Thursday, January 6, 2011

Mr. H started the day by showing us the answers to Chapter 8.2 reading sheet. Next, we moved on to review the blog by Neil. He reviewed the exothermic and endothermic reactions, but added entholpy, which is a form of chemical energy. Entholpy is basically the total energy or heat and is symbolized by a H. In an exothermic reaction, the heat is always on the product side and ΔH is a negative value. In an endothermic reaction, the heat is always on the reactant side and ΔH is a positive value.

Next we moved onto a demo. Mr.H started it off by asking people to feel the beaker, the wood, and the water to prove that this wasn't a trick. He started it off by putting water on the wood. Next, he put the beaker on top of the wood. He then mixed two compounds together to create an endothermic reaction. The mixture inside the beaker pulled the heat from the water and caused it to freeze onto the wood.

After that, we moved on to pg 5 of our packets. We worked on #1 which required us to figure out what kind of reaction it was. The answers are..

a.EX b.EN c.EX d.EN e.EX f.EN g.EN h.EX i.EN j.EX k.EX l.EX

Next, Mr. H showed us this concept on the board.

If A → B + 100J

...then 100J + B → A

...then 2A → 2B + 200J

...then 300J + 3B → 3A

After learning this we moved on to work on problem 2 on pg 5. It required us to find the missing value. The answers are..

a. -2408kJ b. 1204kJ c. 1652kJ d.2.70kJ e. -13.5kJ

Next, we moved onto page 4 #4. This problem was to be a sample problem for our TC3 lab. The answers are

a.Q= -27692.5

b. (Determine the Moles of ice that melted) 3.9111

c. 7.08kJ/mol

Finally, we finished our day by finishing up our TC2 and TC3 labs.

Wednesday, Janurary 5, 2011

Today Mr. H started off class by complimenting Emma on the blog post that she had completed the previous night with a quick review of the previous day using her blog as a guide. To begin class Mr. H started off class by giving us the answers to the reading sheet on pg. 36, but really pages 23 and 24. The answers are:
1-A
2-B
3-A
4-A,E
5-B
6-A
7-False
8-D
9-D
10-A,C
11-B,C
12-B
13-B
14-A,B,C,D
15-False
16-False
17-B
18-B
19-B

As usual, Mr. H walked us through the reading sheet by answering any questions we had or highlighting the key points on the reading sheet. The main focus of the discussion was on endothermic and exothermic reactions. We also learned about the difference between the system and surroundings. Next, we began to work in the packet for the second day and it was pretty exciting. The answers are below in the picture.

The first few questions weren't difficult because they started off distinguishing between basic endothermic and exothermic reactions. The top half of this page discusses this topic. Endothermic reactions are where the system absorbs energy from the surroundings cooling down the surroundings and exothermic reactions release heat/energy from the system heating up the surroundings. The second half of the page was a little more complex because it involved distinguishing what graphs represent which type of reaction. Also, more real world examples were used to help us grasp the concept of endothermic and exothermic reactions. Next, we moved onto pg. 3 of the packet and we got to use our calculators for the first time this unit, SWEEEEET!!

The questions on the following page involved understanding a mathematical concept. The equation that was essential was:

q=m*c*delta-t
q=a measure of the heat flow
m=mass of the substance that you are trying to find the heat flow for
c=heat capacity of the substance
delta-t=change in temperature over the reaction of the substance

We began to work on the problem, which was going to help us work on the lab TC2 that we had begun the day before. We were trying to find the heat flow of water to later find the specific heat of iron. In the equation, m was substituted with the mass of the water which was 100g, c was substituted with the heat capacity of water which is 4.18 and chart with heat capacity numbers can be found in our book in ch. 8. The change in temperature was from 21.9 degrees Celsius to the highest value of 29.5. So the equation looked like this:

100*4.18*(29.5-21.9)=q

Once we found q which was 2758.8 joules were needed to calculate the kilojoules lost by iron in the reaction, so we were able to conclude that the amount of energy water gained was the amount of energy that iron and that was -2.7588 KILOJOULES. The following equation was derived to find out the specific heat or heat capacity of the iron.

