Thursday, April 28, 2011

Thursday April 28th

Todays beautifully short class began with going over the right column of page 5 in the packet. Mr. H. highlighted the water dissociation equation and the fact that the equilibrium of the ions is always 1.0e-14. Also that neutral solutions have equal ion concentrations, acidic solutions have more hydrogen ions, and basic solutions are just written differently and OH dominates. On the back side Mr. H. noted that when discussing powers of ions it really was just referring to the numerical value of the exponent. Below we were introduced to equations we would be using on page 6.
Next we flipped back to page 5. Mr. H. briefly spoke about number 11 which has some useful equations. He also mentioned that on a previous reading webassign there was a study tip regarding those equations:
STUDY TIP:
On a note card, write down the equations/formulas presented in these two sections of reading; these include Equation 13.1, Equation 13.2, Equation 13.3, the equation on the line below Equation 13.3, Equation 13.4, and the first equation on page 356. Keep the note card handy during class, when doing homework and when on FaceBook. Accept it as your friend.
^^^YOU HAVE BEEN WARNED
On page 6 we went over some of the table for number 12. We were able to solve columns 1 and 2 by dividing 1.0e-14 by the given OH or H3O value. The answers were for row1: 1.0e-12 row2: 1.0e-11 row5: 1.0e-7 row6: 2.0e-14 row7: 1.67e-15 row8: 2.5e-15. Next we were easily able to do columns 3 and 4 for 3 and 4. Because the pH and pOH values have to equal 14 row3: 9.00 row4: 3.00. We were also able to do column 3 and 4 for rows 1 and 2. Using our calculator we used the equations from page 26. For row 1 we took the negative log of .o1 with a base of 10 and got 2.00 for pH and 12.00 for pOH. Therefore we knew it was acidic because the pH was less then 7. For row 2 we got 3.00 for pH and 11.00 for pOH and for row 6 we got 13.7 for pH and .3 for pOH. Therefore rows 1, 2, 3, and 4 are acidic and row 6 is basic. That was the end of class.

Wednesday, April 27, 2011

Unit 12,Day 2

In today’s class, we started out by going over Chapter 4.3 and Chapter 13.1 Reading Sheet, found on page 23 of your unit packet. The answers are as follows:
1. C
2. D
3. True, False, False, True, True
4. D
5. A
6. B
7. A, A, A
8. B
9. A, B
10. False because some substances are amphiprotic
11. Amphiprotic means a proton could either be lost or gained, like with H2O, which can become H3O+ or OH-
12. A, A, B
13. B, C

Then we went over Kon’s blog from Monday, which was about the first day of unit 12.
Next, we began learning about Bronsted and Lowry. They said that you can’t have an acid without having a base. This idea is different than what we learned on Monday, where you would have just split the H from its anion, so keep that in mind.
Then we did problem 6 on page 2. Keep in mind the acid is the proton donator and the base is the proton acceptor.
a. HCN(aq) + H2O(l)  CN- (aq) + H3O+(aq)
b. HF(aq) + H2O(l)  F- (aq) + H3O+(aq)
c. HNO2(aq) + H2O(l)  NO2-(aq) + H3O+(aq)
The H3O+ is called the “Hydronium Ion”
d. *Now we are working with BASES! This means that the base GAINS a proton.
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
C5H5N(aq) + H20(l)  C5H6N+(aq) + OH-(aq)

Next we did a few problems on page 3. Keep in mind that Acids donate protons and Bases accepts protons.
Here is the general equation used for an acid-base reaction.
HA + B  A- BH+
(acid) (base) (conjugate base) (conjugate acid)

Note that “the reaction of an acid with a base changes the acid into a conjugate base and changes the base into a conjugate acid.”
a. B, A, CA, CB
b. A, B, CB, CA
f. A, B, CB, CA
g. B, A, CA, CB
F and G are “zingers”. Because HCO3- is and Acid in part F, you would expect it to be an Acid in part G as well. This however is not true, and you have to always look to see what is giving away the proton and what is receiving it.
We ended class by doing Lab AB1.

Tuesday, April 26, 2011

Monday April 25th

Today, we started class by getting our chemistry lab notebooks from the lab tables in the back of the room. Mr. H then went around the room and gave everyone their rubrics for the labs that we did in our notebook. Mr. H then went on to talk about how if you had a bad test grade, then you should go and see him. He mentioned that he won't be available tomorrow, but that any other day he would be available.

