Today's class was comprised of a lab, "Determining the Percent of Acetic Acid in Vinegar." By determining the certain data from the two titrations, we'll figure out the percentage. I'm not too sure how right now, but probably tomorrow. Oh, don't forget to do the Pre-lab, I'm pretty sure we have to do it. For those who weren't present, the lab was quite easy. That's all i guess. (okay this isn't the "best" post, but I tried)
Jennie to scribe for Thursday
Showing posts with label romina. Show all posts
Showing posts with label romina. Show all posts
Wednesday, May 30, 2007
Tuesday, May 1, 2007
THURSDAY'S SCRIBE
Hey guys excuse the late scribe, no net for me. On Thursday, most of you know we had somewhat of a review. In the beginning we were all "reminded" of what equilibrium was. IT IS WHEN THE RATE OF THE FORWARD REACTION = THE RATE OF THE REVERSE REACTION.
Necessities Include;
- must be a closed system
- applies only to a gas and aqueous reactions
- reversible
- dynamic
- rates are equal
That same day, we received a handout on Le Chatelier's Principle. remember he stated that "WHEN A SYSTEM IS SUBJECTED TO STRESS, THE SYSTEM WILL ADJUST SO AS TO RELIEVE THE STRESS." Also, there were various "rules" to follow to figure out if the product or reactant was favoured. Some factors included changing concentration, pressure changes and temperature changes. Don't forget what we were reminded several times, "TEMPERATURE IS THE ONLY FACTOR WHICH WILL CHANGE THE VALUE OF KC" Also, something tricky, adding a catalyst to a system, lowers the activation energy of both the forward and reverse reaction equally thus it will not effect the equilibrium position. Although, if it wasn't at equilibrium the catalyst would make the system reach equilibrium much faster.
Necessities Include;
- must be a closed system
- applies only to a gas and aqueous reactions
- reversible
- dynamic
- rates are equal
That same day, we received a handout on Le Chatelier's Principle. remember he stated that "WHEN A SYSTEM IS SUBJECTED TO STRESS, THE SYSTEM WILL ADJUST SO AS TO RELIEVE THE STRESS." Also, there were various "rules" to follow to figure out if the product or reactant was favoured. Some factors included changing concentration, pressure changes and temperature changes. Don't forget what we were reminded several times, "TEMPERATURE IS THE ONLY FACTOR WHICH WILL CHANGE THE VALUE OF KC" Also, something tricky, adding a catalyst to a system, lowers the activation energy of both the forward and reverse reaction equally thus it will not effect the equilibrium position. Although, if it wasn't at equilibrium the catalyst would make the system reach equilibrium much faster.
Tuesday, March 6, 2007
Missing Scribe For Friday March 2 (sorry)
Well, on Friday, we received worksheets. We had an option to either hand it in on Friday or Monday, but now it would be too late. Well the sheet that was handed in was a review, to check if you know what oxidation and reduction was. The other was about Balancing Equations using a different method, if a redox reaction occured. Here is a guide to interpreting that sheet:
Firstly, find the oxidation numbers, but remembering to use the "Rules for Determining Oxidation Numbers."
Lets use Ag2S + Al -> Al2S3 + Ag
Ag: is reduced because it is gaining one electron. +1 to 0 (more negative number)
S: remains the same, so it isn't oxidized.
Al: is oxidized because it is losing three electrons. 0 to +3 (more positive number)
Multiply the subscripts with the number of electrons lost or gained.
Therefore:
Ag2S: 2(because you have two Ag atoms) X 1 electron (number of electrons that was gained)
Al: 1(because you only have one Al atom) X 3 electrons (number of electrons that was lost)
Now we have to balance. So that we have the same number of electorns gained and lost. We find their lowest common multiple...Hm...2 and 3, 6 goes into both, so lets multiply both to get 6:
Ag2S: 2 electrons X 3 = 6
S: 3 electrons X 2 = 6
Use what you multipled there as your coeffcients: so 3 for Ag2S and 2 for S:
3Ag2S + 2Al -> Al2S3 + Ag
Note: Check both sides are balanced, you may need to balance the rest of the equation. For this case, since you have 6Ag on the reactant side, then you need a 6 in front of the Ag in the end product side. That is all.
Therefore, your answer should be:
3Ag2S + 2Al -> Al2S3 + 6Ag
Firstly, find the oxidation numbers, but remembering to use the "Rules for Determining Oxidation Numbers."
Lets use Ag2S + Al -> Al2S3 + Ag
Ag: is reduced because it is gaining one electron. +1 to 0 (more negative number)
S: remains the same, so it isn't oxidized.
Al: is oxidized because it is losing three electrons. 0 to +3 (more positive number)
Multiply the subscripts with the number of electrons lost or gained.
Therefore:
Ag2S: 2(because you have two Ag atoms) X 1 electron (number of electrons that was gained)
Al: 1(because you only have one Al atom) X 3 electrons (number of electrons that was lost)
Now we have to balance. So that we have the same number of electorns gained and lost. We find their lowest common multiple...Hm...2 and 3, 6 goes into both, so lets multiply both to get 6:
Ag2S: 2 electrons X 3 = 6
S: 3 electrons X 2 = 6
Use what you multipled there as your coeffcients: so 3 for Ag2S and 2 for S:
3Ag2S + 2Al -> Al2S3 + Ag
Note: Check both sides are balanced, you may need to balance the rest of the equation. For this case, since you have 6Ag on the reactant side, then you need a 6 in front of the Ag in the end product side. That is all.
Therefore, your answer should be:
3Ag2S + 2Al -> Al2S3 + 6Ag
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