A 50.00 mL aliquot from a 0.490 L solution that contains 0.470 g of MnSO4 (...
A 50.00 mL aliquot from a 0.490 L solution that contains 0.470 g of MnSO4 ( MW=151.00 g/mol) required 38.0 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO3 ( MW=100.09 g/mol) will react with 1.00 mL of the EDTA solution? mass: mg
Homework Answers
0.8407
mg of CaCO3 will react with 1 mL EDTA solution.
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A 50.00 mL aliquot from a 0.490 L solution that contains 0.470 g
of MnSO4 (...
A 47.30 mL aliquot from a 0.485 L solution that contains 0.465 g of MnSO4 (MW=151.00...
A 47.30 mL aliquot from a 0.485 L solution that contains 0.465 g
of MnSO4 (MW=151.00 g/mol) required 39.4 mL of an EDTA solution to
reach the end point in a titration. What mass, in milligrams, of
CaCO3 ( MW=100.09 g/mol) will react with 1.78 mL of the EDTA
solution?
A 47.30 mL aliquot from a 0.485 L solution that contains 0.465 g of MnSO4 (MW = 151.00 g/mol) required 39.4 mL of an EDTA solution to reach the end...
explanation A 47.90 mL aliquot from a 0.510 L solution that contains 0.485 g of MnSO4...
explanation
A 47.90 mL aliquot from a 0.510 L solution that contains 0.485 g of MnSO4 (Fw 151.00 g/mol) required 35.0 mL of an EDTA solution to reach the end point in a titration. What mass (in milligrams) of CaCO3 (FW 100.09 g/mol) will react with 1.63 m of the EDTA solution? Number Caco, mass 1406.18 mg
A 48.80 mL aliquot from a 0.505 L solution that contains 0.525 g of MnSOg (MW...
A 48.80 mL aliquot from a 0.505 L solution that contains 0.525 g of MnSOg (MW EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCo3 (MW 100.09 g/mol) will react 151.00 g/mol) required 40.8 mL of an with 1.80 mL of the EDTA solution? mass mg
A 48.50 mL aliquot from a 0.475 L solution that contains 0.495 g of MnSO 4...
A 48.50 mL aliquot from a 0.475 L solution that contains 0.495 g of MnSO 4 ( MW = 151.00 g/mol) required 37.2 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO 3 ( MW = 100.09 g/mol) will react with 1.00 mL of the EDTA solution?
A 48.30 mL aliquot from a 0.460 L solution that contains 0.375 g of MnSO, (MW = 151.00 g/mol) required 39.9 mL of an ED...
A 48.30 mL aliquot from a 0.460 L solution that contains 0.375 g of MnSO, (MW = 151.00 g/mol) required 39.9 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO, (MW = 100.09 g/mol) will react with 1.82 mL of the EDTA solution? mass: mg
Please help! Will rate. Cd" forms two complexes with acetate, with the given overall formation constants....
Please help! Will rate.
Cd" forms two complexes with acetate, with the given overall formation constants. 1. Cd2+ (aq) + CH, COŽ (aq) =C(CH.CO2)+(aq) 2. Cd²+ (aq) + 2CH,CO2 (aq) = CaCH,CO2),(aq) R = B = 85 B2 = 1400 Find the stepwise formation constant, K2, for the reaction 3. Cd(CH, CO)* (aq) +CHCO, (aq) = Ca(CH, CO ) (aq) Kg =? You prepare 1.00L of solution containing 1.00 x 10-mol Ca(CIO), and 0.260 mol CH, CO, Na. Calculate the...
EBTA TItration Q3. A 1. volumetric flask. A 50.00-ml aliquot of the diluted solution was brought...
EBTA TItration Q3. A 1. volumetric flask. A 50.00-ml aliquot of the diluted solution was brought to a pH of 10.0 with a NH NH buffer; the subsequent titration involved both cations and required 28.89 ml of 0.06950 M EDTA. A secand 50.00-mL aliquot was brought to a pH of 10.0 with an HCN/NaCN buffer, which also served to mask the Cd? 11.56 mL of the EDTA solution were needed to titrate the Pb*, Calculate the percent Pb and Cd...
A solution contains 1.577 mg of CoSO4 (155.0 g/mol) per milliliter. Calculate a. the volume of...
