what is the formula for acetic acid?
597378
Acetic acid is ethanoic acid, not methanoic.  It is methane carboxylic acid, CH3COOH.  You account both carbons in the name.\n\nhttp://www.jtbaker.com/msds/englishhtml/a0326.htm
633503
Salicylic Acid -> C6 H4 (OH) (COOH)\n\nFor more information, click on the following link.\n\nhttp://www.answers.com/salicylic%20acid\n\nHope it proved fruitful to your question c",)
1070616
mix concetrated sulphuric acid to the sodium acetate, the reaction will produce acetic acid and sodium sulphate.\n\nif you take all the water out of the sodium acetate, you will end up with 100% pure acetic acid.\n\nH2SO4 + CH2COONa => CH3COOH + Na2SO4\n(unbalanced equation)
799608
Vinegar is acetic acid,  CH3COOH\n\nCH3COOH + NaHCO3 ---> CH3COONa + H2CO3 (carbonic acid)\n\nThen H2CO3  gives CO2 + H2O  and CO2 will extinguish flame...Many fire extinguishers are made with compressed CO2.
512442
Vinegar is a solution of acetic acid (CH3COOH) in water. The best vinegar is produced from wine by acetic fermentation. And vinegar is acid with a pH aproximatively 4.5 - 5
1082012
Glacial acetic acid is a trivial name for water-free acetic acid. Similar to the German name Eisessig (literally, ice-vinegar), the name comes from the ice-like crystals that form slightly below room temperature at 16.7 °C (about 62 °F).\n\nThe most common and official abbreviation for acetic acid is AcOH or HOAc where Ac stands for the acetyl group CH3−C(=O)−. In the context of acid-base reactions the abbreviation HAc is often used where Ac instead stands for the acetate anion (CH3COO−), although this use is regarded by many as misleading. In either case, the Ac is not to be confused with the abbreviation for the chemical element actinium.\n\nAcetic acid has the empirical formula C2H4O2. This is often written as CH3-COOH, CH3COOH, or CH3CO2H to better reflect its structure. The ion resulting from loss of H+ from acetic acid is the acetate anion. The name acetate can also refer to a salt containing this anion or an ester of acetic acid.
854131
molecular wt of acetone:  58 g/mol\n\nmass of acetone in bottle:  vol x density = 314 x .788 = 247.4 g\n\nmol of acetone = mass/mol wt = 247.4/58 = 4.27 mol\n\nthere is 6.02 x 10^23 molecules in a mole\n\nso 6.02 x 10^23 x 4.27 = 2.6 x 10^24 molecules
890894
C12H22O11+12O2-------12CO2+11H2O+E
145111
The reaction formula is\n\nCH2COOH                           \nCH(OH)COOH (aq) + NaHCO3 (aq)----> \nCH2COOH                            \n\nCitric Acid\n\n       CH2COONa\n CH(OH)COONa    (aq)+ H20 (l) + CO2(g)\n       CH2COONa
1026749
vinegar : acetic acid CH3COOH\nwashing soda : NaHCO3\n\nCH3COOH + NaHCO3 -> CH3COONa + CO2 +H2O\n\nthere will be a solid (the acetic salt) and water, carbon dioxyde will have escaped if it is not in a closed container
323054
You sure did pick a really long question for explaining.  And I'll have to argue that all of the pH's do not have correct sig figs but the numbers are right.\n\n(Sorry about the numbers in the tables but this place doesn't seem to allow better tables)\n\nA) 25.0mL of HAc(acetic acid) and 10.0mL of NaOH\n\n*This is a reaction so we have to use moles*\n\nHAc : 25.0mL * (0.100mol/L) * (1L/1000mL)= 0.00250 moles\n\nNaOH : 10.0mL * (0.125mol/L) * (1L/1000mL) = 0.00125 moles\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00125      0.00250        0\nChange:   -0.00125     -0.00125      +0.00125 \nEquilib:      0         0.00125       0.00125\n\nHAc and Ac-(NaAc breaks apart) for an equilibrium\n*Use Molarity when no reaction is taking place (best shown by when no volume is changing)*\n\nBoth HAc and Ac- have same molarity:\n\n(0.00125 moles / 35mL ) * (1000mL/1L) = 0.0357M HAc and Ac-\n\nUsing the equilibrium equation:\n\nEquation:  HAc   <--> Ac-     +  H+         Ka = 1.8 * 10^-5\nInitial:   0.0357M     0.0357M  \nChange:    -x          +x         x\nEquilib:   0.0357-x    0.0357+x    x\n\nKa=1.8*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]\nKa=1.8*10^-5 = [0.00357][x]/[0.0357] usually good approx.