Santa in a Snow Storm!!

Check out this page for the extra credit assingment !!

Bubbles

Today, we did the Molar Volume Lab. Here's what happened.
First, we cut a piece of magnesium in half and measured its length because we would not be able to find the mass on a scale. Magnesium is .01085 g/cm. Our piece of magnesium was 2.5 cm. Then, we tied a piece of copper wire around the magnesium ribbon so that the magnesium would not float around in the eudiometer. (http://www.flickr.com/photos/hc1011/5258241891/) The eudiometer is a gas collection tube, measuring gas by water displacement. We poured 10 ml of HCl and 100 ml of water into the eudiometer. The temperature of the tap water was 21.3°C. Room temperature was 22.2°C and the barometric pressure was 30.11 inHg (we have to convert it to mm). We put the cork attached to the copper wire and magnesium at the open end. We then flipped the eudiometer over, safely and carefully, and put the cork end into a beaker with water and clamped the eudiometer so it would stay up. We watched as the magnesium reacted with the HCl and created bubbles of hydrogen. (http://www.flickr.com/photos/hc1011/5258850050/) Slowly, the water was going into beaker and leaving space at the top of the eudiometer. While we waited for the reaction to finish, we recorded the room temperature and the barometric pressure at room temperature. Once the reaction stopped bubbling, we took the eudiometer to the fume hood and measured the volume of the gas. The volume of the gas was 27.4 ml of hydrogen. (http://www.flickr.com/photos/hc1011/5258850358/)
This lab is due Wednesday. Our test for this unit is Friday. Sorry about the pictures, they won't show up. So I included the links to them.
This has been Kaitlyn. (with a Y!) Thank you, thank you very much. :)
The next scribe will be Matt P!

More About Gas Equations and...Ice Cream

We first started class off today by being told by Mr. Lieberman that we will not be doing our lab today, but we will be doing it on Monday instead.

Next we went over our pre-lab. I want to thank Matt P. with some help from Ben A. for explaining to us how to answer each question. Here is how to do them:

1) Make an equation to help find the pressure of hydrogen: 19.8+x=746 and when you solve for x you get 726.2 mmHg of H2.
2)Use your answer from #1 and you will make an equation following the form of PV/T=P2V2/T2. The equation is 726.2(31)/295=760V/273. Solve the equation for the volume of hydrogen at STP and you should get .274 L.
3)This is just a simple stoich problem: .028g Mg x 1 mole Mg/24.3g x 1 mole H2/1 mole Mg =.001167 moles of H2
4) Use your answers from numbers 2 and 3 to get the molar volume of hydrogen at STP: .274/.001167 = 23.5 L/ 1 mole.

Next, after going over the pre-lab, we continued going over notes. The main thing we talked about in our notes is that there are three equations that come from the equation PV=nRT. Those three equations are:
-PV=(g/mm)RT
-mmP/RT=g/v and g/v equals density
-mm=DRT/P~ a way is to remember this is dirty pee...get it..

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Also, a reminder, make sure you are keeping up with your worksheets which are due on the day of the test which is on Friday, the day before break.

Then we did an awesome experiment that we were actually able to eat. We made chocolate ice cream! The ingredients were sugar, milk, whipping cream, chocolate syrup, and what made the magic happen, liquid nitrogen. It was and looked a little clumpy, but it actually tasted really good. And that was the end to another good day in chem.

The next scribe will be......Kaitlin Y!

Gas gas gas

Half of our classmates were gone today...
because of ....... don't know...
Denna wasn't here today so I am the scriber for today

We started the class with a demontration.  Mr. Liberman put some melted dry ice into a styrofoam box and put a manometer,this manometer has a empty metal ball, into that box.  And the pressure went down because the moles which is in that ball starts to move slow.  Through this experiment we could make sure that when temperature goes down then pressure goes down and temperature goes up then pressure goes up.

After Mr. Liberman finish off with this experiment, we wanted him to dump the melted dry ice on the floor.
And he did....Twice :)
The melted dry ice started to move around the floor like when we spill some water on the hot pan.
We were all excited about the movement of  these little particles.
Also Mr. Liberman dumped the ice over my head accidently!
Korri screamed and called 911 immediately
So that's why I am in hospital right now...
hehe I'm just kidding~
Actually he poured vapor over my head. It just feel coooool~

Today, we learned about the Ideal Gas Equation.
The three important gas laws, Boyle's, Charles', and Avogadro's, derived relationships between two physical properties of a gas, while keeping other properties constant.
When we rearrange to a more familiar form we get                    PV = nRT

*In this equation R is the gas constant.

The conditions 0 *C and 1 atm are called Standard Temperature and Pressure (STD).

At STR R = (1 atm x 22.414L) / (1 mol x 273 K)

R = .0821

And experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

If you know the formula with STD, we could convert to different forms.

The important thing is the value of R won't change!

Here is an example

 What is the volume(in liters) occupied by 49.8g of HCl at STP?

T = 0*C = 273K

P = 1 atm

n = 49.8g x (1 mol / 36.45g HCl) = 1.37 moles

V = (1.37 mol x 0.0821 x 273 K) / (1atm)

V = 30.6L

I hope you guys understand what I explained...

If you want more information about the ideal gas equation, I recommended to visit :

http://www.mikeblaber.org/oldwine/chm1045/notes/Gases/IdealGas/Gases04.htm

Today's homework is a Worksheet and Pre-Lab

http://gbs-moodle.glenbrook225.org/moodle/file.php/1814/Worksheets_Unit_6/08_ideal_probs.doc

http://gbs-moodle.glenbrook225.org/moodle/file.php/1814/Labs_Unit_6/lab-molar_volume.doc

The next scribe will be Denna M.

