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The rules of engagement (roe) working group is often used to (select all that apply.)

Salicylic acid is a very important acid. it is used to synthesize the aspirin by treating with acetic anhydride. a 0.2015-g sample of salicylic acid was dissolved in a 100.00-ml volumetric flask, and the solution was diluted to the mark. a 10-ml aliquot of this solution was titrated with standard naoh (0.01130 + 0.2% n) to a phenolphthalein faint pink color end point at 19.81 ml. (a) (calculate the normality of the salicylic acid solution used in the titration. (b) assuming the salicylic acid is pure, what is the equivalent weight of the salicylic acid? practice problems for the final exam (continued) (c) (calculate the inherent error in the determination of the equivalent weight you calculated in part (b). use the following absolute errors in the equipment /glassware when calculating the inherent error. 5.00-ml pipet: + 0.02 ml 100-ml volumetric flask: + 0.08 ml analytical balance: + 0.2 mg 25-ml buret: + 0.03 ml

Ineed someone to see if my answers are correct! if any are wrong let me know what the correct answers would be and how to get that answer! 1. how many moles of sodium chloride are in 28 grams od nacl? a. 265 mole naclb. 856 mole naclc. 479 mole of nacld. 1.2 mole nacl < my choice2. 734 grams of lithium sulfate (li2so4) are dissolved to make 2500 ml of solution what is rhe molaratiy? a. 2.67 mb. 4.56 mc. 3.89 m < my choiced. 1.78 m3. how many grams of cacl2 would be dissolved in 3.0 l of a 0.50 m solution of cacl2? a. 250 g cacl2 b. 166.5 g cacl2c. 113.65 g cacl2d. 98 g cacl2 < my choice4. suppose you had 58.44 g of nacl and you dissolved it in exactly 2.00 liters. the molarity if the solution would be 0.5 mtrue < my choicefalse 5. i would need 22g of naoh to make a 3.0 m solution using 250 ml of solvent.true < my choicefalse6. identify the solute: you have a .0195 m solution made from using 6.5 g of solute and 3 l of solvent. identify the solute by solving for molar weight.a. the solute is nacl because the molar weight is 58.43 g/mol < my choiceb. the solute is h2so4 because the molar weight is 98.06 g/molc. the solute is cacl2 because the molar weight is 111.11 g/mol