Monday, February 28

Today we started off class by going over the Chapter 7.0-1a WebAssign.
The answers are as follows:

1. Ionic bonds normally involve two elements giving or taking electrons and consist of a cation and anion. Covalent bonds share their electrons with another element and can involve a cation and anion or two anions.

2. A
3. D
4. A
5. A
6. D
7. B and C
8. C: 4 F: 7 S:6 Cl: 7
9. Bonds are formed by a pair of electrons being shared among atoms.
10. a. 1, 3 b. 2, 2 c. 3,1 d. 4,0
11. 2, 4, 6
12. a. 8 b. 14 c. 18 d. 10 e. 24 f. 32 g. 14 h. 24
13. It is usually the atom listed first and it is usually the lone atom.

After we went over the WebAssign answers, Mr. H showed us the Lewis Structure Rules:

1. Count # of valence electrons

2. Draw skeletal structure

3. Give each atom 8 electrons (except H-2e-, B-6e-)

4. Count electrons (dots and dashes) if needed, make corrections

Also for 3 and 4 this is trial and error which means if the number of valence electrons you are given don't match up with how many you drew, you must go back and make the necessary corrections.

We then began our lesson for the day on page 5 in our packets.

The first few problems involved basic Lewis Structure questions that were fairly easy to solve. However, the problems became harder and we couldn't solve them with only using single bonds. For CO2 there are 4 V.E. for carbon and 12 for oxygen, 6*2=12, and in total there are 16 V.E. If you do this problem the regular way, you will get an incorrect answer of 20 when the problems states you only need 16. To fix this problem you have to use 2 double bonds bonding the 2 oxygen atoms to the central carbon atom. The result is four V.E. around each oxygen and two double bonds which equal 8 connecting to carbon, and you have 4+8+4=16.

We did a couple more of these types of problems and then went on to problems that broke the octet rule. For XeF4, there are a total of 36 V.E., but when you draw the Lewis Structure, you can only put in 32. To solve this problem you have to satisfy the octet rule for all atoms and put the remaining electrons on the central atom which is usually the larger one.

Once we finished this, Daria told Mr. H to shut up so we could present our skits about bonding. They were all good and provided useful information about some of the bonding rules.

The last thing we did before class ended was go over WebAssign 7.1b and the answers are as follows:

1. A

2. False

3. C

4. To determine which one of the two competing Lewis structures is the more reasonable structure.

5. A

6. B

7. ABCE

Our homework is the 7.2 Reading WebAssign due tomorrow.

Friday, February 25

It was a sad day in chemistry since we did not get to see Mr. Henderson's slick dance moves as seen in class the previous day. We had a substitute, and a new and exciting assignment. The assignment, "Bonding Theatre," is an in-class group assignment. The groups are divided up between lab groups. The assignment is a short 1-2 minute skit that explains a part of either ionic or covalent bonding. An ionic bond is when electrons are transferred (usually between a metal and a non-metal), and a covalent bond is when electrons are shared (usually between two non-metals). Every member of the group must have a role in the skit either as an actor/actress or a narrator. As far as props go, each group can use props that are easy to use. Furthermore, if a person in the group is not there, then the lab group will not make a part in the skit for him. If a person is there, but will not be there Monday, he should participate in the creation of the skit but not give himself a role. So, we brainstormed our ideas for the first 20 minutes of class.

For the last half hour of class, we worked on a WebAssign. This assignment consisted of 13 questions on the topic of bonding (section 8.1). If your lab group had not finished planning for the skit, it was highly recommended that your group sat together. The purpose of this was so your group could quietly talk and finish the planning of the skit. If you finished the WebAssign, then there was no homework for the weekend; if not it was homework.

