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3) The concentration of phosphorous in a stock solution is 10.0 ppm. After the dilution...
Answer 2) To measure the concentration of phosphorous in a water sample, Phosphorous was first converted...
Answer
2) To measure the concentration of phosphorous in a water sample, Phosphorous was first converted into a color complex in the presence of ammonium molybdate and SnClo. The absorbance of a series of standard phosphorous solutions was measured by a spectrophotometer and gave the calibration curve y 1.150 x, where y is the absorbance of phosphorous complex at 690 nm while x is the concentration of P in ppm. An unknown natural water sample was collected. 10.00 mL of...
show calculations please 3. A standard solution of Fe(II) [10.0 ppm) was prepared, the appropriate reagents...
show calculations please
3. A standard solution of Fe(II) [10.0 ppm) was prepared, the appropriate reagents were added, and a series of standard solutions were prepared and the transmittance of the solutions were read out as follows: Vol. of std solution (ml) Final vol. (mL) %T 0.00 100.0 100.0 5.00 100.0 77.7 25.00 100.0 30.2 50.00 100.0 10.4 (a) (3 pts) plot a calibration curve and find out the slope and y-intercept for the linearly related Absorbance with concentration. (b)...
The concentration of a solution after dilution can be found by: where M1 is the concentrated...
The concentration of a solution after dilution can be found
by:
where M1 is the concentrated solution (in
this lab, 0.400 M), and V1 is the volume of the
concentrated solution placed in the test tube. The
V2 is the total volume of
concentrated solution plus water added, and M2
is the concentration of the solution after diluting.
Tube 1 has 2.00 mL of the 0.400 M Ni2+ solution and
8.00 mL of water added. What is the concentration of...
1. Calculate the concentrations of each of your serial dilution solutions. Report the temperature of the...
1. Calculate the concentrations of each of your serial dilution solutions. Report the temperature of the “room temperature” deionized water Show the complete calculations (with appropriate units and significant figures) for the first solution as a sample and the complete Table 1 (below) with the results for all of the solutions. Solution Concentration (M) of crystal violet Absorbance at 590 nm Blank 0.00 0.000 Stock solution 1.00 x 10-5 0.851 Solution 1 0.411 Solution 2 0.203 Solution 3 0.097 Solution...
Electrolytes, Concentration and Dilution of Solutions n this part of the experiment, you will make a dilution of a stock solution of known concentration and calculate the concentration of the di...
Electrolytes, Concentration and Dilution of Solutions n this part of the experiment, you will make a dilution of a stock solution of known concentration and calculate the concentration of the dilute solution. DILUTION OF A STOCK SOLUTION PART C: Measurements and Observations Procedure Check out a 100-mL volumetric flask from your instructor. Fill test tube about 2/3 concentration of the active ingredient from its label including units. Record 1. orlt O " concentration of active ingredient. full of "Magic 2....
1. Serial dilution Questions: a) (5 pts) Suppose you begin with stock solution of dye that has a concentration of 3...
1. Serial dilution Questions: a) (5 pts) Suppose you begin with stock solution of dye that has a concentration of 350.0 mg/L and do the following set of serial dilutions on it: 5.00 mL of stock diluted to a total volume (TV) of 500.00 mL (Dilution 1); 2.00 mL of Dilution 1 is then diluted to a TV of 250.00 mL (Dilution 2); finally, 1.00 mL of Dilution 2 is then diluted to a total volume of 100.00 mL. Calculate...
A stock solution of analyte is made by dissolving 34.83 mg of copper (II) acetate hexahydrate...
A stock solution of analyte is made by dissolving 34.83 mg of copper (II) acetate hexahydrate (fw = 289.73 g/mol) in 25.000 mL of water. A second stock solution of internal standard is made by dissolving 28.43 mg of germanium (I) acetate (fw = 190.74 g/mol) into 25.000 mL of water. These solutions are used to make a series of standards for flame atomic absorption analysis calibration. The standard solutions (each 10.00 mL total volume) should have the following concentrations...
5.00 mL stock 5.00 mL Dilution A 5.00 mL Dilution B + 45.0 mL Water +...
5.00 mL stock 5.00 mL Dilution A 5.00 mL Dilution B + 45.0 mL Water + 45.0 mL Water + 45.0 mL Water O b lo Dilution C Dilution A Dilution B Example #5. Solution A is prepared by diluting 1.00 mL of 0.160 M CaCl2 to 50.0 mL. Solution B is prepared by diluting 10.0 mL of Solution A to 500.0 mL. What is the concentration of CaCl2 in Solution B? O lodno tlo iz obdobimo t 100 babe...
1) Figure below show a 36 uL of stock solution containing 800 ppm of CuSO4 was...
