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Construct a table comparing the similarities and differences of Double indicator, EDTA, and Mohr titration in terms of the:
a.) Reactants
b.) Products
c.) Indicator used
d.) End point
e.) Application / Purpose
a.) Reactants
b.) Products
c.) Indicator used
d.) End point
e.) Application / Purpose
A necklace contains 0.250 mol of silver. How many atoms of silver are in the necklace
a stock solution of 4.0mol/L copper 2 carbonate is available. students need to make a 1L quantity of 0.25mol/L solution. calculate the volume of stock solution needed. then calculate what you need to make 1Lof 2.5×10^6mol/L solution. show the calculations and describe the steps necessary to make the new solutions.
Osmium is completely colored in because
a. It is the element with the highest density on the Periodic Table
b. It is in the center of the Periodic Table.
c. It is the standard for the value of metals on the Periodic Table.
d. It exists for only a short while and it cannot be measured for density.
a. It is the element with the highest density on the Periodic Table
b. It is in the center of the Periodic Table.
c. It is the standard for the value of metals on the Periodic Table.
d. It exists for only a short while and it cannot be measured for density.
If 4.00 moles of gas and his pressure of 5.40 atm has a volume of 122 L what is the temperature
If 500g of marble are decomposed completely and knowing that it contains 90% calcium carbonate, calculate the mass of calcium oxide and the volume of carbon dioxide ( at S.T.P.) produced.
What are merits and demerits of liquid phase synthesis, also exppain different methods used in liquid phase synthesis.(Provide a detailed answer and also give the reference of the provided answer)
A 1.3174-g sample containing chloride salts Mg (24.30 g/mol), Hg(II)
(200.59 g/mol) & Zn (65.39 g/mol) was dissolved in 250.0 mL of soln.
A 50.00 mL aliquot was treated with 10 mL as NH3/NH4Cl buffer at pH
10.0 followed by 25.00 mL of 0.05331 M EDTA. After a few minutes of
mixing, the x’ss EDTA was back-titrated with 11.43 mL of 0.01816 M
MgCl2. The second 50.00 mL aliquot was made basic & treated with
x’ss NaCN, complexing both the Hg & Zn. The Mg in this sample req’d
16.83 mL of 0.005583 M EDTA for titration. The solution remaining at the
end pt. of this titration was treated with x’ss formaldehyde which
reacts with free CN- & Zn(CN)4
The liberated Zn2+ req’d 28.47 mL of the EDTA for titration. Calc. % of
each metal in the sample.
(200.59 g/mol) & Zn (65.39 g/mol) was dissolved in 250.0 mL of soln.
A 50.00 mL aliquot was treated with 10 mL as NH3/NH4Cl buffer at pH
10.0 followed by 25.00 mL of 0.05331 M EDTA. After a few minutes of
mixing, the x’ss EDTA was back-titrated with 11.43 mL of 0.01816 M
MgCl2. The second 50.00 mL aliquot was made basic & treated with
x’ss NaCN, complexing both the Hg & Zn. The Mg in this sample req’d
16.83 mL of 0.005583 M EDTA for titration. The solution remaining at the
end pt. of this titration was treated with x’ss formaldehyde which
reacts with free CN- & Zn(CN)4
The liberated Zn2+ req’d 28.47 mL of the EDTA for titration. Calc. % of
each metal in the sample.
The Cd & Pb ions in a 50.00-mL req’d 40.19 mL of 0.004870 M
EDTA for titration. A 75.00-mL portion of the same sample was made to be
basic & treated with x’ss KCN, masking the Cd as Cd(CN)4
2-. This soln req’d 31.45 mL of the EDTA for titration. Calc. conc. of Cd2+
(112.41 g/mol) & Pb2+ (207.2 g/mol) in the sample in ppm.
EDTA for titration. A 75.00-mL portion of the same sample was made to be
basic & treated with x’ss KCN, masking the Cd as Cd(CN)4
2-. This soln req’d 31.45 mL of the EDTA for titration. Calc. conc. of Cd2+
(112.41 g/mol) & Pb2+ (207.2 g/mol) in the sample in ppm.
An antacid tablet weighing 15.468 g was dissolved in acid
& diluted to 500.0 mL. Then, a 25.00 mL aliquot of the soln was
made basic to ppt the aluminum as Al(OH)3.
The remaining magnesium req’d 16.50 mL of 1.042 x 10^-2 M
EDTA for titration. A second 25.00 mL aliquot was
withdrawn from the 500-mL flask & treated with 50.00 mL
of the EDTA. This soln was made basic & the x’ss EDTA back
titrated with 11.75 mL 5.597 x 10^-3 M MgCl2. Calc. % of
both Mg (24.30 g/mol) & Al (26.98 g/mol) in the sample.
& diluted to 500.0 mL. Then, a 25.00 mL aliquot of the soln was
made basic to ppt the aluminum as Al(OH)3.
The remaining magnesium req’d 16.50 mL of 1.042 x 10^-2 M
EDTA for titration. A second 25.00 mL aliquot was
withdrawn from the 500-mL flask & treated with 50.00 mL
of the EDTA. This soln was made basic & the x’ss EDTA back
titrated with 11.75 mL 5.597 x 10^-3 M MgCl2. Calc. % of
both Mg (24.30 g/mol) & Al (26.98 g/mol) in the sample.
