video analysis- rosalia

Logger Pro Analysis

Logger Pro- Analysis of the Free Fall experiment

We had analysed our free fall experiment with the programm logger pro. Our objective with this is to explain how is the experiment proceded and also to examine one of the videos we had recorded during the task, and with this, showing how the movement of the ball changes and how it is affected by it acceleration and by the height by which the marble is thrown.


MARÍA CASTRO

Conclusion

As we can see in our logger pro analysis, there are two graphs. The first one shows the velocity of the movement of the ball. It shows that as time is increasing, the velocity is decreasing (because when it goes down and it has the effect of gravity, it is negative), so this means that it is getting faster as the time increases. As we can see, the measurements are not very specific with the tendency line, but this is because the first point is when we had thrown the ball, so as the ball is thrown by a person, it is normal that in the first point its not so precised. The last two points of the graph, represent the time when the ball has reached the jelly surface, so it stops. On the graph below, we can see the movement of the ball, the place where it moves. So as the time increases, the ball goes deeper and down to reach the jelly surface that it stops the movement. If we compare our results with my hypothesis, the final velocity had decreased, but I thought in the hypothesis that it was going to increase. However, the velocity has decreased number, but it accelerated the speed. The reason why the numbers are negative is because it is falling down. Also,  the time of the marble to fall had changed, because the time changed the final velocity, which is the one that made the hole to be more deeper. More further is the ball form the surface, the bigger the hole would be, because it will last more time, and in that time, the velocity would decrease, so in this our hypothesis was correct. However, in our experiment we did not use plasticine, but we used the jelly solution, which was easier to measure the depth of the hole of the marble using this substance

Evaluation

I think that in our experiment we could have improved more things, because some data are not very précised. These could be because we could have thrown the ball at a non- précised distance, because maybe we could have made a mistake.  Another common mistake that we could have done is not preparing precisely the jelly solution, which it could have been denser, or less dense than it was expected to be. There is also a possibility that we could have not put the edge of the ruler on the exact place where it had to be, maybe we had put it some milimeters up or down as we expected. To solve this, I think that more people should revise that everything it’s in it correct place, and to make sure the measurements are précised, we could have done the experiment more than once. However, as well as the people, trying it more times to make sure the measurements are correct is also a good solution. Also, I think that with more time and concentration, the experiment would have been better done. 

Free Fall Experiment

Amparo Canto Rodríguez, María Castro Belluzzo and Rosalía Otero Naranjo 10B

Free Fall

How does the height from which the marble is dropped affect the penetration of the depth of the marble into the jelly solution?

Background Information:

A free-falling object is an object that falls because of the gravity force, so it is
in a state of free fall movement. The object has an acceleration of 9.8 m/s. Here there are the free fall/kinematic equations (Physicsclassroom.com, 2014):

In these formulas “d” represents distance, “v_i”   initial velocity and “v_f” final velocity, “t” is time and “a” is acceleration, which as I said before, it is 9.8 m/s².

There are some basic skills that we need to know: When the acceleration goes down, the number is negative (-9.8 m/s²) so a free fall object always has a negative acceleration. Acceleration and velocity are a vector, which is defined as the rate that its velocity changes. A vector is a quantity that describes the motion of objects; it marks the direction of which an object is moved or it can also be a magnitude. If the acceleration goes down, the vector goes down. If an object has been dropped from a certain height, the initial velocity (v_i) is 0 m/s. If we throw an object up straight vertical and then it returns down, the initial and final velocities are 0 m/s and when it goes up, the velocity is positive (for example 20m/s) and when it goes down the velocity is negative (for example -20m/s).

With these basic skills we can solve free fall problems using the equations in the diagram above.

In a free fall, we have the uniformly accelerated rectilinear movement (UARM) which is that in a rectilinear movement, like in free fall, the acceleration is constant. The velocity increases, but the acceleration in this case is gravity, so it doesn’t change.

