Thursday, March 24

We started class today by talking about how we should memorize the equations on page 5 for molality, molarity, mass % and volume % for the test tomorrow. We then reviewed number 9 on page 17. We wwere given some information and had to find the change in freezing point. You have to recognize that hexane is a non-electrolyte so i=1. We are given the constant, 30.0 C/m, so all we need to know to find the change in freezing point is the molarity. We are told that 0.421 moles are dissolved into 500 g CCl4, so we can use that to find the molarity. The molarity would be .421 mol/.500 kg. If you plug that into the m for the equation Tf=i m Kf, you get Tf= 25.3 C. We did number 10 on page 18 which is basically the same problem but wiht different numbers. The answer is 0.430 C.

Next we went to page 13 to do 21 and 22. For 21, the relationship between gas solubility and temperature is inverse becuase as temperature increases, the solubility for a gas decreases. This is because if there were oxygen bubbles in a solution, they would be heading upwards towards the surface of the solution. As you increase the teperature, the oxygen molecules will move faster so they will reach the surface of the solution and get out of the solid more quickly. Once they're out, there's less gas in the solution so the solubility decreases.

For 22, the relationship between gas solubility and pressure is direct. This is because if there are gas molcules right above a solution, as the temperature increases the molecules wil start moving around really quickly. Some of the gas molecules will be forced into the liquid by this so there will be more gas in the liquid, therefore increasing the pressure.

Next we did the colligative properties lab. The goal of this lab was to find the molar mass of an unknown solid by using boiling point elevation. We started by pouring 50 mL of water into 3 seperate beakers. We put them all on the hot plate and when they reached 80 C, we took 2 off. In each of the ones we took off, we put a certain mass of the unknown solid and then put them back on the hot plate. The third beaker was used to find the boiling point of pure water. When the other 2 beakers started to boil, we measured their boiling points. Then you subtracted the boiling point you reocrded of pure water from the boiling points of the solutions and recorded the difference. By using the boiling point elevation formula, we could use the change in temperatures to find the molality. To do this you had to divide the difference by 2x0.52 since the constant was 0.52 C/m and we were told i=3. Once the molality was found, that could be used to find the moles of the solute since molality=moles solute/kg solvent. The final step was to find the molar mass which is mass of solute/moles solute. The mass was the number of grams you were supposed to measure of the solid. Each group had a different mass. The theoretical value was 58 g/mol. Using this, you were supposed to do a percent error calculation.

The homework was to study for the test tomorrow and finish our lab notebooks.

Wednesday, March 23

Today's class didn't have much new content, Mr. H explained at the beginning of class that the last 20 minutes of class would be used to work on the WebAssign homework due Friday. He also explained that the test on Friday would consist of 35 multiple choice questions and 2 work problems, any people who would be absent on Friday should take the test in the Test Center as opposed to waiting 'till after Spring break. Next, we took a look at Matt's excellent blog post. Then we turned to packet page 8 to look at some problems not unlike those featured on Friday's test.

Balanced ionic equations and "Mole Island" conversion factors make a repeat appearance here, a simple if somewhat lengthy review.

After those two problems, we moved on to page 18. Rankings for the freezing point depression/boiling point elevation can be determined by the quantity im. The compound with the highest im has the highest boiling point and the lowest

freezing point, the compound with the lowest im has the lowest boiling point and the highest freezing point.

After two more problems, we spent the rest of class time doing our WebAssign homework on netbooks. All in all, this was a decently productive day. Don't forget to complete the Solubility Lab Write-Up. Oh, and it'd be a good idea to study for this upcoming test instead of, 'yanno, winging it.

Tuesday, March 22

Mr. Henderson started today's day with an argument with himself. He had two voices, one on each shoulder, arguing if the WebAssign due tomorrow should be pushed back to Thursday. He ended up postponing the WebAssign and offered extra credit points to whoever did it by tomorrow at 8 A.M.

We started the bulk of the class period practicing problems on page 7 and 8. These problems were about solution stoichiometry. The answers that we did were:

3a) 0.0182 grams

3b) 0.123 liters

5a) Na2CrO4 + Pb(NO3)2 ---> 2 NaNO3 + PbCrO4

5b) 13.3 grams PbCrO4

These problems consisted of the same concept as did the old stoichiometry problems (mole island). The only major difference was that we had to use a molarity ratio (for example #5b of .4100 M). This means that there are .4100 moles in every 1 liter.

The rest of the class period was dedicated to colligative properties. The four colligative properties are vapor pressure lowering, boiling point elevation, freezing point lowering, and osmotic pressure. Mr. Henderson then explained how this was relevant in every day life. He stated that we add salt to the roads when it snows because it is a colligative property. This means that it lowers the freezing point of water. By doing this it is less likely for the water from the snow to freeze on the roads. This makes driving a lot safer. Hooray for colligative properties! How much lower the freezing point of water gets is only dependent on the solvent and amount of solute.

