I am working on chitosan acetate,how I want to know the amount of acetic acid left in the solution ?
139445
The acetate is working as a counter ion to the amine on the chitosan, you could exchange it with chloride by dissolving the chitosan in NaCl, precipitation with diethylether/ethanol three (or more) times. The chitosan solubility will depend on the molecular weight, degree of deacetylation etc. You could try dialysing the chitosan also. The method used will depend on the subsequent application of the chitosan.
135532
You can either look it up or determine it experimentally.\nI cant get much more specific without additional details.
125497
sciencelab.com has just about anything you'll need!\nhttp://www.sciencelab.com/search.php?q=acetic+acid+-acemetacin+-acetamide+-benzyl
822325
a) Find some Acetic Acid (CH3COOH)\nb) Find some Copper (II) Hydroxide (Cu(OH)2)\nc) Add to form Cu(CH3CO2)2 + 2H2O\n\nWithout the volume or the number of moles, preparation of a 0.530M can be done several ways. \n\n2CH3COOH + Cu(OH)2 --> Cu(CH3CO2)2 + 2H2O\n\nUsing 1L:\n0.530 mol of Cu(CH3CO2)2 means 1.060 mol of CH3COOH\n\n1.060 / 0.5L = 2.120M\n\n0.530 mol of Cu(CH3CO2)2 means 0.530 mol of Cu(OH)2\n\n0.530 / 0.5L = 1.060M\n\nSo mix 500mL of 2.120M Acetic Acid with 500mL of 1.060M Copper (II) Hydroxide.\n\nN.B. Being a weak acid, acetic acid will not fully react with copper hydroxide, and hence the actual molarity requiredmay be higher.
740045
http://www.wellesley.edu/Chemistry/chem120/titrate1a.xls\n\n\nstep by step
326138
the only way to determine the rate of a reaction is to do the experiment yourself.\n\nUse starch as the indicator and plenty of acid.
598258
http://en.wikipedia.org/wiki/Acetic_acid\ndistilled vinegar is approximately 4% acetic acid 96% water\n\nSoap is not a given formula it depends which "greasy guys" you saponify.\nhttp://chemistry.about.com/library/weekly/blsapon.htm\n\nOranges are very complicated, you might want to start with gººgling citric acid.\n\nAmmonia is NH3\nhttp://en.wikipedia.org/wiki/Ammonia\nHousehold ammonia is dissolved in water as ammonia is a gas at standard temperature and pressure.
724819
To determine the concentration of a reagent.
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
479476
You need to know concentrations to determine the amount of substance that precipitates out by using solubility products. In this case the precipitate will be CaSO4, so the solubility product will be given by:\n\nk=[Ca(+2)]*[SO4(-2)]\n\nWhere k is the equilibrium constant (probably a given), and [Ca(+2)] & [SO4(-2)] are the concentrations of Ca(+2) and SO4(-2) in the solution. Any amounts greater than the equilbrium concentration will precipitate out as a solid.
460090
Vinegar: First of all, the kind of vinegar you get at the grocery store is about 95% water, but the vinegary part is acetic acid (also called ethanoic acid or methane carboxylic acid). The formula is written: CH3COOH.\n\ncoke: do u mean cocaine its C17H21NO4\n\nand i forgor about the nail polish remover its like this:\n\n"In chemistry, acetone (also known as propanone, dimethyl ketone, 2-propanone, propan-2-one and β-ketopropane) is the simplest representative of the ketones.\n\nAcetone is a colorless mobile flammable liquid with melting point at -95.4 °C and boiling point at 56.53 °C. It has a relative density of 0.819 (at 0 °C). It is readily soluble in water, ethanol, ether, etc., and itself serves as an important solvent. The most familiar household use of acetone is as the active ingredient in nail polish remover. Acetone is also used to make plastic, fibers, drugs, and other chemicals."\n\n\n\nis it good?!
955333
an easy example you can remember is H2O(water).\nif u want a more difficult one, u can use CH3COOH(acetic acid)!\n\nALL DE BEST!
143234
I think you at least need to know the density of pure acetic acid to solve your problem. We can assume density of water at 1.0 g/mL.\nIf we assume that the density of your 40% w/v solution is 1.0 g/mL then:\n\n    V = volume of 40% solution required would be: \n\n        V x .40  = 0.0225 x 250 \n\n  V = 14.1 mL \n\nYou place the 14.1 mL in a flask and add water to complete 250 mL and voila... You are done. \n\nLet me know if this is wrong.
423892
I'm sure that's what you're supposed to determine in the lab you're working on. Good luck.\n\nI've been able to find a value for the heat capacity of n-hexadecanol (cetyl alcohol) in my copy of _Lange's Handbook of Chemistry, 13th Edition_.  When you've come up with your value from the experiment, post it here and I'll check to see if you're close.
836016
Here is a good guide\n\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Limiting-Reagent.html
932507
Depends on the type of vinegar. But the actual chemical is called acetic acid and it's Boiling point is 118.1°C (391.2 4K) (244.6°F)
802465
carboxylic acid, as represented by -(C=O)-OH. For more info, check the source.