Explanation:
<h3>yes, Radioactive decay involves the emission of a particle and/or energy as one atom changes into another. In most instances, the atom changes its identity to become a new element.</h3>
How much of a 3.75 M KI solution would you need to prepare 252.5 mL of a 0.780 M KI solution?
1 answer:
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Answer:
Average atomic mass = 10.812 amu
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
<u>For first isotope, Boron-10: </u>
% = 19.8 %
Mass = 10.013 amu
<u>For second isotope, Boron-11: </u>
% = 80.2 %
Mass = 11.009 amu
Thus,
<u>Average atomic mass = 10.812 amu</u>
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