Wednesday, December 15, 2010

Today Mr. Henderson started off class with some announcements. First of all, Friday is the test, and he is going to be collecting lab notebooks. Second of all we only have to do 2 out of the three webassigns due Friday. Another reminder, delicious assignment is due tomorrow.

Next Mr. Henderson passed out the lab grading sheet and explained a little about how the lab is going to be graded. Mostly just reviewing whats on the sheet. Later on i will explain more on how to help you finish the lab. After that we went on page 30 and Mr. Henderson gave us the answers to the reading sheet along with some notes: pressure has a direct relationship with the amount of moles of a gas, therefore pressure1/pressure2=moles1/moles2.

The answers to the reading sheet are:

1) A, C

2)A

3)A

4)287

5)570

6)c

7)A).25 B).25 C).50

The next thing we did was page 19. (I will include a picture at the end of this blog and go over number one) first of all we needed to find the vapor pressure from a chart in our book on appendix 1. For this problem since the temperature was 30 C when you look on the chart the v.p. is 31.8 mmHg. Then we need to convert the kPa to mmHg since the other pressure is in mmHg. After that we can conclude that PH2O is 31.8 mmHg since that stands for the water(vapor) pressure. The O2 pressure would be the over all pressure, which we found by converting 120 kPa to 900.1 mmHg, is the total pressure minus the vapor pressure, which is 900.1-31.8=868.3mmHg. Then the next two questions ask for the fraction pressure of H2O which we find by putting H2O's partial pressure untop of the overall pressure, 31.8/900.1 which equals .0353. Next we find the fraction pressure of O2.Which we find the same way as the fraction pressure of H2O except different numbers, so its 868.3/900.1=.9646. The rest of the answers you can see on the picture.

After that Mr. Henderson helped us do most of our lab which the rest were finishing tomorrow. I will show you my data, which you can plug your own numbers in if need be. My data is : .03582g Mg, temperature is 20.5, volume of H2 is 38.2, and by using appendix 1 v.p. is 17.54. So using my grams of Mg i convert, using stoichiometry, to moles of H2. Which i get .00147407 m H2. We are trying to find the volume of H2 at STP. Now using the pressure of the room, which Mr. Henderson gave us, and the pressure of the H2O we can find the pressure of H2 which we need to find the Volume of H2. You subtract H2O pressure from room pressure, 763.52-17.54=745.98mmHg, which is the pressure of H2. Now we use the equations P1V1/T1= P2V2/T2 to find the V2. When plugging in numbers make sure to convert Celsius to Kelvin. (745.98)(38.2)/293.5=(760)V2/273. V2 is 34.87 L. That is all we managed to do in the shortened class periods. As I said we will finish the lab tomorrow.

Tuesday December 14, 2010

Class started off with a reminder that nspire calculators will not be allowed on this Friday's test, Mr. H reminded us that we could bring our own "un-nspired" calculators to class or use the ones provided. He also emphasized that students should spend about 30 minutes a day on WebAssign, the only one we wouldn't be able to do after today's lesson is Chapter 5:5. In addition, the deadline for the Unit 6 Delicious Assignment has been moved back to this Thursday and a point was made to separate the tags "hcp3y1011" "unit6" and "(first name last initial)" with spaces, no commas. The last subject discussed before the lesson was the previous day's post, a good job on Kyle's part.

We began the lesson with a review problem, we turned to packet page 16 and were assigned problem #7 to complete. Part a) was particularly easy, no surprise whatsoever. Part b) involved using converting factors, nothing particularly interesting with the possible exception of the ratio 22.4 Liters in 1 Mole of any gas at Standard Temperature & Pressure. Part c) was where things got interesting: since the reactant was not at STP, it was necessary to use PV=nRT where P=1.20 atm, n=5 Moles, R=0.08206 (a given), and T=65 degrees Celsius=338 Kelvin. Since V=(nRT)/P, the volume of the product here is equal to (5*0.08206*338)/1.20 which yields approximately 1161 Liters of acetylene. It was noted that since R is in the ratio (L*atm)/(mole*K), P should be in atm.

Next was page 15 problem #3. No different than the last; use conversion factors, solve for V using PV=nRT, and done. I'm sure it is vitally important to someone that Mr. H generously gave Kevin one Periodic Table to add to his collection of identical Periodic Tables. Here's page 15 for those who need it, answers are still just app

roximations. Under that is page 17 problems #9-10 for those who need extra practice.

Finally, it was time for a demo. Mr. H took a ping-pong ball with a hole the size of a pinhead and dipped it into a container of Liquid Nitrogen. It filled up inside the ball which was then placed on the table. Gathering up close, the students observed that as the ball spun, the Liquid Nitrogen inside was coming out through the hole and vaporizing instantly, pushing the ball in a continuous circular pattern. Of course, since the majority of developing teens have a small attention span, this got boring after 5 minutes. To shake it up a bit, Mr. H took the ping-pong ball and submerged it in the Liquid Nitrogen for several seconds and, without giving the substance time to escape the ball,

smashed the piece of plastic on the table. All around, the audience was pleased.

"But wait, there's more!" was what Mr. H should have said when he continued with these demos. He brought out the ol' vacuum pump and showed everyone what happens when the atmosphere surrounding an inflated balloon is removed. A loud and satisfying pop resulted. During this time, one student was leaning on the light switch mounted on the side of the demo table and accidentally pushed it down to the "off" setting. Much complaining and laughter ensued. It is not likely that the offender will be identified, though it is known that this student was indeed of the female gender.

Segueing awkwardly out of that tangent, Mr. H finally showed us one last demo. Another unnamed student wondered out loud what would happen if one were to suck the air out of, say, shaving cream. Answering this question, Mr. H sprayed some cream into a beaker and placed it in the chamber connected to the air sucker. Activating the vacuum, we all saw the cream rise up towards the beaker. Almost like The Blob, without getting all "theatrical." Shortly afterwards, the bell rung and class officially ended. Homework is to do WebAssign, Delicious, or both. Just do something and you'll be fine. Till next time!

Monday, December 13, 2010

To start off class today, Mr. H handed us a lab procedure sheet for the Molar Volume Lab. He then told us the purpose of the lab which is to determine the volume of one mole of gas at STP. After Mr. H briefly explained the lab, we got into our groups and started the experiment.

My groups data looked like this:
Barometric Pressure: 763.52 mm Hg
Length of Mg strip: 3.8 cm
Mass of Mg strip: 0.05 g
Density of Mg strip: 0.00995/3.8= 0.0026 g/cm
Temperature of water: 20.1 degrees Celsius or 293.1 Kelvin
Volume of gas: 61.5 mL

After each group finished up the data part of the lab, Mr. H showed us Brooke's blog from Friday which had some really important information about the five equations used for solving gas related problems.

We then started off the lesson on page 15 with Gas Stoichiometry Calculations. Mr. H taught us that Avogadro's Law is used in these types of problems which are very similar to conversions we did in previous chapters. Avogadro's law states that one mole of any gas occupies the same volume when held at the same temperature and pressure.

