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- MTBE, Methyl tert -butyl ether, CH3OC(CH3)3, is used as an oxygen source in oxygenated gasolines. MTBE is manufactured by reacting 2-methy1propeue with methanol. (a) Using Lewis structures, write the Chemical equation representing the reaction. (b) What volume of methanol, density 0.7915 g/mL, is required to produce exactly 1000 kg of MTBE, assuming a 100% yield?Ten mL of concentrated H3PO4 (91.7% by mass, d=1.69g/mL) was accidentally poured into a beaker containing 20.0 g of NaOH. Not all the NaOH was consumed. How many grams of NaOH were left unreacted? The equation for the reaction is H3PO4(aq)+3OH(aq)3H2O+PO43(aq)Gold metal will dissolve only in aqua regia, a mixture of concentrated hydrochloric acid and concentrated nitric acid in a 3:1 volume ratio. The products of the reaction between gold and the concentrated acids are AuCl4-(aq), NO(g), and H2O. The equation for this reaction where HNO3 and HCl are strong acids is Au(s)+4Cl(aq)+4H+(aq)+NO3(aq)AuCl4(aq)+NO(g)+2H2O(a) What stoichiometric ratio of hydrochloric acid to nitric acid should be used? (b) What volumes of 12 M HCl and 16 M are HNO3 required to furnish the Cl- and NO3- ions to react with 25.0 g of gold?
- Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) A. What is the %m/m of the solution? B. What is the molarity of the solution? C. What is the mole fraction of the solvent?Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) 1. What is the % m/m of the solution? 2. What is the mole fraction of the solvent? 3. What is the molarity of the solution? Based from the calculated molarity of the solution in #3, you are to prepare 250 mL of the said molar concentration. If you have an available 2M stock solution of Palmitic acid, How many mL of the 2M palmitic acid stock solution will you need? and; 4. How many mL of solvent do you need to add to the amount gotten from the 2M Palmitic acid stock solution in order to prepare 250 mL total volume of your desired solution?A formula for compounding state qs ad 250 ml. This expression means
- Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) You are to prepare 250 mL of the said molar concentration. If you have an available 2M stock solution of Palmitic acid. How many mL of solvent do you need to add to the amount gotten from the 2M Palmitic acid stock solution in order to prepare 250 mL total volume of your desired solution?Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) You are to prepare 250 mL of the said molar concentration. If you have an available 2M stock solution of Palmitic acid. How many mL of the 2M palmitic acid stock solution will you need?give clear handwritten answer -What is the molarity of nitrate (NO2) in a solution that is 5.42 w/w % iron nitrate (Fe(NO3)3) and has a density of 1.059 g/mL?
- Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) What is the %m/m of the solution? Final answer must be rounded off to 1 decimal place, and shall NOT have any unit.another question The aluminum in a 1.200g sample of impure ammonium aluminum sulfate was precipitatedwith aqueous ammonia as the hydrous Al2O3 · XH2O. The precipitate was filtered and ignitedat 100°C to give anhydrous Al2O3, which weighed 0.2001 g. Express the result of this analysisin terms of % Al.a student prepares sodium acetate by allowing 2.44ml of glacial acetic acid (17.4M) TO REACT W/ 2.5 Grams of solid NaOH. Calculate the theoretical yield in grams of sodium acetate.