Thursday, July 16, 2015

Prompt: Post a picture of your titration setup, and a photo of your analyte at the equivalence point. Write a short summary of your procedure. Calculate the percent ionization of vinegar and describe why it is such a low number.

Titration Set-Up



Equivalence Point:


Summary: In this lab we were supposed to determine at the percent ionization of vinegar. To do so we first cleaned the burette with NaOH to ensure there was no leftover substance from a prior experiment. We then measured 7 mL of vinegar into our flask and added 4 drops of phenolphthalein. We then began to slowly add drops of NaOH. We knew it had reached the equivalence point when the solution remained a light pink color. We ended up having to do the experiment twice to try to get a more accurate result the second time than we had the first. Our second attempt turned out to be significantly more accurate, and in that trial we ended up adding 23.3 mL of NaOH.

Percent Ionization: .45%
Description: Vinegar does not ionize well.

Wednesday, July 15, 2015

Lab 13B

Prompt: Please read the rubric! It is how you will be graded! Include everything listed on the front page of the lab in the manual. Add a photo, too! :) You will rip off the pre-lab and hand it in separately.

Introduction: The purpose of this lab was to identify a solid substance just by using its solubility. This was done by using a graph of solubility's of a few options of what the substance could be, and then testing the solid in water under various conditions. Solubility is the maximum amount of grams of a solute that can be dissolved in a constant amount of solvent without becoming saturated. Saturated is the maximum amount of a solvent that can dissolve of a solute where if it tried to dissolve more a solid would form.

Procedure: To begin with we were told it was one of three salts(NaCl, NaNO3, KNO3) and we were given a graph of the solubility of each of the salts. We began by making a hot water bath at around 30 degrees celsius. We then heated a small beaker of 10 mL of water in the bath and began attempting to dissolve 5 grams of our mystery solid into the 10 mL. We were unable to do so at that temperature, so according to our graph it could not have been NaNO3. We then further heated it to around 50 degrees celsius and the solute dissolved into the water, so we were able to determine that is was KNO3.

Data:
At 30 degrees celsius we were unable to dissolve 5 grams of the salt into 10 mL of water.
At 45 degrees celsius we were able to completely dissolve 5 grams of the salt into 10 mL of water.



Discussion/ Concluding Summary: We identified the substance to be KNO3. We determined this by eliminating the other options. First we added 5 g. of the salt to 10 mL of water at 30 degrees celsius. We knew if it were NaNO3 it would easily be able to dissolve with no solid remaining. However the solution had solid remaining so we knew we could eliminate NaNO3. Next we increased the temperature to around 45 degrees celsius but kept the amount of solute and solvent the same. We knew if the solid was dissolved it would be KNO3 but if a solid was still remaining it would be NaCl. The solid dissolved so we knew it was KNO3. Solubility in solids increases with temperature because the liquid has more energy and it is easier to displace the solute and dissolve it.

Tuesday, July 14, 2015

Lab 12

Prompt: For each case I-IV, describe how you setup the simulator to collect data. Attach your spreadsheet with your data collected and graphs for each experiment. Include answers to questions #3-4a-d in your post.

Setup:
  I: For this case we adjusted the volume, or the size of the box, and the pressure, or the rate of collisions in the box. We did this by changing the volume of the box and collating data on how that changed the pressure. This helped to determine the relationship of volume to pressure as we could see how one responded to the others change.

II: For this case we adjusted the volume, or the size of the box, and the temperature in the box. To do this we changed the temperature and took data on the change in volume as a result and vice versa. This helped to determine the relationship between volume and temperature and how the one responded if the other was changed.

III: For this case we adjusted the temperature inside the box and the pressure inside the box. Pressure is the rate of collisions of particles so to test the pressure we adjusted the temperature and collected data on the results.

IV: For this case we tested the relationship of moles and volume. To do so we changed the number of particles to determine the change in volume and vice verse. This helped to determine the relationship between moles and volume


Questions:
#3:



#4a:
Due to Gay-Lussac's Law, when temperature is reduced pressure is also reduced

#4b:
According to Gay-Lussacs' Law, when the temperature is increased pressure also increases. If the pressure increases inside a confined area, such as a pop can, it would need to escape somehow(i.e. explode).