Mr. H then officially began class by introducing to the class another 50 point web assign that will be due in about 3 weeks. If you want to know more about it, here is the link: http://gbschemphys.com/honchem/index.html
Mr. H did tell us though that in order to get 100% on this web assign, you have to get at least 35/50 questions correct. As an addition to this web assign, Mr. H reminded the class that just like the previous unit, there will be frequent quizzes throughout this unit and a test that will be in three weeks.

Mr. H then began his lesson of the day, which was an introduction to Acids & Bases, by letting us know that there will be a lab tomorrow in class, so be sure to bring your lab notebooks. He then proceeded to write important notes on the board which are essential in understanding Acids and Bases. The Notes are as followed:

Acids and Bases


Definitions:

1) observable properties (Lab AB1)

2) Arrhenius (Early 1800s)

Conductor: it needs to have ions in order to be considered a conductor

^^One of the examples that Mr. H gave about conductors was the light bulb with the two rods. He tested to see if multiple liquids were conductors by putting each liquid in a beaker and then placing the beaker in the two rods. If the light bulb lit up, then we knew that that liquid was a conductor. If it didn't, then we knew it wasn't. The liquids that Mr. H experimented with were HNO3, HCl, HC2H3O2. The first two liquids lit the light bulb really bright, so we were able to conclude that those liquids had lots of ions in them. When we experimented with HC2H3O2, (a.k.a. vinegar), we noticed that the light bulb turned on, but the light was very dim. From this, we were able to conclude that vinegar is a weak acid on the pH scale, and therefore is a weak conductor.

Acids: H___ ---> H+  +   ____-

A= anions

Bases= ____OH----> _____ + OH-

^^ Mr. H continued to use the light bulb as a conductor tester, but this time, he used KOH and NaOH, which are solids, and mixed them in a beaker of water. He once again did the same process as before, and after experimenting with both, the class saw how the light bulb became very bright when both of the solutions were tested on the light bulb prongs. This shows that KOH and NaOH are two very strong bases and have plenty of ions in the solid to become a conductor.


Strong Acids: HNO3------------>  H+    + NO3-    (High K) <--------- lot of reactant, little product
                                 <---

Weak Acids: HC2H3O2<------------ H+    + C2H3O2-     (Low K) <------- lot of product, little reactant
                                                --->

Strong Acids to know:                                                                                        

1)HCl

2) HBR

3) HI

4) HNO3

5) H2SO4

6) HClO4



Strong Bases to know:

1) LiOH

2) NaOH

3) KOH

4) Ca(OH)2

5) Sr(OH)2

6) Ba(OH)2




After writing these notes and copying them, Mr. H had the class open up their unit 12 chemistry packets to page 1 and had us work on problem 1. Before doing problem 1, it was essential to know that:

an acid= a substance which produces hydrogen ions when it dissolves in water

a base= a substance which produces hydroxide ions when it dissolves in water

Knowing this, the class was able to answer question 1. The answers are as followed:

A            B           B
N            B           N
B            A           N

The class then went on to do problems 2, 4, and 5. (#3 was a review)

#2

a) H+  + NO3-  (Acid)

b)Na +    + OH-   (base)

c) H+  + CN-          (acid)

d) Ca2+     2OH-       (base)


#4   ( these answers correspond to a table, so the name goes on the left part of the table and the formula goes on the right part)

hydrochloric acid
HBr
hydroiodic acid
nitric acid
H2SO4
ClO4

lithium hydroxide
sodium hydroxide
KOH
calcium hydroxide
Sr(OH)2
BA(OH)2

#5
a) [H+]  = 6.0M

b) [OH-]= 3.0 M

c) [H+]= 1.0 M

d) [OH-]= 2.0M <------ this is because the product is double the reactant

This concluded our lesson for the day. Tonight's homework: 13.2-3 Rdg Sheet.

The Daily Joke: What do you do with dead chemists?
                         Answer: Barium

Tuesday, April 19, 2011

Tuesday, April 19

Class started off today by Mr. H taking tardies for people who were late (Hannah). He then told us to get out our lab notebooks and write the title and purpose of the lab we will be doing later in the period; Ksp Lab.