It was a very relaxing, yet productive class period. Hopefully on Monday we get to watch another awesome video like this one which we saw on Thursday.

http://www.youtube.com/watch?v=9-V1oW6oENI

Thursday, February 24

Today we began the class by looking at Wednesday's blog. From there we proceeded to review ionic compounds, which was discussed on Wednesdays. We recapped on the difference between ionic, Polar, and Non-Polar bonds. We were also given the answers from the previous days homework, Pg. 3 of the packet.

1.a.- 1 b.-2

c.-6 d.-7 f.-3

2.a- [Ne] 3s1 & it loses 1 electron

b.-[Ne] 3s2 3p4 & it gains 2 electrons

c.-[Ar]4s2 & it loses 2 electrons

d.-[Ar] 4s2 3d10 4p5 & it gains 1 electron

We then began to discuss covalent bonds and the Lewis dot structure. From last nights homework we all know that every element wants to gain or lose electrons so that it has 8 valence electrons (2 in the case of H and He). That is why noble gases are so stable. In an ionic bond, one atom will gain electron wh

ile another atom loses an electron. (ex: in NaCl, sodium gives up 1 electron to chlorine, so that both elements have 8 valence electrons). However sometimes, 2 atoms will form covalent bonds and each element will share atom. To begin diagramming a covalent bond, you have to be able to Lewis Dot structure any given

element given the number of electrons.

For example, if you have Cl, you know it has 7 valence electrons.

Step 1: write Cl

Step 2: place 1 dot around Cl for every electron

-it should look like

Now if you wish to write a covalent, you:

1. begin by drawing the Lewis Dot Structure for each element

2. remove 1 electron and draw an electron to represent a pair

3. it should look like this

Each pair of dots can be represented by a dash. As you can see, the H and Cl are each sharing a dot so that H has 2 valence electrons and Cl has 8. They are both stable.

HW:

Webassign

Honors Chemistry, Period 3: Wednesday, February 23

Honors Chemistry, Period 3: Wednesday, February 23: "Today in class the first thing that we did was get our lab notebooks back. Once we got those back Mr. Henderson gave us our lab grades and..."

Wednesday, February 23

Today in class the first thing that we did was get our lab notebooks back. Once we got those back Mr. Henderson gave us our lab grades and grade sheet with our latest test on it, but he noted that some of the homework has not been recorded. After that he told us that the next unit that we would be studying is chemical bonding. In this lesson we will be learning how to understand the nature of a chemical bond, about the various categories of bonds, what causes 2 atoms to bond together and much more. Mr. Henderson started off by telling us that some people like to think of a bond as a stick that holds things together. He then reassured us that this is not really how it is. The bond is moving to try and find balance, so that a happy state is found between the protons and electrons. It was also made very clear that protons are attracted to electrons. Mr. Henderson also did a little demonstration with 2 magnets(oppositely charges) that he had attached to a track. He noted that if pushed together they repel each other. Then he took out a spring and attached it to the magnets so that they where almost touching. He showed us that the magnets where in a happy state and if pushed too close together they repelled each other and if forced together from a far distance they would do the same. After the demonstration he had us look inside our new unit 9 packets. We first looked at the diagram and saw that as the potential energy decreases and the distance between the nucleus decreases the protons and electrons are brought to a happier state. We then went over ionic and covalent bonds. Ionic bonds are formed with ions; usually occur between metals and non-metals; formed by ions that have opposite charges. There are two types of covalent bond; polar and nonpolar. In polar covalent bond the electrons are shared unequally and in nonpolar the electrons are shared equally. Covalent bonds occur between two non-metals. On page 2 we talked electronegativity, and how the greater the difference in electronegativity , the greater the ionic character of the bond. We then matched bonds between C-H, C-C, C-N, C-O and C-F and how they are portrayed in the pictures. After that looked at a chart and filled bonds. We filled in the information that was needed to find if the bond was nonpolar covalent, polar covalent or ionic. If the difference in electronegativity is > 1.7 then the bond is ionic, < 0.4 it is nonpolar covalent and greater than or equal to 0.4 and less than or equal to 1.7 then it is polar covalent. For our homework we had to do pages 1-3 in the unit 9 packet.