The first problem that involved finding the volume of a certain compound was #4.
Our first step in solving this problem is to write a balanced equation:
C3H8(g) + 5O2(g) ------> 3 CO2(g) + 4 H2O(g)
Then the question told us to determine the volume of CO2 that is produced at STP.
In order to solve this question, you have to make a conversion factor problem:
149 g C3H8 * 1 mol C3H8 * 3 mol CO2 * 22.4 L CO2/44 gC3H8.
The reason I used the number 22.4 L is because if a gas is at STP then there are 22.4 L of that certain gas.
The final answer would be approximately 228 L CO2.

After we did this, we did one more problem on page 16, #6 and the answer was 44.8 L H2.
Then we went back to our lab to record the data measurements and after this class was over.

HW: WebAssign due tomorrow, Test on Friday, Unit 6 Delicious due Thursday, WebAssign due Wednesday, and 3 WebAssigns due on Friday.

Friday, December 10, 2010

The class period started out looking at the blog from the previous night. Mr Henderson then had the class take out the unit 6 packets.

To answer the following problems one of five equations must be used.

1)P*V=n*R*T
(R=0.08206 L * atm/mol * k)
P=pressure=force/area, force=cumulative force from all collisions of particles with container walls.(pounds/inch2=psi, atmosphere=atm, millimeters of mercury=mmHg, torr)

V=volume(liters,mL)

n=number of moles

T=temperature= measure of average kinetic energy or KE (C or kelvin). Kelvin(K)=C + 273

2)V1/T1=V2/T2
(when n and P are constant)

3)P1* V1= P2* V2
(when n and T are constant)

4)P1/T1= P2/T2
(when n and V are constant)

5)[P1*V1]/T1=[P2*V2]/T2

Page#9 number 1

1) V1= 16.9 V2=?
T1=25C (25+273=298), 298K T2=268K

Since the information given contains temperature and volume the next step is to look at which of the five equations contains temperature and volume and the equation is V1/T1=V2/T2

16.9/298=V2/268

V2=15.198 Liters

Page #10 number 7

7)V1=412 mL V2=663
P1=1.00 atm P2=?

Since the information given contains volume and pressure the equation for this problem is P1*V1=P2*V2

412*1=663*P2

P2=0.621 atm

Page #11 number 15

15) P1=2.50atm P2=?
T1=5C(5+273=278), 278K T2=650C(650+273=923), 923K

P1/T1=P2/T2

2.5/278=P2/923

P2=8.30 atm

Page #13 number 26 and 32

26) V1=18.5 L V2=?
P1=85.5 kPa P2=1atm, 1atm=101.3kPa
T1=296 K T2=273 K

[P1*V1]/T1=[P2*V2]/T2

[25.5kPa*18.5L]/296 K=[101.3kPa*V2]/273 K

V2=14.4L

32)V1=5.25L
T1=23C(23+273=296), 296K
P1=8.00 atm
n=?
m=?

P*V=n*R*T n=[P*V]/[R*T]

n=[8.00 atm*5.25L]/[0.08206L*atm/mol*K *296 K]=1.73 mol

m=1.73 mol CO2*[44 gCO2/1 mole CO2]=76.1 g

The class period ended with Mr. Henderson making ice cream for the class.
HW:2 webassigns due Monday and test Friday

Thursday, December 9, 2010

Today we started class off with reviewing Tammy's blog. If you were absent, her blog would be a good review of what we did that day (packet pages 1-3). In her blog, she incorporated the equation: 1 atm=14.7 BSI=760 mmHg=760 torr.

Next, we reviewed the equation: P*V=n*R*T which would used for today's entire lesson. With this equation, the P stands for pressure, V stands for volume, n stands for the number of moles, R stands for the gas law content, and the T stands for temperature.

From that formula, there are other equations that are formed when P, V, or T is constant. The first equation is: V=K*T (assuming that the pressure is constant).This equation is direct which means that as the volume of a sample of the gas increases, the Kelvin temperature of the gas increases also. An example of this would be (#3a, packet page 5): As the Kelvin temperature is increased by a factor of 2, the volume of the gas sample is increased by a factor of 2. When a sample of gas with volume V1 and temperature T1 are changed to a new volume V2 or a new temperature T2, you form a 2-point equation. The theory relating volume and temperature for a two-point equation is called Charle's Law. This equation would be: V1/T1=V2/T2. On packet page 6, number 6 is an example of how to apply the Another equation that can be formed using Charle's Law. For number 6a, you would take the initial temp and the inital volume and plug that into v1 and t1 (5L/300K). Next, you would plug in 600K for t2 and "X" for v2. After solving the proportion (multiplying both sides by 600), the answer would become 10L for the final volume.

Another equation derived off of the forumla PV=nRT is: PV=K (assuming that temperature is constant). This equation is inversely proportional meaning that when the pressure of a sample of gas is increased, the volume of the sample decreases. An example of this would be (#10): As the pressure is increased by a factor of 2, the volume of the gas sample is decreased by a factor of two (or 1/2 of the original value). If a sample of gas with a volume of V1 and a pressure of P1 is changed to a new volume V2 and a new pressure P2, a two-point equation is formed. This theory is called Boyle's Law. The Boyle's Law equation is: P1V1=P2V2. An example of Boyle's law is number 13a on packet page 7. In this problem, you would multiply the p1(1.0 atm) and v1(5.0L). After you get the answer to that, you divide that2.0 atm(p2) to get the final volume(V2 or "X").

The last equation that is derrived off of the formula is: P=KT. This is equation is direct meaning that as the pressure of the gas increases, so does the Kelvin temperature of the gas. If a sample of gas with a pressure P1 and a temperature of T1 is changed to a new pressure P2 and a new temperature T2, you would use a two-point equation. The theory of this equation is called John's Law. The equation is: P1/T1=P2/T2. An example of John's Law is on packet page 8 and number 20a. You would plug in 300K for T1 and .528atm for P1, and 600K for T2 and "x" for P2. After that, you would solve the porportion (multiplying both sides by 600k to get 1.056 atm for the final pressure.

After going over packet pages 5-8, we went to the science computer lab to work on our web assigns. That concluded our day! HW: web assign due Friday

Wednesday, December 8

We started class today by doing page 1 in the unit packet.
The following statements are false:

• Gases are massless matter. Gases are made up of atoms, which have mass.
• Gases will condense when heated. Gases expand when heated.
• Gases are colorless. I2(g) is purple, NO2(g) is brown.
• Gases push on the walls of their container, but ONLY in a downwards direction due to the influence of gravity. Consider a balloon; gas is pushed in ALL directions on the wall
• The volume of a gas is inversely related to the temperature of a gas. They are directly related. As temperature increases, volume increases.
• The particles of a sample of gas are relatively motionless. The particles move very quickly.
• The entire column of our atmosphere applies pressure on objects on Earth; this pressure is so small that it is of little consequence and influence.