We concluded class by performing math problems on this idea. The formula for boiling point elevation equaled (i)(m)(Kb). i is the number of particles per solute, m is the molality, and Kb is the boiling point elevation constant which depends on the solute. i would equal 2 for NaCl, 3 for CaCl2, and 4 for FeCl3 because if these molecules were broken down there would be that many individual particles. The equation for the freezing point depression is the same other than the fact that the Kb turns into a Kf which is a constant as well. We put these formulas to the test and did numbers 7 and 8 on page 17. The answers are:

7) 0.65 degrees Celsius

8) 2.90 degrees Celsius

Homework is to get started on the 2 WebAssigns due Thursday.

March 21st, 2011

We started off today's class reviewing Friday's blog and then going right to the reading sheet corrections, which Mr.H went through in under 2 minutes (classic Mr.H). The answers for pages 23-24 are as follows:
1.a,b,c,d
2.a
3.b
4.b
5.a
6.c
7.Didn't get this one >.<
8. c
9.c
10.a
11.d
12.a
13.c
14a. 0.645 molal
14b.0.225kg
14c.34.45g

After going over these answers, we continued on to Page 5 which is really really really awesome for learning how to set up equations to find Molarity, Molality, Mass Percentage and Volume Percentage. Every equations is given to you in its general form and I highly recommend using this sheet to study for the test coming on Friday. Anyways, we did problems 1,2,3 and 6 on pages 5 and 6. If you got problem 1, you will get the rest of them for sure. But, if you didn't, this is how to go about doing problem 1:
First, get all of your general information out of the way. What I mean by this is calculate moles of CaCl2, kg and L of solvent (water), and mass of solution. The reason that we do this first is because we are looking for Molarity, Molality, and mass percentage in the given solution. If you get all of this general information out of the way, all you will have to do is plug in this information into their respective formulas. Next, we calculated Molarity. By using the chart at the top of the page, we know that this is moles of solute/ L of solution, which, if u followed the steps prior, you will be able to plug in no problem. The answer to the molarity is .284M. Then we calculated Molality. We see that this has the same numerator as the molarity so we can use the same moles of solute that we used for molarity. We also see that the denominator of molality is kilograms of solvent. Once again, we can just plug in the general information that we got at the beginning of the problem. We should get the exact same answer as we did for molarity, but with a molal after it of course (the answer is .284molal). Finally, we calculate the mass %. This is a very simple process in that, even if you didn't get all of the general information we talked about, the answer is pretty much given to you in the question. Just place mass of CaCl2 (given) over mass of CaCl2 + the mass of solvent (100mL or 100g). You should have gotten 3.15/103.15 or 3.05%. If you have gotten this far you deserve a puppy.



(this is Boo. he is happy for you.)
Why the puppy? Well, its because you are about to do question 6 (actually not that difficult) and the Webassign (actually that difficult). When looking at number 6, it is important to assume 1L of solution throughout the question. This will give us a nice, round relative number to work with. Also, it gives you the number of moles of sulfuric acid (H2SO4) which is important. This will give us the mass of solute after we multiply 3.75 mols by the molar mass. You should have gotten 367.75 g of H2SO4. Now, because we assumed 1L of solution, we can just multiply the given density by 1000 to get the mass of solution. Now we can solve the first part of question 6; the mass percent. We will set up just like we did in question , but this time DO NOT ADD THE SOLUTE TO YOUR DENOMINATOR. We don't add it in this case because we already have the mass of solution (not solvent) which has already taken into account the solute and solvent. The answer to mass % is 29.9%. Finally, we calculated molality by placing the given amount of moles over the kilograms of solvent. We find the kg of solvent by multiplying 1230 by 70.1%. We use this percentage because it is what is left over after we calculate the mass of solute in the mass%. So, you should get 4.35 molal.

Congratulations you have finished the "math-intensive part of the Solution Unit. Sadly, you have to do the webassign now which used the same equations that we used on pgs 5 and 6, but with more complicated wording >.<. Good Luck!





(but hey. atleast your not doing this?)

Unit 10, Day 3 (aka Friday, March 18)

We entered the classroom on Friday and immediately went over pages 19 and 20 in our packets. The answers to the 4.1/10.1 Reading Sheet are as follows?:
1. c, d
2. 2.0 M
3. a
4. b
5. d
6. 0.220 mol
7. 0.0209 L
8. c
9. (ask Mr. H)
10. d
11. a - 0.50 M
b - 0.250 mol
c - 0.050 M
12. Mole fraction: a
Mass %: e
Molarity: b

We were then reminded of the WebAssigns due on Monday and Tuesday. Mr. H proceeded to go over the blogs from the last two days. Kendall's encompassed the concept of reading a graph of the solubility curve. Kevin's blog slightly offended Mr. H because it talked about the special St. Patrick's day demo we had the day before. Kevin claimed it was "no biggie" while Mr. Henderson argued that it was indeed "a biggie." Kevin's blog also included information about how to determine molarity as well as a recap of the lab(s) we did that day.
Mr. H then instructed us to turn to page 3 in our packets where we used the equation moles/liters = molarity to figure out the problems. He told us that before we calculated, we had to establish the numerator as moles (which is often converted from grams) as well as the denominator as liters (which is often converted from milliliters). For number 8, we had to convert 35.5 g of HC2H3O2 to moles. Our answer was 0.591 moles. Then we took 88.0 mL and converted it to 0.088 L. With the moles over the denominator, our answer for molarity was 6.72. Mr. H also taught us a snazzy way to write the symbol of molarity for that equation:

[HC2H302] = 6.72 M

Pretty simple. Number 4 was done in a similar manner. 93g of KCl were converted to 1.249 mol and .60 L was already workable. With the moles over liters, the molarity came out to be about 2.1 M.

