GOODLUCKKK TO USSS!!!!! :) :)

thanks Ms.K for, you know, everything!

i'll miss this class...

Electrochemistry Review

Electrochemistry review

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june 8, 2010

Today we did a lab:The Potential of Electrochemical cells.
We skipped steps 8 and 9.
We had to answer both analysis questions.
And for the conclusion omit questions 3 and 4

Monday June 6th, 2010

Today we went over Oxidation-Reduction Assignment 3.

Assignment 4 Question 1 and 2 are due for tomorrow.

Next scribe is annab

Friday, June 4, 2010

Today, we got our tests handed back to us and corrected it together. I don't know if Ms. K is going to upload the slides...

Afterwards, we did a lab on Electronegativity (I think?) [hmmm, you thought wrong. Try electrochemistry!) with the penny, nickel, and an iron nail. I would assume that if you didn't hand that in today, hand it in sometime next week. Perhaps, Monday, if anything.

Acids and Bases Review

Good evening!!

Today we went over chapter 19 study guide Acids and Bases

The answers are up there ^^

Don't forget to study for the Acids and Bases test tomorrow

Good luck everyone~

28 friday

we went over questions 4,5 in the booklet
study guide going over on monday
went over titration questions.
Dave is the next scribe

Percent Dissociation

Tuesday, May 25th 2010

Today, we had a mini refresher, from what we learned on Friday. The Ka and Kb equations. We also, talked about the pH lab we did last Tuesday as well. To review some more, we did this interactive Naming the Acids sort of thing.

We went over questions 1-4 on Acid-Base Equilibria Worksheet 4.

We did example 8, on page 30 in the booklet. We also, learned a new topic in class. It was how to calculate the Percent Dissociation.

The formula to calculate Percent dissociation is

Percent dissociation = Concentration of Dissociation Species

---------------------------------------------------- x 100%

Original Concentration of Acid or Base

We also did examples 10 and 11, as practice. We were assigned to do Assignment 4 in the yellow booklet, the odd numbers. You could do the even questions as practice if you want.

*Grade 12 bbq tomorrow! Don't forget to bring sunscreen. =]

**Our chemistry exam is on Monday, June 21st 2010 in the afternoon! 27 more days till our exam.

Our next scribe is.....annab.

Next scribe will be Jocelyn!

Electrolyte/Nonelectrolyte

Friday, May 21, 2010, we read pages 19 to 29. We learned new terms like electrolyte and non
electrolyte. "conducts electricity" is an electrolyte and "do not
conduct electricity" is a non electrolyte. An example of a non
electrolyte is water (H2O). An example of an electrolyte is sodium
hydroxide (NaOH). Then, Ms. K showed us a demo that determine if a
compound is an electrolyte or non electrolyte. We looked at questions
on page 22. Questions 4, 9, 10 and 14 are identified as non
electrolytes. We added 2 questions #13. H2O and #14. NaCl. Electrolytes
are question #s 1,2,3,5,6,7,8,11,12,13,14.
*all organic compounds are NON electrolytes*
*strong acids are ALWAYS strong electrolytes*
*strong acids are considered to completely dissociate in water*
We also did the 'calculating the dissociation constant' on page 29.
After the booklet, Ms. K assigned us to do Acid -Base Equilibria
Worksheet 4 questions #1-4.
Have a nice long weekend!!!

OOPS NEXT SCRIBE WILL BE JANINE :D !

Logs/Antilogs pH/pOH

Well we had a sub yesterday, but we were given 2 sheets to work on about logs/antilogs related to pH and pOH. There were a couple of formulas in the sheet that we could remember in order to find pH and pOH.

pH = - log[H⁺] pOH = - log [OH^-]
pH + pOH = 14.0 [H⁺][OH^-] = 1.0 x 10^-14
[H+] = anitlog (-pH) = 10^-pH [OH^-] = antilog (-pOH) =10^ -pOH

Today, we went over the worksheet #2 about acids and bases #’s 1 and 2 where we were to identify which one is an acid, base, conjugate acid and conjugate base. Also, to determine whether they were Arrhenius acids/bases or Bronsted-Lowry acids/bases. We also did the Kw problems #’s 1-10 and the Acid-Base Assignment #6 numbers 1 and 2 on the board today.

