Ask your own question, for FREE!
Chemistry 8 Online
kekeman:

More chem please help: https://snipboard.io/bU8BrY.jpg

WhoIsArii:

..

kekeman:

The link don't work?

kekeman:

@tranquility

WhoIsArii:

nope

kekeman:

1 attachment
Tranquility:

Are those the only three answer choices

kekeman:

Yes indeed

Tranquility:

The equation is Co(H2O)6 + 4Cl- <----> CoCl4 + 6 H2O We have two reactants Co(H2O)6 and Cl- We want to see a decrease in either Co(H2O)6 OR in Cl- The way we see decreases are by the sharp straight line and it has to go downwards to signify that it is decreasing

Tranquility:

So I drew out the two possible ways we see a decrease The first one tells us that the other reactant was added into the reaction and so the equilibrium shifts towards the other side and that's why whatever compound it is is being used up to form more products. The bottom one tells us that something was removed and thus the equilibrium shifted towards it. So if you remove a reactant, then it's concentration would decrease sharply (thus the straight line) but now the equilibrium is disturbed and it's going to start forming more of it which is why the line starts to go back up slowly

1 attachment
Tranquility:

Does that make sense?

kekeman:

Ohhhh i see

Tranquility:

Basically, when the equation is in equilibrium all four line should just be a straight horizontal line


But then when you decrease the reactant, the line of that reactant will sharply decrease first and then the equilibrium will shift towards it and cause the reactant to increase again in concentration

kekeman:

Alright so with this information it sounds like its going to be option 1

Tranquility:

So in that drawing I drew, you want to find the answer choice that looks like the bottom one for either Co(H2O)6 or for Cl-

Tranquility:

@kekeman wrote:
Alright so with this information it sounds like its going to be option 1
You got it!

kekeman:

Ohhhhh okkkk i get it nowww

Tranquility:

That's great! We can move on to your next question now

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!