Page four included some concepts about ions. In order to grasp it, we had to turn to page 19, where the marginal notes disclosed information about dissolving molarity. For number 14, Mr. H told us that we didn't need a calculator for part a. The concentration of the ions is 0.60 M. How? I honestly don't know, ask Mr. H. Then for part b we were reminded to get used to the idea that the number of moles = L x mol/1L. So for part b, .1500L x (.60 mol/1L) = 0.090 mol Fe3+.

Number 18 on page 4 is basically impossible. I would recommend you go see Mr. H for that one. I'm clueless. I'm gonna go see him too.

Then, Mr. H handed out some sheets of paper and told us this was the demo. It was a dilution problem that he was prepared to show us. I'm a faliure at scribing cuz I just realized that I left my compsition book with the demo information at school. Well, I'll have it in class tomorrow so if you need to see it, come see me (yes, it is right for once). Then the bell rang...LOL!

Sorry about my mediocre scribing today, Mr. H. I'll do better next time.

Unit 10 Day 3

Today was another CLASSIC day in Mr. H's class. We started by lighting some methane on fire. No biggie. Not that surprising knowing of all the other crazy stuff we've done in this class. Anyways, After that emotional beginning of class, we began to work of some pages in our parquetes.
We began with page 1 number 4. This was a beginning to the mathematical part of this chapter. The main lesson of this day in class was using the equation for molarity which is

MOLARITY = MOLES SOLUTE/LITERS SOLUTION
(molarity=# moles solute per Liter of solution)
So we began with part a. This mainly was made of plugging the numbers into the formula above and then writing it down.
The answers for this section were
a. 2
b. 4
d. 1
f. 1
g. .05
h. 1.5
i. 6
j. .25

^poorly lettered

Next was a new piece of information that stated that the mass of the solute can be related to the mass of solute by the molar mass. We used this in the next problem on the back of page 1 AKA page 2. This page worked on what was previously done in number 4 but added a new element. This was the Mass of solute. All that was needed to get this value was finding the molar mass of the solute and then multiplying by the # of moles.

The Tables answers when like this
a. .5M 1 mol 2 L 40 g
b. 1.5M 3 Mol 2 L 120 g
c. .25 M .5 mol 2 L 20 g
d. .5 M .05 mol .1 L .85g
e. .425 .85 mol 2 L 20 g
f. 2 M 2 mol 1 L 34 g
g. .125 M .5 mol 4 L 8.5 g

we used the rest of the period on a Lab.
That was it
YAY
And now its friday
its the day after thursday
and saturdays after it

Unit 10, Day 3

We started class today by looking over Emma’s blog from yesterday. After that, we turned to page 15 of our packets and continued where we left off on Tuesday. Mr. H introduced the terms saturated, unsaturated, and super saturated to us. To find out which of the following a molecule is, it is necessary to refer to the Solubility Chart. If it is on the line, it is saturated. If it is below the line, it is unsaturated. Lastly, if it is over the line, it is super saturated. We used this information to complete numbers 5 and 6 on page 16. The answers are as follows:

5. a. S

b. US

c. SS

d. SS

e. US

f. S

g. SS

h. S

i. US

6. a. US

b. S

c. SS

d. SS

e. SS

f. S

g. US

After completing those pages, we “took a break” with a YouTube video. The video was very funny and was relevant to what we are learning now. We then tried to do the demo from yesterday, but, unfortunatley, it didn’t work. However, it was still interesting to watch.

Next, we turned to page 1 and Mr. H told us why this unit was so different than the past few we've done....because it has math! He then explained the term “concentration” to us. Concentration describes the amount of solute per amount of solvent in any solution. To make this simpler it is: solute/solvent or solute/solution. With this information, we did problems 2 and 3.

In 2, we looked at the examples to see if it represents the definition we just learned. The answers are as follows:

2. a. Ex

b. Ex

c. Non

d. Ex

e. Non

f. Ex

g. Ex

In 3, we ranked the solutions in order from most concentrated to least concentrated. The answers were C>D>B>A. We figured this out by looking at the amount of white space in the picture.

Today’s homework is to do WebAssigns, although there are none due Thursday. That’s all!

