The next step is to multiply both numerators together to get the numerator of the product and multiply both denominators to get the denominator of the product.

Now that you have your product, remember to simplify your answer if possible.

When we complete questions using heats of dissociation, we must make some assumptions. The first assumption is that, when our bonds break and form, all the energy is retained within the system. If this wasn’t true, our answer would not be accurate. We also assume that all bonds break in our reactants
and reform in our products, even bonds not involved in the reaction. To solve questions in this way is only a matter of adding up all the energy needed to break the bonds in the reactants and subtracting all the energy released when we reform the bonds in the products. Let’s look at an example together:

Step 1: Add up all the energy in the bonds of the reactants by multiplying the number of bonds of each type by the energy associated with that bond in the table above. For example, there are 3 C-H bonds in our reactants, and C-H has a bond dissociation energy of 410 kJ/mol. So, the total amount of energy stored in C-H bonds is 3 x 410 kJ/mol, or 1230 kJ/mol. The total for all the bonds, including the C-O, O-H, and H-Cl bonds, is 2472 kJ/mol.

Step 2: Repeat this process, but instead add up all the bonds in the products. In this example we would add the following bonds: 3x C-H, 2x O-H, and 1x C-Cl. This brings us to a total of 2480 kJ/mol for the products.

Step 3: Subtract the total bond dissociation energy for the products from the reactants. This would be 2472 kJ/mol – 2480 kJ/mol, which equals –8 kJ/mol. The negative tells us that the reaction in question is exothermic, or it releases more energy than it absorbs. A positive enthalpy change means that the reaction is endothermic, or it absorbs more energy than it releases.

There we are! We have completed a question involving heats of dissociation and can now interpret our answer to determine whether our reaction is exothermic or endothermic.

Physics in general can be regarded as the study of how things work and operate in our day to day lives. It can help explain anything as simple as the mechanics and forces involved when you walk to even something as complicated as the origins of the universe. One of the most important and useful applications of physics and kinematics is our ability to predict motion of objects; particularly motion of projectiles. By definition, projectile motion is a type of motion an object undergoes when it’s is launched near the Earth’s surface and travels along a curved path (most commonly a parabolic path). 

Methodology for Solving Projectile Motion Questions

1. Draw a diagram of the problem given
   1. Label and include any sort of information given in your diagram (e.g. distances, velocities etc)
2. Break the problem into x- and y- components
   1. Create a table to better organize your data
3. Determine which variables you need to solve for and using the appropriate kinematic equations solve for the missing data

Solving a Projectile Motion Question

Question: A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

Step #1: Draw and label a detailed diagram

Note: we are taking upwards to be the positive y direction and moving to the right to be the positive horizontal direction

Step #2: Organize all the data into a table

Step #3: Solve for the missing variable using the appropriate kinematic equation

In this question, we are asked to find the initial horizontal velocity therefore, we are looking for  vi  x.  Since we don’t know what the final velocity is in the x direction the only kinematic equation we can apply to the x-components is;

In projectile motion question the time in both the x and y components will always be the same. Hence, for us to solve the equation above we need to find time (t) using the data from the y-component.

Solving for ‘t’ with the data from the y-component we get;

Using the time we just got in the expression previously for the x-component we can now solve for the initial horizontal velocity of the soccer ball.

Therefore, by following these simple steps we found that the soccer ball has an initial horizontal velocity of 16.53 metres/second.

1. Find the number of protons, neutrons and electrons for the atom. The number of protons is the atomic number. The number of neutrons can be found by subtracting the number of protons from the atomic mass rounded to the nearest whole. This is because protons and neutrons both weigh 1 atomic mass unit (amu) and electrons weigh essentially 0 amu. To find the number of electrons you have to compare the charge to the number of protons. A neutral atom will have the same number of electrons as protons. For cations (positive ions), the number of electrons will equal the number of protons minus the charge. For anions (negative ions), the number of electrons will equal the number of protons plus the absolute value of the charge. This is because protons are positively charged and electrons are negatively charged so cations will have more protons than electrons and vice versa for anions.
2. Set up the diagram. To set up the diagram, you will need a circle in the middle. This will represent the nucleus. Here you will write the number of protons and neutrons as shown below in this example using sodium (Na)
3. Add in orbitals and electrons. In the last step you will need to draw circles around the nucleus. These will represent the shells in which the electrons orbit the nucleus. Two electrons will go in the first shell closest to the nucleus and eight can go in each subsequent shell. Using the example of sodium started above, we know that sodium as a neutral atom will have 11 electrons. That means two electrons will go in the first shell, eight in the second and one in the third. That third orbital is our outer shell or valence shell, meaning that sodium has one valence electron.
   
