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