Unit 10, Day 2

Class began by Mr. H briefly explaining the demo we were supposed to do today. He explained that there was a solution in the buret of sodium acetate that was slowly dripping out to form the "stalagmite" looking white substance underneath. However, the solution was not quite hot enough and hadn't reached its full solubilty (50 g of solid in 5 mL of water) so we had to wait. Next we moved on to page 21 (a reading assignment from Monday). Mr H. also noted that in his demo from yesterday he had tricked us because there was a nonpolar substance in the beaker before the water was added so what was really burning was that unknown substance floating on top. The answers are as follows:
1- a
2- c
3- d
4- c
5- a
6- b
7- a
8- Polar solvents dissolve polar (and ionic) solutes. Nonpolar solvents dissolve nonpolar solutes. <- main idea from yesterday 9- d 10- b 11- c 12- a 13- a 14- a Note on side column: Water-soluble vitamins: B's and C

• excreted daily (if not used)
• contain O-H bonds
• must be consumed daily

Fat-soluble vitamins: D A K E

• stored in body tissue
  
• not excreted via urine
• can be toxic

Next we moved on to page 11. First we did number 13 and learned the reason oil and water don't mix is because oil in nonpolar and water is polar so the attraction is not there. Then we moved back up to number 10 and explained after looking back at number 8, that CS2 or CCl4 would be the best options to dissolve the paint because they are both nonpolar like the oil paint and therefore would be able to clean it off. This is why soap is better at getting dirt out then water which is similar to number 14 which again shows that water is polar and oils and dirts are nonpolar so there is no attraction. Before moving on to new material we went over number 15 on the next page. Because the SO4 bond is polar due to the electronegativity the water molecules would be towards the right end. Oppositely the C-H bonds are nonpolar because the electronegativity is right at .4 so the oil molecules would be attracted to the left side. Finally we moved on to the new material on page 15 which was similar to the webassign due Wednesday. We knew that number 1 was no because of how different all the lines were on the graph. For number 2 the answer was that as a gases' temperature went up the gas solubility went down (which was the opposite for solids). After Neil broke the stalagmite class was over and everyone was pissed at Neil. The end.

Unit 10, Day 1

Today’s class started off on a very interesting note. As I walked into class, I saw that Mr. H had lighted something on fire…surprising. But, weirdly enough, it was WATER that was on fire! What a way to start a class.

This demo, obviously, pertained to the new unit we are starting today, SOLUTIONS!

But before I get ahead of myself, let me go over some other stuff that happened early in the period. Mr. H reminded us to look at our HomeLogic accounts to check out our grades for the unit 9 test. He then went on to say that we could come in to see him regarding our tests after school on Tuesday.

Mr. H then began to ease us into our new unit. He pointed out that there would be some major differences from the previous two units, namely that this unit was both conceptual and mathematical. He also let us in on a little secret, there would be an academic stimulus on our webassigns if they were completed at least 24 hours before the due date. He took this chance to say how we had a webassign due this very Wednesday, and to get a head start on it!

Now into the notes:

We opened up our lab notebooks and took some notes on Solubility.
A Solution is made up of: solvent + solute
(Usually more solvent) (Usually less solute)

An Aqueous solution is made of: Solvent (aka H2O) + Solute (aka a Solid)
3 STEPS TO DISSOLUTION:
1. Solute’s intermolecular forces break apart
2. Solvent’s intermolecular forces break apart
3. Attractive forces between solute and solvent are formed

Then we watched a YouTube video:


After that, we went on to page 9 in our new packets. Here are the answers

1. A-B attractions becuase the readily dissolved which means they have attractive forces
2. A-A attractions becuase the didn't dissociate in B, therefore they have no attraction
3. i. Endothermic (uses energy)
ii. Endothermic (uses energy)
iii. Exothermic (Just does it because they are attracted)
we skipped 4
5. Remember that for Polar Covalent, the Delta EN is greater than or equal to .4 and less than or equal to 1.7
a. Polar Covalent b. Non-Polar Covalent
c. Ionic d. Polar Covalent
e. Polar Covalent f. Polar Covalent
g. Ionic h. Non-Polar Covalent

Page 10
6. six is a review from last chapter, about things that we should know how to do
water is Yes/Yes Carbon disulfide is No/No Thiocynate is Yes/Yes
Ammonia is Yes/Yes Carbon tetrachloride is Yes/No Dichloromethane is Yes/Yes

7. Like Dissolves Like:
means that Polar solvents ONLY dissolve in polar solutes (and Ionic) &
non-polar solvents ONLY dissolve in non-polar solutes

8. Seven is used to answer questions from 8
a. polar/ dissolves in water (because water is polar!!!)
b. nonpolar/ doesn't dissolve in water (because CH4 is nonpolar!!!)
c. polar/ dissolves in water
d. nonpolar/doesn't dissolve in water
e. ionic/dissolves in water
f. nonpolar/doesn't dissolve in water
g. nonpolar/doesn't dissolve in water
h. polar/ dissolves in water

And that was our day in CHEMISTRY!

Thursday, March 10

(MY COMPUTER IS ACTING WEIRD AND IS DELETING HALF THE STUFF I WRITE, SO IT WILL SEEM STRANGE READING THIS). Today we began the class by reviewing and doing packet pages 16 and 17. We then proceeded on completing the lab. Here is the information for station 5 and 6.