What it really Boyles down to

Today we started by quickly reviewing Dalton's Law and Boyle's Law, with a couple more gas chamber experiments. he sucked all of the air out of the container (after realizing he needed to attache the hose) and made shaving cream get big, unfortunately he stopped it before it made a mess. He also did an old marshmallow which was rather uneventful.

Then we went on to discuss conversion between temperatures of celcius and kelvin because it is necessary to know for formulas. The reason you need to know the Kelvin temperature is because if you use celcius you may end up with correct calculations stating that you ave negative density, which i hope all of you know is "impossible" according to Mr. Lieberman. To convert celcius to Kelvin all you need to do is add 273 degrees to the celcius measurement.

Then as we continued our notes we went on to Charles' law, which states: The volume of each gas is directly proportional to temperature. V=bT where b is a constant and V1/T1=V2/T2. This means that the original volume over temperature will be equal to the new volume over temperature. To explain this to us he showed us a quick experiment by heating to small flasks with water until they boiled, he then stuck a balloon on top of one which was still boiling and the balloon filled as the gas expanded. he stuck another balloon on top of the other right after removing it fro the heater, the balloon was quickly sucked into the flask because of the rapid change in temperature and the fast reduction of volume.


Next D Liebs quickly explained Avogadro's law which states: Volume is directly proportional to the number of moles of gas. To show this he simply blew up a balloon and let go, letting it soar around and land on Alex's desk. This demonstrates that as moles of air are released there is less and less volume and pressure to keep the balloon inflated, so it eventually collapses. then there is the infamous combined gas law which is P1V1/T1=P2V2/T2 this can be used to apply all forces to one equation instead of several different ones.

There was also a sheet to pick up at the front when we walked in, that sheet is homework and is due come test day. Speaking of which is next friday, the friday right before break. Also, everyone check out the little map thing to the right, people all over the world have seen our blog!

Also since Ben never posted any hilarious pics that im sure everyone wants, here is zoë, looking as happy as a person being vacuum sealed.

And now the next scribe will be... OH MY GOODNESS... Deena M.

The pressure is on

Today in class, we learned about pressure and saw a few demos. We saw Mr. Lieberman lay down on a bed of nails, and then a few of us got to try as well. The reason people didn't get stabbed to death by the nails is that pressure equals force divided by area (P=F/A). He explained that no matter how big or small an object is, the force remains the same. The dependent part is the area. The bigger the object on the nail bed, the less pressure is on them due to the increase in area.
http://www.flickr.com/photos/hc1011/5241587429/

We also learned the different units for measuring pressure, and the conversions between them. They are: pounds per square inch (psi), atmosphere (atm), Torr, or millimeters of mercury (mmHg), and pascal (Pa). The conversions are: 14.7 psi=1 atm=760 mmHg=100 kPa. We also went over how barometers and manometers work, with the atmosphere pushing down to show the pressure.

We went over Dalton's law too, which states that the sum of partial pressures of gases equals the total pressure of the gases when combined. So, if gas 1 is 1 atm and gas 2 is 2 atm, when they are combined the pressure is 3 atm. Boyle's law deals with pressure and volume, stating that the product of the pressure and volume for a gas equals a constant, k (PV=k). So no matter what differences in pressure or volume occur, P1V1=P2V2.

We also got to see Brandon and Zoe get stuffed in a trash bag and have the air sucked out to demonstrate the actual air pressure here on earth, as well as see an inflated balloon expand when the air is sucked out. Happy late arrival tomorrow, and the next scribe is Peter W.
http://www.flickr.com/photos/hc1011/5241588237/
http://www.flickr.com/photos/hc1011/5241588451/

ChemThink gases

Yesterday in chemistry, we went over our new unit, which will end on Friday the 17th with the unit test. After that, we went on the computers to do a ChemThink on the properties of gases. We learned that gas atoms can be affected by temperature as well as size, which changes their speed. We also learned that the pressure of a gas equals its force divided by area. Pressure is caused by gas atoms' impact on the sides of their container. Pressure goes up as the temperature rises due to the fact that there are more collisions, and that they're more forceful. Also, as the number of atoms rises, the pressure rises because there are more collisions. As the volume of the container increases, the pressure goes down because there are less collisions. So overall, we looked at four properties of gases: temperature, pressure, number of atoms, and volume.

Percent Yield

Today, we synthesized all of the concepts we've learned about Stoichiometry into the idea of percent yield. Percent yields are used to compare the amount obtained to the potential that could have been obtained. We learned about the real world usage of percent yields in fields such as the drug industry. It was amazing the impact such a seemingly minute improvement could have on productivity.

The way to determine percent yield is to divide the actual yield by the theoretical yield. First, you have to find what each of those are. Actual yield can only be obtained through experimentation and producing a measurable reaction. Theoretical yield is found through the same stoichiometric formulas we've used before to convert mass of a reactant to mass of a product.

Mr. Lieberman did a demonstration for the class to find the percent yield for. Sucrose was dehydrated using Sulfuric Acid to form Carbon (Graphite) and Water. After the reaction, we realized that the actual yield actually was larger than the theoretical yield. However, there were some flaws in our experiment as there could have still been excess water and acid. From this, we learned that the theoretical yield should typically be less than the actual yield.

Hope everyone had a great break, and there's only two and a half more weeks until the next one!

The next scribe is Ben T

Pumpkin Pie Reactants

I hope everyone had a great thanksgiving!

