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- 1.87 A solution of ethanol in water has a volume of 54.2 mL and a mass of 49.6 g. what information would you need to look up and how would you determine the percentage of ethanol in this solution?Lead = 11.3 g/mL silver= 10.5 g/mL nickel= 9.90 g/mL zinc= 7.14 g/mLSolution 1 is diluted by combining 6 mL of Solution 1 with 64 mL of Solution 2. What is the dilution of solution 1? Report your answer in standard decimal notation rounded to two decimal places. For example, 4 mL added to 16 mL would be a 0.20 dilution. Include trailing zeros if necessary so your answer has two decimal places.
- You have a stock of Rubisco at the concentration of 100 µg/mL. For your experiment you need 500 µL of a working solution of 2.5 µg/mL. What volume of the stock solution do you need? What volume of the diluent (solvent) do you need?At room temperature, table sugar has a solubility of 200. g in 100 mL of water. In boiling water, solubility of sugar increases to 500. g/100 mL water. A solution at room temperature contains 156 g of sugar/100 mL of water. This solution must be _________.What's the differences between two questions? Q1) How much calcium would you ingest by drinking one 8 oz glass of your tap water? Show all calculations. -->Tap water 8Oz = 8 x 0.0296L = 0.2368L Hardness = 66.73ppm = 66.73mg/L CaCO3 1L has 66.73mg CaCO3 0.2368L has 66.73mg x 0.2368 = 15.8017mg MW of CaCO3 = 100g/mol MW of Ca = 40g/mol 100g CaCO3 has 40gf of Ca 15.8017mg CaCO3 has 40/100 x 15.8017mg Ca We would ingest 6.321mg of Ca. Q2) What percentage of the recommended daily dose of calcium (1,150 mg/day) does 1.0 L of your water provide? Show all calculations. --> 66.73mg/1150mg x 100 = 5.80% My Question) Why this calculation is wrong? I think this calculation is same with question 1. Isn't it? CaCO3 = 100g/mol, Ca = 40g/mol 100g CaCO3 has 40g Ca. 66.73 CaCO3 has 40/100 x 66.73mg Ca Ca = 26.70mg 26.70mg/1150mg x 100 = 2.32%
- Give the solution with proper sig figs. Determine the average of the densities of 4.523 g/ml, 4.556 g/ml, 4.821 g/ml and give the answer in sig figs. Sum = 4.523 g/ml + 4.556 g/ml + 4.821 g/ml = 13.9 g/ml or 13.90000000 g/ml Ave = 13.90000000 g/ml / 3 = 4.633333333 g/ml a. none of these b. 5 g/ml c. 4.633 g/ml d. 4.6333 g/ml e. 4.633333333 g/ml f. 4.63 g/ml g. 4.6 g/ml h. 4 g/mlPrepare a solution of NaCl by accurately weighing approximately 1.45g of NaCl into a 100 mL beaker / conical flask. Record the mass of the sample. Add 25 mL distilled water. Mix until all crystals dissolve. Prepare a solution of magnesium chloride by accurately weighing approximately 3.61 g of MgCl2٠6H20 into a 100 mL beaker / conical flask. Record the mass of the sample. Add 25 mLdistilled water. Mix until all the crystals dissolve. 1. Calculate the molality of the NaCℓ and MgCℓ2.6H2O solutions. 2. Calculate the total mass of water (solvent) in kg: a) Mass of crystal water added (kg): b) Mass of water added (kg): c) Total mass of water added (kg): 3. Calculate the molality of MgCℓ2. All questions follow on from each other. Thanks very much.Sucrose solution for experiment 1: (Question a-c)Prepare a solution of sucrose by accurately weighing 9.0 – 9.25 g of sucrose into a 100mLbeaker / conical flask. Record the mass of the sample. Add 25 mL distilled water. Mix until allcrystals dissolve.NaCl solution for experiment 2: (Question d-f)Prepare a solution of NaCl by accurately weighing approximately 1.5 g of NaCl into a 100 mLbeaker / conical flask. Record the mass of the sample. Add 25 mL distilled water. Mix until allcrystals dissolve.MgCl2٠6H2O solution for experiment 2: (Question d-f)Prepare a solution of magnesium chloride by accurately weighing approximately 3.6 g ofMgCl2٠6H20 into a 100 mL beaker / conical flask. Record the mass of the sample. Add 25 mLdistilled water. Mix until all the crystals dissolve
- A 1185 mL sample of drinking water was found to contain 20.5 mg of lead. Calculate the concentration of lead in milligrams per liter. concentration: mg/L MacBook Air トト F11 F12 888 FB F7 F3 F4 24 & * 4 5 6 7 8 Y K2. A serum is diluted 1:20 then 2 parts of the resulting dilution is added to 40 parts saline. Then 3 parts of the last dilution was added to saline to have a total volume of 6 parts. What is the final dilution?How many mL are in 454.6 g of blood plasma? The density of blood plasma is 1.025 g/mL 1 pound = 454 g A) 447.0 mL B) 345 mL C) 1.001 mL D) 0.002255 mL E) 443.5 mL