Where y and x are concentrations of the peptide in the extract and raffinate, respectively in g/L. It is desired to extract at l
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Answer:
the concentration of in the original solution= 0.0088 M
the concentration of in the original solution = 0.058 M
Explanation:
Given that:
The volume of the sample containing Cd2+ and Mn2+ = 50.0 mL; &
was treated with 64.0 mL of 0.0600 M EDTA
Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+
i.e the strength of the Ca2+ = 0.0310 M
Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+
To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :
Volume of newly freed EDTA =
=
= 7.3367 mL
concentration of =
=
= 0.0088 M
Thus the concentration of in the original solution = 0.0088 M
Volume of excess unreacted EDTA =
=
= 8.318 mL
Volume of EDTA required for sample containing and
= (64.0 - 8.318) mL
= 55.682 mL
Volume of EDTA required for = Volume of EDTA required for
sample containing and
-- Volume of newly freed EDTA
Volume of EDTA required for = 55.682 - 7.3367
= 48.3453 mL
Concentration of =
Concentration of =
Concentration of in the original solution= 0.058 M
Thus the concentration of = 0.058 M
Answer : It takes time to produce 3 g of aluminium is, 165 seconds.
Explanation :
As we are given that the mass of aluminium is, 3 grams. Now we have to calculate the total mass of aluminum.
Total mass of aluminum = 3g × 24 = 72 g
Now we have to calculate the moles of aluminum.
Molar mass of Al = 27 g/mol
As, 1 mole of Al has 3 Faradays
So, 2.67 mole of Al has = Faradays
and,
Charge =
Current =
Now we have to calculate the time.
Hence, it takes time to produce 3 g of aluminium is, 165 seconds.