CHEMISTRY COMMUNITY

Kessandra Ng 1K

You know the half-life equation for a second-order reaction is: t_1/2 = 1 / k [A]_o

Therefore, after finding k, you can use the integrated rate law 1/[A] = kt + 1/[A]_o for a second-order reaction to find t.

Kessandra Ng 1K

Yes I think we would need to know. In my discussion section, we went over a question that included labelling the reaction profile so I'd assume it's important! Also, it's good practice to see if you understand the theory behind each of the symbols and what they mean.

Kessandra Ng 1K

Just be sure to know that while the reactant is still in the overall reaction equation, when you write out the rate law, it would be [A]^0 (assuming the reaction is zero order with respect to A) - which equals 1, therefore you wouldn't need to include it in the rate law when you're writing it out.

Kessandra Ng 1K

Hi! Personally I go to the review sessions (I found Lyndon's midterm review session extremely helpful) and do the step-up session worksheets that are posted on Chemistry Community. There's a lot of resources like maths advice on Dr. Lavelle's website so they're good for clarifying things we don't co...

Kessandra Ng 1K

Yep, it's fine though it might help when everyone's gotten their tests back by the end of the week to compare answers and concepts that are unclear!

Kessandra Ng 1K

Test 2 was on electrochem up till the Nernst equation, so I assume that if you do problems about the Nernst equation and after it, it should be fine? I'm not sure though.

Kessandra Ng 1K

Hi! Does anyone know what questions the electrochemistry portion of Test 2 will go up to for the 6th Edition? So what homework problem number for chapter 14.

Kessandra Ng 1K

If the problem asks you about different temperatures, for example, and you had to find ∆H and ∆S in the previous parts, then you would use ∆G = ∆H - T∆S. Otherwise, if they give you the values for ∆Gf for the products and reactants, you would use the other equation instead.

Kessandra Ng 1K

To add onto what the user above said, what to remember is that because it's reduction, you add on electrons onto the left side of the equation.

Kessandra Ng 1K

I would assume these values would be given in any exam questions. In the book, it should be in the appendix tables or given in the question in order for you to actually solve it.

Kessandra Ng 1K

Thanks! I thought that Dr. Lavelle said didn't round anything until the end of the question/the final answer though? Especially since the question asked wasn't specifically telling us to find Qp.

Kessandra Ng 1K

Hi! For question 1A about Q and Kp, why does Q have to be rounded to 1sf (in the midterm solutions it's 2 x 10^-2) even though it's just working out to the final answer?

Kessandra Ng 1K

I personally found Lyndon's midterm review extremely useful - but it's more useful if you attempt his questions beforehand so you're not looking at it for the first time during the session. Also, the UA worksheets are super helpful and Dr. Lavelle's homework problems!

Kessandra Ng 1K

Just to add on, if there's a º to something, i.e. ∆Gº or ∆Hº or ∆Sº, it always refers to "under standard conditions" so you would already know T = 298 K and P = 1 atm.

Kessandra Ng 1K

I think you can use both, and they should give similar (but not necessarily the same) answers. Since you've already found ∆H and ∆S, then it would make sense to use the ∆G = ∆H - T∆S equation.

Kessandra Ng 1K

I think that if the equations are more difficult than combustion reactions (where it's just reacting with O2), Dr. Lavelle should give us the equation in the midterm.

Kessandra Ng 1K

They're just different names for the same process, i.e. a constant volume process. Like the user above mentioned, just make sure to understand the differences between isochoric, isothermal, and isobaric processes.

Kessandra Ng 1K

In Lyndon's review session, he also pointed out that while heat capacity is an extensive property, specific heat capacity and molar heat capacity are both intensive properties.

Kessandra Ng 1K

Hi! Just wondering which chapters/topics we are allowed to do questions on for our homework - are we allowed to do both thermochemistry and thermodynamics?

Kessandra Ng 1K

An important distinction is between closed and isolated systems also, because the bomb calorimeter example seems to come up a lot in class so maybe this is an idea that we need to understand really well. The bomb calorimeter would be closed instead of isolated, as it allows heat to be exchanged and ...

Kessandra Ng 1K

What I usually do for midterm revision is do all the homework problems and Step Up session worksheets and go over my lecture notes if there's a concept that I don't understand - I think this is super important because you can basically do any question once you fully understand a concept!

Kessandra Ng 1K

In the book (6th edition), the exercise questions are split up into topics so you can do all the questions up to the end of the "Entropy" section and before the "Gibbs Free Energy" section, I believe.

Kessandra Ng 1K

We just need to know how changing temperature, pressure (and thus volume), and concentration would affect the reaction at equilibrium, i.e. how it shifts equilibrium position or the equilibrium constant (if applicable).

Kessandra Ng 1K

Dr. Lavelle mentioned in class (I think) that we would be given a table of bond enthalpies for everything we need to know - though like what others said, it would be useful to memorise the more common ones!