Hypothesis:

If a ball is dropped from different heights, the other factors also change. Some of them stay the same, like gravity/acceleration which is always -9.8 m/s² in a free fall motion. In my experiment, the initial velocity (v_i) will also stay the same because the ball will be dropped from a specific height, so the initial velocity will always be 0 m/s. However, most of the factors do change. The final velocity (v_f) will be the one that will change the most because depending in the distance, the velocity increases or decreases. If there is more distance between the ball and the floor, the velocity will have more time to change and the gravity would give more pressure to the ball. If there is less distance, the final velocity will be less because there was less time for the ball to fall, and the gravity gave less pressure to it. Besides, time also changes a lot. If there is less distance, the ball has less time to reach the floor, and on the other hand, if there is more distance, the ball lasts more time to touch the floor. All of these hypotheses can be proved with this equation:

  

In our experiment (free fall) the acceleration stays the same in a rectilinear movement (UARM). The velocity and time change but the acceleration in this case is the same as gravity so it doesn’t hange.

If the ball has more distance, it will make a bigger hole in the plasticine surface. On the other hand, if the ball has less distance it will make a smaller hole in the plasticine surface. Depending on what material the surface is made, the hole can be deeper or not. For example, flour has smaller grains than sand or soil, so the holes would be deeper than the on the other surfaces. A plasticine surface is much harder and dense than flour, so the hole will be smaller. In my experiment I’m going to use plasticine. So if the ball is dropped from a higher space, the hole made in the plasticine would be deeper and bigger. On the other side, if the ball is dropped from a less height, the hole in the plasticine could be smaller; or maybe if the ball is dropped from a miserable distance, it wouldn’t make a very visible hole in the plasticine, or even not do it.

Variables:

-          Independent variable: The independent variable is the distance which is dropped the ball.

-          Dependent variable: The dependent variable will be how deep the hole in the plasticine surface is.

-          Controlled variable: The controlled variables which are the ones that will not change are the acceleration (-9.8 m/s^2)the initial velocity (0 m/s) , the ball (same mass, shape…) and the materials used in the experiment.

Materials:

•       Marble
•        jelly solution (250 mL H20+ 3,5g powder) 
• beaker
• scale
•  Ruler/meter
• Video camera
•   Logger Pro

Data:

3,50 grams of pouder and 250 ml of water

Height (cm)	Depth of the plasticine (cm)
10	1
20	1,9
30	2,5
40	2,9
50	3,2
60	3,5
70	5,1

Method:

1. First, we write down in the notebook the vi and the acceleration, which are always the same.
2. After that, we prepare the jelly solution, adding 250 mL of water to 3,5 grams of powder.

 

1. 3.     After that, we measure the distance from the jelly solution surface and the ball, so the height must be the exact one that I have chosen. The measures between the jelly solution and the ball have to be written on the table of the data (so its what we did). The measures have a difference of 10 cm between each other.

1. 4.      Continually, we drop the marble and when it touches the surface of the jelly solution, we have to measure the depth of the hole made in this solution, with the ruler.
   5.      At the same time, take pictures of it with the video camera.
   6.      Then we write it down in the table. 
   7.      Finally, we do the same with the other measures and we compare the hypothesis with the results.

LAB VISIT 5: CRYOSCOPY

CRYOSCOPY(freezing point)

Objective:

With this experiment our aim is obtaining a graph in which we will show the variation of the melting point of an aqueous solution as a function of its concentration and, also, calculating the molecular mass (mm) of one of the solute from the results that we will obtain.

Background information:

For understanding and developing this experiment we may know that the freezing point of a solution is without exception lower than the one of the pure solvent, with the proportional drop to the molality of the solution:

ATf = Kf * m

Solvent = 5 grams of water.

