CHEMISTRY COMMUNITY

Colin Squire 3B

An anode will oxidize, and the cathode will reduce. I like to use the vowels in anode (a) and oxidize (o) to cement this relationship.

Colin Squire 3B

An isobaric system indicates that pressure is going to be held constant. I like to think of this in terms of the pressure unit, bar, which is in the word. "Iso" meaning kept that same and "baric" relating to "bar" or the pressure of the system.

Colin Squire 3B

An intermediate is a compound that is created as a result of multiple steps in a reaction. However, when it is created, it is consumed in another step, meaning that it will not be in the final reaction or products. It is basically an element that progresses the steps while being consumed.

Colin Squire 3B

k, when it is lower case, refers to the reaction constant. The value of k is directly proportional to the rate of the reaction, meaning that if there is a large k value, the rate of the reaction is relatively speaking, fast.

Colin Squire 3B

A catalyst is generally seen in the reactants of the initial steps, and is seen later in the end of the reaction. It's sole purpose is to lower the activation energy, so it won't interact with what's being formed. It will be seen in the reactants and the product in the same form.

Colin Squire 3B

To find the number of moles, or n, you would need to have the balanced half reactions of the cell, and make sure the number of electrons (coefficient) from each half-reaction is equal to each other.

Colin Squire 3B

The side of the cell where there is oxidation occurring is at the anode, while reduction is happening at the cathode.

Colin Squire 3B

By looking at the total charge of the reactants to the total charge of the products, we can determine how many electrons were transferred in order to find the "n" value. It is also important to note that in some cases (like acidic and basic solutions) water, H+ and OH- can change the overa...

Colin Squire 3B

If I remember correctly, yes, the relationship is proportional. An increase in T will result in an increase in K.

Colin Squire 3B

Activation energy is not able to be replaced with enthalpy.

Colin Squire 3B

Going off of sapling question week 7/8 #20, potential is the driving force to make the electrons move, which is also known, and measured in, voltage.

Colin Squire 3B

Standard Gibbs free energy has the degree sign, or naught, to indicate that the reaction is occurring under standard conditions. The other without standard means that it can be calculated and accounted for in changes in different terms, like temp.

Colin Squire 3B

"n" refers to the number of electrons being transferred in the reaction. To find this, balance the half-reactions and find the total number of electrons used.

Colin Squire 3B

The oxidizing agent is the going to be reduced, and the opposite is also true. The reducing agent will be oxidized.

Colin Squire 3B

Your answers are flipped and the order in which they are placed matters. In the hint after I got it wrong it told me that the chlorine ions would be on the inner most level, closest to the double lines (||).

Colin Squire 3B

E (cell potential) is not a state function given that it is depends on the path taken.

Colin Squire 3B

The gas constant really depends on the units you need to cancel in order to find what you want. This can also mean changing the given value in J to kJ so that we a can minimize the amount of conversions done.

Colin Squire 3B

Finding the value of Q is the same as finding the equilibrium constant K. We divide the partial pressure or the molar concentrations with the products on top and the reactants on the bottom with the coefficients as exponents.

Colin Squire 3B

qrev is the q value or the heat of the reverse reaction.

Colin Squire 3B

In the context of thermochemistry and thermodynamics, I have only seem K used as the equilibrium constant. The only other option I can think of is Kelvin which is also K.

Colin Squire 3B

From what I have seen on the equation sheet, entropy equations (deltaS) is the only one that uses Cp, Cv.

Colin Squire 3B

I was looking for this in the equation sheet as well, but I couldn't find it. In the book, it says R = Cp - Cv, R being the gas constant.

Colin Squire 3B

The integral in the equations is just part of a series of steps to take in order to get the final equation. Essentially it is just part of the derivation of the final equation. I think knowing the conceptual background of the integral in the derivation is key to understanding what we are calculating...

Colin Squire 3B

The way I like to think of it is that the reversible processes are slow because the pressure will need to be slowly decrease, this also results in more work being done. The irreversible expansion is done quickly, in comparison to the reversible.

Colin Squire 3B

Work will be positive as a result of compression, due to the negatives in the work equation will cancel out.

Colin Squire 3B

Vapor and gas are the same thing, they are just said differently!

Colin Squire 3B

Hello, the equation I used was q=-q, or the energy being equal to the energy lost. In the end though, you would want to expand the equations to have q=mC(delta T), and putting the respective values in for each side.

Colin Squire 3B

Heat capacities can't be negative, given the fact that they are the measure of how much heat a system can take before being "raised" to a specific heat.

Colin Squire 3B

External pressure can be looked at as the pressure that is being applied to the system, which in this case, is the pressure the piston is adding.