Kessandra Ng 1K

Dr. Lavelle drew out the Lewis structures of both the products and reactants and highlighted which bonds were added/broken. Though, it would be helpful to do the longer calculation of all the bonds just to check also!

Kessandra Ng 1K

When calculating the concentration of a reactant or product and given that Kc is less than 10^-3, you can disregard the x if it's something like "0.50-x" and approximate it to 0.50 only. When you get your final result, you then use the 5% rule to see if it is a valid approximation.

Kessandra Ng 1K

To add to this, the question would usually give you either concentration values or partial pressure values and when given one or the other, you would know whether to use Kc or Kp.

Kessandra Ng 1K

Weak acids don't fully dissociate in solution, so there is a lower concentration of H+ ion in solution. However for strong acids that fully dissociate, there are more H+ ions in solution which makes the solution more acidic and hence, have a lower pH value as compared to weak acids.

Kessandra Ng 1K

For all coefficients in the ICE table, you would follow the stoichiometric coefficients. I think Dr. Lavelle also explained an example in class which used +2x or -2x also.

Kessandra Ng 1K

I just think of using ICE tables whenever I see a question mention weak acids or bases, when you need to calculate equilibrium concentrations of anything, or when they give you the equilibrium constant.

Kessandra Ng 1K

If the hydronium ion concentration is less than 10^-7, e.g. 10^-9, the solution is considered neutral; a solution producing such a small amount of hydronium ions, which doesn't really affect the pH (hence why it's neutral). Autoprotolysis generates 10^-7 molL^-1 of hydronium ions.

Kessandra Ng 1K

If you're referring specifically to homework questions, I believe most of them will say specifically which K/Kc/Kp they want.

Kessandra Ng 1K

I also thought that you only use Kp if all the the reactants and products in the reaction are gases, but usually they specify whether they want you to use Kp or Kc in the textbook homework questions so I assume Dr. Lavelle will do the same on his tests also.

Kessandra Ng 1K

Hi! I think it's not possible to switch from 14A to 20A, but the other way round is totally fine. You'll be able to catch up on anything we've done so far and to be honest, you probably won't need to catch up on much anyways!

Kessandra Ng 1K

Just remember that if the element can have access to the d-orbital, then it will be able to have an expanded octet as electrons can 'go' into that orbital.

Kessandra Ng 1K

Another good one to remember is that H2O is aqua when naming compounds. I think Dr. Lavelle also mentioned how, for compounds like CN-, you can use either cyano or cyanido.

Kessandra Ng 1K

On Dr. Lavelle's website, he said "Questions will come from the Homework and Online Assessments." and "Final exam covers everything in the Syllabus." - everything is fair game. Also, there's some things we learnt later on in the course that we wouldn't be able to do without knowi...

Kessandra Ng 1K

For VSEPR, just be sure to memorise all the names, bond angles, etc., and be sure you know how to draw Lewis structures (because this is normally how you figure out the rest of the info). For bond lengths and strengths, I think the only thing you need to remember is that the shorter the bond, the st...

Kessandra Ng 1K

Dr. Lavelle's website says: Test 3 covers: End of Bonding from 3.12 (6 Ed.) and from 2D (7 Ed.); and all of Molecular Shape and Structure (see Syllabus and Outlines for details).

Coordination compounds will not be tested!

Kessandra Ng 1K

I think that multiple bonds will always be shorter than single bonds, no matter which elements are involved - i.e. if the elements are the same or different. This is because the more bonds there are, the more electrons that are shared between the atoms, and hence the stronger the bond.

Kessandra Ng 1K

While we won't be required to draw any 3D shapes of molecules, we will be required to know the shapes, bond angles, hybridisation, and how to draw the Lewis structures of molecules. If you start from drawing the Lewis structure, you can get everything else from VSEPR theory so you're good!

Kessandra Ng 1K

The user above is right^. My TA emailed us saying that while there is discussion tomorrow, we would not need to hand in homework until week 9. I guess this is for fairness so everyone hands homework in the same week!

Kessandra Ng 1K

Adding to this post, if you knew the AXE 'formula' of GeF4, you could use that to determine the shape easily also, since all molecules with that particular AXE 'formula' would have the same shape! Therefore from there, you can also get the bond angles.

Kessandra Ng 1K

When we are determining the shape of and bond angles in a molecule, we use the VSEPR theory. By figuring out the steric number of the molecule and drawing out the Lewis structure, you can use the number of bonds and lone pairs (i.e. the steric number) to figure out the shape of the molecule. After y...

Kessandra Ng 1K

VSEPR stands for Valence Shell Electron Pair Repulsion, which should give you an idea of what the theory is for. The principle idea behind this theory is that electron pairs around a central atom will assume a position that keeps them as far apart from each other, as electron pairs and bonds repel o...

Kessandra Ng 1K

I think it really depends, because I was once able to enrol in a class I didn't have the pre-requisite or the co-requisite for (I was taking the pre-requisite along with the actual class). If my.ucla lets you enrol and the prof doesn't drop you, then you're good to go. It's just that you'll be much ...