On Tuesday we turned in our Copper and Silver Nitrate Labs, and our first set of problems in the "HomeFun Problem Set." Mr. Lieberman went over 2 of the homework problems, #67 and #68, before we turned the assignment in. We further reviewed over limiting reactants during class.

Soon after the review we took the Stoich Quiz #3 - Revenge of the Stoich. If you haven't noticed yet all the titles are related to Star Wars in some way.

Over the break make sure you do the Copper Cycle pre-lab.
If you need a copy of the lab here is the link: Copper Cycle

To help you get in the limiting reactants thanksgiving mood, here are the actually ingredients to make a pumpkin pie:
1/3 c. firmly packed brown sugar
1/2 c. sugar
1 tsp ground cinnamon
1 tsp ground ginger
1/4 tsp. ground cloves
1/2 tsp. salt
1/2 tsp. ground nutmeg
2 eggs
1 1/2 c. evaporated milk
2 c. pumpkin puree

You have the following ingredients available:
3 c. firmly packed brown sugar
4 c. sugar
5 tsp ground cinnamon
5 tsp ground ginger
6 tsp. ground cloves
6 tsp. salt
6 tsp. ground nutmeg
6 eggs
3 c. evaporated milk
6 c. pumpkin puree

a. How many pumpkin pies can you make?

b. Which ingredient limits the number of cakes you can make?

c. How much pumpkin puree will be left over?

The next scribe will be Ben W.

Today, we did more practice in stoiceometry with limiting factors. We had the equation "Al + CuCl(2) --> AlCl(3) + Cu". First we had to balance the equation, so we got "2Al + 3CuCl(2) --> 2AlCl(3) + 3Cu". Now that we had the mole ratio, we could figure out which is our limiting reactant with the information of .82g Al and .87g CuCl(2).

To find out which one is the limiting reactant, there are two ways.
The first way, you could find out how much of one reactant you need to react with the other. We used Al for this one. So we take the mass, .82g Al, multiply it by 1 mole over it's molar mass, then find the mole ratio of Al and CuCl(2) (which is 2 to 3), then multiply it by the molar mass of CuCl(2). Since the answer (6.1g CuCl(2) was more than what we have (.87g CuCl(2)), we can conclude that CuCl(2) is our limiting reactant. If out answer turned out to be less than .87g, Al would be our limiting reactant.

The other method is to find out how much of each reactant you need to produce one product. It's always good to look ahead since this question 99.2% of the time will be right after the first: how much of a product you could make. You would follow all of the above steps for each of the reactants and you would choose a product. We used Cu as our finishing product. Remember, whichever reactant produces the least of a product is the limiting reactant. In the end, CuCl(2) produced the least Cu, which turned out to be .41g. Since that's the most that CuCl(2) can produce, that is all that will be produced. You don't do anything with that number!

So after we react this all, how much Al is left? well... to find that there are two ways you could do it, but both are pretty much the same.
First you could take how much of the limiting reactant you have and find out how much of the excess reactant you need to react it, which then you subtract that from how much you start with. To get this, just follow the steps listed in paragraph 2, except for the limiting reactant.

The other way is to take that .41g Cu and see how much of that excess reactant (Al) you need to react with that. It's basically all the same as listed above with different variables.

After we got through all of that, we got to our worksheet, Limiting Reactants, which is part of our homework.
For tomorrow, We need to have our Lab which we recently did and our chemistry question set #1.
1 silver troy ounce = $27.46, I got part of your homework done.
If you're still reading, please come up to me tomorrow and give me a waffle.
We will most likely have a quiz tomorrow, we were lucky not to have one today.


The next scribe will be... Austin W.

Beginning of the End

We ended this week by finishing up the Copper and Silver Nitrate Lab. That is, most of us did. Some people who accidentally dumped their silver before decanting it had to start the whole lab over again. The rest of us used our decanted and dried silver and weighed it. Now that we were all finished with the lab procedure, we got much time to just work on the post lab.

We also had a quiz, although some people haven't taken it yet, so I don't think that I should post the answers. Sorry! Although if you did take the quiz, Mr. Lieberman was super speedy and graded them all before the bell even rang.

The post lab of the Copper and Silver Nitrate lab is due on Tuesday, which is also the last day of the week! Yay!

For other homework, we have to finish the Intro To Limiting Reactions worksheet. If you lost it or misplaced it, you can find it here. If you finished it, you can also find/check the answers here.

We also have the first set of book problems due on Tuesday. The problem numbers are #61-68 on page 70-71. If you want to get ahead, you can do the second set of problems as well, which are due on Wednesday, December first. They are #70-78 on page 71.

Good luck to you guys, and keep working hard. The week's almost over already, and it isn't even Monday.

The next scribe will be Alex K.

How to Make Silver for Pinkie Toe Rings

Today was our lucky day because we didn't have a quiz!! So be prepared to take one tomorrow or Monday.

In class today we got started on the Copper and Silver Nitrate Lab. We were to add 1.4-1.6 grams of silver nitrate crystals into a 50 mL beaker and record its mass. Then we were to fill the beaker with 3o mL of distilled water and stir it until the crystals dissolved. After the crystals dissolved, we were to coil a piece of copper wire around a pencil and then attach the coil to a wooden stick. The coil looked something like this...

Then we put the copper wire into the silver nitrate solution. Once the wire was in the solution, we put three drops of nitric acid into the beaker. We were to let the copper wire sit in the mixture for fifteen minutes and record our observations. The copper wire began to look grey and fuzzy. The liquid began to turn turquoise after about five minutes. The copper wire looked like this...

