Also neutralization of formic acid by NaOH
206555
Aluminum Nitrate, Al(NO3)3, since, in a neutralization you take off any acidic hydrogen's, and any hydroxyls, and you combine the remaining ions in a way that the product molecule has no net charge. So\n Al(OH)3=Al(+3) + 3OH(-) \n HNO3 = H(+) + NO3(-)\nThe H(+) ions and OH(-) ions combine to form water, and each Al(+3) ion combines with 3 NO3(-) ions to produce Al(NO3)3 \n\nFormic acid (CHOOH) and sodium hydroxide(NaOH) produce Sodium formate, CHOONa, for the same reason as above.\n\nNote: The other H in Formic Acid is non-acidic, and so is not replaced by Na
493280
I have a serious feeling that you haven't given us all the info.\nIf you know the density of HCl you can get the first one.\nBut we don't have enough info to even begin to tell you the relevance of the NaOH.   More info, or concentrate on your own homework...
67349
Actually, that is a somewhat simplified reaction,\nin reality, the water and Carbon Dioxide dont form directly, they form after a secondary reaction (Carbonic acid decomposing).\n\n3Na2CO3 + 2H3PO4 --> 2Na3PO4 + 3H2CO3\nH2CO3 --> H2O + CO2\n\nSo combining them,\n3Na2CO3 + 2H3PO4 --> 2Na3PO4 + 3H2O + 3CO2
281912
the proteins are too weak to be titrated directly with alkali,if formalin is added, it reacts with the -NH2 groups to form the methylene-amino(-N=CH2) group and the carboxyl group is then available for titration.\n\nHOOC.CHR.NH2 + HCHO ----> HOOC.CHR.N=CH2 +H2O\n\nHOOC.CHR.NH2(neutral) HCHO(formalin)  HOOC.CHR.N=CH2(acidic)
75283
HCL acid  is acting as a substitution,neutralization or a gas formation reaction on each ion in the compound:\n\n    given reactants :\n\nNaAl(OH)2CO3(s) and HCl(aq)\n\nreactions:\n \n1. Na+(aq) + HCl(aq)->NaCl(aq) + 2H+(aq)\n\n2. Al3+(aq) + 3HCl(aq)->AlCl3(aq) +  2H+(aq) \n\n3. 2OH-(aq) + 2H+(aq)->2H2O(l)\n\n4. CO3-2(aq) + 2H+(aq)->CO2(g) + H2O(l)\n\nadding equations 1 thru 4  and canceling on both sides \nwill give you the overall equation,\n\nNaAl(OH)2CO3(s) + 4HCl(aq)-> NaCl(aq) + AlCl3(aq) + 3H2O(l) + CO2(g)
387581
a. by the proccess of titration.  few drops of universal indicator are poured onto the base. acid is poured  slowly with the help of the burette. when the base is decolourised this indicates that it has been neutralised.\n\nb. vinegar\nc. all of the above\nd. 7
401787
HCL and NaOH are 'strong' acid and base. When they react, the product is neutral. NaOH + HCL --> NaCl + H2O\n\nWhat you need to find is which is the limiting reactant, the other one (what is left over) will effect the pH by its concentration in the otherwise neutral solution.\n\nFor your first problem:\n\n0.0557M HCL = 0.0557mols/liter*0.00542L = 3.02*10^(-4)mol HCL\n\n0.0377M NaOH = 0.0377mols/liter*0.00337L = 1.27*10^(-4)mol NaOH\n\n3.02*10^(-4)mol HCL - 1.27*10^(-4)mol NaOH = 1.75*10^(-4)mols HCL\n\n0.00542L + 0.00337L = .00879L\n\n[(1.75*10^(-4)mols HCL) * (1 L) ] / (.00879L) = .020M HCL\n\npH = -log(.02) = 1.70
515349
acid\ngas naphtha\noxygen  and hydrogen\nacid\nacid
401299
17.13 mL = 0.01713 L\n\n0.01713 L x 0.3150 M = 0.005396 mol\n\nNaOH is a strong base, HCl is a strong acid.\nEach NaOH releases one hydroxide anion.\nEach HCl releases one proton.\n\nTherefore it takes 0.005396 mol of H+ to neutralize 0.005396 mol of -OH.\n\n0.005396 mol / 0.025 L = 0.2158 M\n\nThe concentration of the NaOH solution is 0.2158 M.
452593
Lab: acid and base\nHome: putting ammonia(found in bee sticks) on a bee bite to neutralize the formic acid\nCooking:usng an acid in baking with baking soda
43907
cinnamic acid is reduced to the aldehyde and can be further reduced to the alcohol.\nsodium borohydride is caustic b/c the hydride ion(H-1) reacts with water to produce hydrogen gas and the hydroxide ion.  it acts as a strong base.
415545
Not impossible but Hco3- is considered a very weak acid with resilience towards losing another Hydrogen Ion that's a good explanation for Hydrogen Carbonate formation instead of Carbonate
510285
The strong base NaOH will be fully dissociated in water, and you're adding quite a bit (looks like 5 grams/10 mLs).  The water actually consists mostly of hydroxide anions at this point, and hypochlorite ions (your oxidiser) from the bleach.  The pH is 14 or so.  These are the only conditions you'll find Fe(OH)3 at.\n\nIt Ferric hydoxide is  instable due to its loose bonding (easy dissociation) with the hydroxide ions.
353554
NaOH + NaH2PO4 ===> Na2HPO4 + H2O\n\nH2S + 2KOH ===> K2S + 2H2O\n\n(NH4)2SO4 + 2KOH ===> 2NH3 + K2SO4 + 2H2O\n\nSO2 + Ca(OH)2 ===> CaSO3 + H2O\n\nIn the above equation, further oxidation in air could occur by\n\n2CaSO3 + O2 ===> 2CaSO4
598813
fairly easy, not 100% ionized but increases the concentration of H+, obviously a weak acid ^^
229327
It's the third one.  Formic is stronger than benzoic, which is stronger than acetic, which is stronger than phenol.  The lower the pKa, the higher ionization and the more H+ produced.
304069
The neutralization reaction:\n2NaOH + H2SO4 ---> 2H20 + Na2SO4\nTwo moles of NaOH are required for each mole of H2SO4. \n\nMultiply the molarity and volume (in liters) of the NaOH to find the original number of moles: \n.0346 L * .200 M = .00692 moles NaOH\n\nDivide moles of NaOH by two to get the number of moles of H2SO4:\n.00692 moles NaOH /2=.00346 moles H2SO4\n\nTo find the concentration of sulfuric acid, divide the moles by the volume in liters: \n.00346 moles H2SO4 / .0500 L = .0692 M