R-CHOH react with pcc product will form is?
146289
Primary and secondary alcohols react with PCC ('Pyridinium chloride) to form an alhehyde (primary) or a ketone (secondary). The reaction is a softened form of chromium oxidation chemistry. The reaction was discovered in the group of the famous chemist E.J. Corey. The disadvantage is that PCC is toxic and green alternatives to this reaction are definately desired. (C&E news recently reported a non-Tin Barton reduction of an alcohol to an alkyl group, but I digress)
726370
There should be no reaction.  Provided there is enough water to dissolve the CuCl, the Compounds would separate into their Ion counterparts.\n\nSo, youd end up with Cu+, K+, and Cl- in solution
270274
ur equation is wrong....\nu will get H2O + C...\ncheck again\nif it's correct, u have to find out which one is the limiting reagent by calculating the number of moles of each reactant used...thus figuring out the number of moles of product and multiplying with the molecular mass of methanol...
900859
If the metal not in the compound is higher on the activity series than the metal in the compound then a reaction will occur and the metals will switch spots. If the metal not in the compound isn't higher on the activity series than the metal in the compound then no reaction will take place. You got to be joking if you think I'm going to be doing 20 problems for you. The only way to learn these is to do them on your own. Just so you know in #6 a reaction will not occur.\n\nHere is an activity series:\nhttp://en.wikipedia.org/wiki/Reactivity_series\n\nREACH doesn't know how to do these, don't listen to his answer.
309813
ClOC-(CH2)10-COCl + H2N-CH2CH2CH2CH2CH2CH2-NH2 -------->\nadipoyl chloride         +    hexamethylene diamine\nH2N-(CH2)6-NH(O)C-(CH2)10-CO-NH-(CH2)6-NH(O)C-(CH2)...etc\n\nDifficult to do without drawing, but adipoyl chloride is the acid chloride shown, and hexamethylene diamine is the diamine drawn, the second molecule.  \n\nThe amine has two nucleophilic centres, and the acid chloride, two elextrophilic centres.  These react together to form a long polymeric chain of many repeating units.  There are many forms of Nylon, which are made in the same way using molecules of different chain lengths.  FOr example the first form of Nylon was Nylon 6,6 - made from the same amine as yours, but from an acid chloride containing 4 less methylene (CH2) spacers.  The bonds joining the two molecules together are AMIDE or PEPTIDE bonds.  These are the same bonds which join amino acids together to make proteins, which are formed from DNA.\n\nI hope this helps.
791597
You will get Silver Chromate as the precipitate.  It is red in color.\n2AgNO3(aqueous) + K2CrO4(aqueous) -> Ag2CrO4(solid) + 2KNO3(aqueous)
1131203
G.W. Jr. Shipps, U.P. Spitz, J. Jr. Rebek, Solution-Phase Generation of Tetraurea Libraries, Bioorganic & Medicinal Chemistry \nfound at the site... \nhttp://www1.elsevier.com/cdweb/journals/09680896/viewer.htt?viewtype=keywords&rangeselected=32\n\nlooks like fun stuff :) don't inhale
948592
Kc=2 = conc products / conc reactants\n\nKc = 2 = [CO][H2O]/[CO2][H2]\n\nStarting concs\n[CO] and [H2O] = 0\n\n[CO2] and [H2] = 0.7 moles/10L = 0.07M\n\nEquilibrium concs\n\nFrom the balanced equation, if x CO2 reacts, it reacts with x H2 to form x H2O and x CO\n\n[CO] and [H2O] = x\n\n[CO2] and [H2] = 0.7 moles/10L = 0.07M-x\n\n\nKc=2=(x)(x) / (0.07-x)(0.07-x)\n\nsolve for x and plug back into the final concentration equations.
731401
Correct.\nD. when delta G is positive the reaction is non-spontaneous
865719
its not that the co bonds with the heamoglobin it forms a complex i.e it replaces the nitrogen atoms in heamoglobin to form the complex...CARBOXY HEAMOGLOBIN.. so we dont have the case of ...C reacting..
502725
That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
664298
molecular: \nHCl + K2CrO4 + SnCl2 --> CrCl3 + SnCl4 + KCl + H2O\n\nnet ionic: H+ + CrO42- + Sn2+ --> Cr3+ + H2O + Sn4+\n\nreduction half reaction: \n3 e- + 3 H+ + 5 HCl + K2CrO4 --> CrCl3 + 2 KCl + 4 H2O\noxidation half reaction:\n2 HCl + SnCl2 --> SnCl4 + 2 H+ + 2 e-\n\nBalancing them:\n6 e- + 6 H+ + 10 HCl + 2 K2CrO4 --> 2 CrCl3 + 4 KCl + 8 H2O\n6 HCl + 3 SnCl2 --> 3 SnCl4 + 6 H+ + 6 e-\n-----------------------------------------------------------\n6 H+ + 16 HCl + 2 K2CrO4 + 3 SnCl2 --> 2 CrCl3 + 4 KCl + 8 H2O + 3 SnCl4
15061
K2Cr2O7 + 2Pb(C2H3O2)2 + H2O --> 2PbCrO4 + 2HC2H3O2 + 2KC2H3O2
888737
How about the reaction slugs have to salt?
1165218
you appear to be confused, when a grignard reagent reacts with a carbonyl carbon (such as in RCOOH) the R of RMrBr (the most common grig) is bonded to the carbon. however when the electrons move up the oxygen to form RCO- OH, the remaining MrBr+ ion interacts with this O- forming a salt, i believe this reacts further, eliminating the OH. hope this was clear enough.
125470
Hi pcc,\nNever heard that one before but I'd say not. Good Luck: Blueladybug
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