The acidity of a wine sample is determined to be 13.08 g tartaric acid/L. What is...
The acidity of a wine sample is determined to be 13.08 g tartaric acid/L. What is the mass of tartaric acid needed to prepare an equivalent concentration solution in a 50 mL volumetric flask? (MW tartaric acid = 150.085 g/mole.)
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The acidity of a wine sample is determined to be 13.08 g
tartaric acid/L. What is...
A 40.00 mL wine sample is titrated with a 1.04 times 10^-1M solution of sodium hydroxide....
A 40.00 mL wine sample is titrated with a 1.04 times 10^-1M solution of sodium hydroxide. The equivalence point is determined to be 29.267 mL. What is the acidity of the wine expressed as (g tartaric acid/L, MW tartaric acid 3150.085 g/mole)? 5.71 11.4 11.42 3.38 5.710 The acidity of a wine sample is determined to be 14.21 g tartaric acid/L What is the mass of tartaric acid needed to prepare an equivalent concentration solution in a 50 mL volumetric...
The acidity of a wine sample is 1.05×101, expressed as (g citric acid / L wine;...
The acidity of a wine sample is 1.05×101, expressed as (g citric acid / L wine; MW citric acid = 192.124 g/mole) . The sample is titrated with a 0.118 M solution of sodium hydroxide. The equivalence point is determined to be 39.740 mL. What is the volume of the wine sample analyzed in mL?
A 20.00 mL wine sample is titrated with a 1.04×10-1M solution of sodium hydroxide. The equivalence...
A 20.00 mL wine sample is titrated with a 1.04×10-1M solution of sodium hydroxide. The equivalence point is determined to be 25.926 mL. What is the acidity of the wine expressed as (g tartaric acid / L; MW tartaric acid = 150.085 g/mole)?
A solution of tartaric acid is prepared of concentration 1.2174 g L -1 and it is...
A solution of tartaric acid is prepared of concentration 1.2174 g L -1 and it is found that 50.0 ml of the solution requires 21.48 mL of 3.78 × 10-2 mol L -1 sodium hydroxide for complete neutralization. The formula of tartaric acid is C4H6O6 and its relative molecular mass is 150.2. Calculate how many of the hydrogens in a molecule of tartaric acid are ionised in aqueous solution.
The acceptable range of titratable acidity is 0.7-0.9% grams tartaric acid per 100 mL. After titrating...
The acceptable range of titratable acidity is 0.7-0.9% grams tartaric acid per 100 mL. After titrating 5 mL of wine sample with 0.1N NaOH. You determine that 3 mL of NaOH are needed to fully neutralize the acids present in wine. What is your recommendation for the quality of the product?
14. (14 points) Tartaric acid, H2C4H406, (MW =150.087 g/mol) is a diprotic acid that is naturally...
14. (14 points) Tartaric acid, H2C4H406, (MW =150.087 g/mol) is a diprotic acid that is naturally abundant in grapes. In the process of winemaking, a titration is commonly done to determine the tartaric acid content in the grape juice. a. A particular sample of grape juice requires 40.00 mL of 0.2090 M KOH to completely neutralize the tartaric acid. What mass of tartaric acid was present? b. If the initial grape juice sample had a mass of 75.00 g, what...
1. An aqueous solution is 2.65M in tartaric acid (H,C,H,0). The solution's density is 1.016 g/ml....
1. An aqueous solution is 2.65M in tartaric acid (H,C,H,0). The solution's density is 1.016 g/ml. Calculate the solution's: (a) molality (b) mole fraction of tartaric acid (c) percent by mass
A solution is prepared by adding 0.5630 g of powdered milk into a 2.5E+02 mL volumetric flask and adding 30 mL of 1.0 M...
A solution is prepared by adding 0.5630 g of powdered milk into a 2.5E+02 mL volumetric flask and adding 30 mL of 1.0 M KCL and diluting to volume with DI water. The concentration of a milk sample is determined to be 6.420×10-4 M Ca2+ based on linear regression analysis of calcium standard solutions. What is the concentration of Ca2+ in the milk sample (wt/wt %)? [MW Ca = 40.078 g/mole]
13. (a) Calculate the molarity of the standard base solution if a 5.1362 g sample is...
13. (a) Calculate the molarity of the standard base solution if a 5.1362 g sample is used to prepare 100 ml in the volumetric flask. (Molecular Weight – 121.13g). (b) Calculate the Molar concentration of an acid solution, if 20.74 ml were required to titrate a 10.00-ml aliquot of the solution prepared in 13.
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol....
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
A 40.00 mL wine sample is titrated with a 1.04 times 10^-1M solution of sodium hydroxide. The equivalence point is determined to be 29.267 mL. What is the acidity of the wine expressed as (g tartaric acid/L, MW tartaric acid 3150.085 g/mole)? 5.71 11.4 11.42 3.38 5.710 The acidity of a wine sample is determined to be 14.21 g tartaric acid/L What is the mass of tartaric acid needed to prepare an equivalent concentration solution in a 50 mL volumetric...
14. (14 points) Tartaric acid, H2C4H406, (MW =150.087 g/mol) is a diprotic acid that is naturally abundant in grapes. In the process of winemaking, a titration is commonly done to determine the tartaric acid content in the grape juice. a. A particular sample of grape juice requires 40.00 mL of 0.2090 M KOH to completely neutralize the tartaric acid. What mass of tartaric acid was present? b. If the initial grape juice sample had a mass of 75.00 g, what...
1. An aqueous solution is 2.65M in tartaric acid (H,C,H,0). The solution's density is 1.016 g/ml. Calculate the solution's: (a) molality (b) mole fraction of tartaric acid (c) percent by mass
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
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