C= q/m*delta-t

When the known variables were substituted the equation looked like this:

-2758.8 J/(6.52)*(28.5-86.6)=c

The answer was 7.41 J/g*c.

Next, Mr. H gave us the directions, title and purpose Lab TC3: Heat of Fusion Lab. The goal of the lab was to determine the molar heat of the fusion of ice. We came back to the front of the room for a little bit as Mr. H explained to us how to complete the calculations to the Lab TC2 and TC3. It made it a lot easier to do. Most people didn't finish Lab TC3 and Mr. H said we would finish up the next day from help from him. That was the end of the class and Mr. H reminded us of our Webassign that was due which was a reading sheet.

Tuesday, January 4, 2011

Mr. H began class by greeting us and explaining our agenda for the day. We immeditely were asked to get out our lab notebooks to write down the title and purpose of the TC2 Lab: Specific Heat of a Metal Lab. Purpose: To determine the specific heat of a sample of metal. After going over Kendall's blog we went to the back of the class room to begin the lab. Our lab groups split into two groups to complete to tasks. One group had to Prepare the Metal. Step 1: Measure the mass of a clean, dry, test tube. Record. Step 2: Empty metal (beads) into test tube. Step 3: Measure the mass of the test tube with the beads. Record. [Then subtract the mass of the test tube from the mass of the test tube with metal beads to find m the mass for your equation.] Step 4: Put the test tube into a hot water bath and heat water to about 80 degrees Celsius. The thermometer should be in the water (not the test tube.) Remember the 0th law. The other group had to Prepare the Water Calorimeter. Step 1: Obtain some cool/chilled water. Step 2: Place 50 mL into a dry styrofoam cup. After finishing our preparations we went back to the front of the classroom. Mr. H then reviewed with us our graphs from yesterday. He explained how after the system and surroundings met at thermal equilibrium their temperatures began to go down and would continue to go down indefinitely until the room temperature cooled and then the outdoor air would cool and then the universe would cool and we would all die. But that won't happen for billions of years so were alright. Next we went over Page 1 in our packets. First we were asked to state the zeroeth law of thermodynamics in our own words. I wrote two objects with a different temperature in contact with each other will transfer energy(heat) from high temp to low temp until they have the same temp. Then Alex explained what would happen if a hot piece of dishware was placed in a cold bowl of water. She said heat from the hot dishware would transfer to the cold water until they reached an equal temperature or equilibrium. Then we went over the chart at the bottom of page 1. We were given two situations ans asked to state the system and surroundings, the directional flow of heat, and what happened to the tempeatures (see picture left). Then we moved into the left side of our notebook. Mr. H went over an equation that pertains to our lab (see picture left). Then he went over exothermis process which pertains to our lab. Energy is lost from the system and transfered to the surroundings. The change of T is positive, the q of the surroundings is positive and the q of the system is negative. Next he went over endothermic process. The change of T is negative, the q of the surroundings is negative, and the q of the system is positive. Mr. H also noted through graphing that to being a chemical process there has to be a spark or actication. Finally we went back to the back of the room to finish the lab. We measured the temperatures of the hot water and the cold water then carefully put the metal into the cold water and measured the temperature until we got the highest temperature. Mr. H dismissed us. Our homework was Reading Webassign on 8.2.