After we finished this up, we took a look at Tim's blog from last night which was really classic. "Also, due to the reasons I forgot, the test on Wednesday will be pushed back to Thursday." The wise words of Tim Joo. Anyway, he talked about ICE and some of the Ksp problems which are located on pages 17-19 and 24-25.

Today's lesson focused on continuing with the Ksp problems and learning how to solve for Ksp by using solubility values. The first problem we did was on page 25, problem d. I'm not going to go over it in great detail because we already learned about and I think it is a pretty straightforward topic to understand. But if you want to know the correct answer it is [Cu2+]= 2.29*10^-7 M and [OH-]= 4.58*10^-7 M.

We then moved on to page 26 where Mr. H explained what molar solubility is; a partially souble salt's number of moles which dissolve per liter of aqueous solution. The first problem we solved was 8a and the answer was Ksp= 1.17*10^-10.
Then, we went over to page 27 to solve 8b.
The first step is to write the balanced equation which is Fe(OH)3(s) <---> Fe3+(aq) + 3 OH-(aq). Next, you need to fill out the ICE table. The initial amount for both of these products is 0 because you are given 0 amount of them. The change for Fe3+ is +x because we don't know what the change is and there is no coefficient in front of it. For OH-, the change is +3X because there are 3 moles of OH for every one mole of Fe. The equilibrium amount for Fe is x and 3x for OH. You then have to write the Ksp expression which is Ksp= [Fe3+] [OH-]3, and OH is to the third power because you have to switch the coefficient to a power. but it only matters if the coefficient is greater than 1. After this, you plug in x and 3x for the products values; x*(3x)^3= 27x^4. For x, plug in 1.01*10^-10 and the equation is Ksp= 27*(1.01*10^-10)^4. Finally the last step is to calculate and solve for Ksp= 2.81*10^-39.

After we finished the math portion of the day, Mr. H started off the lab by explaining what we needed to do to complete this lab. The rest of the period everyone worked on the lab.

Homework: Test on Thursday and Lab Notebooks due Thursday.

Monday, April 18, 2011

Monday, April 18

We started our day by getting our quizzes back from Friday. Also, due to reasons I forgot, the test on Wednesday will be pushed back to Thursday.

Shortly, we moved on to page 19 of our packet to work on #7. This was different than the other problems we worked on before because this time, it gave us the equilibrium constant (5.76) and the initial concentrations of H2O (8.2) and CO (8.2). The other concentrations were 0 because it was not given. The mole ratios of the equation were all one, so they all had a change in x. The reaction's equilibrium constant was 8.2-x and the product's equilibrium constant was x. To solve this, all you have to do was set up an equation 5.76= x^2 / (8.2-x)^2. To solve for x, you square both sides, getting 2.4= x/(8.2-x). Multiply both sides by 8.2-x and get 19.68 - 2.4x = x. This equation makes x = 5.78. Remember that the problem is asking for the equilibrium concentration, so the answer is not simply x. The answer would be [H2O]=[CO]=2.42 and [H2]=[CO2]=5.78. Next, we moved on to #8 on the same page. This one was a little bit different because it gave us the concentration on both sides. This is quite similar to #7 except that the direction of reaction must be found out. To find the direction, we have to find Q. If you don't remember what Q is, we did some problems on page 16. The Q value was .5166 and was higher than the equilibrium constant of .175, the equation went to the left. After finding that out, everything was basically the same as #7. The answer to #8 is [HCN]=.85 and [C2N2]=[H2]=.355.

As a reminder, these two type of questions will definitely be on the test. Before we moved on to our next topic, we took a little joke break. I don't really remember any of the jokes so...

We moved on to pages 24-25 of our packet. We worked on a and c. In these types of problems, we were given the Ksp and needed to find the equilibrium ion concentration. In problem a we were given that Ksp=1e-10. Because BaSO4 disassociates into Ba and SO4 their initial is 0. They both have a change in x. The Ksp equation is [Ba]x[SO4], so it is x^2=1e-10. This makes x= 1e-5. The answer to problem a is [Ba 2+]=[SO4 2-]=1e-5. Problem c was basically the same except that one of them had a coefficient of 3. Because of this it becomes +3x and in the equation it becomes 27x^3 because the 3 and the x must be cubed. The rest can be solved with the same method as problem a. The answer to problem c is [OH -]= 3x----> 2.0x10^-9=[Al 3+].

