A 0.5g sample was digested with 10ml of Nitric acid solution
Then diluted to 1:10, 1:3 ration
What is the amount of Mg, Zn, in ppm if given the following: Show all calculation work please step by step
Mg (mg) |
Zn(mg) |
Fe (mg) |
Sample weight (g) |
0.05 |
0.0014 |
0.0017 |
0.5 |
The values 0.05 mg Mg and 0.0014 mg Zn corresponds to diluted solution.
The corresponding values in the original sample can be calculated by considering the dilution ratios.
For Mg
For Zn
This is the amount present in 0.5 g of sample
calculate the amount present in 1 g of sample
For Mg
For Zn
Convert mg/g in ppm
For Mg
For Zn
A 0.5g sample was digested with 10ml of Nitric acid solution Then diluted to 1:10, 1:3...
If you digested 25 mg of pure ferrocene in nitric acid and then diluted the resulting solution up to 500 mL with 1% aqueous hydrochloric acid, what would be the concentration of iron in solution (provide an answer in both mg/L and mol/L)?
4. If you digested 25 mg of pure ferrocene in nitric acid and then diluted the resulting solution up to 500 mL with 1% aqueous hydrochloric acid, what would be the concentration of iron in solution (provide an answer in both mg/L and mol/L)?
A 0.928 g cereal sample was digested in acid and diluted to a final volume of 20.0 mL. It was then subjected to a further 10-fold dilution. When that solution was evaluated using ICP-MS, a particular element was present at 258.9 ppb (wg/L). How many milligrams of the element were present in the cereal sample? Enter the numerical value as a number ( 0.0315 mg as 0.0315). Note: The question is asking for milligrams in the sample not milligrams element...
A 11.17 mg ore sample containing copper is digested with acid, filtered and transferred to a volumetric flask and diluted to exactly 200.00 mL. A 10.00 mL aliquot is transferred to a 100.00 mL volumetric flask containing zincon reagent and a quantity of a pH 9 buffer solution. It is then diluted to the mark and mixed thoroughly. The measured absorbance of the resulting blue solution at 600 nm was 0.326 which corresponded to a Cu(II) concentration of 0.498 ppm...
A. 0.191 g of steel was dissolved in acid and diluted to 100 mL with water. The solution was found to contain 7.7 ppm Mn by atomic absorption analysis. What was the weight percent Mn in the steel? B. A multivitamin tablet with a label value of 13 mg Zn per tablet was dissolved in acid and diluted to 100 mL with water. What is the theoretical Zn concentration of the solution? Answer in ppm.
A 14.8 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid. If 11.5 mL of 1.12 M barium hydroxide are required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture? % by mass
1-5 all please Questions 1. Calculate the pH of a 0.10 M nitric acid solution. How does your calculation compare to your experimental pH? 2. Calculate the pOH, pH, and [H] of a 0.10M KOH solution. How does the pH compare with your data? 3. Use your date to calculate the [H'), POH, and [OH ] of the 0.10 Macetic acid solution 4. Use your data to calculate the pOH, pH, and [H'] of the 0.10M ammonia solution 5. A...
please answer Titration Homework 1. Copper reacts with dilute nitric acid according to the equation 3 Cu(s) + 8 HNO, (aq) + 3 Cu(NO3)2 (aq) + 2NO(g) + 4H20 (1) If a copper penny weighs 3.020g is dissolved in a small amount of nitric acid and the resulting solution is diluted to 50.0 mL with water, what is the molarity of the Cu(NO3)?
Problem 1 - Strontium Hydroxide Acid/Base Question -/1 points A 8.51 mL sample of nitric acid required 13.25 mL of 0.105 M strontium hydroxide for titration. Calculate the molarity of the acid solution. (Hint: It's stoichiometry, you need the balanced equation) Concentration of Nitric Acid M Evaluate Problem 3 - H2Z Molecular Weight -/1 points A solution was made by adding water to 0.22 g of H2Z until the volume totaled 25.00 mL. Subsequent titration required 40.50 mL of 0.11...
Outline of procedure: Dilute bleach solution A 1/10. Prepare analyte for titration by adding 10mL of diluted bleach solution, starch indicator and 5mL of 2 M HCl. Titrate the analyte with 0.05592 M Na2S2O3. Average volume of Na2S2O3 added for Solution A = 31.80 mL Average Volume of Na2S2O3 added for Solution B = 19.40 mL Question: Calculate the percent of sodium hypochlorite in the original bleach bottle using your averaged result. Show the calculation for solution A and only...