Friday, Febraury 19, 2011

Today, class started off by reviewing what we did in class the day before and the blog post put together by Katie. Then, we received the answers to a previously due Web Assign in our packet on page 17 which was a 6.8 reading sheet. The answers are:
1. b
2. d
3. They represent the number of s, s+p, s+p+d electrons....etc.
4. b
5. b,c
6. a,b,c
7. d
8. a
9. d
10. a
11. d
12. b
13. b
14. a, d
15. b
16. d
17. F, Rb

Also, Mr. H showed the class where the highest number of people viewing the blog came from which included interesting places such as Germany, but of course it was mainly from the United States. I would like to say Hi to all the people that are reading this blog not in our class.

After, we moved on to page 10 in the packet which had explainations for the periodic trends of atomic radii, Ionization energy and Electronegativity.

The following is what Mr. H explained in the class as the reason why when moving from left to right on the periodic table the size of the atomic radii decreases: Additional protons led to an increase in nuclear charge=the outer orbitals are pulled inward and shrink.

Explaination for when moving from top to bottom the atomic radii increases: The outer most electrons are entering orbitals that are larger thus the size of the atom increases.

Explaination for why when moving from left to right across the periodic table the Ionization Energy increases: The nuclear charge increases and pulls with more force on the outer electron thus it takes more energy to remove outer electrons.

Explaination for why when moving from top to bottom on the periodic table the Ionization energy decreases: The outer most electrons are entering orbitals that are larger; thus being further away the electrons are pulled away with less force and requires less energy to remove them.

Mr. H said that the explainations for Electronegativity weren't needed, so we skiped that part. The rules though are moving from left to right the electronegatvity increases and going from top to bottom the electronegativity decreases.

Katie's blog has diagrams and graphs that show the properties of all three trends on her blog.

Once we finished learning about the explainations we went to the computer lab to work on upcoming Webassigns or labs that were due.

Homework for the weekend is to study for the test and finish the Webassigns one of which is optional or you could to both to earn yourself an extra throwout at the end of the quarter.

THURSDAY FEBRUARY 17, 2010

We started off class by reviewing question #1 on our Web Assign that is due Friday. The question is: How many electrons in an atom can have the following designations? You were told to state how many electrons could be held in 5px. Since "x" simply is a sublevel of the 5p category, there would only be 2 electrons held in 5px.

Next, we went over the periodic trends which is similar to what we did in the computer lab yesterday. Graphs, diagrams, and explanations can be found on packet page 9. There are three topics discussed on packet page 9: size of atom, ionization energy, and electronegativity.

There are two noticable trends that can be found looking at the size of the atom. As you go down a column, the size of the atom increases. As you go across a period, the size decreases.

There is a diagram provided below to be used as a visual to show these two trends.

When looking at a graph of ionization energy the trends seem to be the opposite to the size of the atom. (note: ionization energy is the amount of energy required to remove an electron from an atom). As you go down a column, the ionization energy decreases. As you go across a period, the ionization energy increases. Below is a graphic demonstrating the trends for ionization energy.

Both of these two periodic trends discussed are discussed and mentioned in lab AS2.

The last graphic we looked at was electronegativity. (note: electronegativity relfects the tendency of an atom to attract the electrons in a bond with an atom of another element--noble gases aren't listed for this graph because they do not usually form bonds) The trends noticed were when you went down a column, the electronegativity decreases. Also, when you went across a period, the electronegativity inceases. Below is a graphic on the trends of electronegativity in elements.

Next, we reviewed #7 on packet page 8 which is similar to the quantum mechanics 3 web assign number 2. The question states: An ion of an isotope has a 2+ charge, an atomic mass of 56.9397 amu, 2 electrons at the n=4 energy level and 13 electrons at the n=3 energy level. From that you were to determine the atomic number, total number of electrons, mass number, and total number of s/p/d electrons. To figure this question out you count the number of electrons stated in the problem. In this case there would be 25 electrons (don't forget to count the electrons in n=1 and n=2)/ Since it has a 2+ charge, the element would have the atomic number of 27. You could then use this information to answer the remaining questions.