Colin Squire 3B

There will be an increase in entropy because during any spontaneous reaction because of the natural tendency for the system to lean toward the more disordered state, which is essentially entropy.

Colin Squire 3B

Work isn't a state function because it matters where the path of the system has taken. I try to think of the mountain example Dr. Lavelle showed us a couple lectures ago.

Colin Squire 3B

Other than the given values from the question, there are multiple values for the gas constant. The reason there are so many fractions is because they are trying to convert atmospheres. The gas constant is given in term of atm, but there is another gas constant in torr that could've been used to avoi...

Colin Squire 3B

When bonds are being broken, there is going to be endothermic properties, shown by the positive value. When the bonds are being created, there is going to be exothermic properties, shown by the negative 4 value.

Colin Squire 3B

The stronger bond being created in the products, the more exothermic it is. If there is a weaker bond in the products, it will be endothermic.

Colin Squire 3B

Essentially, we are observing the amount of energy being put into the system to push it into its next phase. The significance of moving from liquid to gas is that more energy is being added to the system.

Colin Squire 3B

Since strong acids and bases completely dissociate, we know exactly where all of the initial concentration is going. We use ICE tables with the weak acids and bases because the don't completely dissociate and is difficult to determine where the initial concentration ends up. Using the ICE table allo...

Colin Squire 3B

Since the removal is so small that it becomes negligible, we can remove an x from the reactants side. We don't do this to the products side because we have nothing to subtract it to. This can be doubled check to make sure we can do the approx. by finding the % ionization and seeing if it's less than...

Colin Squire 3B

The way I think of it is that if I have an initial concentration, I will need to subtract it to create products. On the same point, -x doesn't apply to the products because nothing can be subtracted from zero in the initial row.

Colin Squire 3B

When the equilibrium constant is less than 1x10^(-3), we can assume that they x value is negligible. Dr. Lavelle also showed us in lecture that if we calculate the percent ionization and it turns out to be less than or equal to 5%, we can use this "short cut."

Colin Squire 3B

A salt can either be acidic, basic, and even neutral. This is all really dependent on the ions of the acids or bases. If they are equal though (like NaCl), they are neutral.

Colin Squire 3B

Temperature is the only factor that can change the K value. When there a purely changes in concertation, the reaction will shift itself either to the left or to the right, depending on what is being added or taken away.

Colin Squire 3B

Since inert gases do not react with gases inside the system, it can be implied that there will be no change in the concentration because they isn't a reaction to change it.

Colin Squire 3B

Molarity (mol/L) is usually the default units to go with but make sure to account for the units given in the question. This also could be put in the context of pressures.

Colin Squire 3B

When the reaction of K is reversed, K becomes the inverse of the reaction, 1/K.

Colin Squire 3B

Total pressure is all of the partial pressures added together. This usually would be a cumulation of all of the the partial pressures in one system.

Colin Squire 3B

The question will generally give you the number of mole (or some sort of way to find them like mass and molar mass) as well as the volume to find the concentration, if not given straight to you.

Colin Squire 3B

P=pressure, V=volume, n=number of moles, R= gas constant, T= temperature

Colin Squire 3B

Assuming you mean the values in between 10^-3 and 10^3, they aren't favored in anyway because the reaction is moving toward the reactants or the products. It is more or less in the middle and not one is being made more than the other. Dr. Lavelle discussed this in his lecture on 1/6.

Colin Squire 3B

Essentially, Q and K have the same equation. Q represents a reaction not yet and equilibrium, while K is used when the reaction has reached equilibrium. The question should tell you whether or not a reaction has reached equilibrium.

Colin Squire 3B

If we go back to 14a, recall the molarity equations, it was the moles divided by the volume. This is essentially the same thing except for being apart of a larger equation.

Colin Squire 3B

The K+ ion will be smaller than the Cl- ion because of the fact that it can pull the electrons in closer because of its + charge.

Colin Squire 3B

One way I like to remember what a Lewis acid is that it an electron pair acceptor (acid -> accepts). From that, I can determine that the base donates. Just by remembering the first one, I then know that Bronsted acids are the reverse of the Lewis acid-base. Bronsted acids donate, bases accept.

Colin Squire 3B

Yes, what you described is correct. One of the simplest examples of this is the formation of NaCl in the products. The strong acid and strong base "cancel each other out" so the salt doesn't affect the pH.

Colin Squire 3B

In class we were told that the oxygens were pulling at the H, making it so that the bonds are weakened. This has to do with the electronegativity. When the bonds are weak, an acid is being made stronger

Colin Squire 3B

What I was was first determine which ones were bases and acids. Bases having OH- and acids having H+. Knowing that NH3 is a weak base, I placed it in the middle and worked around everything else from there, keeping in mind the bond strength for acids.

