CHEMISTRY COMMUNITY

Matthew Choi 2H

This is not a question that requires calculation as much as it is a question that requires conceptual understanding. In a 1st order reaction, the rate of the reaction varies proportionally with the concentration of reactant. But in a 2nd order reaction, the rate will quadruple if the concentration o...

Matthew Choi 2H

The final will definitely cover all of the material that we have learned over the course of this year. You can expect conceptual questions as well as questions that require calculations. I think that most of the material on the final will be on the material that we covered after the midterm, so expe...

Matthew Choi 2H

For many of these types of problems, I think you should first understand the equations that you would use. Once you know which equations are available, you can maneuver the date given to find the answer you need. For this problem specifically, you need the half life equation and the equation that re...

Matthew Choi 2H

You need to find n to find what order that reactant is. The order will determine by what factor the change in concentration of that reactant affects the overall rate of reaction. You need to find k because that is the rate constant that is specific to that reaction that you need to find the rate of ...

Matthew Choi 2H

It doesn't matter which experiment you use because k should be the same. You should definitely check again to see if the k values are different for every experiment because that shouldn't be the case. As long as you are consistent with the values of a certain reaction, then you should be good.

Matthew Choi 2H

You will use any one of the experiments to determine k. Assuming you have the correct rate law for the reaction, you simply plug in the rate, concentrations of the reactants, and solve for k.

Matthew Choi 2H

Instead of looking at the molar coefficients in the reaction, you would only use the exponents of the reactants to determine the overall order of the reaction. So, when you write the rate law of the reaction, H2 and I2 will each have an order of 1 which you add up to get an overall order of 2.

Matthew Choi 2H

If a reactant is second-order, this means that doubling that reactant will result in the rate being quadrupled. This is because the concentration of the reactant is to the second power. Therefore, if you double the concentration (multiply by 2), then the 2 is squared to give you rate that is quadrup...

Matthew Choi 2H

In general, the rule of the thumb for finding the units of k is to find what units will result in the rate having the units of mol/(L*s). First, look at the units that the concentrations give you and make the units of k cancel those concentrations leaving you with mol/(L*s).

Matthew Choi 2H

Normally, you will see that changing the concentration of the reactants will affect the rate of the reaction. However in a zero-order reaction, changing the concentration will not change the rate. Therefore, the rate will equal the k constant. In order for the rate to have the units of mol/(L*s), k ...

Matthew Choi 2H

LOL. The salt bridge provides no other purpose than to allow the free flow of ions from each side of the cell to prevent the polarization of one side that would stop the reaction from occurring. It's not made of salt so it doesn't change the concentration of either solution.

Matthew Choi 2H

When Cl2 is broken up into the constituent Cl ions, then one will obviously react with water to create HClO. The remaining Cl will react with the H+ ion leftover from the creation of HClO to form HCl. But since HCl is a strong acid, it immediately dissociates into H+ and Cl-. This, I believe, is why...

Matthew Choi 2H

I think you just have to figure out the charges on your own based off of other information. In the example Lavelle gave in class, he used the reaction between MnO4- and Fe+2. On the products side of the reaction, you see that MnO4- has a net charge of -1, and since you know that O has a charge of -2...

Matthew Choi 2H

There are two parts to the process in which Br2 becomes 2Br. One part is the breaking of the bond between Br2, and the second part is the vaporization of Br2 to a gas.

Matthew Choi 2H

I think you should just draw the Lewis structure of the molecule and check to see if the formal charges of the atoms are all 0 or close to 0.

Matthew Choi 2H

A lot of these problems involve adding up the steps similar to how you do a Hess's Law problem.

Matthew Choi 2H

If you look at josephyim1H's post in this same thread you will see an explanation for this problem. Basically, the gist of what I said was that the ice will first melt (use delta H) and then increase in temperature to the exact same temperature as the rest of the water that was already in the glass....

Matthew Choi 2H

For phase changes, keep in mind that the temperature of the substance in question does not change. You can see that if you look at a heating or cooling curve. Therefore, you can't use q=(m)(C)(delta T) because you would end up with 0 due to the net change in temperature being 0. Instead, you have to...

Matthew Choi 2H

The key to this problem is knowing that the heat gained by the water is equal to the heat lost by the water. If you think about what will happen to the system logically, then the ice will melt into water while simultaneously cooling down the water already in the glass. All that's left in the glass w...

Matthew Choi 2H

I doubt it because the numbers that we use in the formula usually are at the same temperature.

Matthew Choi 2H

I'm pretty sure q and delta H are the same thing.