After that problem we worked on the Web Assign using netbooks.
The answer to number 2 was eventually given to us, after having complications with answers.
#2:
a) 24
b)6
c)12
d)0
e)26
f) 1s2 2s2 2p6 3s2 3p6 4s2 3d4

HOMEWORK:
--delicious assignment
--web assign

Wednesday, February 16

We started class today by going over the answers to the 6.5-6.6 reading sheet. The answers are:
1.
a. 7
b. 2
c. 3
2. a. 2s
b. 2s
c. 3p
d. 4s
3. a. 2
b. 6
c. 3s
d. 3p
e. F
g. Cl

4. a. 2p
b. 3s
c. 18th
d. 3d
e. 37, 38
f. 5p
g. 13th, 14th

5. a. S, Ca
b. Cl, Zn
c. Xe, Ba

6. Ar, 4, 3, 4, 5
7. a
8. b
9. a. 2
b. 5
c. 1
10. O, Na, Sc, Cl
11. This is a violation of Hund’s rule.

We then reviewed how to write electron configurations and learned new information about quantum numbers. Mr. H put this chart on the board.

n=1 l=0-----> 1s orbital ml=0
n=2 l=0-----> 2s ml=0
l=1-----> 2p (3 of these) ml=-1, 0, 1
n=3 l=0-----> 3s ml=0
l=1-----> 3p (3 of these) ml=-1, 0, 1
l=2-----> 3d (5 of these) ml=-2, -1, 0, 1, 2

The n value is the number of possible energy levels that atom can have. Since there are 3 possibilities for 2p, there needs to be 3 different ml values to distinguish them from one another. To find the ml value, you start with -l and go to l. For example, since n=3, l=2 has 5 possibilites, you need 5 different ml values. You start with -2 since l=2 and go to 2. The vlaues would be -2, -1, 0, 1, 2. We then talked about how when l=0 it is an s orbital (spherical). When l=1, it is a p orbital (pinched cylinder). When l=2, it is a d orbital (dorky). We then reviewed what each of the quantum numbers represent.
n=energy level (size)
l=orbital type
ml=sperical orientation
ms=spin direction. The ms value is always +1/2 or -1/2

We did a practice problem where you had to find the possible quzntum numbers for a 3p electron in Cl. The answer is n=3, l=1, ml= -1, 0, 1, ms= 1/2, -1/2
To find this you can use the chart from above. Since we were given that it was a 3p electron, that indicates that n=3. We can also see from the chart above that the p orbital for any n value has a corresponding value of 1 for l, so l=1. If you go across the chart for n=3, l=1, you will find the possible ml values are -1, 0, 1. Also, ms is always 1/2, -1,/2, so that is what it is for this problem.

After this we went to the computer lab to work on our new lab, It's Periodic Lab. We went to www.webelements.com to look at trends in the covalent radius and first ionization energy of elements. The covalent radius is the size of the atom that forms part of a covalent bond. By looking at graphs of the periodic table, we saw that going down the periods, the covalent bonds increase and going across the groups from left to right, the covalent bond decreases. We then observed the trends in the ionization energy of elements. Ionization energy is a measure of how much energy it takes to remove an electron from a gaseous atom. We saw that going down the periods, the ionization energy decreased. Going across the groups from left to right, the ionization energy increased. The exception to this was that the last element of each group in the p block was smaller than the first element in each gorup of the p block. The lab is due Tuesday.

The homework was our delicious assignments due Friday and 2 webassigns due Friday and Tuesday. There are 3 assignments on webassign, but only 2 are for credit. If you choose to do all 3, one will count towards an extra homework drop.