We then did page 2, #4. The units for pressure are atm, psi, mm Hg, torr, and Pa. The units for volume are L, m3, and mL. The units for temperature are celcius and Kelvin. The answers are:

a) P

b)V

c)V

d)P

e)P

f)T

g)V

h)P

i)P

j)T

We then did #5 on page 2. To do this, we had to use the equation °C+273=K

For a, we were given 100°C to convert to K. The answer was 373 K since 100+273=373.
For b, we were given 400 K to convert to °C. The answer was 127°C since 400-273=127.
For c, d, and e we had to use converting factors.
1 atm=14.7 psi=760 mm Hg
The answers were:
c) 1.03 atm
d) 3.0 atm
e) 684 mm Hg (torr)

We then did page 3, #8 and 9. To determine the pressure, you take the difference in height and add or subtract from 760 mm Hg. For the second manometer in #8 you had becuase P has more pressing power than Hg. For that one, P=700 mm Hg + 760 mm Hg = 1460 mm Hg. In the third one, the gas outside has more pressing power, so you subtract the difference in height. P=760 mm Hg - 560 mm Hg=200 mm Hg.

For #9 we did something similar to measure pressure. For A, we had to convert 75 cm (the differnence in heights) to 750 m Hg. For B, we subtracted the two heights, 82 cm- 25 cm, to get 57 cm which we then converted to 570 mm Hg. For C, we also subtracted the two heights, 125 cm-50 cm, to get 75 cm which we converted to 750 mm Hg.

We then wrote down the answers to the Chapter 5.2-3 reading sheet on page 27 about gas laws. The answers are:

1. a, b

2. b, c, d

3. a

4. b

5. d

6. c

7. a, ,b c, d

8. a

9. d

10. c

11. d

12. b, c, a

13. a

14. b, c

We ended class with some interesting demos. In one, Mr. H put a vlown up balloon in liquid nitrogen. That caused the volume to decrease because the the nitrogen is very cold, so the temperature had decreased. It was so cold that the air turned into liquid air. He also froze a water bottle by putting it in the liquid nitrogen. Another demo was to take the shrunken balloon and quickly put it in an empty bottle. As the temperature increases, the volume of the balloon increases causing there to be a blown up balloon inside a water bottle. At the very end of class, Mr. H poured the liquid nitrogen all over the floor where we watched it seem to disappear as it turned into a gas.

Today's chemistry class had us begin a new unit. We took the unit 5 test friday, so we started unit 6: Gases today. The first thing we did was meet in the chemistry lab to do a chemthink activity.

This link will take you to the chemthink website: http://chemthink.com/chemthink.html

*your login is your school i.d. and password

The Chemthink was on the Behavior of Gases. We started out with the tutorial which was very imformative.

*The first thing we learned was the effect temperature has on gases. As the temperature increases, gas particles begin to move faster, and when the temperature decreases, the particles move slower.

*The next thing we learned was the larger a particle is, the slower it moves. Take helium(He), neon(Ne), and argon(Ar). Argon, the largest particle, moves the slowest, Neon, the second largest, moves medium speed, and Helium, the smallest, moves the fastest.

As we continued through the chemthink, we learned that Pressure=Force/Area. Pressure also relates to Temperature, number of atoms, and volume.

This picture shows the relationship of how pressure varies directly with changes in temperature.

Formula: P/T=constant

This picture shows how pressure varies directly with the number of gas particles.

Formula: P/n=constant

This picture shows how pressure varies inversely with the volume of a gas sample

Formula: P*V=constant

This infomation is an overview of what the Chemthink was, and what the basics of the Gases unit are. After the Chemistry lab, we returned to the room and went over the previous nights webassign homework. The answers for page 27 in your unit packet are:

1. D

2. C

3.B

4.D

5.A

6.A

7.B

8.B

9.860

10. A

11.E

12.C

13.C

Notes from Class Discussion:

Volume (V)- Liters

=Amount of 3D space occupied by an object

*Further information can be found on page 7 in your book

Temperature (T)- Kelvin

=measure of the average amount of energy of motion (kinetic energy) of particles

*further information can be found on pages 7-8 in your book

Number of Moles (n)

1 mole=6.022 x 10^23

*further information can be found on page 55 in your book

Pressure (P)

=Force/unit area

=For gases, pressure results fromparticle collisions with container walls.

*further information can be found on page 104 in your book

Class then ended with Mr.H saying that we would finish going over problems 11-13 tomorrow, and that class the next day would have plenty of demonstrations regarding the new unit!

Thursday, December 2

Today we started off class by reviewing the 3.4 Reading Sheet. After looking over that, Mr. Henderson passed out calculators and told us how to use them more efficiently. There is a button (sto) that can store a numbers in the calculator. For example, if you wanted to store a molar mass because you would be using it in the problem a lot, you would plug that number under sto 1. So when you used that number, you would not have to type it in; you would just have to the recall 1 button. He also said that we should either be comfortable with his calculators or bring your own to use. If you bring your own calculator it cannot be an inspire.

We then went over what would the format of the test would like like.

-14 multiple choice questions

-3 pages of short answers (stoichiometry problems)

-1 question on the lab

To study for these types of problems there are review questions on the website and moodle with answers. A great way to study would be to use your packet. Everyone should have almost a completely filled out packet with correct answers. You could rework a problem that is difficult for you and check your answer with your packet. To prepare for the test in class, we completed review questions 17-19 which simulated what the questions about the lab would look like.

Furthermore, Mr. Henderson lectured us about percent yield in detail. To find the percent yield of a substance you need the theoretical yield (mass of the expected product) and the actual yield (mass of the measured product). You discover the theoretical yield by using the normal stoichiometry conversion factors, and you discover the actual yield by massing what your results are in the lab. The equation for the percent yield is:

Percent yield = Actual Yield * 100

Theoretical Yield

Lastly, we attempted problems involving this percent yield concept on a worksheet Mr. Henderson passed out. The toughest problem was number 5; the answer to it was 709.75 grams. Today was a very productive class period as we got in needed review time for the test. Finally, I would like to wish everyone good luck!

Wednesday, December 1

One-half the class was at the "Aids Day Workshop". Those who were present received a quick review of percent yield problems. We discussed the percent yield equation (see graphic below). We discussed how the yield in a given reaction refers to the amount of product that is produced. When conducting an experiment, the actual amount of yield is often less than the theoretical amount of yield. Spillage and other means of losing materials is the cause of the actual yield being less than the theoretical yield. The theoretical yield of product (i.e., the mass of product that we expect to obtain in a reaction) is related to the mass of reactants via the usual stoichiometry calculations.

We then took time to work on a handout on percent yield. Access Handout on course website.

The answers to the handout are shown below. For the first two columns of the table, think stoichiometry.

We will do more tomorrow on the topic of percent yield and review for the Friday exam. A review sheet for the exam should be up by the end of day on Wednesday.

