Homework Answers
![Ca 2+ + EDTA = (cd EDTA)]2* mn 24 + EDTA - (MM (EDTA)] 2+ let the concentration Cd2+ 8 My 24 is C & C respectively. - from an](http://img.homeworklib.com/questions/96c3e8b0-1714-11ea-b4b8-5b4129b579e8.png?x-oss-process=image/resize,w_560)
Add Answer to:
A 50.0 mL sample containing Cd²+ and Mn + was treated with 43.8 mL of 0.0700 M EDTA. Titration of the excess unreac...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 57.2 mL of 0.0700 M...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 57.2 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 14.6 mL of 0.0150 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN“. Titration of the newly freed EDTA required 11.3 mL of 0.0150 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? concentration: M Mn2+ concentration: M Cd2+
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7 mL of 0.0500M EDTA....
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7
mL of 0.0500M EDTA. Titration of the excess unreacted EDTA required
15.8 mL of 0.0190 M Ca2+. The Cd2+ was displaced from EDTA by the
addition of an excess of CN−. Titration of the newly freed EDTA
required 13.7 mL of 0.0190 M Ca2+. What are the concentrations of
Cd2+ and Mn2+ in the original solution?
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 42.5 mL of 0.0700 M...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 42.5 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 16.8 mL of 0.0120 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN–. Titration of the newly freed EDTA required 26.9 mL of 0.0120 M Ca2+. What were the molarities of Cd2+ and Mn2+ in the original solution? Answer: ([Cd2+] = 0.00646 M; and [Mn2+] = 0.0490 M) The...
A sample containing Cd2+ and Mn2+ was treated with 60.1 ml of 0.0400M EDTA. Titration of...
A sample containing Cd2+ and Mn2+ was treated with 60.1 ml of 0.0400M EDTA. Titration of the excess unreacted EDTA required 16.8ml of 0.013M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN-. Titration of the newly freed EDTA required 10.4ml of 0.013M Ca2=. What are the concentrations of Cd2+ and Mn2+ in the original solution?
Please help! Will rate. A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5...
Please help! Will rate.
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5 mL of 0.0600 M EDTA. Titration of the excess unreacted EDTA required 13.3 mL of 0.0100 M Ca. The Cd2t was displaced from EDTA by the addition of an excess of CN. Titration of the newly freed EDTA required 11.2 mL of 0.0100 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? M Mn2+ concentration: M Cd2 concentration: A...
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10. A 50.0 mL sample of water containing both Ca? and Mg is titrated in another 50.0 mL sample, the Mg* was precipitated as Mg(OH)2...
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10. A 50.0 mL sample of water containing both Ca? and Mg is titrated in another 50.0 mL sample, the Mg* was precipitated as Mg(OH)2 and then, Ca2* was titrated at pH 13 with 9.25 ml of the same EDTA solution. Calculate ppm CaCo, (FW-100.09) and MgCO, (FW-84.31) in the sample. FWca: 40.08; FWm: 24.30
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10....
An unknown solution was analyzed for Ni by an EDTA titration. A 50.00 mL sample of...
An unknown solution was analyzed for Ni by an EDTA titration. A 50.00 mL sample of the unknown was treated with 25.00 mL of 0.2404 M EDTA. The excess EDTA was then back titrated with 8.52 mL of 0.0694 M Zn2+ to reach the equivalence point. What was the concentration of Ni (in unit of M) in the 50.00 mL sample? Please keep your answer to three decimal places.
6. (8) A 50.0 mL sample of hard water containing Ca2 was titrated with 0.0500 M...
6. (8) A 50.0 mL sample of hard water containing Ca2 was titrated with 0.0500 M EDTA standard solution at pH 10 and with Eriochrome Black T indicator. The endpoint was reached when 11.35 mL of EDTA was added a) (6) Calculate the concentration of Ca* in the unknown sample in moles per liter (M). b) (2) What is the pCa of the solution?
Problem: A solution containing 50.0 mL of 0.100 M EDTA buffered to pH 10.00 was titrated...
Problem: A solution containing 50.0 mL of 0.100 M EDTA buffered to pH 10.00 was titrated with 50.0 mL of 0.020 0 M Hg(CIO4)2 in the cell [[ S.C.E || titration solution | Hg Ol]. From the cell voltage E = -0.027 V, find the formation constant of Hg(EDTA)2-. I know how it's done, but I have few questions from the solution below. What does F_EDTA mean? How do you get 0.30 at the bottom when calculating Kf? The following...
In order to determine the hardness of α solution, a small amount of Eriochrome 2. Black T is added to 100 ml of sample. After addition of I1.2 ml of D.01 M EDTA solution, the color of the soluti...
