Photography for sale to fund education

I am selling my photography to help fund my education. I am currently a pre-med student and hope to be going to medical school soon. I am not asking for money for nothing, but I am offering my photography (if you would like to just give money, I won't refuse but I was raised that anything given should receive something in return). I have my photography for sale at http://gmockler1.deviantart.com. I know the name of the site is not that attractive, but it works for the purposes I need it for. There should be plenty of photography for any of your needs. If the photo you want is not in the size you would like, please feel free to contact me and we will work something out.

Rutherford's Gold Foil Experiment

While reading The Making of the Atomic Bomb by Richard Rhodes, I came across the gold foil experiment done by Rutherford in the early 1900’s that is currently used to explain the current model of the atom. Having run across this experiment multiple times in my study and readings I feel it is time to address this concept.

Rutherford and his assistants noticed that when alpha particles are fired at a piece of gold foil the will be deflected by two degrees on photographic paper. They also noticed that some of the alpha particles were shot directly back at the alpha emitter. With this evidence, Rutherford spent a year thinking about the results and experimented with electromagnets suspended so as to make a pendulum to come up with the theory that an atom contains a nucleus of charged particles that repel (like magnets) the alpha particles (which were somehow found to be positive).

Seeing as how the alpha emitter is stationary, it would seem logical that the particles emitted from it would travel in a straight line. Assuming minimal motion from the atoms within the gold foil (it is postulated that solids are still moving but the atoms stay in a relatively fixed position), the alpha particles would be deflected directly back due to the particles having the same electrical charge. We can envision this as a stationary gun shooting at a vibrating target. While the gun is being shot it does not move or alter its trajectory. The vibrating target will deflect the bullets from the gun. These deflected bullets will be shot back towards the gun with very few passing the target to the background medium.

The random few deflections that occurred in Rutherford’s experiment does show that there is a lot of space that allows particles to pass through. What it does not show is that there is a positive centre that deflects these alpha particles back towards the emitter. If we assume that physics for everything in the natural world occurs the same, the gun and target analogy would prove Rutherford’s experiment to be invalid. Even with electromagnetic pendulums being release from the exact same point with exact same trajectory, as Rutherford tried, the deflections would not occur two degrees behind the target but would be deflected back towards the emitter. Without performing this experiment myself, I cannot say whether or not the deflect particles returned in random patterns or if they struck the same point each time.

An experiment could possibly be devised where a recording medium is placed enough of a distance that the emitted particles would be deflected as Rutherford proposed- two degrees to either side of the atom- such that these deflections produce an enlarged image (enough so it can be seen through a microscope, at the very least). If the nucleus is positive and deflects two degrees on either side of the nucleus than this should produce a blank spot on the recording medium, thereby confirming Rutherford’s theory.

An argument that some have proposed is that quantum physics is different than classical physics. This surprises me since many physicists strive to find the “Theory of Everything”. If the macro world works differently than the atomic world, then they will never achieve this “Theory of Everything”. Some may agree with me and many may not, but I feel that this difference in physics indicates that there is a mistake somewhere along the line. Maybe an experiment has been interpretted incorrectly or something was missed. Maybe there was an experimental error that was not taken into consideration.

We must not take for granted that everything done in the past is absolutely correct. We must be willing to revise and revisit these concepts and see if they hold up with current technology and thinking. We must perform these experiments with an open mind, neither trying to prove or disprove lest we skew our results.

Chemistry: Simplicity v Complexity and Moles part 2

Here is a more complicated example that we will use to show both the "traditional" way of solving and then a more simplified way of solving. You will see the experiment as it unfolds.
We are given this problem: "What volume of carbon monoxide gas at STP is produced from 6.14 g calcium phosphate and 8.24 g silicone dioxide according to the following equation:

2Ca3(PO4)2 + 6SiO2 + 10C --> P4 + 6CaSiO3 + 10CO"

The traditional way of solving this equation would most likely use the ideal gas law equation PV=nRT (P= pressure, V= volume, n= the number of moles, R= a constant of .o8206 L-atm/mol-K, and T= temperature). Since we are told that our gas is at STP we know that our temperature is 0 degres Celsius (or 273 K [we need this since our equation use Kelvins instead of Celsius]) and our pressure is 1 atm (we can leave this value alone since we see it used within our equation).
Now, let us look at what values we need to solve for. Here is our equation again PV=nRT. We have pressure (1 atm), we have R because it is our constant (.08206 L-atm/mol-K), and we have our temperature in Kelvins (273 K). The only two variables that we are missing are our moles and pressure. We cannot get our moles of carbon monoxide until we plug in our other values.


6.14 g Ca3(PO4)2 (1 mol/310.181522 g)= 0.0198 mol Ca3(PO4)2

0.0198 mol Ca3(PO4)2 (10 mol CO/ 2 mol Ca3(PO4)2)=0.099 mol CO

8.24 g SiO2 (1 mol/ 60.0843 g)=0.137 mol SiO2

0.137 mol SiO2 (10 mol CO/ 6 mol SiO2)=0.229 mol CO


I think we need to find out which molecule is our limiting reactant. I should have done this first since we had two reactants with their respective weights. We can see that Ca3(PO4)2 produces less CO than SiO2 which means that Ca3(PO4)2 is our limiting reactant.
Now we use the ideal gas formula.