Tuesday November 30th, 2010

Todays class started out by Mr. H informing us that we are not going to continue the copper lab that we had started yesterday. We then started to review what would be on our chapter 5 that is this Friday. The test on Friday will consist of 15 multiple choice questions, 3 pages of problems in which you have to show your work, and out of those three pages one page is lab based, another one is 4th grade stoichiometry, and the last page has problems from 7th, 8th, 9th, and 10th grade stoichiometry. There will be a test review page up on the gbschemphys website by Thursday. Another great way to study for this would be to go through the Period 3, Unit 5 delicious book marks. Mr. H also let us know that since a majority of the class will be on a field trip tomorrow, we will be in a computer lab working on the web-assign that is due Thursday.

We then turned to page 13 in our packets and started work on question number 23, part b.

In part B it says... If 215 g of C6H6 are combined with 590 g of O2, what is the limiting reactant and what mass of excess reactant will be left over?

They have also given you the balanced chemical equation:

2C6H6 + 15 O2 ----> 12 CO2 + 6H2O

now you can find the mass of CO2 once reacted with C6H6 and O2.

215g of C6H6 x 1 mol C6H6 x 12 mols CO2 x 44.03g/mol of CO2

------------- ------------- -------------------- = 727.62CO2

78.06 g/C6H6 2molC6H6 1 mol CO2

LEVEL 9 STOICHIOMETRY

590g of O2 x 1 mol O2 x 12 mols CO2 x 44.03g/mol of CO2

------------- ------------- --------------------= 648.74CO2

32.02 g/O2 15molC6H6 1 mol CO2

O2 is the limiting reactant.

590 g O2 x 1 molO2 x 2 mol C2H2 x 78 C2H2g

---------- ------------ --------- = 192 g C2H2 left over

32 gO2 15molO2 1mol C2H2

215-192= 23g

We were then told to work on page 12 number 18 and page 14 number 24.

18. 2KClO3 -------->2KCl + 3O2

o.550g KClO3 x 1 mol KClO3 x 3 mol O2

------------ ---------- = 0.00673 mol O2

122.55gKClO3 2 mol KClO3

24. B2O3 + 6 HF --> 2BF3 + 3 H2O

a. 85 g of B2O3 x 1mol B2O3 x 2 BF3 mol x 67.78gBF3

----------- ------------ ----------- = 165.43 BF3

69.65gB2O3 1B2O3mol 1 mol BF3

115g of HF x 1 mol HF x 2 BF3 mol x 67.78gBF3

---------- ----------- ------------ =129.97 BF3

19.99gHF 6HFmol 1 mol BF3

129.97 -- HF is the limiting reactant

115g of HF x 1 mol HF x 1 BaO3 mol x 69.65gBaO3

---------- ----------- ------------ = 66.78 BaO3

19.99gHF 6HFmol 1 mol BaO3

Excess: 18.22

PSYCF= Please Show Your Conversion Factors (not cute frogs) (:

Tuesday 23, 2010

Today we started out class by turning in our lab notebooks but, it quickly turned into a very interesting start. Mr. Hendersons computer was broken therefore projecting a magenta color. Some people claimed it was a "purpendicular" color crayola made so we went on a search to find that color. Sadly we couldnt find any proof that color existed. After that we went on the wikipedia page for green, then Mr. H asked girl students and boy students what color they see, guys saw black and girls saw a dark shade of green, therefore when were all married, the men just agree to the womens color likings. Also, a different chemistry teacher came in and informed us that only male turkeys gobble. That was a great start to a chemistry class.

After our great begining we turned to page 8, to review 7th grade stoichiometry in a very picturesque way. We only did number 4. *NOTE* the white circles that are just floating around[theres four] should have a diagonal line going through them so you know which ones carbon, same in your answer key. In the after reaction box we drew in two H2O four H2 and four CO. The limiting reactant was Carbon.

After that we turned into page 9 in our packets. The page was fairly easy as we breezed through it. We skipped question 6 going on to seven which was finding the molar mass. Then we went on to questions 8 and 9. *NOTE*= The limiting reactant is the reactant that produces the least amount of product.In question eight we had to figure out which was the limiting reactant and then the max mass of NH3. In nine it was pretty much the same except we had to list the limiting reactant.

Here are the answers to page 9:

8) A)1 mole, 3 moles, 34 grams/ 2 moles.

B) 2 mole, 6 moles, 68 grams/ 4 moles

C) 5 mole, 15 moles, 170 grams/10 moles

9)A) 2.21 moles, 6 moles, H2, 68 grams./4 moles

B) 1 mole, 4 moles, N2, 34 grams/ 2 moles

On to page ten, on page ten it was mostly just practice of 8th grade stoichiometry. We had a balanced equation and we had to find the limiting reactant from two masses we were given.

Here is all the work for page 10:

HOMEWORK!:unit five delicious assignment due Tuesday! It will be graded!

Two webassigns if you want extra credit are due Saturday and Monday.

Also, test Friday and lab notebooks will be collected again!

Friday November 19th

Today, Mr. H began the class by going over Alex's blog and showing the variety of pictures that were from Wednesday's lab. He then refreshed our memories about the chemical reactions we had done the day before in our packets. (These types of conversions are going to be on tonight's webassign). Mr. H then proceeded to tell us that he had taken each lab group's silver precipitate, and put them in an oven to dry. He mentioned that he would take them out at the end of class so we could find the mass of each beaker that contained pieces of silver.

Mr. H then instructed the class to open our unit 5 Stoichiometry packets to page 6. There, we worked on the second half of the page; it was titled: Decomposition StoiCHEoMISTRY. The class worked on problems 12-15, and then Mr. H showed us the answers so we could correct our mistakes. He went on to answer any questions that students had about these problems. The answers to problems 12-15 are as followed:

12) 2H2O=> 2H2 + O2

13) 364.8 grams O2- this was a 2 step problem ( converting moles of H2O to grams of O2 )

14) 35.67 grams H2- this was a 3 step problem (converting grams of H2O to grams of H2)

15) 1.22 moles O2- this was another 2 step problem (converting grams of H2O to moles of O2)

In order to solve #13, this is how you would show your work:

         22.8 mol H2O * 1 mol O2 * 32g O2= 364.8g O2
                                  2 mol H2O   1 mol O2

Next, Mr. H gave a lecture on "Limiting Reactants." The notes he wrote on the board are as followed:

Limiting reactant- one reactant that produces the least amount of a product

   2 Al(s)             +          3CuCl2              ==>            3Cu +    2AlCl3

0.010 moles                  0.030 mol    =============> 0.03 mol Cu                           
not enough                    too much-precipitate
dissolves                      (excess reactant)
(limiting reactant)==========================> 0.015 molCu
                                                                                                                                                                  

0.030 mol                    0.030 mol ==============> 0.030 mol Cu
too much                      not enough
precipitate                    dissolves
(excess reactant)          (limiting reactant)
|
|
|======================================> 0.045 mol Cu

The reason why you choose the number that is less than the other one is because after you conduct 1 experiment, you'll run out of one of the elements and you won't have enough  of the second element to create another product.