Station 5: Surface Tension:

After completing station 5 my group and I found that if you placed a paper clip on the surface of water, oil, and acetone, the paper clip may sink or float. This depends on the intermolecular forces holding bonds together.

After testing each substance we concluded that a paper clip placed on water will float; a paper clip placed on oil will sink quickly; and a paper clip placed on acetone will sink very slowly.

This is due to the type of bonds holding molecules together. Liquids with the strongest bonds between molecules will hold together very strongly, creating a strong surface tension. Therefore because water was the only liquid to hold the paper clip above the surface, it is safe to say it has the strongest surface tension. Because it has the strongest surface tension, it therefore has the strongest intermolecular forces. Next is acetone. When a pa0per clip was placed on acetone it sunk much slower than when it was placed on oil. Because of this characteristic, acetone has a greater surface tension then oil. This also means acetone has a greater intermolecular force because the strength of intermolecular forces are directly related to surface tension. However, we know that acetone does not have a stronger surface tension than water because acetone was not able to hold the paper clip above the surface.

Surface Tension:

oil

Intermolecular Force Strength:

oil

After completing this assignment, it is clear that the greater the intermolecular force, the greater the surface tension. This is because the stronger the molecules are being pulled together, the less likely an object will pass through them.

Station 6: Beading:

Observing substances bead together is also very closely linked to intermolecular forces. This is due to the fact that if there is a strong intermolecular force between 2 molecules, they will pull towards each other with greater force. The greater the beading the stronger the intermolecular forces.

After completing this assignment, we observed that if water was placed on wax paper it would bead together. Hexane placed on wax paper would not bead together, and oil placed on wax paper would bead only a little. Likewise, if you placed the same substances on the same lab bench, you would get the similar results. Water would have the most beading, followed by oil, and last you would have hexane. The only difference is there would be slightly more beading (for substances that did bead) on the lab bench. This is observed when looking at the beading differences between water on wax paper and water on a lab bench. Because water is attracted to the wax paper there is less interaction between nearby molecules. This is because the attraction to the wax paper is preventing some molecules to attract. Therefore, when water is placed on a lab bench, it does not attract to a lab bench, but instead the molecules pull toward each other with greater force.

This assignment has demonstrated that the stronger the intermolecular force, the more likely that the molecules will bead together. Furthermore, if a molecule repels the surface on which it is rested on, it will bond together with greater strength.

HW:

Test is tomorrow.

Orbitals and More

What is this, you ask? It's a Lewis structure of a methane molecule, or CH4. As you can see, carbon is the central atom, while the hydrogens attach themselves to it. Every atom is happy because each hydrogen gets two electrons, and the carbon gets the eight it requires. But what does this have to do with anything? Simply put, this explains how many electrons each atom in a molecule should get. This also explains why this molecule is non-polar, even though it has polar bonds. Finally, this dot structure reiterates the fact that methane would be composed in a tetrahedral shape, much like this:

We started class today with taking a closer analysis at pages 11-12 in our packets. Not to mention reviewing Lewis structures, we studied a new aspect of looking at molecules: we learned about the bonding type molecules have. All the possible bonding types go as follows:

sp --------------------> AX2

sp2 --------------------> AX3

sp3 --------------------> AX4

sp3d --------------------> AX5

sp3d2 --------------------> AX6

"But I'm confused, Dmitriy. What is it I see above?" Well, I'll tell you what. Those six values are what the layman would call "Hybridization types". The A represents the central atom, in methane's case the carbon. The X represents the surrounding atoms, the hydrogens. The numerical digit would represent how many surrounding atoms there are. As you already know, there may or may not be an E, which would represent unbonded electron pairs.

In short, these hybridization types show all the AX's and which spd configuration they would have.

Next, we introduced intermolecular as well as intramolecular forces on molecules.

"But I'm confused, Dmitriy."

Intermolecular - happening between different molecules, as in molecules acting on each other

Intramolecular - happening within an individual molecule, as in a molecule acting on itself

This is a weak intermolecular attraction. It is occuring between two HCl molecules. "But why is it weak?" Because hydrogen is not bonded with nitrogen, oxygen, or iodine. However, this diagram is very effective at conveying its message because it shows not only intermolecular forces, but intramolecular forces, as well. As you can tell, although the intermolecular forces are weak, the intramolecular forces (covalent bonds) are quite strong.

Furthermore, we discussed London dispersion forces as well as dipole attractions.

London dispersion:

1. temporary, short-lived

2. opposites attract (aka electrostatic)

3. greatest for larger molecules

Dipole-dipole:

1. longer-lived

2. electrostatic (same as above)

3. dipoles align selves

4. NOT experiences by non-polar substances

However, I do understand a picture is worth a thousand words. So I'll give you two thousand. Here are images of both these attractions to demonstrate the validity of the two arguments.

To summarize and conclude, we solidified our knowledge of chapter 9 to be better prepared for the TEST ON FRIDAY

Furthermore, we have instructions to finish our labs. Don't forget to look at the review and do the review WebAssigns!

Here's some chemistry jokes for you:

What do chemists call an iron ring?

A ferrous wheel

Why do professors like to teach about ammonia?

