CHEMISTRY COMMUNITY

RitaThomas_3G

If E˚cell is a positive value, then the overall reaction is spontaneous as written. The standard cell potential can be calculated by using the equation: E˚cell = E˚cathode - E˚anode. The two parts of this equation is calculated by using the half reactions for reduction and oxidation.

RitaThomas_3G

Enaught is equal to 0 when the reaction is done at standard conditions. Thus, the overall equation qe get for E is: E = -(RT/nF)ln(Q). Just be sure to use the correct units according to what the problem is asking for. For question 17, it's asking for in V, so you can use the equation E = 0V - (0.059...

RitaThomas_3G

Hey! So just adding on to what was just stated, because we already know that the Enaught value is 0, the overall equation should look like this: E = (0.0592 V/n)log(Q). You will just need to then plug in the values for n and Q!

RitaThomas_3G

Adding on to what was stated above, the last part is asking for the shorthand notation the electrons moving from the anode (on the left) to the cathode (on the right). The phases are listed in order of which they become in contact, with he double line being the salt bridge.

RitaThomas_3G

For this problem, you can subtract the standard reaction potential that occurs at the cathode by the one that occurs at the anode. Overall, the equation that arises is: Ecell = Ecathode - Eanode. The individual values for the respective half reactions can be seen in the linked chart.

RitaThomas_3G

Hey! So initially I was confused with this concept too, but then I found that it was easiest to see what units I needed for my final answer. Then, I used the R value which allowed me to get the desired result (by cancelling out units that I needed).

RitaThomas_3G

Hi! So looking at the Appendix, we can see that if the delta G value is negative, then the compound is thermodynamically stable. If it is positive, then we can assume that it is thermodynamically unstable.

RitaThomas_3G

If Gibbs free energy is positive, then the compound thermodynamically unstable. If it is negative, then the compound is thermodynamically stable.

RitaThomas_3G

Using the information that many people have stated above, I think its also important to note that if delta G is negative it is spontaneous, and positive is not spontaneous. However, knowing the sign of delta G naught does not prove spontaneity. Because, according to the equation for delta G, there a...

RitaThomas_3G

Hey! So if Q<K, then a forward reaction likely occurs, which means products are favored. When Q>K, a reverse reaction likely occurs, which means reactants are favored. If Q=K, the system is at equilibrium.

RitaThomas_3G

Hey! So looking at your answer, it actually does look correct. One thing that I realized I was doing wrong was putting the "spontaneous" at the bottom sections instead of the top. Maybe try switching these values with the Q<K and Q>K? Hopefully that works!

RitaThomas_3G

Hi! So for the first part of the problem when it is asking for you to solve for the K value, I believe that you have to do so by using the equation delta G = -RTln(K). However, how would you go about solving for K?

RitaThomas_3G

Answering the above equation^, you have to first find the amount of energy supplied in Joules by multiplying the # of minutes * 60s/1min * J/s (this value should be given in the equation). Then, the delta h vap value is calculated by dividing the amount of energy supplied that you just calculated by...

RitaThomas_3G

Hey! So in terms of calculating this answer, you have to take the sum of the Gibbs free energy for products minus the sum of the Gibbs free energy for reactants. Make sure to multiply the individual numbers by the number of moles by each individual value if applicable!

RitaThomas_3G

Hey! It seems like overall the steps that you have taken with ΔS and ΔS are correct. I would just try to double check that the calculations for these individual numbers are correct. For example, with the second example, make sure you take the reverse reaction of the number and multiply it by 2. Then...

RitaThomas_3G

Also, when you are solving for the value, and look at the equations you are using, there is no need to make any calculations using mass because all the necessary values are already given. For example, when you are solving for Ccal = qcal/DeltaT, you already have the q value and DeltaT from the probl...

RitaThomas_3G

Also keep in mind that when you are solving for q for this equation, you will be converting the grams into moles. Then, you can use this value to convert it into kJ. So, in a way moles are accounted for, just not in the equation C = q/DeltaT.

RitaThomas_3G

Hey! So for part a, the overall equation is w = -nRTln(Vfinal/Vinitial). Most of these values are already given in the problem, however you will need to calculate for the number of moles, "n." This will be done by using the ideal gas law, PV = nRT. After you solve for n, you can just plug ...

