CHEMISTRY COMMUNITY

Jana Sun 1I

I'm confused. My answer book does include Cl- in the products and I think that it definitely should be included. Correct me if I'm wrong, but I don't think that being a product of the slow rate affects the fact that it is still a product.

Jana Sun 1I

Oh wow I didn't even notice the -0.59 in the solutions manual until now. I didn't use it at all in my calculations and I still go the same answer (probably because our answer ended up being so large that the -0.59 wouldn't have made a difference). But now I'm really curious about the -0.59. Does any...

Jana Sun 1I

Regarding your first question, catalysts aren't "consumed" in the total reaction because they are formed again in a later step. Technically, they do get consumed in whatever step they're a reactant in, but they always get produced again by the end of the total reaction. A good example of t...

Jana Sun 1I

I understand your overall process, but I don't get why we need to do this step: "To make the problem a bit easier, you can rewrite Ea,cat as a function of Ea,uncat: Ea,cat = 75 kJ/mol and Ea,uncat = 125 kJ/mol Ea,cat = 75/125 kJ/mol*Ea,uncat = 0.60 kJ/mol*Ea,uncat" Why can't we just keep E...

Jana Sun 1I

No problem! Also, in 15.63, we use the same equation but they kept the term as (1/T - 1/T'), so I really don't know why they felt it was necessary to change the it to ((T' - T)/(TT')).

Jana Sun 1I

I agree with Lindsay. The question doesn't really need us to think about the activation energy for the reverse reaction. It just asks us to calculate the activation energy for the forward reaction, which doesn't change at different temperatures. I think we derived the equation we use under the assum...

Jana Sun 1I

That's a good question. I don't know why they changed it around. I solved it by leaving it as (1/T - 1/T') and I got the same answer.

Jana Sun 1I

To understand why the half-life of first order reactants are independent of the substance's initial concentration, it really helps to look at how we get our half-life equation. We start with our integrated rate law ln[A] = -kt +ln[A]o. Since the half-life of anything is the time that it takes for ha...

Jana Sun 1I

That's a good question, I'm wondering about this too. I doubt we would be penalized for drawing out the structures instead of writing them out, but would we lose points if we didn't write them out correctly correctly? In high school, I remember my chemistry teacher just telling us to follow the shap...

Jana Sun 1I

I'm still kind of confused as to why we need this information. The question is just asking us for the time it takes for the concentration of A to decrease from 0.1M to 0.08M. We don't seem to need this information to figure that out.

Jana Sun 1I

Does anyone know why we convert mmol to mol? Is it absolutely necessary? For instance, in 15.17, we're asked to do similar calculations but everything stays in mmol.

Jana Sun 1I

Regarding the first part of this problem, is there a reason why the answer book converts mmol to mol?

Jana Sun 1I

I'm confused about this too. Why do we include H+ in our cell diagram?

Jana Sun 1I

For this problem the cell diagram in the solutions manual divides Ag(s) and AgBr(s) on the anode side with a vertical line. Can someone explain why they aren't separated by a comma even though they're the same phase? This is addressed in this thread: https://lavelle.chem.ucla.edu/forum/viewtopic.ph...

Jana Sun 1I

Why do we use H+ instead of H30+?

Jana Sun 1I

2) I'm still a little confused about this. I previously thought that if work is done by the system, work would be negative. If work is done on the system, work would be positive. In this case, work is done BY the system, so shouldn't our calculated w be negative instead of positive? But if I'm under...

Jana Sun 1I

If Delta G for formation of a product A is negative, we know that the forward process is spontaneous and will "end up" as product A, meaning that product A is stable. In the same vein, lets assume the Delta G for formation of product B is positive. The forward process that forms product B...

Jana Sun 1I

Since a temperature is not explicitly given, I think this is one of the cases where we have to assume that things are happening at room temperature, which we usually assume to be 25C aka 298K.

Jana Sun 1I

If you look at Table 9.1 (page 328) in the book, the last column gives us the standard entropy of vaporization for a bunch of liquid substances (ammonia included!). I think we'd just use that because we're looking specifically at vaporization reactions.

Jana Sun 1I

Bromine is farther down the periodic table than Flourine. Thus, Bromine is larger and contains more elementary particles than F in HF. More particles means more entropy. To compare entropies, look at the states of matter if they are different. Entropy increases from solid to liquid to gas. What exa...

Jana Sun 1I

If I'm understanding you correctly, yes. If 9.15 had asked for the entropy change of the freezing of water not at 0.00C, then yes, you would have to do further calculations, similar to those that you did in 9.19.

Jana Sun 1I

I think this person had the same question: viewtopic.php?f=130&t=26744.

I don't think I got the same question on the test, but for the situation you're describing, work would be done at least (because there's a change in volume and pressure).

Jana Sun 1I

Is there a specific homework problem that uses 7/2R? I'm just a little surprised because I haven't seen it before and I'm not sure when I would know to use it.

Jana Sun 1I

This might be going on a limb, but I thought carbon's most stable form was as diamond. Can someone explain why/how graphite is most stable than diamond?

Jana Sun 1I

In the solutions manual, why is ln (v2/v1) stated as ln (2.4/1.2)? Where does the 2.4 come from? The 2.4 is the final volume, which part b says is the same as part a. In part a, they say that "the gas is allowed to expand through an additional 1.20L." So in addition to the 1.20L already p...

Jana Sun 1I

I had a question on another part of this problem. When we're using q in our calculations, why is our q positive instead of negative (because the question gives us a negative q)? Is it because for the subsequent experiment with HBr and KOH, the temperature rises?

Jana Sun 1I

How exactly does the gas' ability to expand "isothermally" play a role in the problem?

