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  • What is the product of the following reaction? 1)CH_3 OH - Socratic
    These are ostensibly acid-base reactions For ammonium we could write NH_4^+ +HO^(-) rarr NH_3(aq) + H_2O(l) For methanol, the acid base reaction would proceed
  • Is it necessary to have oh group of phenol valways at ortho . . . - Socratic
    No, the "OH" group in phenol is by definition at "C1" > The structure of phenol is You can draw the structure like this: phenol2 However, the structure is not meta-phenol It is phenol because the ring carbon attached to the "OH" group is now "C1" The "OH" carbon can be on any carbon atom of the ring, and the compound is still phenol,
  • Can you give the IUPAC name for the following (CH_3)_3C-OH . . . - Socratic
    So this is a propanol derivative: "2-methylpropan-2-ol" For "isopropyl alcohol", H_3C-CH (OH)CH_3, the longest chain is again three carbons long, and C2 is substituted by -OH, so "propan-2-ol" I think this is right, and I haven't broken any arcane rule Both names seem to be unambiguous
  • Question #54b11 - Socratic
    The balanced equation is "CH"_3"CH"_2"COO"^"-" + "3S" + 4"OH"^"-" → "CH"_3"COO"^"-" + "HCO"_3^"-" + "3HS"^"-" + "H"_2"O" > The reaction is carried out at pH 8, so we must balance in basic solution
  • Question #1a680 - Socratic
    Now, notice that the oxidation state of nitrogen goes from #color (blue) (+5)# on the reactants' side to #color (blue) (-3)# on the products' side, which basically means that every atom of nitrogen gains #8# electrons
  • Question #477c5 - Socratic
    On the product side the Carbonic Acid (#H_2CO_3#) is the Conjugate Acid as it is the hydrogen donor to the Conjugate Base (#OH^-#) as it receives the hydrogen ion
  • Question #5392e - Socratic
    Here's what I get Step 1 Write the molecular equation "Fe" ("NO"_3)_3" (aq)"+ "3NH"_3" (aq)" + "3H"_2"O (l)" → "Fe (OH)"_3" (s)" + "3NH"_4"NO"_3" (aq)" Step 2
  • Question #019dd - Socratic
    Only (C) is fully correct As you can see, steam typically reacts with alkenes by adding H on one double-bonded atom and OH on the other But there are two possible choices for which atom gets the single hydrogen atom and which gets the OH group In some cases like (A) there is a perference (the Markovnikov rule), but the only sure way to get 100% of one product is to have the double bond in
















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