Next, Mr. H introduced us to the next level of stoicheometry and used the notes he gave us in class to solve problem #12 on page 4. In this problem, you have to convert the number of grams of each element to the moles of Ti3N4. This is how you would solve this problem:

162g Ti * 1 mol Ti * 1 mol Ti3N4 = 1.13 mol Ti3N4  <== correct answer
                47.88g Ti   3 mol Ti

79 g N2* 1 mol N2 * 1 mol Ti3N4 = 1.41 mol Ti3N4
               28.01g N2   2 mol N2 

1.13 mol Ti3N4 is the correct answer, because as stated in the notes: you choose the number that is less than the other one because after you conduct 1 experiment, you'll run out of one of the elements and you won't have enough  of the second element to create another product.

Next, Mr. H showed us 7th grade Stoicheometry. This was on pages 7 and 8. This page included practicing finding limiting ingredients and the number of cakes produced. For each problem, we had to look off of the original equation, which was: 6 cups of flour + 4 cups of sugar + 2 tbsp baking powder + 3 cups of milk ==> 2 cakes, and find by how much each ingredient was being multiplied. By doing so, you could find the limiting reactant ( the lowest number) and then multiply that number by the original number of cakes produced from the given equation. The answers for page 7 are as followed:

a) Limiting ingredient: milk               # of cakes produced: 1

b) Limiting ingredient: baking powder            # of cakes produced: 3

c) Limiting ingredient: flour               # of cakes produced: 10

d) Limiting ingredient: baking powder                          # of cakes produced: 100

e) Limiting ingredient: baking powder           # of cakes produced: 100

On page 8, was more practice that involved finding the limiting reactant and the # of products produced. We only did #3 on this page. The answers for # 3 are as followed:

Molecules before reaction             Type and # of molecules after reaction                           Limiting Reactant
3 H2 + O2============>       2 H2O
                                                      1 H2                                                                                   O2

6 H2 + 3O2===========>       6H20                                                                                  none

18 H2 + 4 O2==========>      8 H2O
                                                     10 H2                                                                               O2

Mr. H ended the class by getting each lab groups' beaker from the oven, and then had 1 member of each lab group go and mass the beaker.The students that missed class that day, should ask one of their group members for that information. **Note, there is a QUIZ  on Monday, so be sure to study.

Thursday, November 18

This morning, we walked into class and were notified that there was a lab today. The lab is called ST1 Copper and Silver Nitrate Lab. The purpose is to use careful measurements to determine the mole ratio of copper(II) and silver in the single replacement reaction of silver nitrate with copper. We were instructed to take out our lab notebooks and copy down this purpose onto a clean right-side page. On the left side, Mr. H told us to write the chemical equation for the reaction, for it would be helpful later. It is:

Cu (s) + 2AgNO3 (aq) --> 2 Ag (s) + Cu(NO3)2 (aq)

We all then received a piece of paper describing the lab's procedure. 2 people from the lab group were to carry out the first procedure while the remaining 2 were to carry out the second procedure. The procedures are as followed:

1) Prepare the Silver Nitrate for Reaction:

• Mass an empty 50 mL beaker. Record.
• Add ~1.4-1.6 g of AgNO3.
• Mass beaker plus AgNO3. Record.
• Add 30 mL of water.
• Stir to dissolve solid; rinse stirring rod so as to avoid loss of the AgNO3.

2) Prepare Copper for Reaction and React:

• Obtain ~25 cm of copper wire.
  
• Mass wire. Record.
  
• Loosely coil the wire so as to rest of a wood splint as shown. It should be long enough to hang on the wood splint and dangle in the solution. Yet it should not touch the bottom of the beaker.
  
• Add 3 drops of nitric acid to the solution; avoid contacting the copper wire with the HNO3. Do not stir.

After we did what was required, we were instructed to leave our notebooks at our lab benches and come back to our seats. This time was set aside for the reaction to occur. When we were back at our desks, Mr. H told us to take out our packets and turn to page 6 for

E V E N M O R E S t o i C H E o M I S T R Y!!!!

Mr. H told us to do number 7; write the balanced equation for the synthesis of magnesium oxide from its two elements - magnesium and oxygen. The answer is as follows:

2Mg (s) + O2 (g) = 2 MgO (s)

Then we were told to do number 8; calculate the molar mass of each of the three reactants and products. The answer is as follows:

Mg: 24.305 g/mol O2: 32.0 g/mol MgO: 40.3 g/mol

After, as a class, we went through the next three problems and identified how many steps would be needed (using the Mole Island method, of course). Number 9 was a one step equation. Number 10 was a two step equation. And number 11 was a three step equation.

The answer to number 9 (Determine the number of moles of magnesium oxide produced by the reaction of 3.25 moles of magnesium.) is 3.25 molecules of MgO, simply found by using Mole Island to convert from moles of magnesium to moles of magnesium oxide.

The answer to number 10 (Determine the mass of magnesium oxide produced by the reaction of 5.22 moles of magnesium.) is 210.39 g MgO, again, simply found by using Mole Island to convert from moles of magnesium to moles of magnesium oxide to grams of magnesium oxide.

The answer to number 11 (Determine the mass of oxygen gas which will react with 65.2 grams of magnesium.) is 42. 92 g O2, aaaaaaand again, simply found by using Mole Island to convert from grams of magnesium to moles of magnesium to moles of oxygen to grams of oxygen.

Before we knew it, it was time to check back at our lab. Our whole lab group was to do the third and fourth procedures listed on the sheet of paper:

3) Retrieve Silver Product from Copper Wire:

• Obtain a clean, dry 100 mL beaker. Using a permanent marker, label it with your period and your names.
• Mass the empty beaker. Record.
• Carefully lift the copper wire (with silver attached) from the beaker.
• Hold wire over and into 100 mL beaker and rinse with a forceful stream of DI water. The goal is to knock the silver off the copper and into the beaker.
• Continue rinsing until all the silver crystals are removed.
• Set labeled beaker aside to dry. Once dry (the next class period), mass the beaker with the silver. Record.

4) Rinse and Mass Unreacted Copper:

• Rinse the remains of the copper wire in an acetone bath.
• Being careful not to break or crumble the copper, thoroughly dry it by dabbing it with a paper towel.
• Mass the remaining copper wire. Record.

We did just that. The untouched product (silver still connected in the original beaker) should look like this:The removal of the silver should look like this:

Mr. H then told us to clean up and reminded us that we would be finishing the lab tomorrow.

The homework is the Stoich WebAssign 2.

Wednesday, November 17, 2010

Hello Class!

