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

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

12/13/10

Announcements: Study for test on Thursday.

Homework: chem packet (due Wednesday) , and finish double replacement lab

Today in class we worked on the Double Replacement Lab in class. Mr. Paek explained how to mix the elements to see if there would be no reaction or a reaction. To do the lab you need to set up a data table that shows all the chemicals. Like this.

Then you mix each chemical once with the ones on the the grid and mark an which ones have a reaction, if there is no reaction write NR.

Next scriber: Mike A.

12/8/10

Announcements: do your homework and study! Quiz tomorrow!

Homework: Complete chemthink tutorial and quesitons (chemical reactions)!

On this day (late arrival) we learned about different reaction types:

1) Single Replacement

2) Double Replacement

3) Synthesis

4) Combustion

5) Decomposition

During single replacement, one element replaces another element in a compound. This kind of reaction consists of three elements: AX+Y. In this case A and X attract because they have the opposite charges (+ and -) while Y has a possitive charge. During single replacement, the product will end up being AY+X because now element Y has replaced element X. Example:

Since both copper and iron have a positive charge, they switch.

During double replacement, the positive charged elements and the negative charged elements switch places. This kind of reaction contains 4 elements: AB+XY. In this case A and X have positive charges and B and Y have negative charges. During the reaction, these elements switch pairs: AY+XB. To make it simpler, here is another example. There are two couples: AB and XY. A(boy) decides he doesn't like B(girl) and starts dating Y, and vice versa. Another example:

During synthesis, a 2 elements join together to form a complex compound. This reaction type is much easier to learn. An example: A+B=AB.

During decomposition, a complex compound seperates into two elements. Its the exact opposite of synthesis. An example: AB=A+B

Combustion is the reaciton of oxygen gas with anything. Combustion always and only contains oxygen carbon and hydrogen. It is the easiest reaction to learn because the answer is always CO2 + H2O. After this, all you have to do is balance the equation. Example:

Next scriber... IS PEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEETER I.

11.22.10

Announcements:

1. Lab test tomorrow

2. Quizzes on Monday and Tuesday after break

3. Unit test on Wednesday (12/01)

Homework: none.

Today, we did a lab for most of the time, and then we took a quiz with our lab group. There were two parts to our lab: Part 1 - Solubility and Part 2 - Volatility and Surface Tension.

In part one, we were given six liquids. The six liquids were hexane, ethanol, pentanol, methanol, butanol and acetone. We put each one in their own test tube, and put a few drops of water down the side of each test tube. We had to determine if the liquid would mix immediately with the water and if they stayed mix after shaking the test tube. We also recorded down our observations.

After filling out the table, we answered two questions about this part of the lab. The first question was why doesn't hexane mix in either step 3 or step 5? In step 3, we recorded whether the two liquids mixed immediately. In step 5, we recorded whether the liquids stayed mix after they had been shaken. The hexane did not mix with the water in either steps. The reason for this is because water is polar, and hexane is nonpolar. Previously, we learned that "like dissolves like", and "unlike stays separate". Therefore, because water is polar and hexane is nonpolar, the two liquids did not mix.

The second question that was answered was to explain the difference in our results for ethanol and pentanol. We also had to answer which mixed more easily, ethanol or methanol.

We had several differences in our results for ethanol and pentanol. Both did not mix immediately. After shaking, the ethanol mixed with the water where as the pentanol did not completely mix with the water. We observed that after mixing, the ethanol was clear and kind of yellow, while the pentanol was white and hazy. The methanol also mixed more easily with water because the methanol is more polar than the ethanol.

In the second part of the lab, we had to measure how much each liquid spread out and how fast each liquid evaporated. We put a drop of each of the six liquids on the table. We ranked them in order of 1 to 7 with 1 being "spreads out the most" and "evaporates the quickest", and 7 being "spreadas out the least" and "evaporates the slowest". We learned that there was correlation between whether a liquid was polar or nonpolar, how much each liquid spread out and how fast each liquid evaporated. We answered three questions related to the second part of this lab.

For the first question, we had to explain how the amount a liquid spreads out is related to the polariy of that subance. From observing each of the liquids spread out and evaporate, we recorded data which helped us find the correlations. We found that the more polar a substance is, the less it spread out. For example, water is very polar, and it spread out the least. We ranked it number 7. On the other hand, hexane is very nonpolar, and it spread out the most. We ranked it number 1.

