reduced equation:\n3Ag +  NO3- +  4H+ --> 3Ag+ + NO + 2H2O
831216
Cu (s) + Pb(NO3)2 (aq) ---------> No Reaction. \nCopper will not replace lead.\n\nreaction VI in the attached link.
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.
821643
Copper is more reactive than silver, hence displacement of silver occurs.\nCu (s) + 2AgNO3 (aq) --> 2Ag (s) + Cu(NO3)2 (aq)\n\nSum of the coefficients? 6...\n\nHope this helps :)
136146
The 1st answer is wrong. The reaction is\n\nNa2CO3(aq) + SrCl2 (aq)--> 2NaCl (aq) +SrCO3 (s)
752247
Remember that the halide, which is going to eventually serve as the leaving group, is pulling electrons away from the carbon center, inducing a dipole where most of the electron density is around the halide (leaving group) and less of it around the carbon center.  The incoming nucleophile is thus even MORE drawn toward that carbon center when there is more induction of the dipole.  Therefore, the more electronegative the leaving group, the stronger the induced dipole.  The stronger the induced dipole, the more reactive is the carbon center towards SN mechanisms.
872094
A . 4LiH + AlCl3 → 2LiAlH4 + 2LiCl <-- there's only 1 alluminum so it can't yield 4 in the product\n\nB. Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O <-- there are 1 sn, 4 H, 2 S and 8 O ( multiply the coefficient [2] by the subscript [4])\nit matches in the product\nCORRECT\n\nC. 4BF3 + 3H2O → H3BO3 + 3HBF4< same goes here, just write out how many of each element are in the beginning of the equation and in the product and compare\nCORRECT\n\nD. 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O <-- this one is hard to read, I'm assuming that I is not an L because it's followed by a subscript, so it would be iodine\nbut either way, this one is wrong because there are 5 C/Cl in the product which is wrong
837624
Cassandra C has given you a very good answer.\nYou are interested in a single type of reaction: single displacement. It is a redox reaction so you have to think which compound is more happy having extra electrons (more electronegative).\n\nIn the case of A+BX-> AX + B, B(+) in BX is a cation and is missing an electron. If B is a more electronegative negative element than A, then it will steal away the electron from A and the reaction will occur. If A is more electronegative then B is not "strong" enough to steal the electron and the reaction will mot occur.\n\nIn the case of Y+BX -> BY+X you have the opposite. X(-) in BX has already an extra electron. So if X is more electronegative than Y it will not give away the electron and the reaction will not occur.\n\nThis is an oversimplified reasoning. The proper one is that you have to look at the standard reduction potentials (E0) which give the reactivity series. The half-reaction with the highest reduction potential will be the one favoured or more correctly the reaction will occur if the redox potential of the whole reaction is positive.\n\nFor instance you can find a detailed table at\nhttp://www.northland.cc.mn.us/Chemistry/standard_reduction_potentials.htm\n\nLet's see your reaction\nCu+FeSO4 -> CuSO4+ Fe \nIn this case the half reactions would be\nCu - 2e -> Cu(+2)\nFe(+2) +2e-> Fe\n\nFrom the website we get that for\nCu(+2) + 2e->Cu  E01=+0.342 Volt\nFe(+2) +2e-> Fe E02=-0.447 Volt\n\nE01 is more positive than E02 thus Cu(+2) tends to get electrons from Fe and not the other way around so the reaction will not occur \nOR\nYou find the E of the reaction like this.\nThe second half-reaction is in the form we want it. The first is in the reverse. So we just reverse the reaction and the sign for the potential\n\nCu -> Cu(+2) +2e E01'=-0.342\nFe(+2) +2e-> Fe E02=-0.447 and add\n\nCu + Fe(+2)+2e -> Cu(+2)+ Fe+ 2e with E=E01'+E02= -0.342+(-0.447) \n\nTherefore Cu+Fe(+2) -> Cu(+2) + Fe E= -0.789<0 and so the reaction will not occur.
660043
1.) do your own homework\n2.) this is the computer forum, not high-school chemistry.  Next time post to the right furom.
845374
Use the Nernst equation:\ndE = dE0 - 0.0592 V/n ln [C]^c [D]^d /([A]^a [B]^b)\n\nn is the number of electrons involved, 2 in this case.\nTherefore we have in all cases:\n dE = dE0 - 0.0296 V ln [Sn2+] /([Fe2+]\nFurthermore we use the standard electrode potentials to calculate  dE0 \nSn2+(aq) + 2e- -> Sn(s) -0.14 V\nFe2+(aq) + 2e- -> Fe(s) -0.41 V\nTherefore dE0 = 0.41 - 0.14 = 0.27 V\na) The concentration are 1 M/L and we have dE = 0.27 V\nb) dE = 0.27 - 0.0296 ln 0.01/0.2 = 0.27 + 0.09 = 0.36 V\nc) dE = 0.27 - 0.0296 ln 0.4/0.005 = 0.27 - 0.13 = 0.14 V\n\nOnly in the last case the reaction moves to the left: the concentration of Sn2+ is the highest.\n\nYou can take a look here for explanations and examples:\n\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html