We ended our day with us returning the calculators and Mr.H accusing us of trying to steal them.

Sunday, April 17, 2011

Friday, April 15

Today we started off by flipping our packets to page 23. Mr. H said that this would be somewhat difficult, because we would have to use already learned knowledge in order to be able to work out the problems. He said that we would start out easy. The new topic that we would be learning is Ksp. It is the last topic of the unit and we should know it for the major celebration we will be having on Wednesday. Mr. H noted that if we get stuck we should look at the examples listed in our packet, which are in the book. The first thing that we did on page 23 was some review on Ionic compounds. we had to write the formulas and name the compounds. Then we had some practice with solubility and balancing equations. Mr. H also noted that when he said that there are soluble and non-soluble compounds he meant that they are partially soluble. We had some practice with this on the bottom of page 23. We then flipped to page 24, and this is what we would be putting all of those concepts together for. As I said before, Partially soluble salts dissolve only slightly in water. The dissociation of the salts are controlled by a reversible reaction that has a very small K value (much less than one). This is known as a solubility product constant (Ksp). For 4 on page 24 we had to write the dissociation equation, in which we needed to have the solubility rules. If help is needed reference example 16.1 on page 4332 in the book. After that we did some more practice on page 21 with LeChatelier's Principle. Then at the end of class we had another reading quiz which I am sure everyone did well on.

Wednesday, April 13, 2011

Wednesday, April 13


At the start of class, Mr. H told us to take out our packets and calculators while he was getting Will's blog ready to show to the class. After we went over Will's blog and reviewed the material learned from yesterday's lesson, we turned to page 14.

We began with number 11. It was exactly like yesterday's problems. Step by step: first, you find the ratio of the products to reactants, like always. You plug in the information you have, but we only get the mole values. So, we convert to molarity. That is found by dividing moles by the liters given. Anyways, you plug in the molarity and then solve for x. Simple as that. The answer to number 11 is .00114 for NH3.

Following up with that question, we turned our attention to numbers 13 and 14. The equation is the reciprocal of number 11's equation, so you find the reciprocal of the answer, too. From there, you can figure out the temperature, which is 76.9. Then, Mr. H had us try and figure out number 14. He asked Konstantine and Daniel if the answer to number 14 is the doubled or squ
ared for .0113 or doubled/squared for 76.9. They both said that they think it is not doubled or squared, but kept the same. However, Neil, using his debate skills, countered saying that 76.9 is squared because if you do the ratio for the equation on number 14, it ends up that if you square one side, you have to square the other to keep it equal.

After their debate or whatever, we took the quiz, which was 2 questions. Mr. H said that if we found that easy, we are in good shape.

Finally, we turned to page 15. For number 1, the answer is C. Straight from the packet, for such a situation with a K value greater than 1, the system is described as having an equilibrium which favors the products or lies to the right. So pretty much,
the Kp is a ratio, so the products have to be greater than the reactants if Kp is going to be greater than 1.

When Kp is equal to the ratio, then it is at equilibrium. We learned that Q represents the ratio between products and reactants. When Q is greater than K, there are too much products. If Q is less than K, there is too much reactants. K wants to reach equilibrium, so this is very bad.

Solving is just the same. You find the ratio (products over reactants), raise the power to
the number of the coefficient, plug in the numbers given, and see if it is equal to K (which is given). Like I said before, if Q is greater, then there is too much products (numerator). If Q is less, there is too much reactants (denominator).


We ended with jokes from Mr. H's book. :)

There's a reading assignment due tomorrow so do it!



Tuesday, April 12, 2011

Tuesday, April 12



Today in class we continued to learn about equilibrium constant expressions. Yesterday we learned how to create Kp expressions, and today we finished learning about Kc expressions. We flipped our packets to page 12 and we worked through the page. To form a Kc expression, I will use letter c on page 12 as an example.

c. HF(aq) + H2O(l) <--> H3O+(aq) + F-(aq)

First we get rid of any substances that aren't gases or aqueous solutions. That would mean getting rid of H20(l). So now we have

HF(aq) <--> H3O+(aq) + F-(aq)

So now taking this, we place it into he equation for the Kc.