After fifteen minutes, we took the copper wire out of the mixture and held it over a 100 mL beaker. We held a steady stream of distilled water over it until all of the silver came off of the wire. Once the copper wire was clean, we rinsed it with acetone. Then we massed it and recorded the mass in our data tables. The copper wire was dull because it lost all of its shine. The rinsed copper wire looked like this...

Lastly, we decanted the water from the beaker into a waste flask so that most of the water was discarded. We rinsed the silver with distilled water and then poured the water out about five times. We have to let the solid dry overnight so we were not able to finish the lab today. So far, our beakers looked like this.....

And we will see what they look like tomorrow so we can finish the lab!

Hopefully we will gather enough silver for Ben T. to sell and Kaitlin S. and I can make a pinkie toe ring :)

The next scribe is Aaron G. Enjoy.

Pop Goes the Weasel!

Except sometimes it happens to be a mole. We started off class today with some stoichiometry notes and another one of Mr. Lieberman's crazy experiments. The goal of this one was to make a mole fly out of a tube that Mr. L constructed himself. We were all scared. We took CaC2 + 2H2O -> Ca(OH)2 + C2H2 and created a combustion reaction:
2C2H2 + 5O2-> 4CO2 + 2H2O
From this Mr. L asked us how many grams of H2O will form if .60 grams of CAC2 reacts. This is the formula we got:

We ended up with .17 grams of water. Now it was time to put this formula to the test. Mr. Lieberman loaded up his mole-rocket with the special ingredients from the formula. It was the combination of the CaC2 and H2O that created the Acetylene gas. This gas was the ignition that the rocket needed to make the mole fly. Suddenly there was a BOOM! A little red object shot across the room followed closely by a trail of fire. There were a few screams, particularly from Korri, which was expected. The room then filled with a smell that was immediately placeable: It was a mix of burnt hair and/or that smell from straightening your hair when it is still slightly damp.

Next, we all took our 4 question quiz. It was then graded and handed back. Hope everyone did well! And the next scribe issssss........Mollie M!

Intro Stoich Answers

The answers are posted here

BOOM!

Today we started off our class with Mr. Lieberman announcing that he had a demonstration for us. He filled up a can with methane gas (CH4) through a hole in the bottom and then lit it on fire through a hole in the top. We were told that the can would explode if his lovely assistant Colleen did her job right. Sadly, although Colleen did a wonderful job, Ethan was lacking in spirit and the explosion did not happen. So, Mr. Lieberman left the flame burning and continued on with his lesson.

Today's notes were on Stoichiometry. This unit combines our last two units of moles and chemical formulas and combines them. We can expect to have a quiz every day on what we learned the day before, unless Mr. Lieberman is feeling particularly generous that day.

Today's lesson, aka tomorrow's quiz topic, was about how to do the simplest kind of stoichiometric equation, in which a mole ratio is used to find out how many moles of one substance in a reaction is needed to form a certain number of moles of another substance in the reaction. To do this you must know the mole ratio of the two substances. This ratio is defined by the number of moles wanted/the number of moles given. To find these numbers, all you use is the coefficients from the molecular equation. The example we used in class was:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

How many moles of methane are needed to make 13 moles of water?

To solve this you must multiply the desired 13 moles of H2O by the mole ratio of methane to water: 1 mole methane/2 moles water. The answer came out to be 6.5 moles of methane. These are the types of equations we had on our homework sheet and what we should expect to see on our quiz tomorrow...

BOOM!

I guess Ethan finally got a little faith...the paint can that most of us had forgotten about had exploded. The reason it took so long for the reaction to happen was because the can had to reach the proper ratio of methane to oxygen in order for anything to occur. Once it reached a ratio of 1 part methane for every 2 parts oxygen, it became a combustion reaction resulting in an explosion. (Sorry I didn't embed the video, I'm not great with technology.) Anyways, that was basically our day today, the homework was the sheet he gave us in class, the webassign if you have not already done it, and to be ready to take a quiz on mole ratio equations tomorrow.

ps. the next scribe is....Colleen C. Have fun!

Insolubility and Solubility

Mr. Liberman started off class today by first reviewing the rules of solubility and insolubility. First off, we restated what those words mean: something is soluble if it is able to dissolve in water, and something is insoluble if a precipitate forms in water. Then Mr. Lieberman explained that the lab that we did yesterday was the basis on which we were eventually going to base the solubility of certain compounds off of. This will include compounds that are commonly found around us and whether they are soluble or not, but it is not that black and white. There are some compounds such as silicate (SiO3) that is insoluble unless it contains the ions of K+ and Na+. There are many more examples like this.

To make things more interesting, Mr. Lieberman did the Lead Iodide experiment to show that certain compounds are always soluble (NO3), and therefore, you are able to identify the solids in a reaction. See Video Below:

This reaction is of Lead and Iodine. This solid had to form between these two elements because NO3(nitrate) mixed with any other element is soluble.

Since Lieberman was in a good mood, after explaining all of this, he let us work on the lab and the worksheets we got. Everybody was working until the bell rang and class ended. That was it for the day. The next scribe is....

Kaitlin S.

Lots of Reactions

In class today, we started a lab with many different sodium and nitrate groups, totaling 48 reactions. Why would we do that many, you ask? Well it has to be either:

a) Mr. Lieberman likes to make us do a ridiculous amount of writing and balancing equations

b) To help us formulate rules on formation of precipitates.

or c) All of the above

I'm Gonna go with C.

To start off, we put various solutions containing water and a Sodium Ion in each column. Then, we added a Nitrate solution to each of those, receiving varying results:

Some did absolutely nothing, leaving a clear liquid behind.

Others changed colors, but no precipitate formed.

Still others formed either a cloudy or completely solid precipitate, meaning that a reaction occurred.