1) Figure below show a 36 uL of stock solution containing 800 ppm of CuSO4 was transferred into T1 tube contained with 324 uL of wash buffer. Next, serial dilution was performed from T1 to T5. (CuSO4 molar mass: 159.61 g/mol) 36 ul 120 ML 120 ML 120 L 120 L 800 ppm Cuso T1 Stock solution T2 T3 T4 TS B1 Dilution Factor Final Concentration NA 800 ppm Tube Volume Diluent Total Final Transferred Volume Volume Volume (PL) (wash...
Serial Dilution Questions 1, Serial dilution Questions: a) (5 pts) Suppose you begin with stock solution...
Serial Dilution Questions
1, Serial dilution Questions: a) (5 pts) Suppose you begin with stock solution of dye that has a concentration of 350.0 mg/L and do the following set of serial dilutions on it: 5.00 mL of stock diluted to a total volume (TV) of 500.00 mL (Dilution 1); 2,00 mL of Dilution 1 is then diluted to a TV of 250.00 mL (Dilution 2); finally, 1.00 mL of Dilution 2 is then diluted to a total volume of...
Answer
2) To measure the concentration of phosphorous in a water sample, Phosphorous was first converted into a color complex in the presence of ammonium molybdate and SnClo. The absorbance of a series of standard phosphorous solutions was measured by a spectrophotometer and gave the calibration curve y 1.150 x, where y is the absorbance of phosphorous complex at 690 nm while x is the concentration of P in ppm. An unknown natural water sample was collected. 10.00 mL of...
show calculations please
3. A standard solution of Fe(II) [10.0 ppm) was prepared, the appropriate reagents were added, and a series of standard solutions were prepared and the transmittance of the solutions were read out as follows: Vol. of std solution (ml) Final vol. (mL) %T 0.00 100.0 100.0 5.00 100.0 77.7 25.00 100.0 30.2 50.00 100.0 10.4 (a) (3 pts) plot a calibration curve and find out the slope and y-intercept for the linearly related Absorbance with concentration. (b)...
The concentration of a solution after dilution can be found
by:
where M1 is the concentrated solution (in
this lab, 0.400 M), and V1 is the volume of the
concentrated solution placed in the test tube. The
V2 is the total volume of
concentrated solution plus water added, and M2
is the concentration of the solution after diluting.
Tube 1 has 2.00 mL of the 0.400 M Ni2+ solution and
8.00 mL of water added. What is the concentration of...
Electrolytes, Concentration and Dilution of Solutions n this part of the experiment, you will make a dilution of a stock solution of known concentration and calculate the concentration of the dilute solution. DILUTION OF A STOCK SOLUTION PART C: Measurements and Observations Procedure Check out a 100-mL volumetric flask from your instructor. Fill test tube about 2/3 concentration of the active ingredient from its label including units. Record 1. orlt O " concentration of active ingredient. full of "Magic 2....
1. Serial dilution Questions: a) (5 pts) Suppose you begin with stock solution of dye that has a concentration of 350.0 mg/L and do the following set of serial dilutions on it: 5.00 mL of stock diluted to a total volume (TV) of 500.00 mL (Dilution 1); 2.00 mL of Dilution 1 is then diluted to a TV of 250.00 mL (Dilution 2); finally, 1.00 mL of Dilution 2 is then diluted to a total volume of 100.00 mL. Calculate...
5.00 mL stock 5.00 mL Dilution A 5.00 mL Dilution B + 45.0 mL Water + 45.0 mL Water + 45.0 mL Water O b lo Dilution C Dilution A Dilution B Example #5. Solution A is prepared by diluting 1.00 mL of 0.160 M CaCl2 to 50.0 mL. Solution B is prepared by diluting 10.0 mL of Solution A to 500.0 mL. What is the concentration of CaCl2 in Solution B? O lodno tlo iz obdobimo t 100 babe...
1) Figure below show a 36 uL of stock solution containing 800 ppm of CuSO4 was transferred into T1 tube contained with 324 uL of wash buffer. Next, serial dilution was performed from T1 to T5. (CuSO4 molar mass: 159.61 g/mol) 36 ul 120 ML 120 ML 120 L 120 L 800 ppm Cuso T1 Stock solution T2 T3 T4 TS B1 Dilution Factor Final Concentration NA 800 ppm Tube Volume Diluent Total Final Transferred Volume Volume Volume (PL) (wash...
Serial Dilution Questions
1, Serial dilution Questions: a) (5 pts) Suppose you begin with stock solution of dye that has a concentration of 350.0 mg/L and do the following set of serial dilutions on it: 5.00 mL of stock diluted to a total volume (TV) of 500.00 mL (Dilution 1); 2,00 mL of Dilution 1 is then diluted to a TV of 250.00 mL (Dilution 2); finally, 1.00 mL of Dilution 2 is then diluted to a total volume of...
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