[F-] * [H3O+]
Kc = _______________

[HF]

And that is our Kc value.

I found this picture that shows the value of the Kc.

We moved in class, and added math to Kc and Kp. The only thing we changed from before, is that now the values have numbers to them. Using our knowledge of math, we plugged the equations into our calculators and got our answers. The answers o #'s 4-7 on page 13 are as follows.
4. .222
5. .00379
6. 4.315
7. .025

The next thing we did in class, was doing algebra with the constants of equilibrium. We took our equation as shown above, and we plugged in values and solved for the values we didn't have. The answers for page 14 #'s 9 and 10 are as follows.
9. 4.25 * 10^-4
10. 2.6377

Mr. H then told us that we have a quiz tommorow, and that he won't be tellling us about quizzes ahead of time from now on because we stress out less if we don't worry about it. He said we would finish the day with jokes from his jokebook, but we didn't have time. Maybe tomorrow?

Monday, April 11, 2011

Monday, April 11, 2011

Today, class begin with Mr. H putting up the answers to a reading sheet that was a due a few days ago. It was Reading Sheet 12.1 the answers are as follows:
1.
F
T
T
T
F
F
F
2. b
3.e
4.c
5.d
6.a

As Mr. H explained the contents of the reading sheet, we took down the answers. The 12.1 reading and Mr. H's class discussion provided a basic overview on equations and reactions that are at a state of equilibrium. It was brief overview and discussion that allowed us to complete the lab where we were flipping over red and gray cards in an attempt to reach equilibrium. Now that we were able to sufficiently understand why what happened in the lab we were able to construct the graph and answers the questions on the back. The next 10 min. of class were spent constructing our graphs and answering the questions on the back.

After, we moved into our packet to a discussion on calculation the Kp and Kc expressions of a chemical equation. To construct the expression for Kp you have to take the partial pressure of the products over the partial pressures of the reactants on the bottom. Also, whenever there was an aqueous solution in the equation you are supposed to replace the partial pressure with its concentration. Three basic rules or principles were all that was needed to write express the Kp for gas phase systems.
1). Products always come over reactants in the fraction.
2). Raise the partial pressures of the gases to their coefficient in the balanced chemical equation.
3). When calculation the expressions exclude any liquids or gases.

The following picture shows the answers to the practice problems we did:


Sorry it is sideways.

Remembering those three rules, Mr. H guided us through the process of calculating the Kp constant as we did some problems as a class and most of the packet page(pg. 11) individually. We then moved on to calculating the Kc constant which you needed to follow the three rules for as well. The only difference is that to the equation is the concentration of the products multiplied by each other over the concentrations of reactants. As stated before, the same rules apply.

On page 12 these are the answers to the two problems we did:

1. Kc=[NO2]^2/[N2O4]
2. Kc=[Pb2+]x[S2-]/1

Mr. H ended class by informing us about a change in the grading scale. Homework will no only be worth 5% of our grades. The 15% previously dedicated to homework will be based on small quizzes like the one we received on Friday. Two webassign assignments for homework. A reading sheet due on Thursday and a 40 QUESTION WEBASSIGN due next monday before the test. Mr. H highly advised beginning it to prevent procrastinating.

Wednesday, April 6, 2011

WEDNESDAY APRIL 6, 2011

Today was a late arrival day! We started off by reviewing the answers to the Reading Sheet for Chapter 11.1-2. The answers are:

1. B

2. A

3. C

4. B&C

5. A&D

6. C

7. A

8. B

9. D&C

10. False; M is often not equal to the coefficient

11. The order of this reaction is 1. You determine this by using trials 1&3: [A] doubles as the rate doubles

12. k=about 6.2/s (From trial 1:1.24=k(0.2)^1)