Today’s class was very informative. We started out by first going over Will’s blog from yesterday. Mr. H kept saying what a great job Will had done, and how he really liked one of Will’s graphics, which I have to agree with. Since this graphic is also relevant and helpful with my post, I will also put it up.

We then started doing some practice problems in our Lab Notebooks, problems that are very similar to one of the two webassigns due for homework tomorrow.

Before we go over some problems, just remember these key things:
• All orbitals in a sublevel must have one electron in them before they can have two.
• To write an Abbreviated Electron Configuration, you first go to the previous rows noble gas to use as the starting point
• In the d-block, the rows start with 3d, not 4d as one would suspect
• When dealing with elements that go into the 5d or 6d orbital, ADD THE F-BLOCK ORBITALS IN IT!

Here are some examples of doing Abbreviated Electron Configurations:

Aluminum: [Ne] 3s2 3p1 explanation: Ne is the noble gas in the row above Al. Al fills the 3s orbital, which can hold up to 2 electrons which is why the superscript is 2. Then, Al fills up only 1 electron in the 3p orbital.

Lead: [Xe] 6s2 4f14 5d10 6p2 explanation: Xe is the noble gas in the row above Pb. Pb fills up the 6s orbital, which can hold up to 2 electrons, which is why the superscript is 2. Next, Pb fills up all 14 electrons of the 4f level, which is why the superscript for that is 14. Then, it also will fill up the 5d row. Finally, it fills up only 2 electrons of the 6p level.

Tl: [Xe] 6s2 4f14 5d10 6p1 explanation: Xe is the noble gas in the row aboveTl. Tl fills up the 6s orbital, which can hold up to 2 electrons, which is why the superscript is 2. Next, Tl fills up all 14 electrons of the 4f level, which is why the superscript for that is 14. Then, it also will fill up the 5d row. Finally, it fills up only 1 electron of the 6p level.


*if you want to make sure you did it right, you can add up all the super scripts in the abbreviated electron configuration and add it to the atomic number of the noble gas in the brackets.


Next we went on to IONS
Ions are an atom or molecule in which the total number of electrons is not equal to the total number of protons, which we learned earlier this year. Ions are in the excited state.

Take, for example, a sulfur ion (S2-). To write a abbreviated electron configuration for it all you have to do it write the abbreviated electron configuration for Argon, because they have the same number of electrons.

Ground state vs. Excited state

Ground State for chlorine:[Ne] 3s2 3p5
Excited State for chlorine: [Ne] 3s1 3p6 OR [Ne] 3s2 3p4 4s1
Take note that each of the three configurations has the same amount of electrons, just on different levels. The electrons will eventually give off light and jump down the levels to its ground state.

We ended today in the computer lab working on the webassign due tomorrow.

Monday, February 14

We started today's chemistry class off with a little valentines day fun. Mr. H brought in a small packet, and when he hit it, a foil balloon shaped like a heart expanded inside. He explained that this was caused by the reaction of two substances, and the gas was the product. When he hit it, it reacted the two substances.

But now on to what we learned in class today. We flipped our packets to page 5 and read the first two paragraphs. This explains how electrons are configures within the orbitals of atoms. There are 2 atoms per orbital, and all orbitals in a sublevel must have one electron in them before one can have two.

We then tried some examples on the next page, I'll give you a few answers.

H 1 electron 1s1
He 2 electrons 1s2
C 6 electrons 1s2 2s2 sp1

Above this table is a diagram that shows the order of all the orbitals.

Next Mr. H taught us a shortcut. For elements that are large, and would take a long time to write out their entire electron configurations, you just put the symbol for the closest noble gas that is before the element, and then the electron configurations up to that element.
Example. For Barium.
[Xe] 6s2

Then Mr. H told us how electron configurations explain the orientation of the periodic table. I have a picture here that helps explain it.

The location of the atom determines what the last energy level will be, based on this diagram. Also, this is useful for using the shortcut.