In order to determine the hardness of α solution, a small amount of Eriochrome 2. Black T is added to 100 ml of sample. After addition of I1.2 ml of D.01 M EDTA solution, the color of the solution changes from red to blue. What is the hardness concentration in moles/liter and mg/liter cas CaCO,? Show that [HgOH'] is negligible in Example 5-2 3. 4. A water, pH 5.5, is dosed with 80 mg/liter of alum,C.co,-1x10-4. (a) How much hydrated...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 57.2 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 14.6 mL of 0.0150 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN“. Titration of the newly freed EDTA required 11.3 mL of 0.0150 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? concentration: M Mn2+ concentration: M Cd2+
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7
mL of 0.0500M EDTA. Titration of the excess unreacted EDTA required
15.8 mL of 0.0190 M Ca2+. The Cd2+ was displaced from EDTA by the
addition of an excess of CN−. Titration of the newly freed EDTA
required 13.7 mL of 0.0190 M Ca2+. What are the concentrations of
Cd2+ and Mn2+ in the original solution?
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7...
Please help! Will rate.
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5 mL of 0.0600 M EDTA. Titration of the excess unreacted EDTA required 13.3 mL of 0.0100 M Ca. The Cd2t was displaced from EDTA by the addition of an excess of CN. Titration of the newly freed EDTA required 11.2 mL of 0.0100 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? M Mn2+ concentration: M Cd2 concentration: A...
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10. A 50.0 mL sample of water containing both Ca? and Mg is titrated in another 50.0 mL sample, the Mg* was precipitated as Mg(OH)2 and then, Ca2* was titrated at pH 13 with 9.25 ml of the same EDTA solution. Calculate ppm CaCo, (FW-100.09) and MgCO, (FW-84.31) in the sample. FWca: 40.08; FWm: 24.30
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10....
6. (8) A 50.0 mL sample of hard water containing Ca2 was titrated with 0.0500 M EDTA standard solution at pH 10 and with Eriochrome Black T indicator. The endpoint was reached when 11.35 mL of EDTA was added a) (6) Calculate the concentration of Ca* in the unknown sample in moles per liter (M). b) (2) What is the pCa of the solution?
Problem: A solution containing 50.0 mL of 0.100 M EDTA buffered to pH 10.00 was titrated with 50.0 mL of 0.020 0 M Hg(CIO4)2 in the cell [[ S.C.E || titration solution | Hg Ol]. From the cell voltage E = -0.027 V, find the formation constant of Hg(EDTA)2-. I know how it's done, but I have few questions from the solution below. What does F_EDTA mean? How do you get 0.30 at the bottom when calculating Kf? The following...
In order to determine the hardness of α solution, a small amount of Eriochrome 2. Black T is added to 100 ml of sample. After addition of I1.2 ml of D.01 M EDTA solution, the color of the solution changes from red to blue. What is the hardness concentration in moles/liter and mg/liter cas CaCO,? Show that [HgOH'] is negligible in Example 5-2 3. 4. A water, pH 5.5, is dosed with 80 mg/liter of alum,C.co,-1x10-4. (a) How much hydrated...
Most questions answered within 3 hours.
-
In a study designed to test the effectiveness of magnets for
treating back pain, 35 patients...
asked 13 minutes ago -
Here are summary statistics for randomly selected weights of
newborn girls:
nequals=193,
x overbarxequals=30.5
hg,
sequals=7.3...
asked 3 minutes ago -
Exercise #3:
Create the “MathTest” class. It will have two class variables:
1) a question and...
asked 6 minutes ago -
In epidemiology, how do you calculate the overall incidence of
cure within two groups? What formula...
asked 10 minutes ago -
A 1 liter solution contains 0.357 M ammonium chloride and 0.268
M ammonia. Addition of 0.295...
asked 11 minutes ago -
What are the advantages and disadvantages of using virtual
reality simulations in health care education?
asked 15 minutes ago -
Given input { 66, 28, 43, 29, 44, 69, 19 } and a hash function
h(x)...
asked 36 minutes ago -
A pebble with mass m is thrown straight up with an initial speed
v0 so that...
asked 40 minutes ago -
Let X be a discrete random variable that follows a
binomial distribution with n = 11...
asked 48 minutes ago -
The equilibrium constant, K, for the following reaction is
1.29×10-2 at 600
K.
COCl2(g) --->
CO(g)...
asked 1 hour ago -
It is known that 72% of people have a favorable opinion of their
local police force....
asked 1 hour ago -
A vertical straight wire carrying an upward 26-A current exerts
an attractive force per unit length...
asked 1 hour ago