Is water a solution? If so explain. Im very serious.
2 answers:
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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:
*2Kl+Pb(NO3)2=PbI2+2KNO3: double replacement.
*2Al+3CuSO4=Al2(SO4)3+3Cu: single replacement.
*C2H5OH+3O2=2CO2+3H2O: combustion.
Explanation:
Hello there!
In this case, according to the required, it turns out necessary for us to recall the five types of reactions, combination, decomposition, single and double replacement and combustion as shown on the attached figure.
In such a way, since the first reaction follows the pattern AB+CD-->AD+CB we infer it is double replacement; the second reaction follows the patter A+BC-->AC+B and therefore it is single replacement; and the last one follows the pattern of combustion reaction due to the presence of CO2 and H2O on the products side.
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