Matthew Choi 2H

Can someone explain what a bomb calorimeter does? Thanks in advance.

Matthew Choi 2H

We will usually use atm, but other common units for pressure include kPa, torr, and mm Hg.

Matthew Choi 2H

Inert gases do not react with the reactants nor the products, so it doesn't really have any effect on the concentrations of either. However, when you change the volume, then you can figure out which reaction will be favored by looking at the number of moles on both sides of the reaction.

Matthew Choi 2H

The concentrations of the reactants and the products both don't change when you add an inert gas because inert gases don't react with anything really.

Matthew Choi 2H

You're confusing exothermic/endothermic with a mere change in temperature. Just because they heat the mixture from 600 to 700 doesn't tell you anything about the reaction being exothermic/endothermic. Therefore, you can answer the question by looking at the k values that give you for each temperatur...

Matthew Choi 2H

Notice that C is a solid. Therefore the concentration of C won't change. Then, if you look at the equation again, there are more moles of gas on the products side than the reactants. This is why the reaction will favor the reactants side if you increase the pressure.

Matthew Choi 2H

Typically, you use an ICE table when the question asks for the equilibrium concentrations of a reaction and they give you the initial concentrations. Then when you construct your ICE table, you will fill in the blanks for the initial concentrations but use "x" for the change in concentrati...

Matthew Choi 2H

You should probably convert one of them to the other. And yes, it's usually safe to assume the starting concentration/partial pressures of the products are 0 if not explicitly stated.

Matthew Choi 2H

Using PV=nRT, you can rearrange the expression to read P=(n/V)RT. The (n/V) is essentially concentration so you can convert from partial pressure to concentration this way.

Matthew Choi 2H

a/b) If you start with more moles of O3 then the partial pressure of O3 will increase. You know that the Kp must remain constant so, looking at the equation for Kp = (PO2)^3/(PO3)^2 you can reason that the partial pressure of O2 will also increase to keep the ratio the same. The only way that the pa...

Matthew Choi 2H

The electrons in a lone pair have more energy than bonded electrons because they are free and not restricted to a bond. Therefore, they repel each other more because they are more excited and take up more space. This is why they decrease bond angles and slightly distort the molecular geometry of mol...

Matthew Choi 2H

Like other people said before, count the number of areas of electron density. Lone pairs count as one area of electron density. Single, double, and triple bonds also only count as one are of electron density. It might be weird to think that a double and triple bond count only as a single area of ele...

Matthew Choi 2H

The hybridization of the N is sp. This is because there are two region of electron density: the triple bond with the C and the lone pair. The lone pair counts as one area of electron density. The triple bond also only counts as one area of electron density despite there being 6 electrons involved. T...

Matthew Choi 2H

In total, there are 16 valence electrons. After drawing the skeletal structure for N2O with single bonds stemming from the central N to the other N and O, you are left with 12 valence electrons. The way I was taught in high school was to add the remaining electrons to the outer atoms to fill their o...

Matthew Choi 2H

In a T-shaped molecule, there is one bond going straight up, one bond going straight down, and one perpendicular to both of the ones mentioned before that lies on the equatorial plane. The bond angle between the bond going up/down and the bond along the equatorial plane is 90 degrees. The bond angle...

Matthew Choi 2H

Imagine a molecule with the VSEPR formula, AX4E1. There is a central atom with 4 bonds and one lone pair. The electron geometry for this molecule would be trigonal bipyramidal: one bond going straight up, one bond going straight down, and three on the equatorial plane. However, the formula AX4E1 has...

Matthew Choi 2H

The names that you need to know will most likely include: linear, bent, trigonal planar, tetrahedral, trigonal pyramidal, trigonal bipyramidal, seesaw-shaped, T-shaped, octahedral, square pyramidal, and square planar. The bond angles that you need to know will most likely include: 90, 109.5, 120, 18...

Matthew Choi 2H

In response to the above post, I don't believe it is possible for the model in part (b) to have two lone pairs. If you think about it, if there were two lone pairs, the electron geometry would be a tetrahedral. Therefore, the two bonds would have a bond angle of slightly less than 109.5 degrees. The...

Matthew Choi 2H

Not quite sure either, but I found this explanation online and it somewhat helps: "Benzyne is an neutral reactive intermediate that can be formed by removing two ortho substituents from a benzene ring leaving the p-orbitals with free electrons. Benzyne is an extremely reactive species because o...

Matthew Choi 2H

If you look at the principal levels: the 1st principal level only has an s sublevel, the 2nd has an s and a p sublevel, and the 3rd has an s, p, and d sublevel. In order to achieve more than an octet, an atom must have access to the d sublevel in order to go past their octet. This is why atoms in th...

