Calculate % purity and % yeild. 15mL Tert-Butyl (density= 0.78g/mL) 37mL HCl (density=1.18g/mL) 30mL Sodium Bicarbonate Weight of bottle= 11.21g Weight of final product= 2.275g
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Calculate % purity and % yeild.
15mL Tert-Butyl (density= 0.78g/mL)
37mL HCl (density=1.18g/mL)
30mL Sodium Bicarbonate
Weight of bottle= 11.21g
Weight of final product= 2.275g
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Solved in 2 steps
- 1) 0.1% w/v = ? mg/100 ml? 2) 0.1% = ? ug/ml? 3)What is the concentration of your stock in ug/ml? 4)The molecular weight of tartrazine is 534.4 g/mole. What is the concentration of your stock solution in mM? Be sure to give your answer in milliMolar to TWO decimal places.TITRIMETRIC DATA SAMPLE: CANE VINEGAR % acidity in label: 4.5% %purity of KHP: 99.80% FORMULA WEIGHT of KHP: 204.22 g/mol STANDARDIZATION OF NaOH SOLUTION TRIAL 1 TRIAL 2 TRIAL 3 Weight of KHP, g 0.1012 0.1004 0.09987 Initial volume of NaOH, mL 5.00 10.01 Final volume of NaOH, mL 4.95 9.99 14.93 Molarity of NaOH Average Molarity of NaOH ANALYSIS OF ACETIC ACID IN A VINEGAR SAMPLE TRIAL 1 TRIAL 2 TRIAL 3 Volume of vinegar, mL Initial volume of NaOH, mL 1.00 1.00 1.00 14.98 22.90 30.87 Final volume of NaOH, mL 22.84 30.77 38.75 Molarity of acetic acid Average molarity of acetic acid ANALYSIS OF CARBONIC ACID IN A SODA SAMPLE TRIAL 1 TRIAL 2 TRIAL 3 Volume of soda, mL 20.0 20.0 20.0 Initial volume of NaOH, mL 20.20 22.05 23.93 Final volume of NaOH, mL 22.03 23.89 25.8 Molarity of carbonic acid Average molarity of carbonic acid2. How many mL of a 0.250 M KCI solution must be diluted to 1.000 L so that the diluted solution (density = 1.00 g/mL ) is 400 ppm K* by weight ? (MW KCI = 74.55; AW K = 39.10)
- If 400 mL of a 18% v/v solution of methyl salicylate in alcohol is diluted to 1500 mL, the concentration of the resultant solution is _____ % v/v.1a) 0.1% = ? ug/ml? Only include the number and not any units b) What is the concentration of your stock in ug/ml? Only include the number and not any units c`The molecular weight of tartrazine is 534.4 g/mole. What is the concentration of your stock solution in mM? Be sure to give your answer in milliMolar to TWO decimal places. d)What dilution of your stock do you need to get the first standard (100 ug/ml)? Express as 1:x, where x is the fold dilution needed.Part IV. Preparation of 100 mL 25% Solution X Materials: Solution X, measuring cylinder, distilled water, and parafilm. Show calculation steps. (1) Calculate the volume of Solution X required. (2) Calculate the volume of distilled water required_ (3) Deliver the calculated amount of distilled water in the measuring cylinder. (4) Add the calculated amount of Solution X to the measuring cylinder. (5) Seal the opening of the measuring cylinder with parafilm. Mix well. (6) Transfer the prepared 25% Solution X into a labelled container.
- Cyclohexanol 10 ml Mass of cyclohexanol 9.62 g – 0.053 mol Theroretical yield cyclohexene–7.89g - ? mol Therotical mass of cyclohexene- Actual yield- 6.8ml –4.055 g –? mol Percent yield- 51.04% Boiling point of cyclohexene -83 degree Celsius Theroretical boling point -83 degree Celsius can i get some clarification plsA 500-mg tablet of commercial vitamin C tablet was dissolved in 50 mL water, treated with excess KI solution and 3 drops of starch solution. The solution required 4.8 mL of 0.05 M KIO3 to reach the endpoint. Calculate the % by weight of vitamin C in the tablet. (Molar mass of vitamin C = 172.16 g mol-1) IO3‐ + 5 I‐ + 6 H+ → 3 I2 + 3 H2O (1) I2 + C6H8O6 → C6H8O6* + 2 H+ + 2 I‐ (2) I2 + starch → (I2 - starch) (deep blue color) (3) *oxidized product of vitamin CA 32.5 mL sample of an 5.6 % (m/v) KBr solution is diluted with water so that the final volume is 236.5 mL.
- Hi! I need answers for numbers 2 and 3. My professor said that The unit of solubility is in grams per 100mL, while the solvent used is only 50mL, so a yield of close to 2.0g is impossible. Recalculate using this information. Thanks!CI concentrated X 1. LIAIH₁ 2. H₂O* NaOH 0°CUsing the data below how do you calculate the percent yield Compounds Mass Cyclohexanol used 150 mg = 0.15g Acetic acid used 70 mg = 0.07 g Ultra Clorox (6% NaOCl solution) used 2mL Cyclohexanone obtained 0.124 g