Today was a short but fun-filled day. Late arrival means the class was only 35 minutes long. The first thing we did was reviewing our stoichiometry connected to Mole Island. Mole Island is an interesting topic which has to do with knowing how many "division bars" you have in a conversion equation. (For example, if you're converting moles to moles, you multiply once. If you're converting moles to grams, you do it twice. Finally, if you're converting grams to grams, you do it three times.) See pages 3 to 4 in packet for more info. For trickier problems, such as finding out how many kilograms in a certain substance there are, (like on our WebAssign), just do it regularly but when you're done multiply by 1000. (kilo=1000)

But wait, that's not all!

Then, we started stoiCHEoMISTRY. (You get it? It's like chemistry, but inside another word containing other letters) Basically. . . . . same thing. You look at a reaction, balance it, and then you go on and determine molar mass of every molecule. (By using the PT) Then you go on and find how many grams or moles of a particular substance there is in another substance. Again, see pages 5 to 6.

We ended the class by taking a pop quiz. If you have any questions, see Mr. H or simply refer to his webpage. Also, make sure to continue being proactive when it comes to those WebAssigns.

Bye.

Tuesday, November 16

Today, after we were all seated, Mr. Henderson started off the day by showing us the fantastic job Tim did on his blog. He briefly reviewed what Tim covered in his blog, which is what we learned yesterday. Then we jumped right into working in our packets and working our way through stoichiometry grade school.

Today we learned about using the mole island to help us solve stoichiometry problems. We started on problem 6 of page 2 in our packets. We had to find how much H2 we would be needed to produce 22.8 g of NH3. First we found that the amount of moles in 22.8 g of NH3, which is 1.34 moles. Then we set the equation up like this-

22.8 g/mol NH3 x 1 mol NH3/17 g NH3 x 3 mol H2/2 mol NH3 x 20 g H2/ 1 mol H2 = 4.02 g H2

This shows that it requires 4.02 g of H2 to produce 22.8 G of NH3.
Mole island shows us how to get to a certain value. To get to the value you follow the arrows, and that's how we completed number 6. We went from C island, to moles C, to moles A, to A island. And the number of moles was the coefficient to the substance.

We then worked on the problems in page 3, which I have the answers to here.

1. Ti: 47.9 g/mol N2: 28.01 g/mol Ti3N4: 199.6 g/mol

2. 3.15 mol N2 x 1 mol Ti3N4/ 2 mol N2 = 1.58 mol Ti3N4

4. 2.85 mol Ti x 1 mol Ti2N4/ 3 mol Ti x 199.6 g Ti3N4/ 1 mol Ti3N4 = 189.6 g Ti3N4

And then as well on page 4.

6. 4.91 mol Ti3N4 x 2 mol N2/ 1 mol Ti3N4 x 28.01 g N2/ 1 mol N2 = 275 g N2

7. 26.3 g Ti x 1 mol Ti/ 27.9 g Ti x 1 mol Ti3N4/3 mol Ti x 199.6 g Ti3N4/1 mole Ti3N4= 36.5 g Ti3N4

8. 93.6 g N2 x 1 mol N2/ 28.01 g N2 x 1 mol Ti3N4/ 2 mol N2 x 199.6 g Ti3N4/ 1 mol Ti3N2 = 334 g Ti3N4

If you are able to do these, then you are are in the fifth grade at stoichiometry school!

To finish off the day we had a demo. The demo consisted of Mr. Henderson adding aluminum (Al) to copper chloride (CuCl2) in different amount then that which would make the entire solution balanced. The equation for Al + CuCl2 is balanced to this-

2Al(s) + 3CuCl2(aq) ---> 3Cu + 2AlCl3

Now by adding .02 moles of Al to .03 moles of CuCl2, our end result would theoretically have had all of the atoms reacting in the single replacement reaction. All we would have left is Cu and AlCl3. But Mr. Henderson added only .01 moles of Al in the first reaction. The .o1 was not enough, and we would have leftover CuCl2. In the second reaction, Mr. Henderson added .03 moles of Al to the CuCl2, and then we have leftover Al. In the first reaction, the aqueous solution left in the flask was a darker blue than the pure CuCl2, but it was not as dark as the second reaction. In the second reaction the flask had turned almost entirely brown in color. This was because of the combination of the brownish Cu and the grey AlCl3. This demo showed us that if we do not have substances in the proper amounts in proportion to the coefficients of the equation, then we will have leftover substances.

So in conclusion, today we learned about using mole island for setting up unit conversions, and we also learned about having leftover substances from a reaction.

Monay, November 15

Today we started class with Mr.H handing out our grades. He assured us that even though some of us may not have done well, it is still the beginning of a quarter and there will be more tests to do well on.

After that was over with, Mr.H showed us the Academic Stimulus Program. This gave students an incentive to do WebAssign assignments early to receive extra credit. Here is a chart that shows how much extra credit is rewarded for doing the WebAssign assignments.

Assignment	Stimulus?	Due Date	Conditions of Bonus Points
Stoichiometry - Set 1	No	Tues, 11/16	--
Stoichiometry - Set 2	Yes	Thurs, 11/18	Xtra 20% if 24 hrs in advance
Stoichiometry - Set 3	Yes	Mon, 11/22	Xtra 15% if 72 hrs in advance
Stoichiometry - Set 4	Yes	Tues, 11/23	Xtra 15% if 24 hrs in advance
Stoichiometry - Set 5	Yes	Tues, 11/30	Xtra 15% if 24 hrs in advance
Stoichiometry - Set 6	Yes	Wed, 12/1	Xtra 15% if 24 hrs in advance

We then moved on the the first page to learn the basics of Stoichiometry. We learned that the coefficients in a balanced chemical equation provided information about the relative amounts of reactants and products involved in the reaction. In the equation

N2 + 3 H2 → 2 NH3 we noticed that it had coefficients of 1, 3, and 2 respectively, so for this equation there has to be 1 molecule of N2 and 3 molecules of H2 to create 2 molecules of NH3. The equation must always keep a ratio of 1:3:2. Here are the answers to problem 1.

1 molecule 3 molecules 2 molecules

2 molecules 6 molecules 4 molecules

3 molecules 9 molecules 6 molecules

5 molecules 15 molecules 10 molecules

1 mole 3 moles 2 moles

2 mole 6 moles 4 moles

10 moles 30 moles 20 moles

5.4 moles 16.2 moles 10.8 moles

1.63 moles 4.90 moles 3.27 moles

We then moved on to problem two, which got a little more difficult. In this problem, we determined the relative masses by which reactants combine to form a given mass of product using molar mass. In the same equation N2 + 3 H2 → 2 NH3 N2 had a molar mass of 28g/mole, H2 had a molar mass of 2g/mol, and NH3 had a molar mass of 17g/mol. If there was 28 grams of N2 there would have to be 6 grams of H2 to create 34 grams of NH3. Notice that the sum of the masses of N2 and H2 equals NH3. Here are the answers to problem 2.