It's really basic stuff

What's the name of 007's Arctic cousin?

Polar bond

Problem, Will?

Tuesday March 8, 2011

Today we started off class easy, we went through a reading sheet, finished our lab from a couple days ago, did some stuff in the small packet and then learned something new about HYBRIDIZATION!

Reading sheet answers.

(after the wiggly line Mr. H said the questions were unimportant and we were probably not going to talk about them again.)

1. d

2. 0,1,2

3. c, b

4. c, b

5. a,c

6. c

7. d

8. sp2, sp3, sp3, sp3d

9. sp, sp2, sp3

10. false! 2 electrons are in hybridized orbitals, four electrons are in p orbitals

11. a

12. b

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

13. c

14. d

15. h

16. d

Mr. H then talked about the test on friday. We are also going to turn in our lab notebooks on friday, which inside are going to have one completed lab and not inside is a lab on a seperate piece of paper we are going to do. He then told us the usual how theres a review sheet on moodle as of tomorrow and we should use the webassign study guide to study.

After that Mr. H briefly reminded us about resinence, which is when a molecule has one double bond and one regular bond in a lewis sturcture then in reality its like a bond and a half. Its not a distinct doouble bond and single bond.

Then we opened up the little packet we got to fill in if the moleucles are polar or non polar. Mr. H used a diagram on the board which i copied and will post after this paragraph. He taught us how to figure out if the bonds are polar or non polar. We do this by a set of questions. First we figure out if the bonds polar by numbers, we use the periodic table in our packet to determine the number. If the number is in between and including .4-1.7 then its polar and we have one more question to ask. If not then were done, the molecules are not polar. The next thing we do is figure out the AXiEj notation and use that along with the list of special non polar bonds from our reading guide to make sure the bond is polar.

A little while after Mr. H introduced

hybridization. Which is pretty much a new notation for the s and p orbitals in bonds. I apologize for the bad picture quality but in the margines where we should take notes was two sections that were really important. They showed what the conversions are from AXiEj notation to hybridization notation.

After all that learning we were told about the homework. Tonights homework was a webassign, delicious due thursday, lab notebook friday and if you do the 7.3-.4 webassign today its .25% extra credit SO DO IT.

Friday March 4th, 2011

Today, Mr. H began class by having everyone open their packets to page 26. Once there, he stressed the importance of including for question 17: AX5, AX6, AX2E3, AX4E2. Its important to know that all of these molecules are NONPOLAR. Additionally, he broke down the AXiEj molecule and explained what each variable represented:




A<---------- Central Atom


X<------------------- Terminal atoms


i<--------------------# of terminal atoms


E<-------------------- Non-bonded electron pairs


j<----------------------- # of non-bonded electron pairs


Next, Mr. H went to Grace's blog and emphasized the important parts of the blog. He also stressed the idea about dipoles, especially when grace says "The greater the number in the electro- negativity, the more electrons an element pulls. Cl, has a high number than H, resulting in more electrons towards itself. It is greedy. These are called dipoles. " It is essential to understand the idea behind dipoles, because by understanding dipoles, you are able to understand the molecular geometry of different kinds of molecules. After going over Grace's blog, Mr. H had the class open their packets to page 9 where he continued to talk about the "AXiEj" molecule and explain how to figure out degree angles and types of molecular geometry. For instance, lets take SO2. Here is what it looks like based on the lewis dot structure and its molecular geomtery:

Lewis Dot Structure

Molecular Geometry (Bent)

Based on the Lewis Dot structure, SO2 has 3 total pairs. I was able to figure out the total pairs by focusing around the central atom, which is S, and any bonds or unpaired electrons. So, since we know that the total pairs in SO2 is 3, this limits our answers to either trigonal planar or bent. Next, you count the total number of bonding pairs. Although SO2 has 1 double bond and 1 single bond, the total number of bonding pairs in SO2 is 2 bonding pairs. An important thing to know when counting bonding pairs is that no matter if a bond is a sing, double, or triple bond, it still only counts as 1 bonding pair. So, using that information, you can conclude that SO2 has 2 bonding pairs. Next, you have to count the number of lone pairs. Once again, you focus around the central atom. Since (S) only has 1 pair of electrons surrounding it, that means that it has 1 lone pair. So in the end, the number of electron pairs comes out to be 3-2-1. Once again, this stands for 3 total pairs, 2 bonding pairs, and 1 lone pair. If you are confused on how to find the total number of pairs, a trick that you can use is that if you add the number of bonding pairs by the number of lone pairs, you'll get the number of total pairs. So now since you know the number of electron pairs, you turn to page 9 of your packet and look for the bonding pairs that coincide with what you found. You can find the answer to this problem in the third row of the diagram. when you look across, you can see that the molecular geometry of SO2 is bent. (the picture above this paragraph shows the molecular geometry of SO2). Now, to find out the "AXiEj" molecular set up of SO2, you use the information that i provided at the beginning of this blog and apply it to the molecule SO2. Since "S" was the Central atom, you substitute "S" for A. So far, it looks like this: SXiEj. Next, you substitute "O" for X, because "O" is the terminal atom in this molecule. So now, you have SOiEj. Considering that you had 2 Oxygen atoms, you replace "i" with 2. So now you should have SO2Ej. Next, considering the fact that SO2 had 1 lone pair, you would end up replacing "j" with 1. At the end, you can now see that the molecule was SO2, and it had 1 lone pair electron. Considering the fact that now we know the molecular geometry of SO2,  the next thing to find is the degree angle. If you look at the picture above this paragraph or the picture on page 9, you can see how the bent molecule forms an angle that is bigger than 90 degrees. Now, you have a few answers to choose from. The possible degree angles are: 180, 120, and 109.5. First of all, we can eliminate one of the answers right away, because the picture above doesn't form a straight line, therefore we can eliminate 180 degrees. Next, based on the picture, you would have to conclude that the its 120 degrees because its bigger than 90 degrees, but also less than 180. The choice for 109.5 would be eliminated because the angle formed is not very narrow, instead its much wider. By using all the information i have provided, anyone can find out a molecules molecular geometry and angle measurement.