At the beginning of class Mr. H told us we would use the computers, but we never did. I guess it took a long time to learn this. I hope we get to use the computer tomorrow!

Friday, February 1

We started todays class with the lab that we had started on Thursday. In this lab we were to carefully observe the line spectra of four unknown elements and to identify the four elements by comparing their line spectra to those of known elements. Mr. H told us about a website, http://astro.u-strasbg.fr/%7Ekoppen/discharge.html, where you can compare your spectra with the elements online.

After that we spent a few minutes going over the blog from Thrusday the Brooke did.

Next we started working on page 3 from the packet. Before we began, Mr. H warned us that this unit was going to be on of the most difficult units because it is the hardest to visualize.

The Quantum Mechanical Model of the Atom

Features of the Theory:

· Very mathematical

· Electrons are located in regions of space know as orbitals

· Each individual electron of an atom is described by 4 quantum numbers

· Each orbital holds at most 2 electrons

As we went through the rest of page 3 we added that information to our notebooks as well. The main idea was that each individual e- is described by four quantum numbers, which are n, l, ml, and ms. The first number is the principal quantum number. It is represented by the letter n, describes the energy level and size, and has the possible values of 1,2,3,… The second quantum number is represented by the letter l and determines the shape of the orbital. The quantum numbers n and l are related; l can take on any integral value starting with 0 and going up to a maximum of (n-1). This means that the possible values are 0, 1, 2, …, (n-1). Each principal energy level is has one or more sublevels. These sublevels are denoted by the second quantum number. In general, in the nth principal level, there are n different sublevels. Another method is to use the letter s, p, d, or f to indicate the sublevels l = 0, 1, 2, or 3. Each sublevel contains one or more orbitals, which differ from one another in the value assigned to the third quantum number, ml. This quantum number is used to describe the spatial orientation of the orbitals. For a given value of l, ml can have any integral value, including 0, between l and –l. The fourth quantum number, ms, describes the direction of electron spin. The ms value doesn’t describe the orbital; it describes the electrons in the orbitals. There are only to possible values for ms, +1/2 and -1/2.

After finishing those notes Mr. H instructed us to continue on to page 4.

The ways to remember orbital types:

Smartà spherical

Peopleà pinched cylinders

Don’tà dorky

Failà frog-shaped

(see chart below)

Lastly Mr. H said that we are learning about this stuff because it explains the line spectras and the elements in the periodic table, even the 14 that came after 1926, which Mr. H thinks is cool in a geeky way.

Light and Energy/Bohr Model of the atom TUESDAY, FEBRUARY 10, 2011

Mr Henderson started out the class by having us take out our packets and turn to page#13-14 to give us the correct answers to one of the webassigns due the previous day

(Pages 13-14)The answers to this webassign are:
1)a. Its predictions were inaccurate for atoms having more than one electron.
2)c. The best that one can know of an electron's location is to know the probability that it is in a given region of space.
3)d. Complex differential equations
4)b. Where an electron has a 90% probability of being located
5)b. The value of n cannot be negative AND c. The larger the n number, the higher the energy of the electron.
6)a. The value of l reveals information about the shape of the orbital housing the electron AND c. The value of l cannot be negative AND d. The value of l must be a whole number
7)b. The value of ml indicated the directional orientation of the orbital AND The value of ml must be a whole number
8)b. The value of ms indicated the spin direction
9)b. an l value of 1
10)a. TRUE b. TRUE c. TRUE d. TRUE e. TRUE
11)TRUE
12)b. 2
13)a. For every n value, there are just as many sublevels.
14)a. INVALID b. VALID c. INVALID d. INVALID
15)c. The 3s orbitals are larger than the 2s orbitals.

The class then looked at the blog from the previous day done by Dmitriy and Mr. Henderson had the class watch the video he added of the plane crashing into 12 feet of concrete.