Monday, January 3, 2011

We began class today by passing out our lab notebooks, old papers, as well as a graph and the new unit packet. The next unit is Thermochemistry. We then discussed the last test. The results were unfortunatley not so great and Mr. Henderson gave us the option to come in to review the test. As an opportunity to help our grades, there are bonus points if the Webassign due the 13th is completed before the due date. For the Webassign, questions 1-6 need to be completed by Friday January 7th; 7-10 by Monday January 10th; 11-15 by Tuesday January 11th; and 16-19 by Wednesday January 12th. Tonights homework is a Webassign Reading Sheet on 8.1 due Tuesday.
We then began a new lab (TC1) called Zeroing in On Heat Lab. The purpose is to describe and explain the temperature changes experience by two objects of different temperatures when place din close contact with one another. In order to do this, we marked the temperatures of hot water and cold water every 10 seconds. We made a graph and the times started at 0 and went to 190. The temperatures for hot water (in celcius) were 84.6, 81.6, 75.2, 69.1, 64.9, 62.9, 59.3, 56.0, 51.8, 51.8, 51.3, 50.8, 50.2, 50.0, 49.7, 49.7, 49.4, 49.2, 49.0, and 48.9. The temperature of the cold water was as follows: 7.3, 17.8, 24.7, 28.3, 32.7, 37.8, 39.3, 43.4, 46.1, 47.0, 47.9, 48.3, 49.5, 49.5, 49.7, 49.7, 49.4, 49.2, 49.0, and 48.9. We then graphed our results, which should look something like this:

There is a line through 140 because that is when the hot and cold temperatures were equal. We then made our lab conclusion, which described how the temperature changes over time for both sets of data. My results were that the temperatures are inversely proportional until 140 seconds have passed. The difference between the seconds decreases over time. At the end, the released the same temperature. This constitutes a heat flow (exchange of energy). The hot warms up the cold and continues until they are at the same temperature. The graph should be taped into the lab notebook. Mr. Henderson also mentioned that there is a quest on Wednesday January 12th.

Partial Pressures and Working through a Word Problem

To begin class, Mr. H asked about the webassign that would be due on Friday and whether or not we had any questions on it. The main question was on how to recognize what equations we should use on each problem because, unlike stoichiometry, all our gas equations are word problems that have about 5 key equations. These include:
P1V1=P2V2,
P1/n1=P2/n2,
P1/T1=P2/T2,
PV=nRT (R as a constant),
P1V1/T1=P2V2/T2.
All others can be derived from these core equations and concepts. It is also important to know that if P is divided by a variable, like temperature, then it varies directly with that variable (P/T). If P is multiplied by a variable, like volume, then it varies indirectly with that variable (PV). (look at pgs. 23-24)

After doing a few gas word problems on the white board, we continued on to pg.20. On pg. 20 was a question that seemed very difficult because it incorporated the mixing of two gases and their resulting pressure. We know from the start that the volume of H2 gas is 2L with a pressure of 1.2atm and the volume of the N2 gas is 1L with a pressure of 4.6atm. Temperature and moles stay constant. Although this equations looks very difficult, it is really just a P1V1=P2V2 equation. We begin by solving for 1 gas, H2. First, rearrange the equation using a bit of algebra and make it so that you are solving for P2, our unknown variable. Now, you should end up with P2=P1V1/V2. Next, simply plug in the information we are given to get P2= (1.2atm)(2L)/V2. Because we know that the new volume will be the old volumes added together, V2 is 3L. Now we solve and get that P2=.80atms. Repeat the same process for the N2 gas and you should end up with 1.53atms. Finally, to solve the last problem of the resulting pressure, just add together the two P2 values to get 2.33 atms. The equations should have a diagram that looked something like this:

This is the diagram from the webassign that was due on friday, so no you will not get a free answer!

Finally, we looked at pg. 19, which had similar partial pressure problems (although they are a bit easier). The key to these equations was to use your textbook to look up the vapor pressure of water at certain temperatures. This was given to us by Mr. H. He told us that the V.P. of H20 for problems 1 and 2 where 31.8 and 149 mmHg respectively. This left us with 868.3 mmHg and 151.02 mmHg for the O2 values of questions 1 and 2. Then, to solve parts 3+4 and Q 1-2, we put each pressure value over the total pressure. After converting each kPa pressure to mmHg, the correct answers for Parts 3-4 were 3.53% and 96.46% for Q1 and 49.66% and 50.34% for Q2. The rest of the problems could be solved the same way.

That was all for Thursday's class. Thank You and see you in class!