RitaThomas_3G

Hey! So the equation for q is: q = -Ccal*DeltaT. I think what you were confused about is the equation for Ccal, which is qcal/DeltaT. The q value should be given in the problem, and the DeltaT is correspondent to the first experiment.

RitaThomas_3G

Hey! So just adding on to what Arti said above^, you would have to do (7/2)R - R, which is equal to (5/2)R. Then, from here you just multiply this value by the number of moles and the change in temperature.

RitaThomas_3G

As a response to the previously stated question ^, a combustion reaction is always when one compound interacts with oxygen (as a reactant), and the products will always be CO2 and H2O. This will always occur no matter what the initial reactant compound is. For example, CH4 + 2O2 -> CO2 + 2H2O is a c...

RitaThomas_3G

The main difference, as you can see in the answer choices is the negative sign. This helps to denote between a exothermic and endothermic reaction. From the problem, we can see that it is an exothermic reaction, and thus the answer choice would be that with a negative sign.

RitaThomas_3G

Hey! Also think about using this equation: ΔH = (n) * (ΔH∘), where n is the number of moles. When we look at it, we can see that the value of n and ΔH∘ is given, being 1.31 moles and 358.8 kJ/mol respectively. Thus, using the equation, we can see that we need to simply multiply the numbers to get th...

RitaThomas_3G

Hey! When I approached this problem, I found it really helpful to just put in simply numbers that represented the increasing bond strength (so 1, 2, etc.). By doing this, I found it a lot easier to simply calculate the bond enthalpy for each reaction. If it is positive, it is endothermic, and if it ...

RitaThomas_3G

I actually had the same question! I was thinking maybe it just had to do with the difference between using a strong acids and bases with the salts versus with weak acids and bases with the salts (where you would use Ka and/or Kb values). But also, I wasn't 100% sure.

RitaThomas_3G

Hey! Just to clarify what a state property is (it's definition), we can understand that it is a property that only depends on the initial and final state. Therefore, the path taken does not matter, as long as the other conditions (such as energy) stay the same. Thus, we can understand that this is n...

RitaThomas_3G

Hey! Just adding on to the discussion between the previous responses, I believe that heat and enthalpy can in fact be used to describe the same thing if no work is being done to the system (so heat absorbed or given off is equal to the change of internal energy in the system).

RitaThomas_3G

Hey! Just adding onto what other people have said before, I thought I'd give an example. If you look at the first problem, the products of the reaction are CO2 and H2O. This can be compared to the enthalpy of the combustion of CH4 because, as we have learned before, the products of any combustion re...

RitaThomas_3G

^ @Jaden, you would actually use the same thought process as the other reactions in this problem. An easy way to look at it is as A-A or B-B bonds for A2 or B2 in order to make it easier to look at the chart and compare bond strengths.

RitaThomas_3G

Hey! This is endothermic because the pack absorbs heat from the surroundings in order to make the pack feel cold for us. Therefore, it can be classified as endothermic according to its definition.

RitaThomas_3G

We do not need to include it because it is a strong base, which means that when the reaction takes place it will dissociate completely. Therefore, we don't need to consider it with #7.

RitaThomas_3G

Hey! So one thing I noticed is that, assuming the explanation you gave is all the work that you did, you might have forgotten to find the molar concentration of the other reactants and products as well, other than O2. I see that you divide the moles of O2 by the total volume (4 L), but then not for ...

RitaThomas_3G

Hey! So when approaching this problem, you want to create an ICE table to find the final concentrations of all the amounts, and then solve for Kc using the equation Kc = ([SO2]^2[O2])/([SO3]^2). Just keep in mind that from the given values, you know that the initial concentration of SO3 will be the ...

RitaThomas_3G

That is correct! As many people have stated you could also just take the square root of both sides. Just an FYI, after you solve for x (both ways should give the same value), you just plug it into the final equilibrium concentration for [HI], which is 2x.

RitaThomas_3G

Hey! So when you approach this problem, you can create an ICE table similar to how you would do for the other concentrations we have worked with in this class. So, the equilibrium concentrations of the reactants will both be "x", and for the product it would be "0.0390 - x" since...

