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- 500g of Al2S3 dissolved in H2O to make 15.0 L of solution. What is its normality?(MW: Al = 27g / mol S = 32g / mol H=1g/mol, O=16g/mol)A. 0.11 NB. 0.07 NC. 0.22 ND. 0.04 N The concentration of a 1500mL aqueous KCI solution was increased from 25.0 M to 80.0 M by evaporation. How much H₂O was removed? (MW: K= 39g/mol, Cl = 35g/mol, H = 1g / mol 0=16g/mol)A. 1.03 LB. 0.47 LC. 1968.75 mLD. 4.68.75 ml. Suppose that 15M of NaCl solution with a volume of 15L is heated and loses 2000ml of its water content. What would be the final concentration of your NaCl sample? (MW: Na = 23g / mol , Cl = 35q / mol, H=1g/mol,O=16g/mol)A. 3.75 MB. 0.0375 MC. 37.5 MD. 375.0 MCalcium Chloride solution comes as a 10% solution in a 10 ml vial. How many ml of this solution are needed to add 7 mEq to a 500 ml Bag of D5W?(Mw=110) 1 (Answer to the nearest tenth ml)You need to make 100 mL of 1% Atropine sulfate solution and make it isotonic with sodium chloride Atropine sulfate has a MW of 695 and is 3 ions. (E = 0.12) For 100 ml of 1% atropine sulfate we have 1 g of atropine sulfate 1: How much NaCl is the atropine sulfate equivalent to 2 How much Sodium Chloride would be needed to make 100 mL of water isotonic? 3 How much sodium chloride should be added to the 1% atropine sulfate solution to make it isotonic
- Calculate the final molarity of H2O2H2O2 if 5.9 mL5.9 mL of a 3.0% w/w H2O23.0% w/w H2O2 solution, which has a density of 1.0 g/mL,1.0 g/mL, is added to 5.9 mL5.9 mL of a starch-iodide solution. final [H2O2]=[H2O2]= MM Calculate the final molarity of H2O2H2O2 if 1.6 mL1.6 mL of the 3.0% w/w H2O23.0% w/w H2O2 solution is diluted with 3.4 mL3.4 mL water, then added to 5.0 mL5.0 mL of a starch-iodide solution. final [H2O2]=[H2O2]= MIn the standardization of HCl using pure anhydrous sodium carbonate as the primarystandard for methyl orange as an indicator, 1.0 mL HCl was found to be equivalent to 0.05gof sodium carbonate (MW =106). The normality of HCl is:Calculate how many mEq of sodium (MW 23) are present in an admixture that is prepared by adding a 10 milliliter vial of NaCl (2.5 mEq/mL) to 500 milliliters of 1/2 normal saline. (MW NaCl 58.4; MW CL 35.4).
- Solution A with a mass of 300 g and contains solute A is mixed with 500 g of solution B which is 40% solute B. If MW of solute B is 46 g/mol, what is its molarity after mixing? Sp gr of Solution A = 1.05, Sp gr of Solution B = 1.11Calculate the molar concentration of a thiosulfate solution from the following information:A 40.-mL aliquot of a 0.00653 M KIO3 solution is added to a flask containing 2 g of KI and 10 mL of 0.5 M H2SO4. The resulting solution is titrated to a starch endpoint with 38 mL of the thiosulfate solution.What is the value (g) of the reduced mass of HF? A.)20 B.)0.95 C.)1.05 D.)3.80 × 10^2
- A 500.00 mg vitamin C (MW 176.12 g/mol) tablet was ground, acidified and dissolved in H2O to make a 250.0 mL solution. A 25.00 mL aliquot containing vitamin C, KI and starch was analyzed and titrated with 10.10 mL of 0.009128 M KIO3. what is the % vitamin C in the tablet. Answer = 97.42%A vinegar solution was prepared by diluting 25.00 mL of vinegar to 250.0 mL. From this solution, 50.00 mL portion was taken, and this required 30.00 mL of 0.1000M NaOH solution to reach neutralization end point. What is the percentage (w/v) of HOAc in the sample? Mol wt HOAc = 60.0 g/molA student prepares a stock solution of BaF2 by dissolving 50.00 g of a BaF2 in 1.00 L of water. An additional 2 drops of 37% (w/v%) HCI was added to acidify the solution. How many mL of the stock solution will be transferred to prepare a 100.00 mL solution with 0.10 M BaF2 solution? a. 38.05 mL b. 35.07 mL c. 380.5 mL d. 37. 50 mL e. None of the above