~10 mL 0.100 M Fe(NO 3 ) 3 in 0.2 M HNO 3 ~10 mL 6.00 × 10 -4 M NaSCN in 0.2 M H

ID: 893590 • Letter: #

Question

~10 mL 0.100 M Fe(NO3)3 in 0.2 M HNO3

~10 mL 6.00 × 10-4 M NaSCN in 0.2 M HNO3

~15 mL 0.002 M Fe(NO3)3 in 0.2 M HNO3

~15 mL 0.002 M NaSCN in 0.2 M HNO3

PART A: Obtain about 10 mL of 0.100 M

Fe(NO3)3

in 0.2 M

HNO3

in a small, clean, dry beaker. Obtain about 10 mL of 6.00 ×

104

M NaSCN in 0.2 M

HNO3

in another beaker.

PART B: Obtain about 15 mL of 0.002 M

Fe(NO3)3

in 0.2 M

HNO3

in a small, clean, dry beaker. Obtain about 15 mL of 0.002 M NaSCN in 0.2 M

HNO3

Enter the concentrations of the reactants to be used in each part of the Equilibrium Constant experiment.

Reagent Part A Part B [Fe(NO3)3] Check the number of significant figures.M M [NaSCN] M M

Part A

[Fe(NO3)3]=0.1M

[NaSCN]=6*10-4M

Part B

[Fe(NO3)3]=0.002M

[NaSCN]=0.002M

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