RitaThomas_3G

Yes! I believe that thus far in this course we have only learned that in order to cause a change in K, there must be a change in temperature. Changes in other factors, such as pressure, do not directly change K, but rather how it is balanced.

RitaThomas_3G

Because the volume decreases, but there is more moles on the left, the reaction would shift to the right to create a balance within the overall reaction. To adjust for the change, the reaction therefore simply shifts.

RitaThomas_3G

The way that I would approach a problem like this would be to initially look at the ICE table, where we can see that one of the concentrations is equal to 0.1-x. Logically, we can see that with an x value of 0.126, the concentration would be negative. Therefore, this positive x value would not make ...

RitaThomas_3G

You are correct! When counting the moles, you only include the gas states. Therefore, when you look at this problem, there is only 1 mole of a gas on the left, and 2 on the right. The 2 moles of an aqueous solution would not be counted.

RitaThomas_3G

In terms of the certain equations you would need to choose, it is dependent on which of the equations would give you the desired result (which is shown in the problem). Note that some manipulations to the given equations might need to be made. Also note that, for example, the last equation would not...

RitaThomas_3G

I believe we would need to know how to calculate pH from pOH for problems like we saw from c, d, and e because we are really just applying the other concepts we have already learned. This includes pH = -log([H+]), pOH = -log([OH-]), and pH + pOH = 14. So I would learn how to do it just in case!

RitaThomas_3G

I am also not 100% sure about the correct answer to your question, but from what I understand your reasoning is correct in using cis when the Cl are next to each other, and trans when they are opposite to each other. This way, from the name, we can understand how the molecule is actually shaped, and...

RitaThomas_3G

To answer the question above ^, you can determine if a salt is an acid or base depending on the cation and anions that make up its composition, and the strength they would have in an acid or base. For example, in Na2S, the cation Na is derived of a strong base (NaOH). The anion S, however, is derive...

RitaThomas_3G

Answering the question above ^, I believe that could be the case. However, what I believe is more important than classifying as V1 or V2 is making sure that you put the 200 mL and 0.025 M on the same side. This way, even if you incorrectly classify it as V1 or V2, you would still get the right answer.

RitaThomas_3G

I solved for the [OH-] by doing 10 ^ (- pOH). As we know from the equation, pOH = 14 - pH. In this problem, the pH value can be calculated by doing the -log([H+]). Hope this helps!

RitaThomas_3G

I definitely agree with what has been said previously by other students! However, I think when looking at this specific question, we don't need to overthink it too much. The approach that I found helpful was looking at the molecules that were inside the brackets, not outside. Then, I found how many ...

RitaThomas_3G

A bidente is defined when it is bound to two atoms. If we look at the structure of oxalate, we see that when looking at only one Carbon atom, it is only bound to two Oxygen atoms. Therefore, it is characterized as a bidente.

RitaThomas_3G

Adding onto what everyone else has previously said, it refers to the number of ligands bound to the central atom. If you had a molecule with brackets, you wouldn't consider what is outside of it because it is beyond the coordination sphere. For example, Ba[FeBr4]2, the coordination number would be 4...

RitaThomas_3G

I do think that it would be best if we, more or less, knew the charges of the atoms or molecules. If it is just the atom, we could easily look to the periodic table to see what charge it would be. If it were a molecule, we could calculate the formal charge and find our answer through that method (us...

RitaThomas_3G

One other thing to keep in mind when calculating the coordination number is that you don't need to account for the molecules outside of the brackets, because they are out of the coordination sphere.

RitaThomas_3G

All the different hybridizations are determined by the number of regions of electron density. So, we can see that sp has 2, sp2 has 3, sp3 has 4, etc.

The regions of electron density are shown by a bond (single, double, and triple bonds are all 1 region) or lone pairs.

RitaThomas_3G

An easy way to conclude whether or not there is a delocalized pi bond is to see if there are resonance structures. This means that the double or triple bonds could be in different spots. This indicates that there is a delocalized pi bond.

RitaThomas_3G

Since you have found the lewis structure already, look at how many bonds/lone pairs the oxygen atom has. We can see that it has 4 bonds, which means that it has a hybridization of sp^3.