And a couple did this:

Just kidding. But still, it was exciting. Here are the real results:

And my data table, if you can read it:

And the post lab is the real fun part: we have to write chemical equations for EVERY SINGLE PRECIPITATE! Both molecular and net ionic. For example, the molecular equation would be: 3Na₂SO₃+2Al(NO₃)₃ yields Al₂（CO₃）₃+6NaNO₃ and the net ionic equation would be: 2Al³⁺+ 3CO₃² yields Al₂(CO₃)₃. And repeat. About 20 times. Just a thrilling homework assignment. Oh, and by the way, for those who can't realize it yet: NaNO₃ is a product for every single molecular reaction, and always dissolves.
Thank you for reading my post. I hope you all do wonderful on your labs now. Now, the next scribe will be:.............................................

Ben A.

Keeping Balancing Equations Cool Like They're Going Out of Style

During Friday's class, we went even deeper into balancing equations. So deep that many people would consider it impossible in anyway shape or form. Specifically, we discussed Molecular Equations, Complete Ionic Equations, and Net Ionic Equations.

Molecular Equations

• Molecular Equations show the overall stoichiometry of the equation, but not the actual forms of the substances
• (Example): KCl+AgNO3------------>KNO3+AgCl

Complete Ionic Equations

• Complete Ionic Equations represent, as ions, all substances that are strong electrolytes
• Ionic Compounds dissociate in water
• Spectator Ions are ions that don't react; spectator ions are still included in Complete Ionic Equations
• (Example): K(^+)+ Cl(^-)+NO3(^-)------------>AgCl+K(^+)+NO3(^-)

Net Ionic Equations

• Includes only those ions undergoing a charge (spectator ions are NOT included)
• Net Ionic Equations are only written if a solid is formed
• (Example): Cl(^-)+Ag(^+)------------->AgCl(s)

NOTE: Molecular Equations, Complete Ionic Equations, and Net Ionic Equations are all Double Replacement Reactions

After we deciphered Molecular Equations, Complete Ionic Equations, and Net Ionic Equations, we proceeded to work on our newest worksheet (Net Ionic Equation Worksheet). Here are some tips in completing this worksheet:

1. To find the Molecular Equation, balance the equation given
2. To find the Complete Ionic Equation, seperate ions from compounds (include charges)
3. If a reaction isn't seen, or there is no solid, a Net Ionic Equation isn't needed

We proceeded to work on the Net Ionic Equation Worksheet for the rest of class.

On Monday, we will be doing a series of 48 reactions. The reactions will help establish rules regarding what we talked about on Friday. Furthermore, there is a WebAssign due Monday, on section 4.2 (Precipitation Reactions).

Stay thursty, my friends.

The Scream Heard 'Round the World

Today, class started off with Mr. Lieberman telling everyone to get out their Classifying Chemical Reactions Lab so we could go over the post-lab questions, which asked you to find the equations for some experiments that you did during the lab. People wrote the answers on the board and we cleared up any questions about them. For answers see Brandon L's previous post that described the lab's experiments and their equations.

After we finished discussing the lab, we moved on to the notes for the day. Although we only got through the first part of the notes, we still learned a lot about solutions, and how electrolytes and ions are related. Mr. Lieberman explained that a solution contains a solute, which dissolves, and a solvent, which is what the solute dissolves into. He also explained how water was the most common solvent. As a solvent, water dissociates ionic compounds into its ions. For example if you dissolved table salt (NaCl) into water, then one would see each individual ion separately, because water breaks the Na+ and Cl- apart.

We started to talk about electrolytes, and the difference between strong and weak electrolytes. I general, electrolytes are allowed to pass through a current if there is a sufficient amount of ionization. Strong electrolytes contain enough ions to carry a current efficiently, while weak electrolytes don't completely dissociate and have a small amount of ionization.

To give us a visual of how electrolytes and ions really worked, Mr. Lieberman performed the light bulb experiment. First he used sugar to see if he could light the light bulb. To fake out the class he screamed to make it seem like there was going to be light. Sure enough, the infamous Korri then proceeded to shriek so loudly, that it was heard around the country, and maybe even the entire world because it was as if every students ear drums were being punctured and then completely ruptured. Mr. Lieberman then demonstrated how the sugar, which has no ions, didn't light the light bulb. He then mixed salt and water together. The water acted as the solvent and the salt was the solute. Na+ and Cl- ions were formed and the light bulb irradiated a lot of light.

You can check out this video, as a woman explains electrolytes and currents.

That's it for class on November 4th.

The next scribe will be the oh so special, John A.

Can You Classify Chemical Reactions because I hope I can

Today and yesterday in Lieb's class we started off by talking about the 3 worksheets we need print off of moodle after this we began our class notes on the 5 types of reactions: Synthesis, Decomposition, Single Replacement, Double Replacement, and Combustion Reactions.

A synthesis Reaction is when you are given two reactants that combine to give you one product for example: 2H2 + 02 -----------> 2H20

A Decomposition Reaction happens when compunds are broken into elements or similar compounds. Example: 2H20 ---------> 2H2 + 02

A Single Replacement Reaction is when an element replaces another compound. These only take place in wate. Example: NaCl + F2 ---------> NaF + Cl2

A Double Replacement Reaction happens when one metal replaces another in a compound and the same for the non-metals. Example: AgN03 + NaCl ----->AgCl + NaN03

A Combustion reacion only occurs when a hydrocarbon reacts with oxygen gas. Example: C5H12 + 802 ------>5CO2 + 6H20

Now on to the Classifying Chemical Reactions Lab YES!!!!!!!!!!!