   

Okay, maybe not so simple. But simpler than one would think when they hear the word physics. We will answer the first part of the question first, which is why does a rocket not explode when it is ignited? The secret lies in the bell nozzle of a rocket engine. You can see one in the picture above. The nozzle tends to take up most of the engine, and for good reason. Without the nozzle, the rocket would have no way of turning its fuel into propulsion. This can be explained using forces. Newton’s Third Law states that for every action, there is an equal and opposite reaction. This is also true for tiny particles of fuel present in the rocket. When these fuel molecules collide with the nozzle, they try to expand it, but since the nozzle is rigid, the molecules rush out of the bottom of the nozzle which pushes the rocket in the opposite direction. You can visualize this if you imagine letting go of a balloon full of air. When you create an opening in the bottom of the balloon by letting it go, it shoots upwards because the air molecules are rushing out of the bottom. The bell nozzle is designed to focus the explosion in one direction, away from the rocket. 

Now we know how the rocket focuses the fuel downwards, but this doesn’t answer the question of why the fuel is on fire. This is specifically done in order to maximize the amount of upward force the rocket receives from each bit of fuel. The force received from pushing fuel out the bottom of the rocket is reliant on two things: the number of molecules ejected and the speed at which they leave. We can increase both of these things via the process of combustion. When you burn something, you are combining it with oxygen to turn it into water and carbon dioxide, plus energy. As long as we have oxygen and fuel together, we can turn our fuel into more molecules than we started with which will increase our upward force. Also, the energy created by burning our fuel is used to speed up the molecules leaving the nozzle, creating yet again more upward force on the spacecraft.

Overview

Whether we’re at school or at work, the one thing we tend to take for granted is our home. Not our physical houses, but our Home – the Earth. So let’s take some time to learn more about our amazing Home and the many layers to it. The Earth has a radius of 6,378 kilometres (3,963 miles) measured from the centre of its core all the way to the ground (where we stand). Between the 6,378 kilometres, the Earth can be fundamentally broken down into 3 main parts; the Crust, the Mantle, and the Core. Let’s start our exploration with the Crust.

The Crust

The Crust is the outermost layer of the Earth. The crust is completely solid and made up of rocks and various kinds of minerals. It is actually much thinner than you might expect it to be, as it only makes up roughly 1% of the Earth’s total volume. With a depth of 40 kilometers (or 25 miles) it can generally be broken down into two distinct types. The first type of crust is a thicker, continental crust and the second being a thinner oceanic crust. 

Continental Crust

The continental crust is primarily composed of many different types of granites. Aluminum and silicon are the most abundant elements found in this type of crust, hence, geologists also refer to the continental crust as “sial”. Sial stands for silicates and aluminum respectively. In comparison to the oceanic crust, the continental crust is much older in age.

Oceanic Crust

The oceanic crust extends beneath the ocean floors by approximately 5-10 kilometers (3-6 miles). This sort of crust is mainly composed of different types of basalts. The most abundant elements that make up the oceanic crust are silicon and magnesium; as a result, geologists refer to the oceanic crust as “sima”. Sima stands for silicates and magnesium respectively. The oceanic crust is much younger in age in comparison to continental crust because of the process of subduction. Due to plate tectonics, many areas where two or more oceanic plates can cause one plate to sink beneath the other; grinding away at both plates. As a result, the oceanic plates tend to erode and recycle often leading them to always be much younger in age.

