Month: September 2016

How to solve oxidation-reduction reactions

How to solve oxidation-reduction reactions

09/21/16

“How can we solve these complex oxidation-reduction reactions?”

Oxidation reduction reactions can be quite difficult to grasp for many new students in Chemistry due to the long process often involved. However, I will show you how even through working through these problems might be convoluted at first, they are actually very intuitive if you think about how it works from a systematic point of view.

Let us start with an example oxidation-reduction reaction, CrO4 -2 + I- → I2 + Cr +3. The first thing we must do is split the problem into two half reaction, I- → I2 and CrO4 -2 → Cr+3. Now that we have isolated these problems into separate forms, we now have to balance the equation with the exception of hydrogen, oxygen, and charges! (Which we will come to later). So our new half reactions will be 2I- → I2 and CrO4 -2 → Cr+3. Now, you are probably curious about how oxygen is present on the left side of the second equation but not on the right! But before we give up, it is possible to add hydride ions to both sides. So what we do is take hydride, add enough to balance out the charge of the oxygen, and then add the corresponding amount of H2O on the other side. Now that we have done this, we have to account for the extra charges on both side. To solve this, we simply add extra electrons to the more positive sides of the reactions. To illustrate, the first equation will become 2I- → I2 + 2e-, and the second one will become CrO4 -2 → Cr+3 + 5e-. Now that both equations are fully balanced, it would seem logical to just combine them again, right? Well, not quite yet. If we would do that, there would be an imbalance in electrons. To solve this, all we have to do is multiply each side by a numerical constant to the lowest common factor of the electron charges to balance the electron charges. Since we have 2 electrons for the first equation and 5 for the second, the lowest common factor between them will be 5 and 2 respectively. So then our equation will be 2(CrO4 -2) + 5(2I-) + 10e- →2(Cr+3) + 2(4H20) + 5I2 + 10e-, and after crossing out the electrons on both sides, our final answer will be 2(CrO4 -2) + 5(2I-) →2(Cr+3) + 2(4H20) + 5I2

Once you have completed this process a few times, you will realize how logical and systematic it can be!

O-rings

O-rings

09/20/16

“How can we apply our knowledge of gaskets to tubular geometries?”

Pipes and tubes are some of the most ubiquitous materials used to transfer fluids. And since they often do it at such high pressures, safety must be a central focus of their design. We already know that gaskets can be utilized to ensure the secure transfer of fluids, so how can we take our knowledge of these parts and apply it to tubular machines? Well, let’s use our engineering mindset to figure this out.

First of all, let us analyze the geometry that we are working with. This problem specifically has a uniform circular nature. So any fluid that might escape will do so in a direction perpendicular to the surface. So wouldn’t it be logical if we covered all critical points with a ring like material? Well, this is the operating principle behind an O-ring. O-rings can be found in all of sorts of applications, ranging from everyday uses such as car engines and even to extraordinary achievements of human through such as space shuttles! (in fact, the primary reason behind the infamous challenger failure was due to an O-ring failure)Since O-rings come in a diverse range of shapes and sizes for their diverse uses, these items are typically defined by the diameter of the inner circle and outer circle. All in all, O-rings are a salient example of how a simple part can have be utilized in far more complex ways.

Gasket

Gasket

09/19/16

“What is a simple way to ensure that fluids can be transferred without leakage while under compression?”

In modern machinery, fluids are often transferred when under compression. However, this action often leads to fluid leakage, so how can we design an apparatus that allows us to safely transfer this fluid . Well, let’s think about at using our engineering mindset. Well, often times it is the compressed space surrounding the mechanical transferring parts that causes the leakage. So wouldn’t it be logical if we just added a new part in between this space to prevent any unwanted interference? Well, this is the operating principle behind a very fundamental piece of technology called a gasket. The design of gaskets are usually made out of soft materials such as asbestos, aluminum and copper. The design and material makeup of gaskets is contingent upon the substance needing to be sealed, the operating temperature, and the geometry of the conjoined parts. Gaskets are often prone to be worn out after continual use, and must be replaced. Gaskets have many ubiquitous applications, most notably in car engines. In some applications, gaskets are being phased out in favor of sealants.

Guide to the chemical hazard diamond

Guide to the chemical hazard diamond

09/18/16

“What does that multi-color diamond near each chemical symbol represent?”