So, to begin with we went over the Pre-Lab in Question 1 we had to summarize the chemical reaction of sodium bicarbonate, carbon dioxide, and water. This turned into

2NaHC03-----> C02 + H20 + Na2C03. When balanced.

We then had two days to go through the 7 different stations and do all the experiments. In reaction 1 We Burned Magnesium ribbon. The Magnesium ribbon lit on fire and got chalk white after it burned. The chemical Reaction that occured was 2Mg + 02 ----> 2Mg0.

In the first picture we see the product of the Magnesium ribbon turning pure white and in the second it is lighting on fire.

In reaction 2 we put hydrochloric acid in a test tube with Magnesium Metal, and then we lit a wood splint and put it in the tube where it was extinguished. These turned into

2HCl+ Mg ----> MgCl2 + H2

The answer above has been balanced. In this picture we see the hydrochloric acid and Magnesium Metal Ribbon mixing and bubbles occuring.

In reaction 3 we put ammonium carbonate in the test tube an burned for 30 seconds then we put a piece of litmus paper at the top and an odor was released. This odor was amonia next we put a piece of wood flint on fire into the tube which was extinguished. The reaction for this was (NH4)2C03---> C02 + NH3 + H20 when balanced.

In this picture we see the mixture of amonia being released.

In reaction 4 we put calcium carbonate in a test tube with hydrochloric acid and once again lit a wood splint which was extinguished.

This reaction when balance became 2CaC03+ 2HCl-----> 3C02 + 2CaCl + H20.

In the second picture we see the wood splint being extinguished.

In the first picture we see the Calcium Carbonate and hydrochloric acid mixed.

In reaction 5 we mixed copper (II) Chloride and mossy zinc together and the mozzy zinc change color from silver to read. CuCl2 + Zn-------> ZnCl2 + Cu. When balanced.

In this picture we see the zinc changing color.

In reaction 6 we added copper chloride with sodium phosphate and a precipitate formed. When this reaction was balanced it became

3CuCl2 + 2Na3P04------> Cu3(PO4)2 + 6NaCl

In this photo you can see the precipitate forming.

In reaction 7 we mixed sodium hydroxide and phenolphthalein and the hydrochloric acid. During this reaction it at first became pink and then back to clear liquid color whise. In this reaction when balanced it became NaOH + HCl-----> H20 + NaCl As you can see in the pictures below it demonstrates the pink on the right and then clear again on the left

Well that was our class in a nut-shell and the coveted spot of scribing for Thursday goes to my man Josh D CONGRATS!!!

Intro to Balancing Equations

We started off the class today by going over the notes

Balancing equations

View more presentations from gbsliebs2002.

two very important rules to remember, and that Mr. Lieberman was very persistant in reminding us was:

1. DO NOT CHANGE THE SUBSCRIPTS

2. There are no stead fast rules to balancing equations, so the best way to master this skill is to practice, practice, practice.

After showing us these liberating notes he demonstrated a couple experiments for us. Both of which come with an entertaining story.

1. The Water from hydrogen and oxygen eperiment

It all started when Mr. Lieberman was taking a strol in deserted mountains with no one around and no water. Why he was walking in the heat without water or people in general is a mystary to us all. Well, while he was on this hike he got thirsty, but without any water and no connection to civilization he was stuck in the middle of nowhere without a source of water. This caused a problem for Mr. Lieberman, but it just so happened that a hydrogen spring was there and he was carrying two rulers, a balloon and a match. Attaching the balloon, which he had filled up with hidrogen from the spring, to one of the rulers he was ready to make water. His idea was that if he caused a reaction to occur he could make H2O out of the hydrogen and the oxygen in the air. He ran the experiment and low and behold, there was water (or at least some form of it). He demonstrated this awesome experiment in our class:



Along with this riviting story, came the story of his pet elephant Stampy:

Now Mr. Lieberman owns a pet elephant and this elephant is known as Stampy. Stapmy has needs just like any other elephant, and this includes brushing his teeth. Mr. Lieberman being the thoughtful man that he is makes his own elephant toothpaste, just for his Stampy. He demonstrated the making of this special toothpaste for us in class:



These stories made the class interesting and educational, depicting scinerios that use the topics we are learning of in class. Don't forget to write up the prelab including any tables needed for the procadure. If you don't do that you will get no credit for the WHOLE prelab, as I have had to learn from experience.

And now your next scribe will b.......Brandon L. Congrats

Friday's Class

Today in class was great because we got our tests back from the day before. This test was all about moles, but not this type of mole:



this type of mole:



We used this on the test and many people did very well. After answering all of the quetions about the test, we got time to go over to the computer lab and work on the ChemThink that is due on Monday. Make sure you finish that if you have not done so already. The next scribe will be....Mollie M!!!

Magnesium Oxide Lab

Hey everyone! Today was a really fun day in chemistry because we did the Magnesium Oxide Lab.

At the beginning of class, Mr Lieberman checked in our pre-labs (don't forget that means a data table as well) and we got started on the Magnesium Oxide Lab. We began by massing an empty crucible and lid, and then massing the crucible, lid, and a 25 cm magnesium ribbon together. We recorded these two values in our data tables and then proceeded to place the covered crucible on the clay triangle. We lit the burner and removed the lid every three minutes for a total of fifteen minutes. After the flame had been carefully turned off and the crucible cooled down, we massed the crucible, crucible lid, and magnesium oxide product.