Mrs. K assigned numbers 3 and 4 today and this was to be handed at the end of the class.

Dont forget the equilibrium law can be written as:

Kw = [H_3O⁺][OH^-] Where the value of Kw is constant at 1.0 x 10^-14.

Conjugate Acids and Conjugate Bases

Friday may 14, 2010

We went over the booklet Acid & Bases pg 3- 12. Then we were given a worsheet to work on the first page - Acids & Bases Project Assignment I, indentify the acid, base, conjugate acid(ca), & conjugate base(cb).
Answers:
1.a) acid + base <---> ca + cb
b) acid + base <---> ca + cb
c) base + acid <---> ca + cb
d) base + acid <---> ca + cb
e) acid + base <---> cb + ca

2. write the equation for the reaction of each of the following with water.Indicate wether the ion or molecule is an acid and base. Which one of these are arrhenius acid/bases and which are Bronsted-Lowry acids/bases?
e.g. HBr + H2O <--->H3O+ + Br- , Bronsted - Lowry & Arrhenius Acid
Answers:
2. a) <--->C3H3O2- + H3O+ Arrhenius Acid & Bronsted - Lowry
b) <--->HS- + OH- Bronsted - Lowry Base
c) <--->HSe+ + OH- Bronsted - Lowry Base
d) <---> H2CO3 + OH- Arrhenius Acid & Bronsted - Lowry

Homework for this weekend:
- finish the Next page of the woorksheet Acids & Bases Woorksheet 2 Question 1 & 2, hand in for monday.
- also the foldables (Arrhenius, Bronsted - Lowry , Lewis) follow the rubric for foldables to get a perfect mark this is also hand in for monday.

& the next scribe will be: julien:)

Foldables

So today in chem, we made foldables about the different types of models. Such as the Lewis, Arrhenion and the Bronsted Lowry Model. This assignment will be due Monday. Don't forget to be creative with this and add every little bit of detail to each model.
And our next scribe will be bernadette.

hi there guyss! On monday, we corrected the work sheet called "solubility equilibria assignment #3. And also, we looked at pages 36-37. We learned how to determine if a solution is saturated or unsaturated.

Remember guyyys!:
- Q-Ksp, the solution is just saturated and no ppt( no precipitation)
-Q> Kso, the solution is saturated and a ppt occurs.
- Q< Ksp, the solution is unsaturated and no ppt formed.

GOODLUCK ON THE TESST' DUN DUNN DUUUUUNNN......
the scribe for thursday will be arjel:)

May 07, Friday

I wasn't late today, so I volunteered to do the blog. :) :)


First, we did the questions 3 & 4 in the Solubility Equilibria Assignment #1. Obtain the answers from Ms. K. Note: in letters b and d you have to change the unit to moles/liter before you can solve for the Ksp.


We also went over the pages 31-35 on the Chemical Equilibrium booklet. In that, we learned a new lesson, determining ion concentration from Ksp. For all of you who hates ICE, you'll life will be easier. It's no different from the ICE method we've been doing, except that:

• the initial concentration of the ions (products), before dissolving, is zero.

• reaction is consists of a saturated solution as reactants and ions as products so you get to IGNORE the saturated solution, the solid one, in your ice table.

Example: The Ksp of magnesium hydroxide is 8.9 x 10 ⁻12. Determine the solubility of Mg(OH)₂.

• Mg(OH)₂(s) <=> Mg2+ (aq) + 2OH⁻(aq)

I...............___.................0................0
C.............___................+x................+2x
E..............___...............+x................+2x


To solve for x, use the formula, Ksp = [products] and substitute the values. We'll find that the solubility of Mg(OH)₂ is 1.3 x 10-4.


That's determining ion concentration using Ksp in PURE WATER. Another thing we learned is determining ion concentration with an ION IN COMMON WITH THE COMPOUND.

