Hydrochloric acid NF contains approximately 37 g of hydrochloric acid (HCl) in 100 g solution of HCl in water. If the Specific gravity of HCl NF is 1.18, a 100 g solution is equivalent to ___________ mL, and the amount of HCl contained in 250.0 mL of HCl NF is _____________ g.
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Hydrochloric acid NF contains approximately 37 g of hydrochloric acid (HCl) in 100 g solution of HCl in water. If the Specific gravity of HCl NF is 1.18, a 100 g solution is equivalent to ___________ mL, and the amount of HCl contained in 250.0 mL of HCl NF is _____________ g.
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- A solution of sulfuric acid was made by dissolving 20.0g of 98% sulfuric acid in a 250mL of water. (H2SO4 = 98.078 g/mol; H2O = 18.015 g/mol)Additional information for 35-37 What if the solution has a total volume of 262ml? 1. Express the concentration of this solution in % by weight.a. 7.3%b. 4.4%c. 8.0%d. 7.8% 2. Express the concentration in molaritya. 0.10 Mb. 2.00 Mc. 0.76 Md. 0.93 M 3. Express the concentration in normalitya. 2.0 Nb. 1.5 Nc. 1.0 Nd. 0.76 N 4. Express the concentration in molalitya. 0.80 mb. 0.77 mc. 1.00 md. 1.50 mConsider a 1.000-L stock solution of 9.69 M Na3PO4. A 2.000-mL aliquot is taken from this stock solution and diluted to a final volume of 1.000 L. A 5.000-mL aliquot is then taken from this new solution and further diluted to make a new solution with a final volume of 750.0 mL. Calculate Na+ in the final solution.1. How many milliliters of concentrated hydrochloric acid, 38% (wt/wt), specific gravity 1.19, are requiredto prepare 1 L of a 0.100 M solution? Assume density and specific gravity are equal within threesignificant figures. (H-1.008 amu, Cl-35.45 amu) 2. A 0.456-g sample of an ore is analyzed for chromium and found to contain 0.560 mg Cr2O3.Express the concentration of Cr2O3 in the sample as %w/w.
- When performing the gram stain, the Crystal Violet solution is composed of 20 mL of 95% ethanol in a final volume of 100 mL. If only 70% ethanol is available, how many mL of ethanol should be added to achieve a final volume of 1,000? Report your answer to 2 decimal places. Perform all calculations in a single step using Excel to avoid accumulated rounding error.Read the Limit Test for salicylic acid in aspirin below then answer the question below. Remember: Aspirin = acetyl salicylic acid (ASA), and salicylic acid = SA Limit Test for salicylic acid: Extract powdered Aspirin tablets (i.e., active + excipient) containing 0.4g acetylsalicylic acid (i.e., active) with 4 mL of 95% ethanol. Add water to give a final volume of 100.0 mL and mix. Transfer 50.0 mL of the filtrate to a Nessler Cylinder and add 1.0 mL of acidified ferric ammonium sulfate solution. Mix and allow to stand. Prepare the standard solution used for comparison as follows: To another Nessler Cylinder add 3.0 mL of freshly prepared 0.01% (wt/vol) salicylic acid, 2.0 mL of ethanol, 1.0 mL of acidified ferric ammonium sulfate solution and sufficient water to give a total volume of 51.0 mL. Mix. A violet colour produced by the sample should not be more intense than that of the standard, used for comparison. Example of "Limit" calculations: Based on the above limit test, the…By experiment, the normality of H2SO4 was found to be 0.5172. If 39.65 ml of this acid neutralized 21.74 ml of NaOH, What is the normality of the base?a) 0.0095433Nb) 0.9433Nc) 9.533N 0.200grams of pure Na2CO3 requires 35.5ml of vinegar for titration. Calculate the normality of vinegar.a) 1.06 Nb) 0.106Nc) 10.6Nd) 106N
- Given that liquid alum is used as a coagulant. Specific gravity of alum is 1.33. One gallon ofalum weighs 11.09 pounds (5.03 kg) and contains 5.34 pounds (5.42 kg) of dry alum.Determine: (a) mL of liquid alum required to prepare a 100 mL solution of 20,000 mg/L alumconcentration, (b) the alum concentration, (c) the dosage concentration of 1 mL of stocksolution in a 2000 mL Gator jar sample.• Determine alum concentration in mg/L• Prepare 100 ml stock solution having a 20,000 mg/L alum concentration• Calculate mL (y) of liquid alum to give 2000 mg• Find 1 mL of alum concentration (z) in 2000 mL sample (jar)NaOH Concentration (M): 0.09669M Trial 1: Inital Volume of NaOH: 0.55mL Final Volume of NaOH: 28.25 mL Trail 2: Intial Volume of NaOH: 18.1 mL Final Volume of NaOH: 46.1 mL Concentration of HCl Solution Subtract the initial buret reading from the end point reading. Convert the readings to liters by dividing by 1000. Solve this equation for the concentration of HCl: (CHCl) (0.01000 liters ) = (CNaOH) (volume of NaOH in liters). CNaOH is the normality of the NaOH you recorded from the bottle. Average the two HCl concentrations.1.) Use the line of best ! Tables: Data Port 1. Synthesis of Asprin \table[[Mass of galicylic Acid used 2.024gt to calculate the concentration of the salicylic Acid in your asprin saple solution (in 50.0 ml volumettic flask) 2.) Use M1V1 = M2V2 to determine the concentration of salicylic acid in your 100.0 mL solution before dilutuon to 50.0mL for analysis. 3.) Determine the mass of salicylic acid present in the 4g sample of synthesized asprin tht was dissolved in the 100.0ml, volumetric flask 4.) Determine the percent of salicylic acid and percent asprin in the 4g sample of synthesized asprin7 5.) Calculate the theoretical yelld of asprin for your synthesis reaction. Assume that the salicylic acid used in the salicylic used in the synthesis reaction is the limiting reactant. 6. Determine the percent yild of the asprin for your synthesis. For the 1st cacization i got 1.340x10^-4 M C7H603 For the 2nd calculation i got M167x10^-5 Please check my calculations and explain the remaining
- Your 50 mL beaker has a mass of 6.65 g. You add 10.00 mL to the 50 mL beaker and now the total mass is 17.45 g. You originally used 5.14 g of NaCl to make the solution. What is the percent composition of your solution? These calculations are described in the lab procedure in detail.Solution 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.Cedrick and Astrid a 20.00 ml aliquot of grapefruit juice with a 0.165 M NaOH solution to the end point. The initail buret reading was 1.72ml and final buret reading was 15.51ml. They calculated that there was 0.1457g of citric acid present in the juice sample. What is the amount mg of citric acid present per ml of juice? the answer is 7.29mg/ml. (can you please show me how to calculate this answer)