RitaThomas_3G

I definitely get what you mean! I was also really confused when initially looking at this problem just because there's so much going on, but what helped me was to just focus on the Phosphorous atom. By looking here, I saw that there was 3 bonds and 1 lone pair, which meant it was sp^3.

RitaThomas_3G

You would have to consider each region of electron density. So, this means that lone pairs would count. In terms of bonds, single, double, and triple bonds all account for one region. Therefore, whether it is a single or triple bond, it would only count as one region.

RitaThomas_3G

As stated by the answers above, I believe 104.5 would be within the range of answers possible. Maybe just make sure to say it is approximately that angle to show that it could be slightly more or less that number.

RitaThomas_3G

The see-saw shape has 4 bonding pairs and 1 lone pair. For example, SF4 has a see-saw shape.

RitaThomas_3G

Trigonal pyramid, because there are 3 bonding pairs with Oxygen, and 1 lone pair on S.

RitaThomas_3G

Hello! I believe XeF2 could also be seen as linear. Given the number of electrons it has, 22, there would be a F atom to each side of Xe, and a lone pair of electrons.

RitaThomas_3G

I don't think we would have to know the exact approximation, more just the approximate range that the angle would fall in. As said in the previous answer as well ^.

RitaThomas_3G

I would assume so since that would be the most accurate. Also, since the midterm would be multiple choice and there are several correct structures, I would definitely try and go for the "best possible" answer.

RitaThomas_3G

Hi! I'm not 100% sure but I believe I saw someone else on another discussion post say that for right now, he will give us the molecular formula. However, I do think it might be beneficial to know some of the more well-known one's just in case.

RitaThomas_3G

I think the best way to go about this problem would be to see which number the actual bond length is closest to, rather than focusing on the terminology of the problem.

RitaThomas_3G

Hey! So in terms of structure it looks correct. However, for C, the charges are incorrect. C should not have a charge, and both O should have a negative charge. The N is correct! This will bring the overall formal charge to -1.

RitaThomas_3G

Nitrogen is from Period 2, meaning it cannot have an expanded octet. However, looking at your diagram, you have two double bonds and 1 single bonds. To fix this, try putting only 1 double bond.

RitaThomas_3G

Hey! Resonance simply means that there are different ways to draw the Lewis structure for a molecule. This might mean changing where a double bond is located. The most stable structure is the one where the formal charge is closest to 0.

RitaThomas_3G

Double bonds have more electrons and are more strongly negatively charged, which makes the attraction to the positive nuclei stronger.

RitaThomas_3G

Yes! Thinking about it logically, it also does make sense because it is asking for which structure is more "plausible." The experimentally determined values are from structures that have occurred, and therefore it makes sense for the plausible values to be close to this.

RitaThomas_3G

Also, after memorizing all the trends, I highly recommend immediately writing them on the periodic table once the exam starts. This way you don't have to worry about possibly forgetting about them later on in the test.

RitaThomas_3G

In terms of memorizing them, I am personally a visual learner so I like to use arrows indicating where the specific value increases. It makes it easier to see which way it increases, and then afterwards make the necessary changes. Also, whenever I start a chemistry exam, I will write all the arrows ...

RitaThomas_3G

I also looked a bit more into this after he said it and I believe the equation is 2(n^2), where n is the shell (principle quantum number). We can see that this equation does work in the following examples: n = 1, 2 electrons - 2(1^2) = 2 electrons n = 2, 8 electrons - 2(2^2) = 8 electrons The equati...

RitaThomas_3G

Although theoretically there could be more possible orbitals, because the question specified exactly which ml value it is (+1), there is only 1 possible orbital.

RitaThomas_3G

Just adding a bit more information that helps understand the concept a bit more! s - subshell: 1 orbital p - subshell: 3 orbitals d - subshell: 5 orbitals f - subshell: 7 orbitals n = 1 shell: 1 orbital (s subshell) n = 2 shell: 4 orbitals (s and p subshells) n = 3 shell: 9 orbitals (s, p, and d sub...

RitaThomas_3G

When you first approach a part of this problem, you will want to convert the element into its electronic configuration. This is because you know that when you remove an electron, it starts from the outermost shell. With the electronic configuration, it will be easy to find which shell this is becaus...