The Mantle

Venturing further down into the Earth, after the crust, we have the mantle. The mantle is a mostly-solid portion of the subsurface of the Earth which has a depth of about 2,900 kilometers (1,802 miles). In comparison to the measly 1% of Earth’s volume that the crust makes up for, the mantle accounts for a staggering 82% of the Earth’s total volume. The rocks that commonly make up the mantle are silicates. Some silicates (which are compounds that have both silicon and oxygen molecules) that can be found in the mantle are; olivine, garnet, and pyroxene. Another common type of compounds/elements found in the mantle are; magnesium oxide, iron, aluminum, calcium, sodium, etc. The expected temperature in the mantle falls between the range of 1000°C (1832°F) to 3700°C (6692°F).

The Core

Finally, we’ve arrived at the Core. The core can be broken down into two sections; the Outer Core and the Inner Core.

The Outer Core

The outer core is the portion of the core that is the closest to the mantle of the Earth. The outer core has a thickness of 2,200 kilometers (1,367 miles) and is composed of liquid iron and nickel. The alloy that is formed, NiFe, has a very high temperature. The outer core has a temperature range of 4,500°C (8,132°F) to 5,500°C (9,932°F). The liquid metal of the outer core has a very low viscosity, meaning that it deforms very easily and is malleable.

The Inner Core

The inner core is a hot, dense sphere composed of iron. The sphere has a radius of about 1,220 kilometers (758 miles) in size.The temperature of the inner core is approximately 5,200°C (9,392°F). In comparison to the outer core, on average the inner core is much hotter than the outer core. 5,200°C (9,392°F) is far beyond the melting point of iron. Despite that, you might expect the inner core to be liquid like the outer core but due to the inner core’s intense pressure (and the entire rest of the planet and its atmosphere) the iron is prevented from melting. In short, the pressure and density are simply too great for the iron atoms to change state to a liquid. Because of these uncommon characteristics, most scientists prefer to interpret and understand the inner core not to be solide by as plasma.

Another unique characteristic of the inner core is that it rotates eastward, like the surface. The inner core makes a rotation about every 1,000 years. The Earth’s magnetic field is also theorized to be as a result of the inner cores rotation

The story of the atom begins way back in the days of the Greek philosopher Democrotus (470-380 BCE) who proposed that all matter was made up of smaller particles he dubbed “atomos”. 

It wasn’t until 1808 before an English school teacher, named John Dalton, would use experiments to elaborate on these Atoms. He created the first Atomic theory which consisted of these main points:

1. 1. All matter is made up of tiny, indivisible particles he called atoms. He imagined them as hard with a distinct mass and movable. They could neither be created nor destroyed. 
   2. All atoms of the same element are identical. Identical in terms of mass and also properties. 
   3. Atoms of different elements are different. 
   4. Compounds are formed by combining atoms of different elements in fixed whole numbered ratios. 
   5. Chemical reactions occur when atoms are rearranged. 

In 1897 J.J. Thompson proved Dalton wrong when he discovered electrons. This showed that the atoms were not indivisible as Dalton had suggested. Thompson did experiments using Cathode Rays which were thought to be beams of atoms at the time. He noticed that the beam travelled further through air than their size would suggest. He was able to estimate that the beam was made up of particles that were 1000 times lighter than a Hydrogen atom. Further experiments using electric fields to deflect the ray showed that the particle was negatively charged and the same in every type of element. With these results he proposed the “Plum Pudding” model of the atom. This model suggested that atoms were a large positive sphere with negative electrons embedded into it. 

Earnest Rutherford discovered protons and the nucleus in 1911. In his experiment he shot a beam of positive particles at gold foil. He found that most of the particles would go straight through the particle but strangely a couple would be deflected. He proposed that rather than the atom being a large positive sphere, as Thompson had suggested, the atom had a small dense center (nucleus) with positive charges. The positive charges were surrounded mostly by empty space with electrons orbiting around the nucleus. Neil Bohr expanded on this idea by proposing that the electrons orbited the nucleus in levels which he called shells. 