Example: The Ksp of magnesium hydroxide is 8.9 x 10 ⁻12. What is the solubility of magnesium hydroxide in a 0.10 M solution of NaOH.

Step 1: Determine the concentration of the common ion, OH ⁻.

NaOH(s) <=> Na+(aq) + OH -(aq)
[OH⁻] = [NaOH] = 0.10 M

Step 2: Set up an ICE table for Mg(OH )₂.

• Mg(OH)₂(s) <=> Mg2+ (aq) + 2OH⁻(aq)

I...............___.................0................0
C.............___................+x................+2x
E..............___...............+x................0.10 +2x

***You have to include the concentration of OH - from NaOH(s) in the table because when we dissolve Mg(OH)₂ in the solution, OH ions are present. According to Le Chatelier's Principle, adding more OH ions to a saturated solution would increase the overall concentration of the products shifting the equilibrium to the left. This would result in more solid formed and a decrease solubility.***


Step 3: Substitute values into the solubility product expression.

Ksp = [Mg2+][OH-]2

8.9 x 10⁻12= (x)(0.10 + 2x)2

**at this point, you can see that you'll have a quadratic equation if you solve further, so, ignore the x of the ion in common, OH. Ms. K said that it's too tiny to matter.**

8.9 x 10⁻12= (x)(0.10)2

8.9 x 10⁻12= 0.01x

8.9 x 10⁻10M = x

Let's check if the value makes sense. The solubility of Mg(OH)₂ in pure water is 1.3 x 10-4 M and in a 0.10 solution of NaOH is 8.9 x 10⁻10M. The solubility decreased, just as we predicted.



I dont know if I have to pick for someone to do the next scribe... if so, i'll pick Ericka! :) :)

Common Ion Examples

Common ion

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Ksp Examples from Notes

Ksp problems

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May 4 AND 5

SOOO pretty much what we did on May 4 is go over the study guide questions which Ms. Kozoriz already put up the answers to. We studied the rest of the period because test was on MAY 5th!

I really don't know what else to put... and I feel like I should scribe for May 6 since this post is quite short and sad.

Ch. 18 Chemical Equilbrium

Ch18equilibrium

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Thursday, April 22 and Friday April 23

On Thursday we went over and corrected the Equilibrium Project Assignment, pages 1 questions 1-4 and page 2 questions 5-11. After that we worked on Lesson 3 Homework Assignment and did questions 1-5.

We also were given a assignment that is to be handed in Determining Equilibrium.


On Friday we did our Equilibrium & Le Chatelier Analogy Lab which should be handed in today.


That wraps up what we did before the weekend.

Wednesday, April 14

We went over yesterday homework (page 32, 33 and part B on page 34). We have to do Kinetics assignment #4 to hand in on Friday.
Tomorrow, we'll go straight to Ms. Truong's room at the beginning of class.
Test is on Friday on Chemical Kinetics. Study!!
Erick is to scribe next

Rate Law

Helloooooooooooo everyone,
Today we learned about rate law, reaction order, and calculating rate order. We did example 2 on page 28 in the blue booklet. For homework, we have to do page 32, 33, and part B on page 34.
That's it.
Chào!
p/s: that's how i say bye in Vietnamese. ^^
And nest scribe is............................................Socola. ok? :D

DOUBLE DOWN!

Kay so this is a double blog post, cause I thought someone else posted but they didn't and then some things happened, it involved some elephant and a penguin... but I digress. Anyways, don't mind my memory lapse and this late blog.

On Thursday we were taught how to do average rate, you know
Rate = [(final concentration of A) - (initial concentration of A)] / [(final time) - (initial time)]

In other words:
Rate = [A]f - [A]i / Tf - Ti

Or in other words besides that:
Rate = Δ[A] / ΔT

We also learned about instantaneous rate: The rate at a specific time. The example is shown on page 12 and is answered in steps on page 13.