28g 6g 34g

56g 12g 68g

2800g 600g 3400g

45g 9.64g 54.54g

32.1g 6.88g 39.98g

41.53g 8.9g 50.43g

November 10, 2010 Wednesday

Today we started out the class by reviewing for the chemistry test on Friday. Mr. H gave the class ways to review for the test. He told us about a review worksheet on the gbschemphys website. After this we turned to page 13 in our Unit 4 packets and reviewed what we had done the day before.

We know that when an ionic compound dissolves in water, it dissociates into ions. These are the balanced chemical equations which show the dissociation of ionic compounds in water.

a. sodium chloride

NaCl(s) ---> Na+(aq) + Cl-(aq)

b. aluminum nitrate

Al(NO3)3 (s) -----> Al3+(aq) +3 NO3-(aq)

c. barium chloride

BaCl2 ----> Ba2-(aq) +2 Cl-(aq)

d. sodium sulfate

NaSo4(s) ----> 2 Na+(aq) + So4 2-(aq)

We also know from our lab that that Na and NO3 are soluble, but there are more things that are soluble in water. On page 13 of the packet there is a table of solubility rules. The rules for compounds that have high solubility are...

High Solubility

1.All alkali metal and ammonium compounds

2. Hydrogen compounds containing H+

3. All nitrates, acetates, and chlorate's (AgCH3COO only moderately)

4. All chlorides, bromides, and iodides EXCEPT those containing Ag+, Pb2+, and Hg2 2+ (moderately soluble in warm water)

5. All sulfates EXCEPT those containing Pb2+ or the 4 heavier ions of Group IIA ( Ca2+, Sr2+, Ba2+, Ra2+)

Low solubility

6. All sulfides EXCEPT those containing alkaline earths and rules 1 and 2

7. All hydroxides EXCEPT those containing Sr2+ and Ba2+ and rules 1 and 2

8. all phosphates, carbonates, and sulfites EXCEPT for rules 1 and 2

We then turned to page 14 and finished up problem number 3. This was about writing net ionic equations. First you have to present the ions that were present in each solution, and then you identify the solid that are being formed, and then indicate the ions remaining in the solution. And then at the end you write the net equation of the solution.

example a.

Na2SO4 + BaCl2

-ions: Na+ Cl+

-solid: Ba SO4

net ionic equation: Ba2+(aq) + SO4 2-(aq) ---> BaSO4(s)

example b.

AgNO3 + NaCl

-ions: Ag+ Cl+

-solid: Ag+ Cl+

net ionic equation: Ag+(aq) + Cl-(aq) ----> AgCl(s)

We then went on to finish the rest of the page. In problems 5-8 you had to analyze the major species in the solution and write the net ionic equations for the solutions.

5. Pb(ClO4)2(aq) + NaI (aq) ---> PbI2(s) + NaClO4(aq)

Pb2+(aq) + 2I-(aq) ----> PbI2(s)

6. Ca(OH)2(aq) + H3PO4(aq) ---> H2O(l) + Ca3(PO4)2(aq)

Ca2+(aq) + PO4 3-(aq) ----> Ca2(PO4)2

7. CdBr2 + Na2S ---> CdS(s) + 2 NaBr(aq)

Cd2+(aq) + S2+ (aq) ----> CdS(s)

8. NaOH + Fe(NO3)3 ----> Fe(OH)3 + NaNO3 (s)

Fe3+(aq) + 3H9aq) ---> Fe(OH)3 (s)

For problem 9, you are given a sentence which has the names of the solutions and you have to write them out and then write the net ionic equation.

9. Silver nitrate is mixed with sodium chloride solution

AgNO3(aq) + NaCl(aq) ----> AgCl(s) + NaNO3(aq)

net ionic equation: Ag+(aq) + Cl-(aq) ---> AgCl(s)

Some notes to keep in mind for the test....

-Aqueous solution of ionic compounds contain the compound in ion form.

-Ion charges can be found from....

.....the location on the periodic table for ions of the main group elements..

....the Roman numeral used in the name for ions of transition metal elements..

* when two aqueous solutions of ionic compounds are mixed, a precipitate can be formed; an ion from one of the solutions reacts with an ion from the other solution to form a solid precipitate.

Monday, November 8

Mr. Henderson started class today by examining what the rest of this week looks like. We will be learning a new topic for the next couple of days preceding the test, followed by the test Friday. He told us that the test Friday will be composed of: 14 multiple choice questions, 1 double-sided page of writing and balancing equations, no math work, and a section on identifying the reaction type. Mr. H said that what is going to make this test difficult is that it requires a lot of unit 2 work -- writing equations and such. But, we have a very resourceful tool to help us review, this blog! Mr. H walked us through Brooke's blog and talked about how helpful it will be to study for the test. To review, we can also use the Delicious websites created by the class (which, reminder, are due on Thursday!).

We then moved on to the new topic of today, Dissociation and Hydration. He begun this by doing page 11 in our packets. The first small section of this page was identifying a compound as being either ionic or molecular. An ionic compound is one which consists of a cation (metal) and anion (nonmetal), while a molecular compound consist of, normally, two anions. A good, common example of a molecular compound are those hydrocarbons Mr. H keeps talking about. These are the answers to the first part of page 11 ("i" for ionic and "m" for molecular):

1.

a. NaCl I

b. CO2 M

c. NaNO3 I

d. D2H5OH M

e. NH4Cl I

f. C6H12O6 M

If you need extra help distinguishing between the two types of compounds, this is very helpful: http://www.buzzle.com/articles/ionic-compounds-vs-molecular-compounds.html

The following section of the packet introduces a new topic to us, dissociation/hydration. We were supposed to find what the compound looked like after it had dissolved in water (after being broken up into individual ions by the water). The first few problems/answers are:

a. KF(s) -> K+(aq) + F-(aq)

b. CaCl2(s) -> Ca2+(aq) + 2Cl(aq)

c. Na2SO4(s) -> 2Na+(aq) + SO42-(aq)

One of the most difficult things to remember is that Uncle HONClBrIF does not apply to these ions because they're not considered to be "by themselves", but "with" water.

As should be obvious by now, all we are doing is splitting the compound into the ions of which it is made up. But then it gets a little more trickey; we are asked to perform the same type of equation, they give us the name and expect us to derive the formula from that. For example the first question asks what happens when you put lithium phosphate in water:

First, we figure out that the formula for lithium phosphate is Li3PO4

Next, we figure out that the ions of which it is mad up are Li+ and PO43-

Then, we balance the equation to get something which looks like this -- Li3PO4(s) -> 3Li+ + PO43-

After doing a couple more problems like this one, we moved onto yet another new topic. This is the topic of electrolytic (E) and non-electrolytic (NE) solutions. What makes a solution electrolytic is if it is created by ionic compounds, a cation and an anion.