The second question asked us to explain how the rate of evaporation is related to the polarity of that substance. We found that the less polar a substance is, the faster it evaporates. For example, water is very polar, and it took the most time for it to evaporate. But hexane is very nonpolar, and it evaporated the quickest.

Our third question was to list other factors (other than polarity) that might influence the rate of evaporation. I think an important one is temperature. If we spill water, and it's not very humid or humid outside, it takes a lot longer for the water to evaporate than if it's hot and humid. So I think that the temperature plays an important role in how long it takes substances to evaporate.

After we finished our lab, we took a short quiz together with our lab group on things that we have learned recently. Tomorrow, Mr. Paek will try to find the "weakest link" in each group, and ask that person to explain how we got our answers.

11.18.10

Announcements:

1. Tomorrow is Chem-day in the Lyceum.

2. On Tuesday, the day before break, we will have our lab test for this unit.

AFTER BREAK

3. On Monday 11/29/10, we will have a quiz.

4. On Tuesday we will have another quiz.

5. On Wednesday, December 1, we will have our unit test.

Homework: none.

Today in class we first got pages 25 through 30. Then we officially begin the class by going over the worksheet that we worked on Wednesday with the substitute. There were two major problems that we worked on which were C2H2 and C2H6O. For C2H2, the correct Lewis structure was: H-C=-H. (The double bond is suppose to have another line on top, so after the C is three lines.) The second problem that students were confused about was C2H60. The Lewis Structure for this molecule is a bit too complicated, but it not something you should worry about so much.

Next we went over polar and non polar. As a reminder, polar means unequal or different, and non polar means equal or same. Make sure that you understand those two terms completely so that the future will be easy for you. After we reviewed a little bit about non polar and polar, we did page 25. Then we started a lab.

This lab starts on page 27 and is called "Polarity Olympics: The Trials". First on the page, we decided what the charge of each molecule was in H2O. The two Hs' have an electronegativity charge of 2.1 The oxygen molecule has a charge of 3.5 and because it has a higher electronegativity, it has a negative charge. The two hydrogen atoms have a positive charge. After doing the first page together as a class, we then finally began our lab.

(Page 27)

In the lab, we used a penny, water, hexane,watch glasses, capillary tubes, markers, and two cotton swabs. For the first part of the lab, we put many drops of water onto a penny until it overflowed. We recorded the number of drops and then drew a side view of the penny. We did the same procedure twice a second time, but instead of water we used a liquid called hexane. From our results, water had the most drops than hexane. We observed that hexane evaporated a lot faster than water did when it touched the table.

For the second part of the lab, we used capillary tubes and watch glasses. We put 10 drops of water onto one watch glass and hexane onto another. The we took the capillary tubes and touched them with each liquid separately. The capillary tube that held the most liquid was hexane.

The third part was easy just like the previous two parts. The point of this trial was to clean two types of marker from the glass. On one watch glass we drew a line with permanent marker and on the other, we drew on it with a transparency marker. Then we put water on a cotton swab and wiped the transparency marker. The transparency mark went away and the cotton swab turned green. With the same swab, we wiped the permanent marker, but it didn't go away. Next we put five drops of hexane onto a new cotton swab and wiped the transparency mark, but it didn't erase. When we wiped the permanent mark with the hexane filled swab, the mark went away. From our data, we concluded that water erased the transparency mark, but not the permanent mark and hexane erased the permanent mark , but not the transparency mark. The water and hexane were opposites.

This was the lab we did today and it was pretty easy. We ended the class by cleaning up our mess. Also our lab test is most likely going to be somewhat similar to this one! If you have any questions about anything, be sure to ask Mr. Paek for help!

Next Scriber: Petrina Z.

11.09.2010

ANNOUNCEMENTS: 1. There will be a total of 6 quizzes, one everyday 2. If you're absent before a quiz, that quiz will be the one that's dropped 3.You can drop one quiz and retake one quiz 4. We taped in pages 9 to 15

HOMEWORK: Finish pages 11 to 13

Today in class we learned how to make Lewis dot structures for formulas like CH4 or F2 and others like the examples below:

Before drawing the actual structure, Mr. Paek suggests doing N, H, S and B every time because it makes it a lot easier. These letters stand for Need, Have, Share and Bonds. 

• NEED: you put the number of electrons needed to be stable, for most elements it's 8, but for Hydrogen it's 2. Then add all of it together.
• HAVE: the number of valence electrons then add it all up. 
• SHARE: subtract the number of valence electrons from the sum of the electrons needed. 
• BONDS: you divide the number shared by 2.