RitaThomas_3G

From the problem, you are given the value of n = 2. This means that you are going to be dealing with the shell that has s and p orbitals. Then, to find the answer, you just need to add the number of orbitals in s and p. Respectively, the answer is 1 and 3, and if you add this together you get the an...

RitaThomas_3G

Hey! I definitely had the same problem when trying to understand the question, but then I realized that I could just use Avogadro's number. By this I mean that initially find the molar mass of the diatomic oxygen, which is 32 g. Then, divide it by Avogadro's number. Then, you must also convert this ...

RitaThomas_3G

Hey! So for this part of the problem, I like to look at it like dimensional analysis when solving it. This is because you solve it by dividing the energy from part a (units: Joules) by the total energy given in part b (units: Joules/Photon). You can notice that when dividing the two values, the Joul...

RitaThomas_3G

Hey! So these are the steps that I went through to do this question: Step 1: Decide which equation to use. Since I am solving for wavelength, and seeing the given values in the problem, I decided to use De Broglie's Equation: \lambda = \frac{h}{mv} . Step 2: Write down given values. First, we know t...

RitaThomas_3G

Hey! So to find the energy of the electron, first think about what equation to use for energy. In this case, it would be KE = \frac{1}{2}mv^2 . After looking at the problem and the given information, you can notice that you do not have a value for "v" yet. But, you can also notice that you...

RitaThomas_3G

Hey! Similar to what other people are saying on this post, I also solved for this problem in a similar way: first I used Debroglie's equation to solve for v, and then used the KE formula and converted the units. Maybe just make sure to double check all the numbers you're inputting, with the correct ...

RitaThomas_3G

Hey! So when I would be solving a problem like this, I would initially start out with what I am solving for, which is wavelength. So, I would start with this equation: \lambda = \frac{h}{mv} . Then, after looking at the equation, I see what values I already have. I know that h is Planck's constant, ...

RitaThomas_3G

E=hv is used to find the energy of a photon using Planck's constant and frequency of light. E=pc is used to find the energy of a photon using the speed of light and the momentum of a photon. I'm pretty sure we only use E=pc when momentum of the photon is provided. Hey! Just adding on to what was sa...

RitaThomas_3G

Hey! So the first part of this question asks you to solve for the work function. After you do that, you get an answer with the units J/photon. In the second part when you are solving for the number of photons, you can notice how this unit (photons) can cancel out with the photons from the units of t...

RitaThomas_3G

Hey! Just adding onto what was stated previously, Max Planck was trying to show that when energy is being exchanged, it occurs in "packets" of energy. This constant helps to show that given the frequency of a wave, it can only carry a certain amount of energy.

RitaThomas_3G

Hey! So for Question #7, I understand that you need to use the equation \lambda =\frac{h}{mv} since they give you the average speed of the molecule. However, i'm not too sure how you would calculate the "m" to solve for wavelength? Is it simply 32g for Oxygen molecule, or do you have to us...

RitaThomas_3G

Yes! If I am understanding what you are saying correctly, then you certainly can use mol/g or g/mol depending on the situation. However, you must always be careful to keep the ratio the same. For example, 1 mol NaCl/ 58.44g NaCl could also be used as 58.44g NaCl / 1 mol NaCl depending on the situati...

RitaThomas_3G

Hello! When we see in the problem that it is a combustion reaction, the only thing we are guaranteed to know is that the products will be CO2 and H2O. However, this does not necessarily mean that the moles of each molecule will be 1. This is because the reactants of the reaction can always be differ...

RitaThomas_3G

Hey! So looking at the work you did, all the steps seem to look correct. However, maybe one place you might have made a slight mistake is rounding between steps, so maybe your final answer was off by a few digits. One thing that I like to do when solving these types of problems is to do all of the c...

RitaThomas_3G

Hey! First, I just wanted to say that yes you are correct in stating that each of the bonded atoms represent the amount of moles in total. This is because the definition of a mole is the "amount of a substance (page F38 of chem textbook)." This means that in the drawing, since there are 2 ...

RitaThomas_3G

Hey! So I was also having a similar problem with understanding this question as well, because I was confused as to how to go from "x" moles of CO2 to percent composition of C, since I didn't initially think we could use CO2 again to find the moles of C. However, it became more clear to me ...

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