PV=nRT or since we are solving for volume we can use V=R(nT/P)

1 atm(V)=0.099 mol CO(0.08206 L-atm/mol-K)(273 K) solve algebraically

V=2.22 L CO

Earlier I plugged this into a chemistry calculator from http://www.chembuddy.com and came up with 2.21 L CO so this checks out.
This was quite the ordeal to figure out and took me roughly two hours and refering to two different textbooks to figure out how to solve this. I am sure you chemists out there would have found this somewhat easy to figure out.

Now that we have done this the traditional way, I would like to try my experimental way. I have yet to work out a problem like this using percentages versus moles and then converting to volume. Let us experiment!

We are solving for carbon monoxide in this equation and because we are using percentages and the law of conservation of mass we only need to solve for carbon and oxygen. Once again our balanced equation is

2 Ca3(PO4)2 + 6 SiO2 + 10 C --> P4 + 6 CaSiO3 + 10 CO

Assuming that the atomic weights are correct we figure out the percentage of oxygen in each formula and then make sure we have the same amount on the product side of our equation.

Ca3(PO4)2 weighs 310.181522 g so we divide the weight of eight oxygen to figure out the percent oxygen, 3 calcium for percent Ca, and 2 phosphorous for percent P.

15.9994(8)/310.181522= 41.3% oxygen

40.078(3)/310.181522= 38.8% calcium

which means that phosphorous should be 19.9% of the molecular weight.

SiO2 weighs 60.0843 g and we divide the weight of two oxygen to figure out the percent of oxygen and one silicon for the percent silicon.

15.9994(2)/60.0843= 53.3%

which means that silicon takes up 46.7% of our molecule.

Since we are given the weights of two reactants that contain oxygen we can find the weights of each of the elements in each formula.

6.14 g Ca3(PO4)2 will contain 2.54 g oxygen, 2.38 g Ca, and 1.22 g P

8.24 g SiO2 will contain 4.39 g oxygen and 3.85 g Si.

According to the law of conservation of mass, we should have a total of 6.93 g of oxygen, 2.38 g Ca, 1.22 g P, and 3.85 g Si on the product side. Since carbon is all by itself we can assume that it will weigh 12.01 g per unit which will mean that we should have 12.01 g C on the product side of our equation.
We can neglect phosphorous and just figure out what the weight of oxygen is in CaSiO3 by taking the weights of Ca and Si from the reactant side, multiplying them by their respective proportions, and adding them together and then figuring out their respective percentages of weights to solve for oxygen. Actually, all that stuff above was just for fun. We only really needed to figure out the final mass of oxygen and then figure out its percentage within CO and the rest of the equation.
We notice that our carbon monoxide contains ten of the twenty-eight oxygen atoms that we have. That comes out to be roughly 36% (35.7%)oxygen. So, let us multiply 6.93 g oxygen by 35.7% to figure out the weight of oxygen in our carbon monoxide molecule.

6.93 g (35.7)=2.47 g oxygen

Now that we have this weight we can figure out what percentage of the carbon monoxide molecule is made up of oxygen.

15.9994 g O/(28.0101 g CO)= 57.1% O which means that carbon makes up 42.9% of
the weight of carbon monoxide.

Let us use proportions to figure out the weight of carbon in this particular case.

2.47/57.1= x/42.9

x= 1.86 g C

Let us check to make sure this is correct by dividing the respective weights of our molecules into the total weight of the molecule.

2.47/4.33= 57.0% (57.0438%)(this is within experimental error)

1.86/4.33= 43.0% (42.956%)(this is also within experimental error)

It is safe to assume that 2 Ca3(PO4)2 + 6 SiO2 + 10 C --> P4 + 6 CaSiO3 + 10 CO with the respective weights of 6.14 g calcium phosphate and 8.24 g silicone dioxide produces 4.33 g CO.
Now comes the challenging part: changing weight into volume. We need to find a formula that use both weight and volume to figure out what our volume will be. I think our density formula would work nicely for this: density= mass/volume, but we need it modified for our purposes. To solve for volume we will use volume= mass/density.
The density of carbon monoxide if we look at wikipedia (http://en.wikipedia.org/wiki/Carbon_monoxide) is 1.250 g/L at STP. Now, we just plug in our known values into our equation.

Volume= 4.33 g CO/1.250 g per L (grams will cancel out leaving us with
litres)
Volume= 3.46 L

This is a rather interesting result. I think the only way to verify the results is to actually perform this chemical reaction.
I may have made a mathmatical error somewhere, but I feel this is fairly accurate considering that we used the law of conservation of mass. As long as the law of conservation of mass and the atomic weights are correct, this should be a correct result. We may lose some mass in the form of energy. The final result may actually be somewhere between the 2.21 L that moles and the calculator gave us and the 3.46 L that using percentages and the law of conservation of mass.