\nKa = 1.8 * 10-5 = x = [H+]\n\npH = -log [H+] = -log [1.8*10^-5] = 4.745 (3 sig figs as only after the decimal counts for logs)\n\nB) 25.0mL HAc and 20.0mL NaOH\n\n*See A for HAc moles and double the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00250      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:      0            0          0.00250\n\nCalculate molarity of Ac-:\n\n(0.0250moles /45mL) * (1000mL/1L) = 0.0556M Ac-\n\nUsing the other equilibrium equation (Kb):\n\nEquation:  Ac- + H20   <--> HAc     +  OH-         Initial:   0.0556M   -        0\nChange:    -x        -        +x        + x\nEquilib:   0.0556-x  -       +x         + x\n\nThis gives us the Kb equation\nKb = ([HAc][OH-])/[Ac-] = with approx ([x][x])/[0.0556]\nso Kb = x^2 /0.0556\n\n**but we need to calculate Kb\n**Kw = 1*10^-14 at STP I'll assume this for this problem\n**Kw = KaKb so Kb = Kw/Ka\n**Kb = (1*10^-14)/(1.8*10^-5)= 5.56*10-10 \n\nso x = 5.56 * 10^-6 = [OH-]\n\npOH = -log [OH-] = 5.253\nPH = 14 - pOH (at STP)= 8.747\n\nC)25.0mL HAc and 30.0mL NaOH\n\n*See A for HAc moles and triple the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00375      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:   0.00125            0       0.00250\n\nAfter the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.\n\nCalculate NaOH molarity\n(0.00125moles/55.0mL)*(1000mL/1L)= 0.0227M NaOH\n\nStrong Bases dissociate 100% so [OH-] = 0.0227M\n\npOH= -log[0.0227] = 1.644\npH = 14-pOH = 12.356\n\nThis should show how they calculted the half-neutralization point (A), the neutralization point (B) and after the neutralization point (C).\n\nIf you have questions on this, feel free to send me a message.  I taught this a couple weeks ago so I'm still pretty fresh on the subject.
551403
the acids are carboxalic acid and  carbonic acid
559772
vinegar is a solution of acetic acid (9% in water). Acetic acid (CH3COOH) is a weak acid that disolve rust (FeOOH according ones or Fe3O4 according another ones)\n\nSo, a soluble salt is obtained: (CH3COO)3Fe
974122
White vinegar - mild acid.
1104461
Synonyms: Acetic acid ethyl ester; Acetic ether; Acetoxyethane; Ethyl Acetic Ester; Ethyl ethanoate \nCAS No.: 141-78-6 Molecular Weight: 88 Chemical Formula: CH3COOC2H5 \n\nPhysical and Chemical Properties\nAppearance: Clear liquid. \nOdor: Fruity odor. \nSolubility: 1 ml/10ml water @ 25C \nSpecific Gravity: 0.902 @ 20C/4C \npH: No information found. \n% Volatiles by volume @ 21C (70F): 100 \nBoiling Point: 77C (171F) \nMelting Point: -83C (-117F) \nVapor Density (Air=1): 3.0 \nVapor Pressure (mm Hg): 76 @ 20C (68F) \nEvaporation Rate (BuAc=1): 6 \nhttp://acad.udallas.edu/chemdept/hendrickson/MSDS/ethyl_acetate_msd.htm
28707
C20 H30 O is the chemical formula for retinol (more commonly known as Vitamin A). It's got 20 carbon atoms, 30 hydrogen atoms, and one oxygen atom in one molecule.
314832
First of all carboxylic acid  is a group.\n\nA carboxylic acid is an organic molecule with a -(C=O)-OH group. The group is also written as -COOH and is called a carboxyl group. The hydrogen on the -COOH group ionizes in water; carboxylic acids are weak acids. The simplest carboxylic acids are formic acid (H-COOH) and acetic acid (CH3-COOH).