If 45.0 mL of a 0.0500 M HNO3, 10.0 mL of a 0.0500 M KSCN, and 30.0 mL of a 0.0500 M Fe(NO3)3 are combined, what is the initial
1 answer:
Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
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⭐ But the last electron of Zn , Cd , Hg and Cn enters in the s-oribital of their respective ultimate shells rather than the d-oribitals of their respective penultimate shells . Therefore, these elements cannot be regarded as d-block elements .
☃️ But properties of these elements resemble to the d-block elements rather than s-block elements .
☃️ Therefore, to make the study of periodic classification of elements more rational, they are studied along with d-block elements .
✍️ Thus <u>on the basis of properties</u> all transition elements are d- block elements, but <u>on the basis of electronic configuration</u> all d -block elements are not transition elements .