In 1920 Rutherford realized there was a discrepancy between the mass of the atoms and the mass between the combined protons and electrons. He thought that there must be another particle that was neutral in charge which he would call a Neutron. He asked James Chadwick to find evidence for this Neutron. Later in his life, Chadwick became interested in experiments where Beryllium was bombarded by Alpha particles. The hit Beryllium would emit radiation with strange properties. The strange beam was able to knock loose protons. Chadwick, after confirming with his own experiments, concluded that the beam was made up of a neutral particle with the same mass as a proton. 

Combining all these previous discoveries and proposals we can see the model of the atom as the Bohr-Rutherford model.

In modern days, there have been even more new developments on the model of the atom. For example, we know that the nucleus is extremely dense and tiny compared to the rest of the atom and that the majority of the atom’s mass lies within the nucleus. Also, rather than in rings, the electrons actually move around the nucleus in electron clouds. However, electrons are so small that this is mostly empty space. Because the electrons are so small, it is also impossible for us to tell where an electron is at any point of time. Instead, we can only determine the probability of an electron to be in a specific area. 

As we learn more about the atom, the atomic model may change again and again. We are constantly learning new things in science and tweaking the models used to represent our current knowledge.

Tell us about yourself

My name is Susanna Talley. I am a sophomore at Simpson University, where I study Psychology. Originally I am from St. Louis, Missouri, but now I reside in Watsonville, California. I am passionate about helping others, my faith, and working with children. In my spare time, I enjoy eating and drinking coffee, playing soccer, and watching Netflix.

What has been your best experience in school?

My best experience in school was taking the AP language class my high school offered. The teacher was legendary across campus for being the most difficult, and it was definitely a challenging first semester. However, I was grown and stretched as a writer which ended in so much growth as a a student and learner.

What influenced your choice in major?

I love people, and want nothing more than to enter a profession in which I can help others. I have a special interest in childhood disorders such as anxiety and depression. I have seen my friends and family be affected by mental health issues, and I want to make the lives of people like them better. Furthermore, my high school counselor made a positive impact in my life and I would love to do the same for other students in the future.

How do you envision yourself in 5 years?

In five years, I hope to be pursuing a graduate degree. Although I have not decided whether it will be in Social Work or School Counseling, I look forward to figuring out my next steps. In the mean time, I hope to continue finding meaningful experiences from my education, including internships. Because I am graduating a semester early, I would love to travel abroad by myself at some point.

What do you hope to learn from your school experience?

Of course, learning job skills and obtaining information is important to me. Yet, I feel that college is bigger than that. I hope to learn how to interact with diverse groups of people, learn better communication skills, and invest in my community. These skills are more important to me than any textbook information.

What extracurricular activities do you enjoy?

Apart from my rigorous academic schedule, I enjoy working as an Admissions Counselor Assistant at my school. I also enjoy eating, reading good books, journaling, and spending time in nature. During my summers, I spend eight weeks working as a camp counselor at a local sleep-away camp. I also love to spend quality time with my family and friends.

The First step in Naming any organic compound is to find the suffix functional group (if one exists). In this compound, There exists both an ester group, and two pairs of double bonded carbons. Since We don’t include double bonds in the suffix (along with triple bonds, and halogen groups), this compound is named as an ester.

Since all esters have a name of the form “alcohol-yl acid-oate”, we need to determine the alcohol and carboxylic acid that condensate to form this ester.

Naming the alcohol group is fairly easy, as it’s a simple two-carbon chain. Thus, this compound’s name will begin with ethyl.

The acid group of this compound is a little harder, but as it’s a single straight chain of 10 carbons (deca) with 2 double bonds (dien), we know that the root of its name will be decadienoate. The next step is to number the carbons of the double bonds, and record which ones are in the Cis (Z) or trans (E) configuration. In doing so, we want to choose the direction that minimizes the suffix group’s number, meaning that we have a trans double bond on carbon 2 and a cis double bond 4. As such, we can say that the compound’s name will end with (2E, 4Z)-decadienoate.