Materials:

• Distilled water (H20)
• Sugar (C12H22O11)
• Test tubes
• A digital thermometer
• Ice
• Salt (NaCl)
• Balance
• 250 mL beaker

Table:

Grams of sugar	Tf (ºC)	ATf (ºC)	Molality (mol/kg)
0.0	0.6	-----------------------	0.0
0.5	0.3	0.3	0.29
1.0	-1.9	2.5	0.58
1.5	-2.5	3.1	0.87
2.0	-4.5	5.1	1.16
2.5	-6.8	7.4	1.46

Calculations (molecular mass, moles and molality):

Sucrose : C12 H22 O11

Carbon: 12  Hydrogen: 1 Oxygen: 16

Molecular Mass=  12 (12) + 1 (22) + 16 (11) = 144 + 22 + 176 = 342 g/mol

Moles: mass/molecular mass

Molality = moles / kg of solvent

0.5 grams: (0.5 : 342)/ 0.005 = 0.001/ 0.005 = 0.29

1.0 grams: (1.0 : 342)/ 0.005 = 0.002/ 0.005 = 0.58

1.5 grams: (1.5 : 342)/ 0.005 = 0.004/ 0.005 = 0.87

2.0 grams: (2.0 : 342)/ 0.005 = 0.005/ 0.005 = 1.16

2.5 grams: (2.5 : 342)/ 0.005 = 0.007/ 0.005 = 1.46

Graphs:

Conclusion:

As we can see on the table, as the amount of sugar (in grams) increase, the freezing point is colder, so you need a colder temperature to freeze the solution. As we lower the temperature, the molecules have less energy to move. Adding sugar makes it harder to slow the molecules down, so the temperature has to be lower than pure water to freeze. Adding more sugar it takes more energy out. The sugar slow down the molecules and the temperature is lower than water. This way it makes more energy going out. We can see that in the first graph the results of the molality is increasing, when mass and molality increase relatively while in the second graph we can see how mass increase and the freezing point decrease. The second one is inversely proportional. Molality is higher because its related with the sugar, not with the temperature. In the molality column, we can see that as the amount of sugar increases, the molality increases, because in the formula the number of moles changed depending on the amount of sugar, that in this case was decreasing. In the other graph, its inversely proportional because its the relation of the grams of sugar and the freezing point, making it descent because as we first said, the sugar makes the freezing point be lower.

Lab session 3-2-2014

LAB SESSION 3-2-2014
PRECISION MEASUREMENTS MASS VS. VOLUME

Materials:
- Pipette
- Measuring Cylinder
- Test Tubes
- Test tube holder
- Scale
- Beaker
- Spatula
- NaCl
- Water
- Cyclohexane

PROCEDURE:

What happened? Why?
A) 10,0 mL of H2O have a mass of 10,0 grams.
B) 2,5 grams of NaCl have a volume of 1,0 - 1,5 mL.
C) 2,5 grams of NaCl in 10 mL of water shows a volume of 11,0 mL.

A) EXPERIMENT 1: 10,0 mL of H2O have a mass of 10,0 grams.

1) First, we measured the wieght of the empty measuring cylinder, which was 76,3
2) Then, we pour 10 mL of water with the pipette in the mesauring cylinder.
3) After that, we weight the measuring cylinder with the water in it.
4) Finally, we calculated the difference between the mass of the measuring cylinder with water and the mass of the empty measuring cylinder.

In our case, first we did it but the difference was 6,9 grams. The mass of the measuring cylinder with water in it was 83,2 grams and the measuring cylinder without water weight 76,3 grams. So the difference, that as firstly said was 6,9 grams was not correct because logically it needed to be 10 grams. So we had to do it again and on the second time, the mass without water was of 76,5 grams. The mass of the measuring cylinder with water was of 86,4 grams. This time, the difference was of 9,9 grams.

First time : 83, 2 - 76, 3 = 6, 9
Second time : 86, 4 - 76, 5 = 6, 9

EXPERIMENT 2:

Then, we did the same experiment but instead of water we used salt (NaCl)
1) This time our measuring cylinder weight 70, 8 grams.
2) Then we put 2,5 grams of NaCl in the measuring cylinder.
3) After that, we measured the mass of the measuring cylinder with the salt (NaCl) and weighted 73,4 grams.
4) Finally we calculated the difference (73,4 - 2,5) that was 2,6 grams, so that meant that the experiment was well done because the difference was very approximated to the grams of salt that we put in the measuring cylinder.