Next the class took out their calculators and turned to page #1 to work on problem 4a. and 4b.
These problems will probably be the only time calculators will be needed to use for the rest of this unit.
The equation needed to remember for these equations is E=h*f=h*c/wavelength. f stands for frequency.

4a...red photon (f=4.80*10^14 Hz)

E=h*f=(6.62*10^-34)(4.80*10^14 Hz)=3.18*10^-19 Joules

4b...green photon (wavelength=5.43*10^-7)

E=h*c/wavelength=(6.62*10^-34)(2.998*10^8)/5.43*10^-7=3.66*10^-19 Joules

We continued the class period by turning to page #2 and started the problems in the chart at the bottom of the page.
Photon(Energy in Joules) use the equation from page #1
a. 3.03*10^-19
b. 4.09*10^-19
c. 4.55*10^-19
d. 4.85*10^-19
e. 5.00*10^-19
f. 1.64*10^-18
g. 1.93*10^-18
h. 1.06*10^-18
i. 1.55*10^-18

For these next problems use the equation En=Rh/n^2

• Rh is the constant called the Rydberg constant and its value is -2.180*10^-18 Joules.
• replace n with the number given in the problem.

E Initial(in Joules)
a.-2.42*10^-19
b. -1.36*10^-19

E Final(in Joules)
a. -5.45*10^-19
b. -5.45*10^-19

Change in E(in Joules)
a. -3.03*10^-19
b. -4.09*10^-19

The class was then given glasses that when put on revealed the true colors from white light included in roygbiv(red,orange,yellow,green,blue,indigo,violet)

We then also looked at different types of lighting which revealed different strengths of the colors from roygbiv.

At the end of the period the class ended the day by beginning a lab in the back of the room. The lab will be continued tomorrow

HW:next webassign not due till the 14th.

Waves that don't have anything to do with the ocean

Good morning students!

We began class today by talking about our test, and with Mr. Henderson saying how our scores were very spread out, unlike some other tests we may or may not have taken. Getting to the point we started the class with a discussion about atom models, and how said atom models were changed over history.

Firstly, let's start with John Dalton.

Dalton did not know how atoms looked, and didn't even bother trying to explain it to others. At least he wasn't a bragger. Dalton published his works during the early 1800s. Before him, only two men attempted to take a stab at the mystery of atoms: Socrates and Democritus.

Although this wasn't discussed in class, Socrates took a piece of cheese and cut it in half. Then he cut that in half. Then he cut that in half. So on and so forth, until he could not cut anymore. He called that the "atom".

Democritus had a slightly different approach.

He took some stuff, put it in a pestle, and crushed it up with a mortar. He called that "atoma" for indivisible. What these Greeks didn't know however, is that matter gets much smaller than even this! Stupid Hellenics. Sorry, Konstantine.

Back to the topic, we looked at Thomson next. He devised the "plum pudding" model. A sea of negatively charged particles inside an atom that looks like pudding. Yum.

Finally, Rutherford. He concluded that an atom is mostly empty space, simply because he fired alpha particles at a sheet of gold foil, and only a few bounced back.

We then proceeded to move on to waves, and how light consists of waves. Mr. Henderson brought up the class guinea pig to demonstrate. Will stood with one end of the coil to his cheek, and Mr. H interacted with the other end. The result was Will got hit in the face, and because of this, Mr. H showed us that you can get someone's attention simply by using energy. Wow.

Finally, what we did was we discussed different wavelengths. We went from the weakest to strongest, radio to gamma, and in between was the visible spectrum of color. We spoke about UV and how they settle in your skin, X-rays and how they settle in your bones, and gamma and how they go right through you. Oh and by the way, don't put your cat in the microwave. Its H2O molecules will start dancing.

Remember that video Mr. H talked about? That plane crashing into 12-feet of concrete video? Here it is.

http://www.youtube.com/watch?v=xM8E-CogkYE

There you have it. At that, we concluded our day.

Fact: Nukes contain gamma rays.