** by the way - the use of [A] can be replaces by any other variable [B], [C], [...]

We also went over:
Kinetics Assignment #1 Answers q. 1-3

On Friday (April 9th) We learned about Rate and Stoichiometry. The example on page 14 shows:
2NO2(g) -> 2NO(g) + O2(g)
We gave 2NO2(g) the value of 4g/min To find out the values of 2NO(g) and O2(g) we use ratios. 2NO2(g) and 2NO(g) have a ration of 2:2 or 1:1 so they have the same value, however O2(g) has a different concentration or ratio of 2:1 so it's 1/2 the amount of 2NO2(g) so it has 2g/min... get it?

We also created / answered some questions on Reaction Rate graphs. (Chapter 18 Worksheet)
(Sorry picture was too blurry... just google it or do your work :D !)

We also did all of the questions on page 15
Received Chapter 18 Worksheet or Reaction Rates Worksheet
Answered Questions 4 and 5 on Kinetics Assignment #1
Was assigned and corrected Questions 1-7 in Kinetics Assignment #2

** AND I THINK WE HAVE A CHEM TEST TOMORROW (April 12) ... I MAY BE WRONG AND IT MAY BE ON WEDNESDAY! ... SO YEAH! **

Yeapp, that's pretty much it... I think I mixed up some events that occurred on those days, but ... whatever they're bound to happen ha. Now a parting gift:


The KFC Double Down - Inspired the post.


Next scriibe is HYCharang :P

KINETICS

hi there!! its me jeck. I'm suppose to scribe yesterday but I totally forget it even Ms. K reminds about it. I'm sorry about that and my apologies to my grammars.

Anyway, yesterday Ms. K shows as how catalyst work on the solution. Catalyst define as something that initiates or causes an important event to happen. Originally a term used in chemistry for the volatile (active) chemical in a formula. (http://dictionary.reference.com/browse/catalyst). In simple explanation it speeds up the reaction of the two aqueous solutions. As I said earlier Ms. k shows us how it works in the solution. Well it really works but there was a little problem happen. The point is catalyst really works on the solution.
Yesterday we did our lab too, the part 3 and also we answer the questions in the lab sheet.

Sorry guys but I'm really bad at this thing (blogging). I work till 11:30 so I totally forgot about it.

April 05 and April 06 2010

Hey there! I am scribing for two days:


One is the day after the Spring Break, Monday, April 05, 2010.

In class, we continued our studies on Chemical Kinetics. We answered the assignment assigned to us before Spring Break which was to do the questions on page 10 of our booklet in Chemical Kinetics. You have to read the previous pages to understand the answers. Here are they: (note: the answers are my own words, Ms. K was too lazy to write hers on the board...)(oh yeah, it said that you have to show diagrams.. but I don't know how to show mine here so figure that one yourself. :)

1.) The increase of concentration of the reactants the faster the reaction to occur therefore, increasing the reaction rate.

2.) a.) Increase in temperature causes the molecules to move faster, therefore more collision. More collisions/reactions --> increase in reaction rate. Decrease in temperature does the opposite --> slower reaction.
b.) The larger or more exposed the particle means more surface area. More surface area means greater chance of collisions --> increase in reaction rate. Less surface area, you know, slower reaction.
c.) More pressure means less travelling space. Less travelling space means faster collisions among the molecules --> increase in reaction rate. Less pressure means more travelling space and the collisions won't occur as fast.

3.) Catalysts provides an easier path for the reaction to proceed. It lowers the activation energy of a reaction, so, more particles are able to surpass the activation energy required. These means more collisions and an increase in the reaction rate.

4.) a. ) i. is faster because in a solution, reaction will occur more quickly to the ions because they are free to move.
b.) ii is faster because Ag is hungrier for elections than Cu. This brings us back to our previous unit, Atomic Structure. Ag has a higher Electronegativity value than Cu, so it can attract more electrons than Cu... I think.
c.) i is faster because all elements in the solution are ions.
d.) ii is faster because all of the elements are in the same state, gaseous form.

By the way, we only spent half of the class; we were called down to the gym for a meeting with the principal.


The second day is today, April 06, 2010.

We did a lab... and that's pretty much it. It was about how elements involved, concentration of the elements, temperature, a catalyst and particle size affect the reaction rate. We only did two parts of the lab sheet, so expect that we will do more of it tomorrow.