A solution contains 1.577 mg of CoSO4 (155.0 g/mol) per milliliter. Calculate a. the volume of 0.007840 M EDTA needed to titrate a 25.00-mL aliquot of this solution. Volume = mL b. the volume of 0.006901 M Zn2+ needed to titrate the excess reagent after addition of 50.00 mL of 0.007840 M EDTA to a 25.00-mL aliquot of this solution. Volume = mL c. the volume of 0.007840 M EDTA needed to titrate the Zn2+ displaced by Co2+ following addition...
A solution contains both NaHCO3 and Na2CO3. Titration of a 50.00 mL portion to a phenolphthalein...
A solution contains both NaHCO3 and Na2CO3. Titration of a 50.00 mL portion to a phenolphthalein end point requires 20.38 mL of 0.1068 M HCl. A second 50.00 mL aliquot requires 46.8 mL of the HCl solution when titrated to a bromocresol green end point. Calculate the molar concentration of NaHCO3 in the solution.
3. a) A standard ZnCl2 solution is prepared by dissolving 0.6483 g of Zn in an...
3. a) A standard ZnCl2 solution is prepared by dissolving 0.6483 g of Zn in an HCl solution and diluting to volume in a 1.00 L volumetric flask. An EDTA solution is standardized by titrating a 15.00 mL aliquot of the ZnCl2 solution, which requires 16.12 mL of EDTA solution to reach the end point. Determine the concentration of the EDTA solution. (2 pts) b) A 1.5146 g sample of powdered milk is dissolved and the solution titrated with...
A 47.30 mL aliquot from a 0.485 L solution that contains 0.465 g
of MnSO4 (MW=151.00 g/mol) required 39.4 mL of an EDTA solution to
reach the end point in a titration. What mass, in milligrams, of
CaCO3 ( MW=100.09 g/mol) will react with 1.78 mL of the EDTA
solution?
A 47.30 mL aliquot from a 0.485 L solution that contains 0.465 g of MnSO4 (MW = 151.00 g/mol) required 39.4 mL of an EDTA solution to reach the end...
explanation
A 47.90 mL aliquot from a 0.510 L solution that contains 0.485 g of MnSO4 (Fw 151.00 g/mol) required 35.0 mL of an EDTA solution to reach the end point in a titration. What mass (in milligrams) of CaCO3 (FW 100.09 g/mol) will react with 1.63 m of the EDTA solution? Number Caco, mass 1406.18 mg
A 48.80 mL aliquot from a 0.505 L solution that contains 0.525 g of MnSOg (MW EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCo3 (MW 100.09 g/mol) will react 151.00 g/mol) required 40.8 mL of an with 1.80 mL of the EDTA solution? mass mg
A 48.30 mL aliquot from a 0.460 L solution that contains 0.375 g of MnSO, (MW = 151.00 g/mol) required 39.9 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO, (MW = 100.09 g/mol) will react with 1.82 mL of the EDTA solution? mass: mg
Please help! Will rate.
Cd" forms two complexes with acetate, with the given overall formation constants. 1. Cd2+ (aq) + CH, COŽ (aq) =C(CH.CO2)+(aq) 2. Cd²+ (aq) + 2CH,CO2 (aq) = CaCH,CO2),(aq) R = B = 85 B2 = 1400 Find the stepwise formation constant, K2, for the reaction 3. Cd(CH, CO)* (aq) +CHCO, (aq) = Ca(CH, CO ) (aq) Kg =? You prepare 1.00L of solution containing 1.00 x 10-mol Ca(CIO), and 0.260 mol CH, CO, Na. Calculate the...
EBTA TItration Q3. A 1. volumetric flask. A 50.00-ml aliquot of the diluted solution was brought to a pH of 10.0 with a NH NH buffer; the subsequent titration involved both cations and required 28.89 ml of 0.06950 M EDTA. A secand 50.00-mL aliquot was brought to a pH of 10.0 with an HCN/NaCN buffer, which also served to mask the Cd? 11.56 mL of the EDTA solution were needed to titrate the Pb*, Calculate the percent Pb and Cd...
A solution contains 1.577 mg of CoSO4 (155.0 g/mol) per milliliter. Calculate a. the volume of 0.007840 M EDTA needed to titrate a 25.00-mL aliquot of this solution. Volume = mL b. the volume of 0.006901 M Zn2+ needed to titrate the excess reagent after addition of 50.00 mL of 0.007840 M EDTA to a 25.00-mL aliquot of this solution. Volume = mL c. the volume of 0.007840 M EDTA needed to titrate the Zn2+ displaced by Co2+ following addition...
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