How Do You Formulate Potassium Laurate?
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CH3(CH2)10CO2-K+\n\nCH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CO2-K+\n\nIt is the potassium salt of dodecanoic acid.\n\n"Dodeca" means 12 in greek, showing there are 12 carbons in the chain.  The "ate" part at the end indicates it is a carboxylate salt or carboxylate ester, but because of the potassium you know you have a salt.
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Well, I do not know that syllabus.  You will need to sift useful answers.\n\nI would use:\n\ndifferent solubility in solvents\ndifferent boiling points\nrecrystallise one out of a mixed solution\n\nYou might be at too junior a level to know about ion exchange resins, and other chromatography techniques.
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KCl(s)  ---->  K+(aq)  +  Cl-(aq)
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use the equation  pH= -lgKa+ lg( salt/acid)\n(this equation only applies for buffer system)
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This compund has an octahedral geometry and is coordinated with 3 bidentate oxalate (C2O4) ligands.\nReaction of potassium oxalate hydrate and ferric chloride\n\n3K2C2O4ˇH2O + FeCl3 --> K3[Fe(C2O4)3]ˇ3H2O + 3KCl\n\nWater molekules do not act as ligands and potassium ions act as counter-ions
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Potassium Hydroxide and Hydrochloric acid.\n\nKOH + HCl = KCl + H2O
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The answers are KCl, CaF2, and Na2O. Potassium has one electron on the outer ring while chlorine has seven. To make chlorine's ring "full" potassium gives the one electron to chlorine. They are "happy" now. :) That's what my science teacher told me. He's weird.
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Aluminium is Al3+\n Chlorine is CL1-\n Potassium is K1+ \n\nHahaz, I think should be correct. Hahaz, I did your homework.
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1.  Oxidise toluene to benzoic acid using strong oxidising agents.  KMnO4 (aq) is strong enough to do this.\n\n2.  Convert you benzoic acid to benzoyl chloride, using (COCl)2 - oxalyl chloride/DMF(cat.) or SOCl2 or Ghosezs reagent or PPh3/CCl4 or PPh3/NCS etc - the former two being by far the most common methods.\n\n3.  Nitrate your malonic ester in the 2 position using isoamyl nitrate in the presence of a base, ie NaH or NaOMe.\n\n4.   Reduce your nitro to an amine (H2-Pd/C)\n\n5.  Acylate your amine with your prepared benazoyl chloride (DIPEA as a base, DCM as a solvent for example)\n\n6.  Saponify both your esters with 1M NaOH.\n\n7.  Acidify your sodium carboxylates to the free carboxylic acids and heat for a couple of hours to promote de-carboxylation.  This is a standard procedure so will work well\n\n8.  Isolate your prepared hippuric acid!\n\nThere will be many other methods - especially to prepare your methyl 2-aminomalonate.  You don;t need to form the acid chloride also.  You could just perform an amidation between your amine and benzoic acid in the presence of a coupling agent such as DIC/HOBt, CDI, HATU, PyBrOP etc etc - there are too many to mention.\n\nYou could also brominate your methyl malonate with Br2, to give methyl 2-bromomalonate.  Treat your benzoyl chloride with ammonia gas to give benzamide.  Then treating your benzamide with NaH to de-protonate your amide, and then treat with your methyl 2-bromomalonate will generate the same product as described in step 5 above.  Following the rest will give you your product.\n\nThere are more methods but hopefully this has given you some useful information.
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Well use a series of steps.\n\n1st step add Calcium Carbonate. This will fizz in the case of the vinegar, as acid plus carbonate gives salt, water and carbon dioxide. U can even test the gas to be CO2, as it turns lime water milky. U will eliminate the vinegar from further tests.\n\n2nd is separate the alcohol from the salt and fresh water. Do this by testing with Cobalt chloride paper. The paper changes from blue to pink if any water is present. The sample which doesn't change the paper would be the alcohol.\n\n3rd is to separate the salt water from the fresh water. Do this by adding silver nitrate followed by aqueous ammonia. Salt contains sodium chloride, so the presence of the Cl- ion will cause the formation of a white precipitate that dissolves in excess alkali.\nSo, the one that produces no ppt is therefore pure water.
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Nutrient broth ingredients:\n\nMeat extract 10.0 gm\nPeptone 10.0 gm\nSodium chloride 5.0 gm\nDistilled water 1000 ml\nMix the ingredients and dissolve them by heating in a steamer. When cool, adjust the pH to 7.5-7.6.\n\nTo make it agar add:\nTo the ingredients as in nutrient broth, add 15 gm agar per litre. Dissolve the agar in nutrient broth and sterilize by autoclaving at 121oC for 15 minutes. Prepare plates and slopes as required.
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Add 50mL of acetic acid to 50mL sodium acetate. The pH is the pKa, unless unequal amounts of reactants are used.
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Cl2 + 2KI --> 2KCl + I2\n\nsingle replacement
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Synonyms: Alum; Aluminum potassium sulfate, dodecahydrate; Kalinite; Potassium alum; Sulfuric acid, aluminum potassium salt\n\nSynonyms: Sulfuric acid, aluminum salt (3:2), octadeca hydrate; Cake alum; Patent alum \nCAS No.: 10043-01-3 (Anhydrous) 7784-31-8 (Octadecahydrate) \nMolecular Weight: 666.44 \nChemical Formula: Al2(SO4)3.18H2O \nProduct Codes: \nJ.T. Baker: 0564 \nMallinckrodt
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SUGAR-FREE “SPORTS DRINK”\n\n1 packet or container sugar-free drink mix (to make 2 qts)\n1/4 tsp table salt\n4 potassium tablets (99 mg each)\n2 quarts water\n\nTake about 1/4 cup of the water and put in a cup with the potassium tablets. Heat in the microwave until the water gets close to boiling. Stir to dissolve the tablets. Add the salt and stir. \n\nPut the drink mix in a pitcher, add water, and stir until the drink mix is dissolved. Add the salt/potassium solution and mix thoroughly. Chill.\n\nPer 8-oz serving: 5 cal, about 75 mg sodium, 50 mg potassium. ZERO POINTS.\n\nIf you’d like less sodium, use a 1/8-tsp measure, and do one level measure plus 1/2 of another. This will cut the sodium to about 55 mg/serving instead of 75. (FYI, 1/4 tsp of salt contains 590 mg sodium.)\n\nThis drink tastes just fine… it has the characteristic slight saltiness that one associates with the commercial sports drinks, but it’s not objectionable at all.
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Cl2 + 2 KI → I2 + 2 KCl
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KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)