Kessandra Ng 1K

This means you can determine the shape by looking at the number of atoms the central atom bonds to, and also the number of lone pairs of electrons on the central atom. Those are the 2 main factors in determining shape.

Kessandra Ng 1K

Here, the thing to remember is that even though we normally think of ionic bonds as between a non-metal and a metal, as in BeBr2, we have to consider the relative electronegativity of the atoms involved in the bond and the electronegativity difference between them. Usually, the electronegativity dif...

Kessandra Ng 1K

The distinction between the 3 shapes you named are the number of lone pairs on the central atom, as the electron-electron repulsion by the lone pair of electrons will cause the other atoms to shift closer to each other; thus, the number of lone pairs on the central atom affects the shape of the enti...

Kessandra Ng 1K

If the moles of each atom are basically the same and near an integer value, then you can just go off those to find your empirical formula. If the moles of each atom don't seem to follow any obvious ratio, you'll just have to trial and error different numbers to multiply the mole ratios by to get nea...

Kessandra Ng 1K

Does this mean things like periodicity trends, i.e. electronegativity trends down a group and across a period, won't be tested because that's section 3.12 in the 6th edition of the textbook?

Kessandra Ng 1K

Dr. Lavelle said that "All material covered in Fundamentals, Quantum, Bonding to end of 3.11 (6 Ed.) and to end of 2C (7 Ed.)"

For the 6th edition, does this mean up to question 11 in chapter 3? I'm kinda confused because I feel like we've learnt the content for questions past question 11!

Kessandra Ng 1K

Resonance hybrids are just the 'sum' of resonance structures one can have for that molecule. As such, the benzene resonance hybrid would only be valid for benzene itself. Other molecules and ions that have several resonance structures can have resonance hybrids too, like the carbonate ion.

Kessandra Ng 1K

I don't think we're expected to list the exact ionisation energies in our answers unless we're given them in a table or something. I would assume not because we're only given the basic periodic table in our tests anyways! Just as long as you get the order of elements right.

Kessandra Ng 1K

I think it's hard to figure out what ions form from the transition metals, so for example copper can form 1+, 2+, 3+, etc. ions. I don't think we'd need to know the ions that the transition metals would form for the test other than the really simple ones? All we need to know is that the electron is ...

Kessandra Ng 1K

So Dr. Lavelle mentioned that he would have several important announcements on Monday, so I'd assume he'd be giving us information about the midterm then. I'd recommend starting reviewing everything we've done so far, though I'm not sure if we have time to go through the entirety of chemical bonds b...

Kessandra Ng 1K

I don't think you would need to know exactly which electron is +1/2 or -1/2; just know that the electrons in the same sub-shell must have opposite spin so one of them must be +1/2 and the other -1/2.

Kessandra Ng 1K

This is because the 4s sub-shells are of a lower energy than the 3d ones so that's why the electrons fill the 4s ones up first.

Kessandra Ng 1K

ms-1 is metres per second, so a measurement for the speed/velocity of something, while ms is usually milliseconds, which is a measurement for time.

Js is joules multiplied by a second, so a measurement for the action of a system over time, while J is just joules, which is a measurement for energy.

Kessandra Ng 1K

Effective nuclear charge is basically the attraction the valence electrons around the nucleus feel from it, so what Dr. Lavelle is saying is that the inner electrons 'shield' the amount of attraction the outer electrons feel from the nucleus, hence why it feels a lower electrostatic attraction.

Kessandra Ng 1K

I think Dr. Lavelle said that everything from The Quantum World could be tested?

Kessandra Ng 1K

If the wave has frequency 300Hz, you can use wavelength = speed of light/frequency to find the wavelength. To calculate energy, you can use energy = Planck's constant x frequency.

Kessandra Ng 1K

Hi! I'm going to guess that you need to first understand the properties of light before you can fully understand the nature of an electron (through understanding how the photoelectric effect works), and subsequently the nature of an atom.

Kessandra Ng 1K

You could look at either of the three to distinguish between different 'types' of waves. I think looking at the wavelength is the easiest, but whatever method you prefer works too!

Kessandra Ng 1K

If it's something large like 2.38, or as long as it's not very near the whole number, I think you have to find another number to multiply it by. So if it's like 1.01 or 4.99 then it's fine to treat it as 1 and 5 respectively, but if it's 1.2 or 4.8 then you should definitely find another number to m...

Kessandra Ng 1K

A solution is prepared by dissolving 0.500 g of KCl, 0.500 g of K2S, and 0.500 g of K3PO4 in 500. mL of water. What is the concentration in the final solution of (a) potassium ions; (b) sulfide ions? I know how to do these questions when it's only one solute, but since there's three solutes here how...

Kessandra Ng 1K

Hi! I think concentration and molarity and basically the same meaning and you can use both of them, but while concentration is more general as the measurement of a certain quantity of solute in a certain unit of solution, molarity specifies that it's a measure of the number of moles of solute per li...

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