Jana Sun 1I

For reversible reactions, you would use work= -nRT(ln(Vf/Vi)) or work= -nRT(ln(Pi/Pf)). For irreversible reactions, which we'll probably need to know better for the first exam at least, we use the work equation described above aka work= -(Pext)(delta V).

Jana Sun 1I

Also just a heads up, in my discussion section today, my TA mentioned that we don't really need to know/use the reversible equation on the first test, but it's fair game for later tests/midterm.

Jana Sun 1I

I'm still a little confused. The answer book gives our answer in kJ without dividing our standard reaction enthalpy by any mol value. For instance, in #63 part a, we're looking at the production of nitric acid. If we wanted our standard reaction enthalpy in kJ, wouldn't we need to divide by 2 (becau...

Jana Sun 1I

Technically, I think it can still be negative. The negative sign would just represent that x amount of KJ would be released per liter, so you could just clarify that if you choose to write a negative sign. However, my TA mentioned that in real-life applications, people typically just represent entha...

Jana Sun 1I

One key piece of information is that the standard enthalpy of formation is the standard enthalpy of reaction for the formation of one mole of a substance. Therefore, the units on a standard enthalpy of formation will always be in KJ/mol. You can also refer to this previous post addressing the differ...

Jana Sun 1I

On a test, we would probably be given the standard enthalpies of formations we would need to solve the problem so with that information, we could probably deduce which ones are zero. In terms of understanding though, you bring up a good point with your water example. My guess is that it is due to st...

Jana Sun 1I

Generally, to determine whether or not to omit the x in the concentration of the reactant depends on x. If the calculated value of x is less than 5% of the initial concentration of that product, you make generally omit x. However, I find this to be kind of confusing as well. You don't really know if...

Jana Sun 1I

Although we might not HAVE to memorize them, I've found that in past chemistry classes it oftentimes makes problems easier to memorize them. In addition to understanding their characteristics (as mentioned in the previous post), it might be easier to just know a few strong acids (HCl, Hbr, HI, etc.)...

Jana Sun 1I

I am confused about part b. It is asking for the conjugate acid of NH2NH2. Both nitrogens have a lone pair, so could the hydrogen from H2O attach to either one? Yes, you're right. The hydrogen could form on either nitrogen. NH2NH2 will only accept one hydrogen to create a conjugate acid and I'm not...

Jana Sun 1I

Also remember if water is outside the brackets, we use "hydrate".

Jana Sun 1I

I think you just have to memorize polyatomic ions and know how their nomenclature works/changes. I'm not sure if there's an easier way to tell, sorry!

Jana Sun 1I

Maddie Hong 1I wrote:Also, I would note that water is sometimes written as OH2 when in the brackets, because the oxygen of the water is what connects it to the central atom.

Would water outside the bracket ever be written in OH2 form?

Jana Sun 1I

I agree with the post above; I think we're just supposed to memorize the charges of the ligands. As for the formula for calculating the oxidation number of the metal, I personally don't write everything out in the formula. I just know that the charges of the molecule should add up to the overall cha...

Jana Sun 1I

However keep in mind that due to the differences in the shapes of cis- and trans- molecules (described by previous comments), cis molecules are often polar while trans molecules are often nonpolar!

Jana Sun 1I

The term "angular" also came up in 4.5. In the answer key, it says that the shape of SOCl2 is angular but the arrangement is trigonal planar. I'm a little confused because I thought "bent" or "angular" and "trigonal planar" are both shapes...so how can SOCl2 b...

Jana Sun 1I

It's easy to spot a radical because of the number of odd electrons. However, is there an easy way to spot biradicals? Thank you!

Jana Sun 1I

I was also confused with the notation in this problem. How would you draw the compound under a single bracket then?

Jana Sun 1I

Actually, since nitrogen has a perfectly half filled p-orbital, it is more stable than carbon. Meanwhile, carbon only has two half filled p-orbitals out of three. Therefore, it wants another electron more than nitrogen does in order to fill half of its p-orbital completely; so, its electron affinity...

Jana Sun 1I

Like people have noted before, we can first use the noble gas notation [Xe] to notate all the electron configurations up to Xe. Then, we write 4f^14 to account for all the electrons in the f block (starting with elements 57 and all the way to element 70). We get the number 4 because n-2 gives us our...

Jana Sun 1I

Just a quick note: my high school chemistry teacher taught us a neat trick to memorize the m sub l quantum numbers associated with each s-, p-, d-, and f- orbitals. You can create a number pyramid like below: 0 (s-) -1, 0, 1 (p-) -2, -1, 0, 1, 2 (d-) -3, -2, -1, 0, 1, 2, 3 (f-) The numbers on each l...

Jana Sun 1I

I can confirm that a constants and equations sheet will be provided! Going back to the original question though, I was confused by the last part: "report the difference as a percentage of the wavelength of a neutron". Are we supposed to just calculate both wavelengths, find the difference ...

Jana Sun 1I

Would this also apply to Planck's constant? Sometimes I see 6.626*10^-34 Js but other times it's shortened to 6.63*10^-34 Js.

Jana Sun 1I

I agree with the post above. It might depend on the problem. Some might ask specifically for you to convert m to nm, but if it doesn't specify, I don't think you would lose points or anything if you kept it in m. However, it just might be easier to interpret a result if it's in nm.

Jana Sun 1I

This is a really small detail, but what if we started with KNO3 in the beaker and then mixed in water rather than starting with water in the beaker and then mixing in KNO3? Does the order make a difference?

Jana Sun 1I

You're on the right track! After you figure out the limiting reactant, you can start with the amount of that limiting reactant you're given, whether that's moles or grams, then use dimensional analysis with molar masses and molar ratios to determine the amount of product produced. In this problem, i...

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