For example: The formula SeF2

• N= 24 (Se needs 8 electrons and F2 needs 2 sets of 8 because of the subscript "2", so you do 8+8+8=24)
• H= 20 (Se has 6 valence electrons and F2 has 7 and 7 because of the subscript "2", then you do 6+7+7=20)
• S= 4 (N-H or 24-20=4)
• B= 2 (4 divided by 2)

Once you're done with all that, you have to draw the structure.Using the same formula SeF2, Se would be in the center with two F's on either side, it doesn't really matter what side it's on as long as there are two F's. The bonds represent the lines from Se to the F's. Each line representing the number of electrons shared, which in this case is four, so two lines represent four electrons shared. The picture also has to have 20 total electrons in it because of the have. Since the lines already represent 4 electrons, you need 16 more, so you draw dots around the Se and the F's. I know this may sound confusing, but the picture below should make it easier to understand. The formula I used for this example is the first one below:

NEXT SCRIBER: Seena K.

11.8.10

Announcements- 6 QUIZZES this chapter, Chemthink & Webassign MUST be done before class. We picked up the calender and pages 1-8 in class today.

Homework- Pages 1-8, except 3(most finished in class)

Today in class we learned about covalent bonds. We observed the “potential energy curve” with covalent bonds.




Using a graph like this, we explain that when two atoms are moved far apart to close together the potential energy of the two atoms decrease, making the stability (how stable the atoms are) increase. This would make the electrostatic force a proton-electron attraction meaning that one proton from one atom attracts an electron from the other atom.

When the two atoms move from close together to far apart the potential energy of the two atoms would increase, making the stability (how stable the atoms are) decrease. This would cause a proton-proton repulsion meaning that the protons repel as they get too close to each other.

Next in class we talked about naming covalent bonds. Covalent bonds are ONLY nonmetals. You don’t need to worry about figuring out the charges. When naming the first element NEVER has a mono, but the second one will.

We use these prefixes:
1-mono 2-di 3-tri 4-tetra 5-penta 6-hexa


If the element starts in a vowel and the prefix ends in a vowel take out the prefix vowel, for example Carbon Monoxide.

• Here are a few examples of naming covalent bonds:
CO- Carbon Monoxide
CF3- Carbon Tetrafluoride
N2O3- Dinitrogen Trioxide
AsCl3- Arsenic Trichloride

*Remember to still use –ide at the end of second element





When figuring out if it is an Ionic or Covalent bond look for these:

• Polyatomic Ions (SO4-, NO3-, NH4+, etc.)
• Transition Elements (Cu, Fe, Pb, etc.)
• Metals (Ca, Al, Na, etc.)

Here are some examples on figuring out ionic (I) or covalent (C) bonds and writing the formulas:

• Nitrogen Dioxide C NO3
• Copper(II) Phosphide I Cu3P2
• Sulfur Trioxide C SO3
• Chromium(III) Bromide I CrBr3








Next Scriber- Gennah L

11. 1. 2010

• At the beginning of class, Mr. Paek collected any quizzed people needed to take home on Friday and the extra credit unit 4 packets. He also check in the homework that was assigned over the weekend which was pages 9-13 in the journal.
• We went over the homework (pages 9-13) and if people had any questions on the homework they were answered.
• Mr. Paek taught the class a short cut on writing the formulas for compounds. For example if the compound is cobalt (III) carbonate, since cobalt has a charge of +3 and carbonate has a charge of -2, in order to write the formula you can just switch the charge numbers, ex. Co2Ca3
• For the majority of class we worked on our journal pages alone and asked questions as needed.
• At the end of class we took a short quiz testing out ability to write compound formulas and name compounds etc.

Don't forget to read pages 253-258 in the book and answer question number 7 on page 258.

Also we have 2 web assigns.

10.29.10

At the beginning of first period, Mr. Paek tough us a little on ionic formulas and how to figure out the formulas for different compounds. For example, if we had a K+ and a O2- then we would need to get potassium's charge the opposite of oxygen and to equal zero. Here are some other Examples of ionic formulas:

Ex:
Co3+ and Cl- = CoCl3-
K and N= K3N



Mr. Paek also talked about Polyatomic Ions and their symbols. Polyatomic ions are ions that such as Sulfate, Nitrate, Phosphate, Etc... Mr. Paek also went over how to name the formulas Such as CaCl2 would be calcium chloride. He also noted that if there was a roman numeral in front of the first named element that it was the charge. After we walked everything trough we took a quiz on a few things we've already talked about its nothing to worry about. after the quiz Mr Paek assigned some homework.