13. 2

14. With respect to A: 1 With respect to B: 2 Overall order: 3

After that, we spent the rest of the class period review problems that have to do with reaction rates. We reviewed how to write the rate law equation. We then learned how to calculate the rate constant, otherwise known as K. To determine the value of K, you simply divide the initial rate by all of the other individual concentrations in that trial multipled together (don't forget to multiple each concentration by it's order of reaction number). After you calculate the rate constant (K), you are able to predict the initial rate when given the concentrations which is another new topic learned in class. The equation for predicting the initial rate is: Rate=K(concentration1*order of reaction)(concentration2*order of reaction)(concentration3*order of reaction). An example of all of these topics learned can be found in our packet on page 2 #4. The problem is listed below:
After we did problem number 4, we turned to page 3 to try a problem on our own. The problem lists a balanced equation. Below the balanced equation, there is a chart of collected data which shows the different trials listed with the concentration of the compounds along with the Rate in mol/L/s. You are to determine the order of the reaction along with the overall order of the reaction which is found by looking for patterns in the chart like we did in class. Once there is a pattern found, you are able to determine the power for that compound which will be the order of the reaction. After you find the order of the reaction for all of the compounds, you add them together to calculate the overall order of the reaction. Next you are to write the rate law, which goes along with the order of the reacitons. An example of a rate law is shown below (Rate=K*[F2]*K[ClO2]^2). After you write the equation for the rate law you are to calculate the rate constant. Again, the example for the rate constant is shown below (K=1.20x10^-3/(.10)(.010^2). The constant for this problem is calculated to be 120. Lastly, you are to calculate the rate of the reaction at the tim when [F2]=.010 mol/dm^3 and [ClO2]=.020 mol/dm^3. To figure out how to do this, you simply take the rate constant (120) and multiply it by (.010) and (.020^2) to get 48.80x10^-4 as your rate.

That was all we had time for today, remember there is a Web Assign due Friday!


--Katie Jennings

Tuesday, April 5, 2011

Tuesday, April 5

Mr. Henderson began the class period having us take our lab notebooks out to copy down the title and purpose of the lab we would be working on at the end of the class period. Notes were given the previous day: Unit 11 Kinetics and Equilibrium chapter 11(reaction rates) consider 2 N2O5 --> 4 NO2 + O2 1 liter container Initially 10 moles/L 0 0 change -2 moles/L +4mol/L +1mol/L sometime later 8 moles/L 4mol/L 1mol/L (1hr later) Rate= (-change[N2O5])/(2 change t) Rate= (change[NO2])/(4 change t)= (change[O2])/(change t) In General: aA + bB --> cC + dD Rate= (-change [A])/(a change t)=(-change[B])/(b change t)=(change[C])/(c change t)=(change[D])/(d change t) We then looked back onto page one which we finished monday in class. 1) a. Rate= (- change[NH4])/(change t)= (-change[NO2-])/(change t)=(change[N2])/(change t)=(change[H2O])/2 change t b. Rate= (-change[N2])/(change t)= (-change[H2])/(3 change t)=(change[NH3])/(2 change t) 2) a. As the concentration of NO is doubled while keeping the Cl2 concentration constant, the rate of the reaction is increased by a factor of 4. This can be seen while comparing rows 1 and 2. b. As the concentration of Cl2 is doubled(while keeping the NO concentration constant), the rate of the reaction is increased by a factor of 2. This can be seen while comparing rows 2 and 3. c. Cl2: 1+ NO: 2 = Overall: 3 Mr. Henderson pointed out that in this specific question, the order of reactions for Cl2 and NO happens to be the coeficients. The overall reaction is simply determined by adding the order of reactions of both of the reactants which in this case is Cl2 and NO. So you then add Cl2's oreder of reaction,1, and add NO's order of reaction, 2, to get 3. d. Rate= k[NO2]^2 Rate=k[Cl2]-->Rate=k[NO]^2*[Cl2] k is the rate constant. e. consider row 3: 9.12=k(1)^2*(1) 9.12=k 3. a. As the concentration of I- is doubled while keeping the S2O8^2- concentration constant, the rate of the reaction is increased by a factor of 2. This can be seen while comparing rows 1 and 2. b. As the concentration of S2O8^2- is doubled(while keeping the I- concentration constant), the rate of the reaction is increased by a factor of 2. This can be seen while comparing rows 1 and 3. d. Rate=k[I-] Rate=k[S2O8^2-]--> Rate=k[I-]=[S2O8^2-]^2 We then were given time at the end of the class period to work on Lab KE1 Feeling Blue_No Longer.