Colin Squire 3B

Hello, "en" is just short for ethylenediamine! When doing the question don't forget that it is a bidentate! This messed me up a little too. Hope this helped!

Colin Squire 3B

Hello, I believe what you are missing in the answer is the "trichloro" in the name. Since in is inside the bracket and has a subscript of 3, it will be to the right of the transition metal, and be denoted by the prefix "tri." hope this helped!

Colin Squire 3B

I kinda minimized the steps of naming the compound into a couple words if it helps. I wrote: 1. cation, 2. complex(ligands+transition metals), 3. anionic ligands, 4. prefixes, 5. ligands in alphabetical order, 6. transition metal with oxidation state, 7. anion. Just remember the subtle rules in betw...

Colin Squire 3B

The prefixes I wrote down from my discussion section include: di-, tri-, tetra-, penta-, and hexa-, which are all respective values starting from 2, 3, 4 etc. If there is a polydentate, when then use the bis-, tris, tetrakis, pentakis- prefixes.

Colin Squire 3B

The transition metals are the elements from group 3 to 12. They have the d orbitals in them. Some examples include Cobalt(Co) and Fe(iron).

Colin Squire 3B

Since there are three domains of electron density, one of which being a lone pair, the shape will be bent (or angular) since the lone pair pushes down on the hydrogens.

Colin Squire 3B

Double bonds include both 1 pi bond and 1 sigma bond, while single bonds just have 1 sigma bond. This would mean that there are a total of 13 sigma bonds, and 3 pi bonds.

Colin Squire 3B

Lone pairs are considered areas of electron density, meaning that there would just be a total of three domains. This includes the two bonding pairs and the single lone pair. Three domains of electron density refer to a trigonal planar shape, which has a bond angle of 120 degrees.

Colin Squire 3B

In order to find the hybridized orbitals of the oxygen, you should first look to see how many electron domains it has. Since there are two bonding pairs and two lone pairs, there are 4 domains of electron density, which refers to sp3.

Colin Squire 3B

Since the question is only asking for the hybridizations of the phosphorous, we only need to look at the electron domains around them. Since there are 3 bonds and 1 lone pair, for a total of 4 domains, the hybridization is going to be sp3.

Colin Squire 3B

Due to the three domains of electron density, two of which being bonded pairs, and the other being a lone pair, the overall shape of H20 is bent. This is partly due to the fact that the lone pair pushes the two bonded pairs down, if that makes any sense. Hope this helped :)

Colin Squire 3B

In the lectures, Dr. Lavelle told us that the axial bonds are the ones that make 90 degree angles with respect to the equatorial bonds. I believe there is a question in the Sapling HW that goes over this in the solutions tab of the question. I'm sure that will give you stronger understanding of the ...

Colin Squire 3B

The result of two bonding pairs and a total of three lone pairs is a linear shape. I have found it useful to reference a VSEPR chart to familiarize myself with the specific geometries with their respective lone pair and domains of electron density.

Colin Squire 3B

See-saw shape is created when there are 5 domains of electrons density and 1 lone pair. For me, it helps to reference a VSEPR chart so that I can get comfortable with all of the different configurations of molecules.

Colin Squire 3B

They are experimentally determined and based off of relative bond angles. That being said, there is no way to calculate out what the bond angle will be. The best way to determine it is to account for the shape and the lone pairs, then make an assumption from there.

Colin Squire 3B

Sulfur has access to the d-orbitals since it is in the third energy level. Because of this, it can hold more than 8 electrons, breaking the octet rule.

Colin Squire 3B

I found it best to write down the actually trends on the periodic table and go from there. It think it also helps to apply why there is a trend so there are two ways of remembering them.

Colin Squire 3B

Since the Ga is in the 13th group of the periodic table, the charge of it is +3. This also applies to other elements like aluminum and boron.

Colin Squire 3B

If you draw out the Lewis Structure of the molecule you will see that since hydrogen is all the same surrounding the carbon, the charges all cancel and it will be non polar. I'm not too sure if this works for all cases but I look if the molecule is symmetrical to see if its nonpolar.

Colin Squire 3B

By adding different bonds to the same Lewis Structure, without going over the total number of electrons, you can change the formal charge of any of the molecules (usually as close to zero as you can get) to get them more stable. Essentially, by finding the different structures of the same molecule, ...

Colin Squire 3B

What I did was write down all of the trends on my periodic table and went over them from there. I think it is also important to know why those trends occur, like why the radius decreases as we move from the left to the right of the periodic table. Combining these too have helped me.