The next thing we did after going over page 9, was a lab. The lab we did was CB1-Molecular Geometry and Polarity Lab. Purpose: To determine the Lewis structure AXiEj notation, molecular geometry and polarity of the following molecules: CH4, NH3, OH2, CO2, SO2, XeF2, SF4. Mr. H color-coded each element with a certain piece in a box of pieces which would be used to create models of molecules. In this lab, you and another partner built models of each molecule, recorded the necessary data, found the molecular geometry, found AXiEj, but didn't find its polarity. (that is to be completed later on). After spending about 10 minutes on this lab, Mr. H passed out a quiz to the entire class and we took a quiz for the last 20 minutes.

Tonights homework: Webassign 7.4 Rdg Sheet. Due: Tuesday March 8

  

Thursday, March 3

We started off the day by making fun of Tim and his inability to speak because he had food in his mouth, which Mr. H does not like in the classroom. We then took out our packets and Mr. H gave us the answers to the reading sheet 7.3. The answers are as follows:

1. ADE
2. False; it is nonpolar, e- are shared equally
3. A
4. C
5. a.) H-F b.) H-F c.) H-O
6. C
7. ABC
8. A
9. a.) Cl-Cl b.) C-N c.) C-O d.) H-O
10. -
11. C
12. If a diatomic molecule has a polar bond, then it is polar.
13. ABC
14. BC, IO
15. B
16. C
17. The polarity of the bonds and the molecular geometry.
18. ADE,

*Mr. H added AX5, AX6, AX2E3, & AX4E2. These will be important for a later time in this blog.

After that, Mr. H reviewed the basics of polarity and went over Tim's blog. After that, he reminded us about the pop quiz tomorrow, Friday, March 4. It will cover the basics of ionic and covalent bonds, the Lewis structure, and the molecular ge

ometry. You will have to memorize the numbers corresponding with the type of geometry the molecule h

as, but Mr. H will provide us the names. It will be multiple choice, along with some Lewis structure drawings and the molecular geometry.

We then turned to the other packet that Mr. H gave to us on Tuesday, and worked on SO2. Like always, you would count the number of valence electrons, do the skeleton structure, and do trial and error. When you have the right Lewis structure, you will then imagine the shape of the molecules without looking it up in the packet. We then did numbers 11 & 13. The answer to number 12 is: 3; 2; 1; Trigonal planar; Bent; 120 degrees. These types of questions will be on the pop quiz.

We then turned back to pages 25, 26, and 10 in our main packet. This will be on the test, not pop quiz. Mr. H explained to us that on the electronegativity chart (page 10), F, which is on the top right corner has a high electronegativity. This means that it is very greedy this its electrons. On the contrary, Fr has a low electronegativity, which is located on the lower left corner. Let's have Cl and Na. One would take the numbers for both elements on the electronegativity and then subtract it. If the difference is greater than 1.7, the elements would just give an electron away, which is an ionic bond. However, if you take Cl and H, Cl and H can share because the difference is less than 1.7. But, Cl would get partial charge. The elements would not share the electrons equally. This is called polar covalent bonding. Number 8 on the reading sheet (7.3, page 25)shows the difference between each element bond. The electronegativity chart shows how polar or non polar an element is.

Number 13 shows this: H -------->Cl. The arrow points over to the element that the majority of the electrons are getting pulled. The greater the number in the electronegativity, the more electrons an element pulls. Cl, has a high number than H, resulting in more electrons towards itself. It is greedy. These are called dipoles. We then turned our attention to the top of page 26, where we take notes on the class discussions. It says, "Some molecules have polar bonds but are overall non-polar molecules." Mr. H explained this idea with the example of SF6. SF6 has F atoms that are opposite of each other in the molecular geometry. On one side, the electrons will pull, but on the other side, it will pull the opposite directions, leading to the cancelling each other's polar bonds. this would make it so that it is non-polar. Same with XeCl2. Non-polar molecules are mostly gases. H2O is bent, so it is polar. Always go to the molecular geometry to determine if this idea is used for that molecule.