For the post lab:
1. we are first asked to calculate the mass of the magnesium metal and the mass of the product, using the law of conservation of mass to calculate the mass of oxygen that combined with the magnesium. (Law of conservation mass: the mass of substances in a closed system will remain constant, no matter what processes are acting inside the system). This should be done by subtracting the mass of crucible, lid, and ribbon from the mass of the crucible, lid, and magnesium oxide.
2.Then, we are asked to calculate the percent composition of magnesium oxide. To do this, start by calculating the formula mass, and dividing each component mass by the formula mass and multiply by 100.
3. We are then asked to use molar mass to calculate the number of moles for each reactant. (see class notes)
4. calculate the ration between the number of moles of magnesium to the number of moles of oxygen and find the empirical formula of magnesium oxide. Look at Elim's post if you still need help with empirical formulas!
5. convert mass to moles to find theoretical yield (this shouldn't be too complicated)
6. calculate your percent yeild

Hope this helps! Tonight's homework is to complete Avagadro's Crash Activity. If you still need help on empirical formulas I found the following video very helpful http://www.youtube.com/watch?v=r2Log6-voWo! Also, work on the lab (due Thursday), and don't forget to begin studying for Thursday's test! Tomorrow's scribe is Katie I!

Empirical vs Molecular Formula

Today, we started class by getting back our quizes from last week.  I didn't take the quiz so I have no idea what happened in the beginning of the class.  I am sorry~  Anyway we went over the quiz about 15 minutes.
Then we started to learn what empirical formula is, and how this Empirical formula relates with Molecular formula.

Molecular formula = (Empirical formula)n

* n is an integer

Once you get Empirical formula, then easy to get Molecular formula.

The Empirical formula is the simplest form of the Molecular formula.
For example :
Molecular formula - C6H12
Empirical formula - Divide by 6 - CH2
Here is a sample we did in the class:

Determine the Empirical and Molecular formulas for a compound that gives the following %.
 - 71.65% (Cl)
 - 24.27% (C)
 - 4.07% (H)
The molar mass is known to be 98.96g/moles

First, assume a 100g of the sample
 - 71.65g (Cl)
 - 24.27g (C)
 - 4.07g (H)

Second, Convert gram to moles
 - 71.65g  x  1moles / 35.5g  =  2.018 moles of Cl
 - 24.27g  x  1moles / 12g  =  2.02 moles of C
 - 4.07g  x  1moles / 1g  = 4.07 moles of H

Third, divide each moles by smallest number of moles
 - Cl  :  2.018 moles / 2.018 moles  =  1 mole
 - C  :  2.02 moles / 2.018 moles  =  1 mole
 - H  :  4.07 moles / 2.018 moles  =  2 moles

Finally we can get the Empirical formula like this :  CClH2

Now, we can find the molecular formula by finding the mass of the Empirical formula and setting up a ratio:

CClH2 → C2Cl2H4


*If you want more information about Empirical formula
I recommend you to visit http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm


After we had finished the note, Mr. Liberman introduced an activity.  This activity is called Crash of Avogadro Airlines Flight 1023.
The goal of this activity is finding a murderer.
On the first and second side of the sheet, there are 3 tables which help to find the horrible murderer.
In this activity each chemical can be used once or no time.

Tonight's homework is to finish the hydrate lab, do a Magnesium Oxide pre-lab and Avogadro Airline activity which is due on Wednesday.
Just a reminder, the unit 3 test is on Thursday.


The next scribe is Ellen Hirsch

Hydrate Lab Day!

We started class with finishing the back sides of our quizzes for 10 minutes. Then we started the pre-lab for the Hydrate Lab we did in class. We first discussed three rules to follow when operating a Bunsen Burner. The first rule is to not leave it on unattended, do not touch it while hot, and keep all hair tied back and loose sleeves out of the way. The next question we discussed was to find:

a. The mass of the hydrate

b.the mass of the water lost during heating

c. the percent water in the hydrate

We were given the mass of:

Empty Test Tube: 18.42 grams

Test Tube and Hydrate (before heating): 20.75 grams

Test Tube and Anhydrous Salt (after heating): 20.41 grams

To calculate these you had to:

a. 20.75 g -18.42 g= 2.33 g

b. 20.75g- 20.41 g =.34g

c. .34g/2.33g=14.59% (round for significant figures)=15%

After this question you had to complete the table given on the Hydrate Lab sheet.After this, we got together with our lab groups and began the Hydrate Lab here are the steps:

1. We took the mass of the pyrex test tube.
2. We added around 2grams of blue hydrated (with water) copper (II) sulfate.
3. We then massed the test tube with the copper (II) sulfate in it.
4. We then set up the clamp and tilted the tube slightly down so the evaporating water can leave the test tube.
5. Then then focused the heat from the Bunsen burner on the copper (II) sulfate and heated it evenly until it turned white. We tried to avoid burning it. (turned brown if burned)
6. Then we wiped out the excess water droplets and massed the tube with the anhydrous salt.

So that was our Hydrate Lab! The next scribe is... Ellen Hirsch!!

This was Korri Hershenhouse! Don't forget to do your How Much Are You Worth Project for monday! Also don't forget to complete the webassign and do the post lab also for monday!

Who's Worth More: Your Lil Brother or Lil Wayne?

We began class today by learning about percent composition in our notes packet. A percent composition compares the mass of each element present in 1 mole to the total mass of the compound.

