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Topic: Chemistry 12 Exam  (Read 15389 times)

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Offline ghostanime2001

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Re: Chemistry 12 Exam
« Reply #15 on: March 12, 2009, 01:44:49 AM »
when you say this :

Quote
Looks OK.

Are you suggesting there is a better way of solving the problem ? or you didnt think thats how i would solve it ?

Offline Borek

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Re: Chemistry 12 Exam
« Reply #16 on: March 12, 2009, 03:40:30 AM »
I mean that the solution seems to be correct.
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Offline ghostanime2001

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Re: Chemistry 12 Exam
« Reply #17 on: March 12, 2009, 07:01:57 PM »
A couple questions:

1. Why cant i use mol ratios from Al2(SO4)3 to Al(OH)3 and then break up that into its ions and calculate # of moles of each ion once this ionic compound breaks down and then divide my 1.125 then sub that into my Qsp expression?

Al2(SO4)3 ---(mole ratios)-->Al(OH)3---(break apart into Al3+ and OH-)--->Calculate # of moles of each ion divide my 1.125 L solution----(sub molarity values into Qsp)-----> compare ppt or no ppt

Offline Borek

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Re: Chemistry 12 Exam
« Reply #18 on: March 12, 2009, 07:20:31 PM »
I am not sure what you are talking about, but intuition tells me you are again trying to use stoichiometrical coefficients from the reaction equation to calculate concentrations. If so, you will get different concentrations of reactants depending on the reaction, while reactant concentration depends only on amount of teh substance dissolved. Amount of products, that's different thing.
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Offline ghostanime2001

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Re: Chemistry 12 Exam
« Reply #19 on: March 12, 2009, 07:29:22 PM »
yes i am trying to use stoichiometric coefficients to calculate concentrations only for Al(OH)3 from the stoichiometric coefficient of Al2(SO4)3 then breaking up Al(OH)3 into ions and and calculating mols of each ion divided by the volume of NaOH solution cuz the ions are in solution and at equilibrium with the undissolved solid.

Offline ghostanime2001

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Re: Chemistry 12 Exam
« Reply #20 on: March 18, 2009, 03:05:34 PM »
Are you saying that stoichiometric coefficients only involve the amount of substance a person has in his/her hand while concentration means the amount of dissolved in a solvent. Okay, i sort of get it. Concentration of a substance might not always mean ALL of the substance dissolves in the given solvent,, maybe only a part of it. But the stoichiometric coefficient tells you the ACTUAL amount of substance in the lab or in your hand correct?

Offline ghostanime2001

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Re: Chemistry 12 Exam
« Reply #21 on: March 20, 2009, 04:17:33 PM »
I know the idea kinda seems twisted and confusing but in a nutsell this is what im asking. Whats the difference and meaning of stoichiometrical coefficients and concentration. Also, why cant i use one with the other ?

Offline ghostanime2001

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Re: Chemistry 12 Exam
« Reply #22 on: March 20, 2009, 04:45:51 PM »
When a insoluble compound on the products side of an eq rxn has its ions in two different compounds on the reactant side. What type of precipitation problem do we call this ?

For example the balanced rxn is:
Al2(SO4)3(s)     +     6NaOH(aq)            3Na2SO4(aq)    + 2Al(OH)3(s) 

What type of precipitation problem is this called ?     

Offline ghostanime2001

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Re: Chemistry 12 Exam
« Reply #23 on: July 04, 2009, 07:16:52 PM »
A quick question:
1. Which [OH-] do we use ? [OH-] = 0.015 M as was stated in the problem or calculate [OH-] from mole ratios of Al2SO4 ? and also explain why because i dont understand

Offline Borek

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Re: Chemistry 12 Exam
« Reply #24 on: July 05, 2009, 03:41:13 AM »
[OH-] = 0.015 M

This one. That's what you have in the solution when you mix everything.

If the precipitate forms, final conentration of OH- will be different, but you are not asked to calculate it.
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Offline ghostanime2001

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Re: Chemistry 12 Exam
« Reply #25 on: July 05, 2009, 10:52:46 PM »
my question is why can't I calculate [OH-] from Al2(SO4)3 ?

Offline Borek

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Re: Chemistry 12 Exam
« Reply #26 on: July 06, 2009, 04:06:32 AM »
Please elaborate - no idea what you refer to.
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