Have you ever noticed how this a multi color diamond near many chemical containers? And have you ever wondered what it represented? Well, it turns out that that geometric object is simply a chemical hazard diamond. A chemical safety diamond is an indicator of the types of hazards that are found in the contained chemical. The yellow diamond on the left indicates the reactivity, the red diamond on the top indicates the flammability, the blue diamond on the left indicates the toxicity, and the white diamond indicates the specific hazard. The first three use a 0-4 numbering system to indicate the danger level, 0 being the lowest and 4 being the highest, while words and symbols will be written on the bottom white diamond to attest to it’s specific danger. Manufacturers by law are required to print the diamond on each container to inform the users about the safety precautions they must make.

Aqueous solution

Aqueous solution

09/17/16

“What makes solutions in water so unique?”
During one’s study of chemistry, one will come across the fact that many substances dissolve in water. However, how can Chemists classify this, and how are they unique? Well, this phenomena is usually termed an aqueous solution (Aqueous coming from aqua-, Latin for water). However, what is truly special about aqueous solutions is that allow for the substances to be dissolved into ions within it. This gives rise to electrolytes, which can make water conductive, and this is where acid and base reactions take place. In fact, this ability to dissolve substances into ions is one of the reasons that makes water the prime factor for life, as different substances can be transformed and reacted within it.

Chemical precipitation

Chemical precipitation

09/16/16

“Why does a solid sometime forms when you mix two chemicals?”

Have you ever wondered why sometimes when you mix two aqueous solutions together, a solid will sometimes form? Well, the explanation for this very simple. When two aqueous solutions mix, their atoms will dissociate and recombine into new chemicals. Sometimes one of the products might be insoluble, meaning that it will not be dispersed into the surrounding substance. As a result, a solid will form. Chemical precipitations are a very common phenomena, for example, did you know that kidney stones are actually the results of chemical precipitation, and also pipe clogging?

Clamps

Clamps

09/15/16

“How can we use a simple tool to prevent two objects from separating?”

When doing engineering work, holding two discrete objects together might be a necessity. There are many ways to accomplish this, but how can one do it using a simple mechanical device? Well, luckily for us, engineers have already devised a tool known as a clamp. To make a clamp, simply create a solid “c” shape part, then drill a hole through the bottom curve, and then put an adjustable screw through the hole. If you put an object through the air gap of the “C” shape, and then adjust the screw to touch the object, you can apply enough pressure to entrap it in place.

Tidal locking

Tidal locking

09/14/16

“How can one face of a celestial object always face the same side of the object that it is orbiting?”

We all know that objects in space revolve around other objects. The moon revolves around the Earth, the Earth around the sun, and the sun around the center of the galaxy. These objects also usually have their own spin. For example, this spin gives rise to the days and nights on Earth. But what if an object’s rotation was in sync with it’s orbit, so that one side always faced the object it was orbiting? Well, not only is this phenomena possible, but it is also happening in our very own backyard, with our very own moon exhibiting this! When the moon revolves around the Earth, the gravity from our planet will cause the moon’s shape to be slightly from solid tides making the moon distorted to give it an almost (American) football shape to it. This distortion means that there will alway be a portion of the moon closer to the Earth then the rest, which results in that side experiencing a greater force being “hooked” to the Earth, causing it to stay on one side. Astronomers and Astrophysicists have termed this phenomena tidal locking as a result of the solid-tide induced locking on the moon.

Equivalent forces

Equivalent forces

09/13/16

“How can we simplify force diagrams?”

When working in physics or engineering, we all have to work with forces. Sometimes, we will have a multitude of forces, all going in different direction. However, how could we simplify all of these different elements of a problem to get the big picture and streamline our solution process? Well, let’s think about it. First we should think of our objective, and that is to see what happens when all of these forces are combined. So how about we take the components of each of the separate forces and moments, add them together, and find the equivalent force for all of these values? For example, let’s suppose that we had a bar of length 6 meters, with one force of 20 acting on the far left from the top and another one of 20 newtons acting at the far right coming in from the bottom. When we do all of the calculations, the equivalent force in the x direction will be 0 Newtons (Since none of the forces have an x component), The net force in the y direction will be 0 Newtons (since both forces are going in opposite directions, they will subtract each other) and the net moment will be 135 N-m (Since they are both in the same moment direction, 3m*20N+3m*25N=135N-m). With the use of equivalent forces, we can analyze an unlimited amount of problems, ranging from structural engineering to electrodynamics