73, 4 - 70, 8 = 2, 6

B) 2,5 grams of NaCl have a volume of 1,0 - 1,5 mL.
In this experiment we used NaCl mixed with cylohexane (C6H12)
1) First, we took 2,5 grams of NaCl (measuring it first on the balance) and pour it on the measuring cylinder that weight 70,8 grams
2) Then, we pour 3,0 mL of cyclohexane (C6H12), measuring it first with the measuring pipette.
3) Finally, we calculated the final volume that was 3,0 mL.

In this experiment, the cyclohexane and the salt couldn't dissolve.

C) 2,5 grams of NaCl in 10 mL of water shows a volume of 11,0 mL.
In this third experiment also used NaCl mixed with water.
1) We put 2,5 grams of NaCl in the measuring cylinder.
2) Pour 10,0 mL of H2O.
3) The final volume was of 12,5 mL
4) The total mass was of 75, 6 grams.
In this experiment, the water and the salt could dissolve.


CONCLUSIONS:
In our results, we could prove that water could dissolve with sodium chloride (NaCl) but cyclohexane couldn't dissolve with NaCl. Water could dissolve with the salt (NaCl) because they have a positive and negative, so they are both polar substances, so they are ionic compounds. Polar substances attract each other, so in this case, they are easier to dissolve. On the other side, we could prove that hexane could not dissolve with NaCl because they are not polar substances, so they are not negative and positive charges. NaCl and cyclohexane are covalent bonds so its hard to dissolve them.
In this experiment, we have seen different changes in the volume; sometimes the volume stayed the same, however we had difficulties to prove that it had to be the same, for example in water. In the second experiment, the volume of the cyclohexane is 3,0 mL. When we mixed water with NaCl the final volume was of 12,5 because the amount of water, that first was 10,0 mL had dissolved with NaCl that wa 2,5 grams (the same as 2,5 mL) so the total volume was of 12,5. In the other case, cyclohexane and NaCl could not dissolve so it just rested the amount of cyclohexane, that was of 3,0 mL.

In this experiment, the measurements are not very accurate so this is because we could have had some problems during the experiment, such as not calculate well the volumes, or not do it very precisely so some measures were not very correct, as we had proved in the first experiment. However, they are realistic values and we proved them. To improve it we could do it more precisely with more time, concentration and ability. We could also impriove it by taking more pictures, that can complete our lab report.

Seeing the characteristics of Butyl acetate

Here is the bottle of Butyl acetate

Butyl acetate has the following formula: C6H12O6

It´s atomic mass therefore is the following: C= 12 H=1 O=16

Molecular mass= 72+12+96=180g/mol (A mol is 6.02x10^23 particles)

Here is it´s atomic model:

 Blue: Oxygen
White: Hydrogen
Black: Carbon

Pressure and Temperature relation

In this experiment we have seen Gay-Lussac´s law comparing pressure at different temperatures. As we may recall pressure and volume are proportionally related.

Initial pressure/Initial temperature=Final pressure/Final temperature

We did the experiment with logger pro and took the result of the pressure

Results with logger pro (KiloPascals)

15º C: 6,3 kPa
20ºC : 10,1 kPa
25ºC: 15,2 kPa
30ºC: 21.7 kPa

Table:

Temperature (ºC)	Pressure (KPa)
15.0	6.3
20.0	10.1
25.0	15.2
30.0	21.7

Graph:

CONCLUSION
As we can see in our results, as the temperature increases, the pressure also increases. Gay Lussac's law states that when there is more temperature, higher is the pressure because the particles move faster. This is because the temperature give energy to the partices, so when the temperature is higher, it gives ore energy to the partciles to move faster, causing more pressure.

Still, this results are not very accurate since when we took of the vacum pressure started increasing cause air was entering due to the fact that our equipment is not perfect. To make sure the changes in pressure were not only for air entering after doing all 4 temperatures we put it back in 15ºC and it dropped down to 10,5 kPa which shows that Temperature and pressure are directly related.
Here are some images from the experiment:

 Here we are checking pressure at 30ºC with Maria´s bath. Data are being taken to the Logger Pro program

This is the material we used to take out the air so our initial pressure was 0 torrents