Question

Design a separation scheme for Cu2+, Fe3+, Mn2+, and Ni2+

Answer #1

Let us observe the separation scheme for the cations provided. For the background information, is group 2 cation and the remaining three cations are of Group 3. The scheme is given below-

Now as CuS is separated out, we would be left with the solution containing . Now there is no feasible way of separation of these three cations in a mixture. The only way out here is that, as we already know we have the three cations, we simply divide the sample! Let us see the scheme below-

Design a seperation scheme for the following cations: Al3+,
Ba2+, Cu2+, Mn2+

what color are cu2+, ag+, co2+, ni2+, fe3+ after development and
after ammonia in paper chromatography?

Determine the chemical equation for the reaction between
Fe3+ and Cu2+ with NH3

1) Balance the reaction Fe2+ + MnO4– → Fe3+ + Mn2+ in acidic
solution. What is the correct coefficient for Fe2+ when the
smallest whole number coefficients are used?
2) Balance the reaction Fe2+ + H2O2 → Fe3+ + H2O in acidic
solution. How many electrons are exchanged when the smallest whole
number coefficients are used?
3) Balance the reaction Cu + HNO3 → Cu2+ + NO in acidic
solution. What is the correct coefficient for NO when the smallest...

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is
treated with 25.00 mL of a 0.03828 M EDTA solution. The solution is
then back titrated with 0.02192 M Zn2 solution at a pH of 5. A
volume of 15.73 mL of the Zn2 solution was needed to reach the
xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2
solution is fed through an ion-exchange column that retains Ni2 .
The Cu2 that passed through the...

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is
treated with 25.00 mL of a 0.05231 M EDTA solution. The solution is
then back titrated with 0.02324 M Zn2 solution at a pH of 5. A
volume of 20.98 mL of the Zn2 solution was needed to reach the
xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2
solution is fed through an ion-exchange column that retains Ni2 .
The Cu2 that passed through the...

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is
treated with 25.00 mL of a 0.04127 M EDTA solution. The solution is
then back titrated with 0.02003 M Zn2 solution at a pH of 5. A
volume of 16.44 mL of the Zn2 solution was needed to reach the
xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2
solution is fed through an ion-exchange column that retains Ni2 .
The Cu2 that passed through the...

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is
treated with 25.00 mL of a 0.04728 M EDTA solution. The solution is
then back titrated with 0.02103 M Zn2 solution at a pH of 5. A
volume of 19.11 mL of the Zn2 solution was needed to reach the
xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2
solution is fed through an ion-exchange column that retains Ni2 .
The Cu2 that passed through the...

A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is
treated with 25.00 mL of a 0.03146 M EDTA solution. The solution is
then back titrated with 0.02115 M Zn2 solution at a pH of 5. A
volume of 16.97 mL of the Zn2 solution was needed to reach the
xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2
solution is fed through an ion-exchange column that retains Ni2 .
The Cu2 that passed through the...

Find the calculated voltages for each cell. Use equation
Eo(cell)= E0(cathode) + Eo(anode)
Cell 1: Cu/Cu2+ Zn/Zn2+
Cell 2: Cu/Cu2+ Fe/Fe3+
Cell 3: Cu/Cu2+ Ni/Ni2+
Cell 4: Zn/Zn2+ Fe/Fe3+
Cell 5: Zn/Zn2+ Ni/Ni2+
Cell 6: Ni/Ni2+ Fe/Fe3+
Find the calculated voltage using the equation Eo(Cell)=
-(0.0592/n) log (dilute/concentrated)
Cell A 0.0100M 1.00M
Cell B 0.0100M .100M
Cell C 1.00M 0.100M
I don't understand anything, please help.

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