Friday, February 4

Class began with a review of Chris's blog from Tuesday. Then, Mr. Henderson announced that, as a result of the snow days, the test was pushed back to Monday. He also mentioned that lab notebooks will also be turned in on test day. Mr. H then provided us with suggestions on how to study. He reminded us of all the study materials we have available; such as the Moodle Review, Delicious bookmarks, unit packet, WebAssigns, and the textbook.
For a review, we did some problems in the packet. We started on page 22 with #9: For the enthalpy (delta H) of the reaction, the products are added and then the reactants are subtracted. When plugging in the standard entropy for just an element, it will always be zero. So for problem 9, the enthalpy should look like 2(o)+1(o)-2(-286)=572KJ. For the entropy (delta S), it should be 2(131)=1(205)-2(70)=372J/K. After finding this, the answer needs to be converted to KJ, so it should be .372. The next part of the problem is to find the temperature range using the equation G=H-T(S). The equation should be 0=572-T(.327) and the final answer should be 1749K and any temperature greater than that is spontaneous.
We then went to page 11 and did problem #1 to practice Hess's Law. The final answer for this problem was -221. To get this, the first equation was multiplied by 2 and the second was flipped.
Next, we did another Hess's Law problem. This one was slightly more difficult. The first equation was multiplied by 2, and the second and third equations were flipped. The final answer was -168.6KJ.
We then finished the Hess's Law Lab. The procedure was to first find Q, using the equation Q=mCT, then to find the moles, and lastly to find the enthalpy by dividing those results.

Tuesday February 2, 2011

Today began with the usual Web Assign reading sheet review, the answers to chapter 17:3-5 on page 35 were presented and explained in class. It is interesting to note that, according to Mr. H, J. Willard was A. a snazzy dresser, B. a real hunk, C. a true scientist, and D. a great uncle of Mr. H's. Unfortunately, the lack of any "select all" option posed a real challenge to students; this particular student happened to choose option C.

After that attempt at humor, Mr. H went over the previous blog post. Very helpful with explaining the conce

pts of entropy. The plan for today was to explain Gibbs Free Energy (ΔG), finish lab TC10 Reaction 2, and go over the Chapter 9 Quiz. Basically, ΔG is to be used to determine whether or not a reaction is spontaneous, non-spontaneous, or borderline (neither spontaneous nor non-spontaneous, the reaction can occur in either direction). The formula for finding Gibbs Free Energy is ΔG=ΔH-T*ΔS, where both ΔH and ΔS are in kJ and the temperature is in Kelvins. If ΔG was negative the reaction is spontaneous, if ΔG is positive the reaction is non-spontaneous, if ΔG was zero then the reaction is borderline. After that q

uick lesson, we turned to page 21 to practice using this formula.

Problem number 2 was a review of how ΔH and ΔS related to the spontaneity of the reaction. If both are positive the reaction is spontaneous at higher temperatures, if both are negative the reaction is spontaneous at lower temperatures. Moving on, it is important to always check that the units are the same before calculating ΔG. That is, T should be in Kelvins and ΔS should be converted into kJ.

On problem 6. c., although ΔG technically is a negative value, the reaction is classified as borderline. One's reasoning will vary from person to person, but in this case the ΔG value is so close to zero that the reaction would be classified as a borderline example regardless.

Next, we preformed Lab TC10 Reaction 2. Not all to different from the previous day's lab portion, for the most part the processes were identical. Since the data has been acquired, the conclusion/discussion will be done in class. After that little bit of fun, the Chapter 9 Quizzes were handed back to the students. Due to a grade calculation error, the multiple choice portion was graded out of 26 points instead of 22 points plus a 4 point written portion. Thankfully this error was fixed using some common sense and a rather fat Sharpie pen.

As for homework, the deadline for the Web Assign has been pushed back due to the snow day. Similarly, Delicious Bookmark assignments are due the next class day. That's all for now, enjoy a day without school!