Answers to the last section:

a. NaCl E

b. SO3 NE

c. C2H5OH NE

d. KF E

e. BaCl2 E

f. C12H22O11 NE

He demonstrated this concept through a conductivity test. If, when the wires were placed in the solution, it caused the lightbulb to light, then the solution is electrolytic. This also implies that it contains an ionic compound. He used a video (from the blog) to demonstrate this:

http://g.web.umkc.edu/gounevt/Animations/Animations211/StrongWeakNonElelytes.swf

After this, we moved on to Lab CR2, the Chemical Dropout Lab. We mixed various substances, observing whether or not they each created a precipitate. If a solid developed, or the mixture became cloudy, we knew that a precipitate had been formed.

This picture shows perfectly what precipitates look like.

Overall, today was a very intense day. We were introduced to many new concepts, but luckily Mr. Henderson is explaining them very well and we have many review tools at our fingertips.

Tuesday, November 9

As always, Mr. Henderson started off with a recap of what we did yesterday, and a preview of what will happen in the class. He told us to get out our lab notebooks and page 12 in our unit packets. We first went over what PPT meant in our lab from yesterday, called Chemical Drop Out. PPT means precipitate.

He then went and showed us a video on the Chemistry website, found here. It corresponds with question 5 on page 12. In the video, the first substance that was dissolved in the water was NaCl, sodium chloride. Mr. H explained that the "Mickey Mouses" (H2O) were connecting with sodium chloride. The positive of H2O, H, forms a bond with the negative Cl ion. Then, the O, the negative ion in H2O, bonds with the positive Na. This is an example of dissociated ions.




That is why the light bulb lighted in the video; it is an example of a strong electrolyte. He then explained the next two examples. The second example is called a weak electrolyte because there are more associated ions (ions that did not break off and make bonds with the water) than dissociated ions. This causes the light bulb to light very dimly. When methanol is added to the water, it does not dissociate into ions. This is called a nonelectrolyte.

For Question 6, we had to decide what type of electrolyte it was by the picture. The first picture is SE, strong electrolyte. Someone in the class explained that since there were no bonds and just molecules in the picture, it means that it completely dissociated and therefore is a strong electrolyte. The second picture is a non-electrolyte because there are no floating ions that are separated. Finally, the last one, you guessed it, is weak electrolyte. This is because there are some associated bonds, but there are other dissociated ions, too.

We moved onto question 7, now. We had to find the major species that would be in water by the compound given. He went through A, B, C, D, and F together with us. The answers are as follows:

A.) Compound: NH4Cl
Major Species: NH4+, Cl-, H2O
B.) Compound: C2H5OH
Major Species: C2H5OH, H2O
C.) Compound: H2SO4
Major Species: H+, SO4 2-, H2O
D.) Compound: NaC2H3O2
Major Species: Na+, C2H3O2, H2O
F.) Compound: HF
Major Species: HF, H2O
Minor Species: H+, F-

Note:

• All soluble ionic solids are strong electrolytes.
  
• Certain acids are strong electrolytes and goes as follows: HCl, HNO3, H2SO4, HBr, HI and HClO4
• Other acids are referred to as weak electrolytes.
• Molecular compounds held together by covalent bonds do NOT dissociate and are considered non-electrolytes.

Mr. H told us to refer to this for the WebAssign, due Wednesday, November 10.

PAGE 13

For 1a and 1b, refer to Hannah's blog on how to do them. For C and D, the steps are as follows to solve:

1. Get the formula of reactants
2. Write formula of product AND charges
3. The state that it is in (solid, gas, liquid, or aqueous)
   
4. Balance

The rules of solubility is written in the box on top of page 13.

LAB FROM YESTERDAY/PAGE 13

Mr. H made us take out our little data sheets from the lab, Chemical Drop Outs. He explained that each row (horizontal) had sodium (Na) and the columns (vertical) had nitrates (NO3). We had page 13 out, to refer to the box on the top. It has the solubility rules. For the rows, focus on the ANION, not cation (Na). Similarly, focus on the CATION, not anion (NO3). Mr. H had an example on the board, but he said that most carbonates are insoluble. Insoluble, meaning that there are more PPT's written in the boxes in the row/column. For the lab Conclusion/discussion, we had to write a sentence about each row and column. For example, for the row of Na2CO3, most of my group's boxes are written with PPT. So, this means that carbonate is insoluble. So, my sentence for that row would be carbonate is insoluble with exceptions of aluminum, ammonium, and calcium. The EXCEPT is there because not all of the reactants had a precipitate. Those would be the NR in the boxes. If a row or column has more NR's than PPT's, you would write that that row/column is soluble with exceptions of the boxes with PPT in it. Every group is different, which would result in different sentences.

Remember that the homework for tonight is a WebAssign and a Delicious bookmark!

Friday, November 5

Mr. Henderson started out the period by looking at the scribe from the previous day by Katie then began to work on problems #33, 34, and 35 on page #10.

*when working on these problems and those similar to them...
1) write the reactants
2) write the products
3) balance the equation

PAGE #10

33.Butane gas (C4H10) is burned in a gas lighter.

C4H10 + 13/2O2 ---> 4CO2 + 5H2O ... need to have coefficients that are whole numbers so multiply all coefficients by 2

C4H10 + 13O2 ---> 8CO2 + 10H2O

34.Zinc reacts with silver(I) nitrate in a single replacement reaction

Zn + 2AgNO3 ---> 2Ag + Zn(NO3)2

35.Fluorine reacts with sodium bromide in a single replacement reaction.

* UNCLE HONClBrIF*
F2 + 2NaBr ---> 2NaF + Br2

NOTE:
~Whenever something is burned, combustion occurs and the result is the oxide of the element.
~Hydrocarbons are most commonly burned.
>octane(car or bus ride to school)
>methane CH4(to heat home or cook food)
>propane C3H8(grill)
-result in carbon dioxide CO2 and H20(water)
~CO2 causes environmental problems
(greenhouse gas -> greenhouse affect -> global warming)
-sunlight can get through and heat the earth but cannot escape well

DEMO:
SOLUBLE SOLID-
Potassium iodide in beaker(white solid)
solution=solute (present in least amount)
water=solvent(present in greatest amount)
-end result is the solid is gone(dissolved in water)
KI(s) ---> KI(aq)

TO BE ABLE TO TELL IF A SOLUTE DISSOLVES...
1. boil water(about 30 minutes)
~result would be KI in solid state (white)
2. taste it(shouldn't just taste like water)

Copper Chloride
-dissolves in water(changes color to a greenish blue)

Lead Nitrate(heavy)
-dissolves in water(whitish liquid)
Pb(NO3)2(s) ---> Pb(NO3)2(aq)

ALL LIQUIDS MAKE THE LIGHT BULB LIGHT UP BECAUSE THE SOLIDS BROKE DOWN INTO CHARGED IONS.

SUGAR DOES NOT LIGHT THE LIGHT BULB BECAUSE NO IONS ARE PRESENT.

HW:
*TEST MOVED TO NEXT FRIDAY
*webassign due Monday
*delicious assignment probably due Wednesday(tags- hcp3y1011 unit4 first name, last initial. do not use commas to separate the tags just separate by spaces and keep the tags in this order.
*lab notebooks will be collected Friday