Sample Problem:

Penicillin F has the formula C₁₄H₂₀N₂SO_{4. What is the mass percent of each element?}

To begin this problem, we calculated the total mass of the compound:

14 x 12 + 20 x 1 + 2 x 14 + 32 + 4 x 16 = 312 grams/mole

To determine the mass percent of Carbon, we multiplied the amount of carbon in the compound (which we used to determine the total mass of the compound) by 100 and divided by the total mass of the compound. We repeated this process with hydrogen, nitrogen, sulfur, and oxygen to find the rest of our results. We found the following:

Carbon had a mass percent of 53.8 %

Hydrogen had a mass percent of 6.4 %

Nitrogen had a mass percent of 8.9%

Sulfur had a mass percent of 10.3%

Oxygen had a mass percent of 20.5 %

After going over the notes, Mr. Lieberman introduced us to the 'How Much are You Worth?' project. The project asks us to calculate the cost of ourselves, a friend, family member, or favorite actor, although, of course, we are all priceless! We are to show all work in a factor-lable style, and are required to include a picture of our person. A maximum of 5 extra credit points can be given for creativity in your presentation style. Mr. Lieberman showed us some examples of cool projects that have earned students extra credit in the past, but remember that going above and beyond is not required.

Then we took our review quiz which we will have some time to finish in class tomorrow.

The % Composition Bubble Gum Lab is due tomorrow.

Web Assign is due Monday.

'How Much are You Worth?' project due Monday.

The Unit Exam is Thursday 10/28.

The next scribe is...Korri H!

'MOLE'ten Lava

Honors chemistry period six is now on round two for these scribe posts because we had to skip Elim.

Mr. Lieberman began the class by letting us correct and ask questions about the worksheet "Mole Problems 2." The worksheet was more practice, that we all needed, on converting particles, moles, and mass. For those that missed it, 6f should be scratched out.

We also corrected and asked questions about the second converting sheet, the one with no name.

Like Kaitlyn stated in her post, the conversions are:

For grams to moles, divide by molar mass

For moles to grams, multiply by molar mass

For particles to moles, divide by Avagadro's number

For moles to particles, multiply by Avagadro's number

We, then, moved on to the lab, "% Composition of Bubble Gum," which we did not do a pre-lab for. Mr. Lieberman told us to fit all of the lab on one page, because it is a relatively short one. The procedure was as follows:

1. Take the gum and mass it with the wrapper (record)
2. Unwrap and chew gum
3. During the chewing process, mass the wrapper (record)
4. Place chewed gum on wrapper and mass it (record)

Amazingly, the mass of the gum dropped drastically, but that's because the sugar in the gum was consumed by the chewer of the gum. Our lab group's gum had a mass of 6.13 grams and the sugar within the gum had a mass of 3.97 grams. From here we had to answer 5 post-lab questions:

1. Calculate the mass of sugar in the bubble.
2. Calculate the % composition of the sugar (by mass of the bubble gum).
3. Calculate the moles of the sugar and gum using the molar masses given the procedure.
4. Using your answer from question number 3, determine a possible empirical formula for the bubble gum.
5. Researchers have found that the ideal formula for the gum is GS2, where G is a fictional elemental symbol for gum and S for sugar. How does your gum compare to the ideal? What might be some sources of error?

An empirical formula can be made by the following procedure from http://chemistry.about.com/od/workedchemistryproblems/a/empirical.htm

1. Assume you have 100 g of the substance (makes the math easier because everything is a straight percent).
2. Consider the amounts you are given as being in units of grams.
3. Convert the grams to moles for each element.
4. Find the smallest whole number ratio of moles for each element.

There is a mole quiz tomorrow, and this lab is due friday. Don't fall behind on all this work, it is alot to do. Also, there is a "mole day" party on saturday at 6:02 AM, but Mr. Lieberman did not give us alot of information on it.

-Emilio I

The scribe for tomorrow's class is: Zoe S.

Insert Mole Title Here

So first off, I'd like to congratulate Kaitlyn on a wondeful posting job.
Now for class, we had kind of a work day in store for us. We got more practice with moles so we could better understand what it is and also we got more practice with unit conversion.
First off, we went to our notes and finished two "Learning Check!". First problem we had was:
Question: How many atoms of K are present in 78.4 g of K?
Work: Take the amount (78.4 g), multiply it by (1 mole of K over 39 (molar mass) grams of K), then multiply that by (Avogadro's number (6.02 times 10^23) over 1 mole of K).
Answer: 1.21 times 10^24.

Our second Learning Check! was:
Question: What is the mass (in grams) of 1.20 X 10^24 molecules of glucose (C6H12O6)?
So this question is actually working the opposite way of our other problem. We've got to find the mass (instead of molecules) and we start with the molecules (instead of the mass).
Work: 1.20 X 10^24 times (1 mole of glucose divided by Avogadro's number) times (the Molar mass divided by 1 mole of glucose).
To find the molar mass for this problem, we must take all the subscripts, multiply them by the mass of their individual atomic masses, then add all of those together. In this problem, the molar mass would be:
6 (subscript of Carbon) X 12 (atomic mass of Carbon) + 12 (hydrogen) X 1 + 6(oxygen) X 16 = 180 grams
Once we plug in the molar mass, our answer should be 359 grams per mole.

So, in order to really understand how we got this, you must already know about moles and molar mass. If you do not understand these fully, please either look back at your notes (you can find extras in moodle), or just look below at Kaitlyn's post.

After we finished those two problems, we had the rest of the period to work on our mole workshop. This compromised of 5 questions all of which we should have done by tomorrow with our partner. An extra copy of this worksheet can also be found in moodle under unit three worksheets (unfortunately the answers are not there).

Homework: Finish the Mole Workshop as well as the second mole worksheet. Both were given out before class.
Important dates: 10/20 (tomorrow!) review quiz on unit conversion and moles.
10/28 Unit exam, remember to have your homework problems done

Random fact of the day: Black whales are born white.
Also, I'd like to point out we had our first visitor from Japan.

Next Scribe will be Elim J