Asked by Rina
Chemical Reaction System Unit Test
Use the dissociation reaction to answer the question.
Fe SCN2+(aq) = Fe3+(aq) + SCN- (aq)
in the reaction. Fe SCN2+ ions are red, Fe3+ ions are pale yellow and SCN-ions are colorless A solution at equilibrium is yellow-orange. The concentration of SCN ions is increased by a moderate amount Which color will the solution most likely become? 1 point)
o orange
O yellow
o red
o colorless
Use the dissociation reaction to answer the question.
Fe SCN2+(aq) = Fe3+(aq) + SCN- (aq)
in the reaction. Fe SCN2+ ions are red, Fe3+ ions are pale yellow and SCN-ions are colorless A solution at equilibrium is yellow-orange. The concentration of SCN ions is increased by a moderate amount Which color will the solution most likely become? 1 point)
o orange
O yellow
o red
o colorless
Answers
Answered by
heyo ❄
i think its B: yellow
Answered by
DrBob222
@ Heyo. You must have guessed. You had a 1 in 3 chance of getting it right but didn't.
I have two things. The first is the questions. I have seen many many solutions of SCN^- and they are colorless. They are NOT orange. This is a Le Chatelier's Principle question and not a very good one at that. Le Chatelier's Principle says that if we do something to a system at equilibrium that the system will try to undo what we did to it. SO, if we add SCN^- it will try to use up the added SCN^-. How can it do that? The equilibrium can shift to the left which will use SCN^- and use Fe^3+ and will form more FeSCN^2+. So the end result will be a solution that is more concentrated in FeSCN^+ and less concentrated in Fe^3+. What color will that be? Who knows? Enough SCN^- and the end result will be red. Less than that and you will have red color mixed with the yellow color of the Fe^3+ to give an orange color. So it all depends upon how much SCN^- is added. My best guess is that a MODERATE amount of SCN^- means some Fe^3+ is left and the mixture of yellow and red will produce an orange color. My best guess as to answer the teacher wants is "red". Why? Because the SCN^- is erroneously cited as being orange in the problem and a shift to the right (the wrong way) will definitely produce an orange color. Sorry for getting on my soap box.
I have two things. The first is the questions. I have seen many many solutions of SCN^- and they are colorless. They are NOT orange. This is a Le Chatelier's Principle question and not a very good one at that. Le Chatelier's Principle says that if we do something to a system at equilibrium that the system will try to undo what we did to it. SO, if we add SCN^- it will try to use up the added SCN^-. How can it do that? The equilibrium can shift to the left which will use SCN^- and use Fe^3+ and will form more FeSCN^2+. So the end result will be a solution that is more concentrated in FeSCN^+ and less concentrated in Fe^3+. What color will that be? Who knows? Enough SCN^- and the end result will be red. Less than that and you will have red color mixed with the yellow color of the Fe^3+ to give an orange color. So it all depends upon how much SCN^- is added. My best guess is that a MODERATE amount of SCN^- means some Fe^3+ is left and the mixture of yellow and red will produce an orange color. My best guess as to answer the teacher wants is "red". Why? Because the SCN^- is erroneously cited as being orange in the problem and a shift to the right (the wrong way) will definitely produce an orange color. Sorry for getting on my soap box.
Answered by
Conexuss is grosss
@DrBob222 Nerd.
Answered by
Conexuss is grosss
but thanks:D
Answered by
Helloooo
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