is this an E1 or SN1 reaction?
971292
as it involves leaving group at primary carbon so the carbocation if formed will be highly unstable ......so neither  E1 nor SN1.\n\nIt will be SN2,.......because ..EtO-  is a good Nu- and also Br- is a good leaving group.......and also C--O bond which will be formed will be stronger than C--Br bond ,so energetically also favourable.
109708
if i am not wrong..presence of one component( i forget the name)..makes its positive...n negative..no need to worry..cheers
928582
A change in color would indicate some sort of reaction.  xenon is a noble gas and is nonreactive, so no.
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.
707224
E1 reactions proceed through carbonium ions. Carbonium ions resulting from particular carbon skeletons/derivatives are unique. Once the carbonium ion forms, it has a characteristic tendency to eliminate to an olefin, or to hydrate or solvate to an alcohol or an ether. So you can't control how much olefin, alcohol, or ether you'll get.\n\nOn the other hand, if the substrate is really cheap, like cyclohexanol, then an E1 reaction is the way to go, because you'll get cheap cyclohexene regardless of what ever else you have to put up with.
890894
C12H22O11+12O2-------12CO2+11H2O+E
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
871220
The second one:\nAF + H --> AFM + TM\n\nActivation E: +10KJ/mol is moderate\nHeat of Rx= +5.2 KJ/mol is positive, making this an endothermic reaction (requires the input of heat or energy to accomplish).\n\nSee the site below:
270316
it's not.  one is a chemical reaction and the other is an auto-immunological response.
791347
1. Methyl iodide: would involve a primary carbocation, which do not exist. Methyl iodide only reacts by an SN2 mechanism.\n2. 2-iodopropane: secondary carbocation\n3. 2-chloro-2-methylpropane--i.e. t-butyl chloride: tertiary carbocation\n4. 2-iodo-2-methylpropane--i.e. t-butyl iodide: tertiary carbocation. Iodide is a better leaving group than chloride, so this substrate is better for SN1.
898657
Answer is d..\n\nThe reaction is as follows:\n\nR-O- + R-X → R-O-R + X- \n\nwhere X is an halogen and R should not be cyclic.
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
858619
The easiest way to approach problems like these, my friend, is to look at your alkyl halide first and then look at your nucleophile or base.\n\nT-butyl bromide is a tertiary halide.  Tertiary halides undergo E1 and SN1 reactions because of their ability to form a stable carbocation intermediate.  If you recall, tertiary carbocations are more stable than primary and secondary.\n\nYour -OCH3 is a strong base known as methoxide, the conjugate base of methanol.  Strong bases favor elimination reactions and thus the elimination product is your answer.
760435
You are looking at a displacement reaction, in which the more reactive metal (Cu) is displacing the less reactive metal (Ag) from its salt solution (AgNO3).\nCu (s) + 2AgNO3 (aq) --> 2Ag (s) + Cu(NO3)2 (aq)\n\nHope this helps. :)
946990
FeCl3 + NaCl -> FeCl2 + NaCl3\n\n\nFeCl3 + NaCl -> FeCl2 + NaCl3\n\nthese reactions are not possible \nas this reaction can take place if iron(fe) was more reactive than sodium or fe was less electropositive than na\nhence \nthese reactions are not possible
136146
The 1st answer is wrong. The reaction is\n\nNa2CO3(aq) + SrCl2 (aq)--> 2NaCl (aq) +SrCO3 (s)
518533
first one: no, your missing an Mn on the right.\nSecond one: yes\nthird one: no, you're missing 5 ammonia on the left\n\nthe rest as standard metathesis reactions, use the ones u've done as a guide