Colin Squire 3B

The charge symbol on the outside of the brackets essentially shows the overall charge of the molecule. For example, for sulfite, SO3 2-, you would have your Lewis structure [SO3] and the charge, indicated by the 2-, goes on the outside of it. It'll look like this, [SO3]2-.

Colin Squire 3B

There are more protons in each element as we move to the right in the periodic table. This means that the pull they have on the electrons are stronger, which then the radii are going to get increasingly smaller.

Colin Squire 3B

Hello! I think its implied that we already know that double bonds are shorter than single bonds, so I don't think we will need to draw them out shorter. The Lewis structure is a representation of what it would simply look like!

Colin Squire 3B

I could be wrong but what I did was add all of the points I earned out of the total possible points to get a percentage. So like adding 4 weeks worth of chem community points, midterm, and homework points.

Colin Squire 3B

The Rydberg equation is final n state minus the initial n state! :)

Colin Squire 3B

The v in the equation E=hv is actually the Greek letter nu. In the equation sheet it seems like it is slightly italicized so that might be one thing to look out for. For velocity, it is just the regular letter v.

Colin Squire 3B

Since the SI unit for distance is meters, nanometers would need to be converted. This would just make it easier to compute using the de Broglie relation because the speed of light units is in meters per second.

Colin Squire 3B

When using the equation, c="lambda"*v, and solving for frequency (c/lambda), the speed of light is value is divided to the Rydberg constant when it is 3.289e15. This gives the value 1.097e7. Essentially the equation is rewritten to equal 1/lambda.

Colin Squire 3B

I would say to double check the SI units of the given information! nm to m, and the mass from g to kg.

Colin Squire 3B

I'm not too sure about the other temperatures you're referring to, but the conversion from Celsius to Kelvin is on the review sheet near the middle of the page on the left side. It's written as "0 Celsius = 273.15K."

Colin Squire 3B

Yes, the decimal does add to the significant figures!

Colin Squire 3B

There are a set of conversions that are given in the equation sheet, but just to be sure there are some conversions that Dr. Lavelle gave us when he went over fundamentals in week one. There weren't too many but they were there. I'd say, just to be safe, get familiar with them.

Colin Squire 3B

The midterm will be multiple choice so I'm sure your answer doesn't need to be exactly on the dot. To be safe though, I'd recommend using the given values on the equations sheet we were given.

Colin Squire 3B

If there are multiple answers that I have to solve for in order to get an final answer, I only apply significant figures to the final answer. The initial answers that help me get to the final one aren't rounded.

Colin Squire 3B

For hydrogen, yes, it will start at the n=1 energy level. But for other elements, there are other energy levels, like the 1s and 2s energy level which are further in the periodic table.

Colin Squire 3B

When an electron is excited, it is being brought to a higher energy level, for example, from n=1 to n=2. When an electron is being ejected, it is going to a lower energy level, this is because of conservation of energy. The extra energy needs to go somewhere else. Hope this helps!

Colin Squire 3B

I would think that it is fine to leave constants in their respective variables, but I usually just write in the values so it makes computing in the calculator easier for me. :)

Colin Squire 3B

That is the equation to kinetic energy (KE).

Colin Squire 3B

You should round to the smallest number of the given sig figs, so yeah your example is correct.

Colin Squire 3B

The job of significant figures is to keep the solution as precise as possible without giving too much precision (if that makes any sense). The answer can only be so precise with the given information. That's why at the end of the problem we look at how many sig figs the given info has and round the ...

Colin Squire 3B

Kelvin (K) is the SI unit for temperature. I wouldn't be too concerned with conversions to Fahrenheit, I would just be familiar with the conversion from Kelvin and Celsius and vice versa, which isn't too difficult. I'd assume that for most of the questions we get will already be given temperatures i...

Colin Squire 3B

The steps I took for this question was to first find out what M was equal to. I did that by finding the molar mass of the hydroxide (OH)2, which was 34.08g/mol, and subtracting it to the given molar mass, 74.10g/mol. The missing molar mass would be 40.08g/mol which is the exact same as calcium. For ...

Colin Squire 3B

I'm just pulling from my past knowledge of AP chem. The theoretical yield will always be 100%. When we were asked to solve for the actual yield it was in a lab setting where we physically had to measure it out, otherwise I'd just say it would have to be given information for like a homework assignme...

Colin Squire 3B

There isn't a strict procedure to balancing equations, there is some trial and error. What I tend to do is balance the least occurring element first. If carbon appeared once on the reactants and once on the products, I would most likely balance the carbon. I've also seen that doing this naturally he...

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