Homework:
Worksheets handed at beginning of class. pages 11, 12, 13, and 14.

10.13.10

Wednesday: late arrival! :) 35 min period

At the beginning of the period, Mr. Paek showed us a few vidoes on metal reactivity. He explained how the periodic table consists of different families:
Row 1(downward) - Alkali metals
Row 2(downward)- Alkaline earth metals
D Block(sideways)- Transition metals

The vidoes consisted of experiments about the reactions of the alkali metals and water. The farther down the row in the periodic table, the stronger/more intense the reaction. In one of the videos, it included an experiment with a glass. The metals were put in the water one by one:
Li- Floated on the water while giving off hydrogen
Na- The same as Li, but a more vigorous reaction. Na gave off much more hydrogen
K- Produced sparks and a small fire
Rb- Produced bigger sparks and bigger fire
Cs- Completely shattered the glass cup

After watching the vidoes, Mr. Paek introduced a short lab. The metals that were being tested were calcium, magnesium, and aluminum. We placed these metals into well plates and used water, HCL, phenolphthalein indicator (PHTH) (liqiud) to see how they would react with the metals. Her is the data:
Appearance:
Ca- little rocks
Mg- thin metal sheets
Al- Flat rocks

PHTH:
Ca- water turns pink
Mg- magnesium sheet turns pink
Al- water becomes cloudy

Water:
Ca- sizzles and dissolves
Mg- nothing
Al- nothing

HCL:
Ca- sizzles and disolves a little
Mg- sizzles and gives of hydrogen
Al- nothing

PHTH and HCL
Ca- turns pink
Mg- nothing
Al- nothing

NO HOMEWORK!!!
Next Scriber: Sal :)

Thursday 9/30/10

part one of the lab
part two of the lab

data collected from lab completed in class

Today in class, we were given six sheets, two for the lab we did in class, the other four we didn't get to yet.

The lab todaywas the Rutherford Simulation Lab, in the lab we took a piece af paper with six circles on it, and six small squares at the center of each circle. We layed a piece of carbon paper on top, and with a partner, bounced a marble on top of the paper 110-120 times. After we finished, we counted the number of dots: in the circles, in the squares, and out side of the circles.

Using the collected information we found various data, like the percentage of dots that landed in the circles, for example me and my partner got 69 dots in the circle, out of 120 dots. to get the percent divide the number dots in the circle to the number of total dots: 69/120=.575 or 57.5%.

then using the total area of the paper, 93.5in. sq., and the percent of dots that landed in the circle, we found an estimated area of all 6 circles. We did this by multiplying the area of the paper times the percentage in decimal form: 93.5*.575=53.76in. sq.

Using the information we just found, we divided by six to find the next question, which was, what is the area of one circle: 53.76/6=8.96in. sq.

Then we solved to find the radius. To do this divide bothe sides by pi, then the square of your answer. To solve for diameter, multiply the answer you just got by 2. 8.96/3.14159=2.85 then the square of 2.85= 1.689in is your radius. 1.689*2= 3.378in is your diameter.

For problem 7, use a similar process to find the area of a square, like finding the area of a circle. First divide the number of dots in the squares by 120, i got 2/120=1.56, then divided by 6 to get one squarer alone. 1.56/6=.259in. sq.

to find the length of a side, take the square root of the area of one square.

Question Answers:


What does the paper represent?

the gold foil

What does the circle represent?

the atom

What does the marble represent?

the alpha-particle

What does the square represent?

the nucleus of the atom

The true diameter of each circle is 3.2 inches. find your percent error. to do this subtract your value minus the true value, then divide by the true value, and multiply by 100.

>>> (3.378)-(3.2)=.178/3.2=.055625*100=5.5625%

What could account for your own error in this lab?

>>>answers may differ. ex: not enough data collected.

Homework: Finish Question Four of the Lab, Test next Friday, October 8th, 2010

Next Scriber: Lauren

Wednesday 9/29/10

Modern View-electrons can be in ground or excited states and can jump between levels. When electrons drop levels (high to low) they give of a beam of light! Its like bees in a beehive and bees are electrons moving around the nucleus like a cloud. Similar to the picture above

Niel Bohr from 1914 discovered that electons(-charge) "ORBIT" around positive nucleus in different- ORBITALS(levels). Electrons can jump up and down these levels. As seen above.

