Tuesday, March 15, 2011

Announcements: noneHomework: answer the questions from the lab, finish the graphs

Today we started Unit 12, we did a lab on the heating curve. For the lab we took down the temperature in degrees Celsius every 30 seconds until our ice melted and boiled, then continued to take the temperature for another 3 minutes after boiling.

My data for the lab was:

After we got our data we had to graph what our heating curved looked like. An example of a heating curve looks like this:

After the lab, we cleaned up our stations, and worked on our hw as mentioned above.

Wednesday 2.9.11

Announcements: Instead of having 7 quizzes we'll only have 6. We took a quiz today, and we're taking another  quiz tomorrow
Homework: pg 32-34 due tomorrow, Web assigns, and text questions due before the test 2/15


Today we worked on part one of our lab. We did all of the following steps listed in the picture below. To sum it up, we basically just put copper (II) chloride in water and stirred it until  the copper (II) chloride dissolved. Then we massed the nails and recorded data. After that we put the nails in the solution. We will come back to doing the lab tomorrow.

After we finished part one of the lab, we took a quiz on limiting reactants. When we finished the quiz, we used the rest of the class time to work on our homework. (pg 32-34, mentioned in bold above)

Next Scriber: Maddy M.

2/7/11

Announcements: Study for upcoming quizzes. if you missed Friday's(2/4/11) quiz, make that up as soon as you can!

Homework: journal page 27 #2, 29, 30, & 31

Today we learned more about limiting reactants; what they are, how to find out which one is the limiting one between two reactants, and also, a little more on how to draw the reaction with the product and its excess.

A limiting factor is something that you only have a limited supply, and then have a greater amount of something(s) else, so you can only make a certain amount of the product.

To find the limiting factors, its exactly the same as Stoichiometry, you are just take it one step farther because you are searching for something different in the end. but as a review, you must convert two different reactants into one common reactant. So to start, you will sometimes or usually have one reactant in its state of mass, you then convert it into moles, followed by moles of the common reactants. Sometimes, from here you can convert to the mass, but it really depends on the following question. You then do the same thing to the second reactant, so you should have two equations in the end. Finally, you get your two usual outcomes, just like Stoich, but with those two answers you decide which one is smaller, and the reactant that you started that equation, with the smaller answer, is the limiting reactant.

Below is a picture of an example that shows this process with actual reactants.

Since 0.065 is smaller than 0.5, that makes O2 the limiting factor

Here are some more examples of limiting reactants we did in class.

For drawing out a limited reaction, you start with the equation you are given and the two reactants that you need to determined which is the limiting reactant. You draw out your given amounts of each, and then pair them up for the out come. Which ever one does not have excess, that is the limiting reactant.

This is a picture to help show a drawn out version of limiting reactants.

NEXT SCRIBER: Alyssa P

2/1/11

HAPPY TWO DAY SNOW DAY EVERYONE!

Announcements: There will be a quiz tomorrow reviewing Stoich with Density.
**Remember** If you get 100% on 5-7 quizzes, you will get extra credit points so study hard.

Homework: Finish page 23 and 24

In class we reviewed Stoich. Page 20, "Stoich 3" was our homework but it was kind of hard because it involved density. So in class we also started to learn Stoich with Density (page 23). Mr. Paek taught us how to convert using density like milliliters per grams. A good example of a tough problem that might be on the quiz is shown below (since I can't upload pictures, I just typed it up).


// = FRACTION BAR

EQUATION: C2H6O + 3O2 ---> 2CO2 + 3H2O

2. c) If the density of O2 is 0.00143 g/ml, what mass of ethanol will be needed to react with 950 ml of O2?

950 ml * .00143g of O2 // 1 ml of O2 * 1 mol of O2 // 32g of O2 * 1 mol of ethanol // 3 mol of O2 * 46g of ethanol// 1 mol of ethanol = .65g of ethanol


*** Things to remember:
1. If it says g/ml, that means you start with one of them (in ^^ that case, ML) and put the other unit (^^ Grams) on top.
2. Make sure your units always cancel out. If you have Grams on top, you have to have grams on bottom in the next one.
3. Use the equation to look at how many moles each unit will need to react with the one above or below it
4. Always write the elements name (for ex: 3 mols of >C2H6O<) after how many units it is, UNLESS you want to lose points on a quiz :)


You can refer back to previous posts for more examples on how to do normal stoich without density.

1.27.11

Announcement: None

Homework: None

Today we started with taping in about 30 or more sheets on the new unit. After that we took the quiz on the mole equations. Once everyone was done, Mr. Paek introduced the unit of stoichamerty, but we didn't really get through much. We learned that it's basically like the mole equations, but instead of moles you would switch it for atoms.

Example worksheet:

You can refer back to the previous post to better understand whats on the worksheet.

Next Sciber-Brandon

1.26.11

Announcements: Study for tomorrows mole quiz and the other upcoming quizzes.

Homework: Journal pages 7-8

Today we started with a new mole lab, testing the mass of different elements, how many moles are in the molecule and how many molecules are in the mole. For each of these ways you have to use conversion factors.

For example finding how many moles in krypton.

2.7 x 10^14 x 1mol/6.02 x 10^23

=4.49 x 10^-14

Also finding how many atoms are in an mol.

Ex. Aluminum atoms

89.35 x 6.02 x 10^23/1mol

= 5.38 x 10^15

Next Scriber- Peter

1.24.11

Anouncements: We will be starting to take a variety of quizzes starting this Thursday on the Mole unit. This unit is crucial so if you do not understand go see Mr. Paek or the TLC!!

Homework: Finish the Pair, Dozen, Mole Lab!! (3 pages)

Today in class we started the new Mole unit and learned the concept of a mole in order to solve mole problems. We learned that 1 mole= 6.02 x 10E23 (10 to the 23rd power).

In order to fully understand the concept of a mole, we did a Mole Lab and started by reviewing the fact that a pair is always 2 and a dozen is always 12. Like those, a mole is always 6.02 x 10E23.

During the lab, we practices converting problems involving the atoms mass in grams. Each elements atomic mass on the periodic table is the amount of mass measured in grams per mole. For example, Al has an atomic mass of 26.98.

26.98g of Al = 1 mole of Al

Another way of saying this is that for every 26.98g of Al, there is one mole of Al.

We also were able to add up the masses of two different atoms such as (H2O) two Hydrogen and one Oxygen so that we were able to figure out that there are 18.02g in a single H2O molecule. Like the concept Al's mass of 26.98g being a mole, 18.02g of H2O is equal to 1 mole as well.

Next Scriber: Dan

1.25.2011

Homework: journal pages 1-6





Today we started class by going over the lab from yesterday. We also went over the fact that to convert grams to moles ( or vice versa), you have to convert it to moles first. For example, if you wanted to find the number of molecules in 0.0908 grams of NiCl2 you need to go through at least to steps to solve. First you need to cancel out grams:


1 mole


0.0908 g x ---------


129.59 g





Then, you need to cancel out moles to find your final answer:


1 mole 6.02 x 10^23


0.0908 g x --------- x ------------


129.59 g 1 mole





This equals 4.22 x 10^20 molecules of NiCl2








Next Scriber: Cyril