Mockler Manuever: A Method to Get Rid of Hiccups

I have found a technique that works really well when my students have hiccups. It is based off the theory that hiccups are caused by the diaphragm experiencing a muscle spasming and that pressure will cause the muscle to relax.

Just a word of caution: Due to the placement of the xiphoid process, anyone attempting this should use caution. If there is any question about how to do this, leave it to a physician. Also, I am not a licensed physician and am providing this information only so that it can be investigated further by researchers and physicians. I cannot be held liable for your actions should you harm another person while attempting this.

As I said before, this works really well on my dance students when they have hiccups in class. I came up with this idea based off the fact that a muscle can be released by pressure and that some hiccups are caused by the diaphragm moving in the opposite direction than it should while breathing.

Placing two fingers on either side of the linea alba (usually the thumb and first finger) just medially to either the 6th or 7th rib, you press in and up to get under the rib and into the diaphragm. With the other hand, you press on the posterior side of the thoracic cavity at the same level. You will want to hold this for at least ten seconds if not longer (especially if the hiccups are spaced at greater intervals). Occasionally, I have felt the diaphragm trying to initiate another hiccup but it is of lesser magnitude and I can release the patient (dancer) just after that.

I would like to see if this works with severe cases of hiccups, and I hope that it can be useful to health care professionals in a medical setting.

Chemistry: Simple v Complex and Moles

It has been my understanding that science has been all about explaning the world around us in the simplest of terms so that we can all understand it. Personally, I love biology and it follows this formula. Life is complex but this complexity is made up of simple systems.
First, I must say that I am not a genius but neither am I a simpleton. All throughout my college education I received A's and when I didn't get an A in the class it was a B because I just didn't try. I have a graduate degree and am currently back in school pursuing medicine. That being said, let me start my rant.
As I mentioned earlier, life is complex. No one will disagree with that, but life, when broken down, is made of many simple components all put together to form something complex. Chemistry just does not fit into this category. Maybe it has been the professors that I have had or the books I have read, but chemistry seems to start extremely complex only to become more complex. Also, once again it might be my professors and/or the books I've been assigned, it seems that in the 200+ years that chemistry has been around that no one has challenged the interpretations and beliefs since they have been made. Finally, a lot of chemistry cannot be observed, and if I remember correctly, science is all about observation and what can be observed. It just seems that there is a lot of conjecture, "interpretation" of experiments, and unobservable phenomena.
According to the Law of Conservation of Mass, "Mass cannot be created/destroyed, although it may be rearranged in space, and changed into different types of particles. This implies that for any chemical process in a closed system, the mass of the reactants must equal the mass of the products." Thanks Wikipedia! This brings me to my first major issue with chemistry besides the complexity issue and the fact that it seems chemistry has not been challenged only expanded upon for 200+ years.
I understand that a mole is the "collection of Avogadro's number of objects." That's from my chemistry textbook. (We won't go into my rant about Avogadro's number today) Basically, the way I understand a mole is that it is one "unit" of whatever object that is there. Let's say I have sodium chloride (NaCl). It, in and of itself, is one mole, but it is also composed of one mole of Na and one mole of Cl. That being said, let's go back to the conservation of mass and how chemists compute weights of reactants and products.
Using the example of salt: NaCl we have a chemical formula of Na + Cl --> NaCl. One mole of Na + one mole of Cl produces one mole of sodium chloride. We all have that and understand that. Now, let's say that I have 39.0 grams of NaCl, how many grams of Na do I need to complete this reaction. Chemists at this stage would find out the molar mass of the 39.0 grams of NaCl and then multiply it by the ratio of the reactants in the formula.

39.0 grams NaCl (1 mol NaCl/ 49.45 grams)= 0.789 mol NaCl

0.789 mol NaCl (1 mol Na/1 mol NaCl)= 0.789 mol Na

0.789 mol Na (14.00 grams/ 1 mol Na)= 11.0 grams Na

If you were able to follow that, what it tells us is that we will need 11.0 grams (three significant figures [which we won't go into in this discussion]) of Na to make 39.0 grams NaCl. That was a pretty easy example.
Now, if the Law of Conservation of Mass is correct, we shouldn't need to convert to moles to figure out the weight of Na needed to get 39.0 g of NaCl. In theory, we should be able to use the percent of the weight Na takes up in NaCl.

NaCl (if atomic weights are correct) weighs approximately 49.45 g. Na weighs 14.00 g.

14.00 g Na/ 49.45 g NaCl= 28.3%

28.3% (39.0 g)= 11.0 g

Now wasn't that so much easier? I realize that this is a relatively simple example but it proves my point. Why do we need to convert everything to moles? Most students do not even understand what a mole is. It is difficult to visualize what a mole is and how it can be used.
I have yet to test this theory on molarities and volumes but I am pretty sure that it will work there. If you are a chemistry person, test it out and let me know how it goes.