We turned our attention to number 18. Mr. H told us that this helps with tonight's WebAssign. I'll use AX2E3 for an example. A represents the central atom of the molecule. X is attached to the A (a bonded pair) and E is the lone electron pairs. The numbers next to X and E are the number of X's and E's there are in the molecule. We ended with writing the Purpose of Lab CB1 and started to work on the data section individually.

Homework tonight is a WebAssign and to study for the pop quiz.

Wednesday, March 2

We started class by picking scribes. Mr. H also figured out that Grace ditched yesterday to avoid being the scribe. Afterwards, he went on to review yesterday's material by going over Mathein's blog.

After all the reviewing, we studied page 9 and the packet we received on Tuesday. He showed us the bond angles of the different joined atoms. He told us that a linear can only have a 180° angles, a trigonal planar can only have 120° angles, and a tetrahedral can only have 109.5° angles. Trigonal Bipyramidals, however, were different. They could have any combinations of 90°, 120° and 180°. After explaining the angle bonds, we moved on to the packets for some practice. We worked on numbers 5, 7, 9, 16, and 17. To help us on these problems and our future webassigns, Mr.H recommended that we look at page 9 in our packets and pg177+181 in our book.

For number 5 we were required to figure out the information of the compound SF4. The lewis dot diagram looks like the diagram on the right. To figure out the number electron groups we looked at the central atom. Sulfur had a total of 5 electron groups because it made 4 singles bonds and had a pair of non bonding pairs. Because it had 4 bonding pairs and 1 non-bonding pair it was a trigonal bipyramidal. Its molecular geometry was a seesaw and had bond angles of 90°, 120°, and 180°. We then continued to work on problems 7, 9, 16, and 17. The answers are...

7) PH3= 4 electron groups, 3 bonding pairs, 1 non-bonding pair, tetrahedral, trigonal pyramid,109.5°

9) SO3= 3 electron groups, 3 bonding pairs, 0 non-bonding pairs, trigonal planr, trigonal planr, 120°

16) ClO2- = 4 electron groups, 2 bonding pairs, 2 non-bonding pairs, tetrahedral, bent, 109.5°

17) ClF2- = 5 electron groups, 2 bonding pairs, 3 non-bonding pairs, trigonal bipyramidal, linear, 180°

The First Of March

To begin, mr H went over the reading sheet that was assigned the previous day on web assign. The Answers For this are as follows
1.B
2.C
3.B
4.B
5.A
6.C
7.A
8.C
9.A
10.a.2,2
b.3,1
c.2,3
d.3,0
11.a. Bent
b.Trigonal Pyramid
c. Linear
d. Trigonal Planar
12.a. Linear
b. Linear
c. Tetrahedral
d. Bent
13. A
14. C
15. AX2E2



Next, Mr. H went over therules that will assist us while drawing Lewis Dot Diagrams. The rules were:
1. Always count the total number of valence electrons and use that number as the number needed of electrons in the diagram.
2. Hydrogen always forms the duet rule
3. Second row elements C,N,O,F will always satisfy the octet rule.
4.Second row elements Be and B will usually have less than the octet rule
5. 3rd row and heavier elements may exceed the octet rule

These rules assisted us with the drawing of some more dot diagrams.

We first started with the compound SF6. This was the simplest of all of the following compounds. We began by counting the valence electrons of both S and F. We got these from the PT. After counting, we ended up with 48 electrons. Next we drew the skeleton with S in the middle and the 6 F's surrounding it. We then gave all the F atoms 8 electrons and ended up with the right amount of total electrons. But there was one problem, the S atom had 12 electrons. But if we refer to the rules above it is acceptable for the S to have more than 8 electrons because it is in the 3rd row.

The next problem was I3-. This was a relatively easy problem after being informed on a new piece of info. When you end up with a ion. You must add electrons if it is a negative ion and subtract electrons if it is positive.

We next did the following problems, BCl3, CN-, O3 and NO3. They all behaved in similar ways and according to the rules above.

After this Semi-review we moved onto the new material for the day. This brought u sthe the Valence Shell Electron Pair Repulsion Theory or more appropriately named the "VESPR".
This chart described how electrons arranged themselves in a compound. The Chart above is the arrangement of the electrons in different arrangements. On page 9 of your paquete you can see a chart containing the corresponding geometry of the pair to the number of electrons in the pairs. After examining this table, we moved onto doing some problems in the mini-paquete that Mr. H gave us. we did the first 3 problems.

For CH4 we began by drawing the Dot Structure of the atom. If you cant do this refer to previous blogs and above. Then we used the diagram to count the number of electron groups which were the number of groups of atoms that were surrounding the central atom. Then we counted the number of bonding pairs. The bonding pairs were counted generally by the number of bonds that were in the diagram. For this problem we got 4 for both the number f groups and the number of bonding pairs. Then we went back to page 9 in the paquete and looked for the corresponding geometry with the table. We found that it was a tetrahedral.

Then we proceeded with the next 2 problems which were the same as the first.

That was the end of the class. YAY