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D-Flaim
11/8/2011 08:50:39 am

When I did my empirical formula I got that there was 2 mols of oxygen, but oxygen is diatomic so would I write that (O2)2? But those 2s would be subscripted.

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Rourke
11/8/2011 09:17:03 am

Flaim Test: there is one mole of O2 which is 32 grams. Does that sound right? If you had 64 grams, then you'd have 2O2 not (O2)2. Got it?

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jake
11/8/2011 09:31:58 am

uhh dont we just use the one formula with the train tracks u gave us a example wont that just work out for the magnesium lab we did with this one or will it we different?


.05g mg/1mol Mg/2moles MgO/MgO 40.3g MgO
/24.3g Mg/ 2 moles Mg/ 1 mole MgO

were (.5 X 40.3g)
----------
24.3

= 0.8g MgO will that example work with this one to?

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jake
11/8/2011 09:38:19 am

ALSO I FORGOT ANOTHER QUESTION can time be used as an error for like we had original data but was lost when we ran out of time and couldent complete the data. So we had to obtain different data that effected the results we would of gotten?

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jake
11/9/2011 06:03:16 am

how will i explain the train tracks?

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Jake.b
11/9/2011 06:47:54 am

I suggest if u still have ur old lab reports to look back at them they may be helpful still in this experiment.

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Renae
11/9/2011 07:28:09 am

Jake, Use an excel document to write out the train tracks or summarize it into an equation. But make sure to explain it fully in the Data analysis section.

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Brittany Groen
11/9/2011 09:07:08 am

Rourke: My partner and I had to do a different procedure so when we are doing the post lab assignment do we have to answer the questions in the lab that deal with the procedure that was on the worksheet?

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Brittany
11/9/2011 09:29:04 am

I do not get question 7 on the post lab assignment. When was nitrogen present in this lab??

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Ashley C
11/9/2011 11:39:07 am

I dont know y bout our magnessium lost grams :( anyone else get that?

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Michael Patrick
11/10/2011 12:00:16 am

It may be because your magnesium ribbon may have gotten caught on fire and it burned some of the Mg into the atmosphere if you had your cover too open.

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Michael Patrick
11/10/2011 04:06:57 am

Rourke,
do you want the reaction formula stated in the equation format, As if it were a mathematical function?

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Michael Patrick
11/10/2011 04:34:59 am

Brittany, what it saying is that the majority of air is nitrogen gas, so some of it had to have been bonded into the heated Mg ribbon. look here:

http://www.elmhurst.edu/~chm/vchembook/143Amgoxide.html

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Rourke
11/10/2011 09:02:53 pm

Michael: Sure, if you put the chemical equation into the lab report in the Equation format that would look nice.

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Caleb
11/19/2011 09:40:52 pm

what were the little glass sticks that we used to decant called

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Michael Patrick
11/20/2011 01:48:23 am

Rourke, What were the glass decanting rods called?

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Anyssa
11/20/2011 02:51:31 am

SINGLE REPLACEMENT LAB

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brittany and anyssa are awesome.(;
11/20/2011 02:53:03 am

The rods were called glass stirring rods. And there was approx. 30 mL of solution in the glass bottle before reacting.

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Michael Patrick
11/20/2011 03:16:37 am

Rourke, I have a higher actual yield than my theoretical. My math is correct, but I have .94 actual and .512 theoretical. This is not supposed to happen right?

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anyssa
11/20/2011 04:07:32 am

On all the websites I looked up, the equation for the percent yield was actual/theoretical.

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mianna
11/20/2011 05:49:01 am

ya i saw that too anyssa. what is everyone putting in data analysis? so far i just have why the reaction occurs and why the actual and theoretical are different.

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Michael Patrick
11/20/2011 05:52:46 am

That is for a ratio between the two, not an actual percentage. the one rourke gave us is the correct one. either way I am getting a negative percentage because of my messed up numbers.

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Michael Patrick
11/20/2011 05:53:37 am

I think that is it Mianna. I can't think of anything else.

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