Matthew Choi 2H

My TA said that sometimes the problem won't ask you to draw all of the possible resonance structures even if there are multiple resonance structures. This is because (1) they might not want to give away the fact that there are resonance structures in the specific problem or (2) they aren't asking fo...

Matthew Choi 2H

For these types of questions, first find out how many valence electrons the unknown element adds to the total amount of electrons in the molecule. Take the total amount of electrons in the molecule and subtract the number of valence electrons of the known elements. For this specific problem, you end...

Matthew Choi 2H

As you travel along a period from left to right, the number of protons and electrons both increase by one. The proton is added to the nucleus and the electron is added to the same sublevel. Since the electrons are added to the same shell, the distance between the charges doesn't increase, but the ch...

Matthew Choi 2H

Electron affinity is described as the propensity for an atom or ion to gain an electron. The way that you can tell that C has a higher electron affinity than N is by looking at the 2p sublevels of a C and N atom. C has one electron in two of its three orbitals. N has one electron in each of its orbi...

Matthew Choi 2H

Ionization energy is defined as the amount of energy required to remove an electron from a gaseous atom or ion. In order to answer this question, you need to look at the orbitals of both N and O. If you look at the 2p sublevel for N, there is one electron in each orbital. If you look at the 2p suble...

Matthew Choi 2H

You can tell if an atom is in an excited by looking at the end of the electron configuration. If you find an electron that is in a higher energy level than where it typically should be, it is most likely in an excited state. Take neon (Ne) for example: The normal electron configuration would look li...

Matthew Choi 2H

The Einstein equation only accounts for the energy of a single photon of light, so if you had no other information, you wouldn't really be able to answer the equation. All you need to find the total number of photons is the total energy of the light source. You take the total energy and divide that ...

Matthew Choi 2H

Don't worry if you get a super small or large value for your delta v or delta x. As long as you plugged in the correct numbers then you should get the correct value in the end. The size of the value only determines how big the range of uncertainty is. So, if you end up getting a super small value fo...

Matthew Choi 2H

Electrons DO exhibit wave particle duality. It was proved through an experiment by Davisson and Germer. It's kind of hard to explain so I'll just give you a link to a video I watched to better understand it.

https://www.youtube.com/watch?v=Ho7K27B_Uu8

Matthew Choi 2H

If you look at a typical spectrum with the range of visible light, you will find that many spectrums use wavelength to differentiate between colors. However, since wavelength and frequency are inversely related, you can say that frequency also determines the color of light.

Matthew Choi 2H

I used the chart that was found a few pages before the questions to identify which event correlated to which wavelength. I'm not sure if we have to memorize those sections though.

Matthew Choi 2H

I know that the ml quantum number is the magnetic quantum number, and that it can be px, py, or pz, and it represents the orientations of different orbitals of a subshell. But what do those things mean? Thanks in advance.

Matthew Choi 2H

In order to balance this equation, you first need to recognize the reactants and products. The reactants are C7H6O3 and O2, and the products are CO2 and H2O. So first, write the equation as follows: C7H6O3 + O2 ---> CO2 + H2O Now comes the part where you balance this equation. Since both the C and t...

Matthew Choi 2H

This chart contains most of the common polyatomic ions that I've needed to memorize in high school! I'm not sure how much more we'll need to know for this class though. https://www.google.com/search?q=list+of+common+polyatomic+ions&safe=strict&source=lnms&tbm=isch&sa=X&ved=0ahUKE...

Matthew Choi 2H

First, you need to find the molarity of the original KMnO4 solution. In order to find molarity, you need moles of solute and liters of solution. Use the molar mass of KMnO4 to convert grams to moles, and divide that number by .15000 L. You should end up getting .211 M KMnO4. Second, you need to use ...

Matthew Choi 2H

In high school chemistry, I typically used 6.02*10^23 as Avogadro's number, but in lecture I remember Lavelle saying that we should use 6.022*10^23. I guess we can use that extra sig fig now that we are grown college students.

Matthew Choi 2H

In your first multiplication calculation (1*12.011), you will end up with 5 significant figures. In your second multiplication calculation, (2*1.008), you will end up with 4 significant figures. However, in your last calculation, you adhere to the addition/subtraction rule for significant figures.Wh...

Matthew Choi 2H

Whenever you are doing a multi-step problem, it is always good to round at the very end. If you round after each step, your answer may come out to be very different and inaccurate. Try your best to keep all of your answers in your calculator so you can go back and use the exact answers in all subseq...

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