explanation A 47.90 mL aliquot from a 0.510 L solution that contains 0.485 g of MnSO4...
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...
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
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
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?
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...
Name Section Experiment 28 Advance Study Assignment: Determination of the Hardness of Water 1. A 0.3946 g sample of Caco, is dissolved in 12 M HCl and the resulting solution is diluted to 250.0 mL in a volumetric flask. a. How many moles of Caco, are used (formula mass = 100.1)? moles b. What is the molarity of the Ca2+ in the 250 mL of solution? c. How many moles of Caare in a 25.00-mL aliquot of the solution in...
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...
Given the following information: Buret contains .01 M EDTA; Both Solution A and B started with .042 g Calcium Magnesium Acetate; Solution A contains both magnesium and calcium; Solution B contains JUST magnesium (Ca was precipitated out); EDTA reacts with both Mg2+ and Ca2+ in a 1:1 ratio (if you use .0005 moles of EDTA, you have .0005 moles of metal ion). Titration of Solution A: Buret reading 1: 18.53 mL, Buret reading 2: 49.12 mL Titration of Solution B:...
Co. A 50.0 mL aliquot of water (assume contains only Ca?") required 4.08 ml. of o.0100 M EDTA to titrate all of the Cr" cation present what is the ppm concentration of Ca2+ (molar mass: 400 g/mol).