By putting these two together, we can now say that this compound is Ethyl (2E, 4Z)-decadienoate.

In this compound, we have either an ester, or an alcohol as our potential suffix group. Since the ester group is the more reactive of the two, we choose that one as our suffix.

As before, the alcohol portion of this ester is rather straightforward, Methyl.

For the acid, we must first find our parent chain. We can see that the carboxyl group is attached to a benzene ring. We call this benzene group with an additional carbon a Benzyl group. This Benzyl group is modified however, it has a hydroxyl group attached to it. To locate the hydroxyl group on the benzene ring, we assign the number 1 to the carbon that connects to the carboxyl, and chose the direction that minimizes the hydroxyl group’s distance from carbon 1. After all this, we can say that the acid part of this ester is 2-Hydroxybenzoic acid.

When we connect the two, we get Methyl 2-hydroxybenzoate, which is the preferred IUPAC name, however, this compound goes by a few more familiar names. Early botanists isolated samples of 2-hydroxybenzoic acid from willow trees, naming it Salicylic acid after the trees’ genus: Salix. As such, This compound can also be called Methyl Salicylate.

The suffix group of this compound isn’t too complicated, the most reactive groups are carboxylic acids, and there are 2 of them, making this a Dioic acid. The parent chain is 4 carbons long, and no matter which direction you start from, the carboxylic acids are on carbons 1 and 4, and the double bond starts on carbon 2.

Since a carbon can only hold a maximum of 4 bonds, any carbon participating in the carbon-carbon double bond can’t also have the 3 bonds necessary to be in a carboxyl group.
Therefore, if we state that double bond is on the 2nd and 3rd carbons, we don’t need to say that the carboxyls are on carbon’s 1 and 4, its Implied.
As such, This compound’s IUPAC name is But-2-enedioic acid.

As in many fields, the manner in which we approach a problem In mathematics can greatly affect how we go about solving it. By finding the factors of a polynomial, we can more easily identify roots, divide polynomials, and simplify rational expressions. These are just a few examples that you may see in the classroom, but the applications of polynomial factoring are far-reaching.

All forms of polynomial factoring take advantage of the distributive property:

a(b+c) =ab + ac

Where a, b, and c can represent constants (3), variables (x), or even entire algebraic expressions (3x² + 2x + 1).

By applying it in reverse. Unfortunately it is often difficult to identify possible values of ‘a’, and a fair bit of guesswork is required. This guesswork doesn’t need to be random however, as these simple tricks will help you narrow down your list of possible factors.

Tip 1. Start with easy guesses.

A constant factor (a), has one coefficient to determine, A degree-one factor(ax + b) has two A degree-two factor (ax2 + bx + c) has three, And so on. Larger degree factors become harder to determine, and to check, and most teachers aren’t mean enough to give a polynomial with no degree-one factors.  

Tip 2. Look at the highest and lowest terms for guidance.

Let’s use a generic third order polynomial.

(r₁ x +s₁) * (r₂ x +s₂) * (r₃ x +s₃)

= (r₁ r₂ r₃)x³ + (r₁ r₂ s₃ + r₁ s₂ r₃ + s₁ r₂ r₃)x² + (r₁ s₂ s₃ + s₁ r₂ s₃ + s₁ s₂ r₃)x + (s₁ s₂ s₃)

Before you run away screaming, notice that the first and last terms seem much simpler than the other two? In fact, this expansion tells us that the 3 ‘r’ values multiply to make the first term’s coefficient, and the 3 ‘s’ values multiply to make the last term.

Knowing this, we can save some time by starting with factors that make up the first and last term.

Tip 3. Synthetic division.

Another way that we can save time in the guess-and-check process is by reducing the time it takes to check our guesses. Synthetic division helps cut down on the time and space required for polynomial division by only writing coefficients, leaving the x’s out and focusing on the divide-multiply-subtract-bring over method. As always, if the remainder term is zero, you’ve found a factor.

Even with these tips, the most important key to success in polynomial factoring is perseverance. It may take a few tries to successfully factor a polynomial, but If you keep at it, it is a skill that you can easily master.