#4c:
The pressure will be reduced. Nothing will happen with the volume, however, as it is a rigid container.

#4d:
The moist heat will not help as pressure increases with heat, so adding heat to the abscess will increase pressure instead of the intend lessening.

Data Graphs:









Monday, July 13, 2015

Lab 11B

Prompt: Post a brief summary of the lab. Include images of your data tables. Answer the four questions listed in your lab manual.

Summary: The goal of this lab was to determine the number of Calories in a variety of different foods. The foods provided in this were a cheese puff, a pecan, and a cashew. My partner and I, however, were not able to test the cheese puff due to time constraints. We were able to determine number of Calories in our two foods by burning them over water and testing the temperature change of before and after the heating. This determines Calories as a Calorie is the amount of energy it takes to heat 1 gram 1 degree.

Questions:

#1: We measured a temperature change in the water.

#2: We measure the amount of energy gained by the water to determine the amount of energy released by the food

#3: It's absorbed by the air or the tin can.

#4: I was surprised that such small pieces of food could actually have an impact on the water and that they were able to raise the temperature that significantly.



Lab 11A

PromptPost a brief summary of the lab and how you were able to identify your unknown metal. Indicate your calculated specific heats for the metal. Provide the identification of the metal you tested.

Summary: The goal of the lab was to determine metal by finding the specific heat. This was done by measuring the temperature change of the rock in boiling water to that same rock in room temperature water and then calculating the specific heat. This helped us to become more comfortable identifying and working with specific heat.



Data
Mass of styrofoam cup: 7.90 g
Mass of cup + water: 300.75 g
Mass of water: 292.85 g
Initial temp of water in cup: 20.5 degrees celsius
Initial temp of metal in boiling water: 96.5 degrees celsius
Final temp of metal and water: 26.5
Mass of metal alone: 249.52

My partner and I concluded that the metal we worked with was steel.

Sunday, July 12, 2015

Lab 10

Prompt: In your post, include an image of your pre-lab table. Include an image of your data table. Answer questions number 2-4 from your lab manual in your post.

Pre-Lab Table

Data Table

#2: Different intermolecular forces and different molar masses contributed to the varying differences in temperature. All of the substances had at least one hydrogen bond, but they each had different molar masses. If a substance has a high molar mass it is harder to for it to evaporate. Water and Glycerin were slightly different in that they respectively had two and three hydrogen bonding sites which made it even more difficult to bond.


#3: Methanol has a molar mass of 32.042 and Ethanol has a molar mass of 46.069. Methanol's change in temperature was 17.9 degrees while Ethanol's was 12 degrees. Ethanol has a lower change due to Ethanol having a higher molar mass. A higher IMF requires more energy so it would take longer to evaporate, limiting the amount the temperature would have dropped.

#4: OH- bonds create hydrogen bonds, which are the stronger than Dipoles and LDF's. They increase the IMF of the compound, which means it would take longer for the substance to evaporate.

Wednesday, July 8, 2015

Lab 7

Prompt: On your blog post, include a brief summary of the lab, the answers to the pre-lab questions from your manual, a photo of one of the flames you test, the identity of the two unknown flames, and how you know the identity.

Summary: The purpose of this lab was to identify chemicals based solely on the reaction that occurs when they are heated. We did this by testing various known chemicals and then trying to determine unknown chemicals based on our observation of known ones.

Pre-lab questions:
1. Excited electrons have absorbed energy and jumped to a higher energy level.
2. "Emit" means to give off or produce.
  3. In this experiment, atoms get their excess energy from the high temperatures.
4. Different atoms give off different colors of light due to their different chemical composition.
5.It is necessary to clean the wires before each test to make sure they are not contaminated with the previous chemical that was tested.

Unknown Chemical:
1.LiCl
2. KCl

We determined the unknown chemicals by comparing how they reacted with the heat to the chemicals we had already tested.The unknown chemicals behaved very similarly to ones that had already been tested so it wasn't too difficult telling which was which.