Yay! Jeck, it's your turn. :)

Kinetics

Today, we started the 3rd unit which is "CHEMICAL KINETICS". Ms. K explained the important terms through pages 3-9. She also did a lab experiment which a Magnesium ribbon was individually dropped to 3 test tubes each containing 1M of HCL, 6M of HCL and 12M of HCL. When a Mg ribbon was dropped into 1M of HCL, it took 4 mins for it to dissolve completely. In 6M and 12M of HCl, it took 7 secs for it to dissolve. After the lab experiment, we were assigned to do page 10 questions 1-4.

Have a nice SPRING BREAK!!!!!!!!

The next scribe will be jen2x

march 23 - Anna B's scribe

for today’s class, we just corrected the answers from yesterday and answered some of the questions from the booklet

hand out answers:

using the periodic table:


1.
a. 8 or 0
b. 1
c. 3
d. 2
e. 6
f. 7
g. 5
h. 4

2.
a. 4
b.2
c. 7
d. 1
e. 3
f. 5

3.
a.[he]2s¹

b.[he]2s²2p⁵

c.[ar]4s²3d¹⁰4p³

d.[kr]5s²

e.[xe]6s²5d¹⁰6p³

4.

a. group 2A period 1 block S

b. group 3B period 3 block P

c. group 1A period 4 block S

d. group 3B period 4 block D

e. group 6A period 6 block P

Electrons In Atoms

Section 5.1 light and quantized energy

1. Energy

2. Wave

3. Light

4. Speed

5. Wave length

6. Amplitude

7. Frequency

8. Hertz

9. A and C

10. B

11. 2 waves/s or 2 hertz

12. C

13. B

14. D

15. B

16. F

17. T

18. T

19. T

20. F

21. F

22. T

Section 5.2 Quantum theory and the atom

1. Ground state

2. Frequencies

3. Lower

4. Higher

5. Electron

6. Energy levels

7. Atomic emission spectrum

8 and 9 doesn’t need to be answered

10. C

11. A

12. D

13. B

14. bohrs model treats electrons as particles traveling in specific circular orbits while the quantum mechanical model treats electrons as waves and doesn’t describe the path of an electron around the nucleus

15. do not

16. two

17. spherically shaped

18. n

19. electrons

20. three

21. 2s and 2p

22. nine

5.3 Electron configuration

1. electron configuration

2. lowest

3. stable

4. ground state electron configuration

5. aufbau principle

6. pauli exclusion principle

7. spins

8. hunds rule

Element	Atomic number	Orbitals	Electron configuration
9. helium	2		1s2
10. nitrogen	7		1s22s22p3
11. neon	10		1

12. atomic number :32 electrons :32

13. noble gas notation uses the bracketed symbol of the nearest preceding noble gas atom in the periodic table. Using noble gas notation allows you to represent the complete electron configuration of an atom with many electrons in a short hand form <

14. [Ar] 4s²3d¹⁰4p²

15. C

16. B

17. D

18. B

19. A

20. A

page 36 in the booklet

periodic trends assignment

1. O, C, Sn, Sr

Because there is a decrease in atomic radii as it moves from left to right across the periodic table and there is an increase in atomic radii as you move it moves down a group

2. Ar, Al, Cs ,Na

Because there is a decrease in ionization as it moves from left to right across the periodic table and there is an increase in ionization as it moves down a group

3. Mo, Te, I

Because Mo is a metal which is a good conductor of heat while Te and I are non metal

4. Additional orbitals between the nucleus and the outer electrons are occupied which causes the shielding of the outer electrons from the pull of the nucleus which increases the distance off sets the greater pull of the increased nuclear charge

Back page

3. largest: antimony smallest: nitrogen

4.

a. Fluorine

b. bromine

c. bromine

d. fluorine

5. group 6A period 3 block P

8. right, because non metals gain electrons

9. >1.7 ionic

< 1.7 covalent

10.

a. covalent

b. covalent

c. ionic

d. covalent

the next scriber is janine D.