Rutherford discovered all atoms have tiny positive centers(nucleus).

Ernest Rutherford from the early 1900's shot alpha particles(+ charge) through gold foil as shown in the picture above. Most of them went through but some deflected and these some were - charged particles!

Joseph John Thomson or J.J. from 1897 studied cathode ray tubes. He would excite the gas in the tubes with electricity. He would hold a magnet to the sides of the tube noticing a bend in the light (+ side attracted beam, - side repelled beam). He discovered that atoms must have a negative(-) charge and if that, then with overall neutrality, they must also have a positive(+) charge. This is demonstrated in his model above.John Dalton was an english schoolteacher from the early 1800's. He thought that each element was mad of elements, an atom of one element would be identicle to an atom of the same element. Also atoms do not include the existence of the nucleus, dosen't explain the existence of ions or isotopes, and dose not talk about subatomic particles (electrons, protons, and neutrons). His view of the atom looked something like the picture above.

Today we started off class by finishing the notes from yesterday on the Atomic scientists (Dalton, Rutherford, Thomson, and Bohr). You may look at yesterdays scribe for what they look like and foe more help.

After that Mr. Paek gave a lecture on the four scientists, their major contributions, what they discovered about the atom, and their actomic model. As shown above^ NOTE: start from the bottom picture and read up!!

no homework! lab test due by next friday!

Next Scriber: Maddy M.

Tuesday 9/28/10

Black container that contained the bead. There were 12 different containers with differents line patterns inside of them.
(scroll down all the way for more details)

Black box lab

We did this sheet in class. The sheet had four atom scientists' names and we had to fill out information about them. The information can be found on the colored sheets pictured below.

These are the colored sheets of paper with the information on  four atomic scientists. (Dalton, Rutherford, Thomson, Bohr)

2nd part of the black box lab

Today in class, we started a new unit. We began studying the atomic theory. We did the Black Box Lab. In this lab, we had to go to several stations with black circular containers. Each container had a bead inside of it. We would rattle and move the container to determine what type of shapes/lines were inside of the canister. We could determine the shapes by feeling how the bead bounced off the surfaces, and how the bead sounded when it bounced off certain spots of the container. We weren't allowed to open up the container to see the shapes.Then, we would sketch what we guessed was in the container, on our lab sheet. It was sort of like the concept of trying to figure out what atoms looked like, back in the days of Dalton, Rutherford, Thomson and Bohr, who were atomic scientists.

We got four colored pieces of paper with information about 4 different atomic scientists. The information on the sheets of paper can be used to fill out the Atomic Scientists worksheet that we got to do in class.

Next Scriber: Grace Ridge

No homework. Test next Friday, October 8th, 2010

9.27.10

Today in class we did not do anything complicated. It was an easy day and we reviewed the test that we took last Thursday. We also talked about the lab test that we took last Wednesday.

First of all, we started off by getting our journals back from Mr. Paek. He was supposed to check them in for points, but he did not yet because he was a bit too busy. To get an A for the journals just make sure the pages don't fall out and that they should be in a similar order to Mr. Paek's journal. We also got four new pages for our journal for unit two. (The new pages are shown at the bottom.) Please leave a blank page after the first two pages to have a page titled, "Text Questions." He also crossed out the names of people who were scribers, so if you were a scriber already, make sure you tell Mr. Paek.
Afterwards, Mr. Paek said that there were good news and bad news. The bad news was that the class did not do a good job on the lab test. Since our class did poorly, he is giving us an opportunity to get half the points back from the lab. In order to get the points back you can go see Mr. Paek during 5th period, 6th period, before school, or after school to work on a worksheet based on the lab test. You have one week to get points back from the lab test, so the option to do the lab worksheet is open until next Monday October 4th. The good news is that the test results were great! We got back our tests today and went over a few problems. After getting back our test, we also got a grade report. As a reminder, remember that labs/projects are worth 20% and tests/quizzes are worth 50% of your chemistry grade.

--The four new pages:

This is all what we did during class today. If you have any questions make sure to ask Mr. Paek. Also, don't forget to get the four new journal pages! There is no homework tonight!

-Next scriber: Alyssa P.
****When you're done with writing your own scribe post, please write the name of the next scriber that you choose.