Asked by Mary
college chemistry-absorbance of an equilibrium mixture
Problem: A student mixed 4.00mL of 1.02x10^(-1) M Fe(NO3)3 with 100.00mL of 1.98*10^(-4) M KSCN, using 5.0*10^(-1) M HNO3 as the solvent for both solutions, and found the absorbance of the resulting equilibrium mixture to be .253. Given these data, calculate the following:
1. [Fe*]
2. [SCN*]
3.[Fe*] + [SCN*]
4.[Fe*][SCN*]
5.A([Fe*] + [SCN*])
6.A / ([Fe*][SCN*])
7. (A([Fe*] + [SCN*]))/([Fe*][SCN*])
I would like the process explained to me, not the answers! (please)
thanks.
Problem: A student mixed 4.00mL of 1.02x10^(-1) M Fe(NO3)3 with 100.00mL of 1.98*10^(-4) M KSCN, using 5.0*10^(-1) M HNO3 as the solvent for both solutions, and found the absorbance of the resulting equilibrium mixture to be .253. Given these data, calculate the following:
1. [Fe*]
2. [SCN*]
3.[Fe*] + [SCN*]
4.[Fe*][SCN*]
5.A([Fe*] + [SCN*])
6.A / ([Fe*][SCN*])
7. (A([Fe*] + [SCN*]))/([Fe*][SCN*])
I would like the process explained to me, not the answers! (please)
thanks.
Answers
Answered by
Anonymous
ClO32- + cl- = cl2 + ClO2 balanced net ionic equation
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