CHEMISTRY COMMUNITY

AnnaYan_1l

Half reactions are usually given as reduction reactions.

It's the difference when you're keeping both anode and cathode $E^{o}$ 's at their reduction potentials

It's added when you flip the anode reaction (making it go from a reduction to oxidation) and thus flip $E^{o}$ .

Hope that helps!

AnnaYan_1l

State functions are path independent. They are: Pressure, density, temperature, volume, Enthalpy (H), Internal energy (U), Gibbs free energy (G), and entropy (s). Lyndon's trick to remembering is: when i'm under PRESSURE and feeling DENSE, all I want to do is watch TV (temperature and volume), and g...

AnnaYan_1l

Catalysts reduce the activation energy between reactants to products, meaning it takes less energy to make the forward reaction occur

AnnaYan_1l

Hi, it was just upload last night!

AnnaYan_1l

I have the exact same question! When I was doing the test, I assumed that the gibbs free energy would be greater than 0. However, is it less than 0 (spontaneous) because combustion involves purely adding oxygen and doesn't require heat? (and, would it be correct to say that the BBQ being made from t...

AnnaYan_1l

Yes, it is. K = the concentration of products over the concentration or reactants, which is also used in acids and bases

AnnaYan_1l

Correct, there is zero order reactions do not have molecularity and do not have a rate law

AnnaYan_1l

The Van't Hoff Equation can be used to calculate K at different temperatures, if you know the standard change in enthalpy.

AnnaYan_1l

Jchellis 1I wrote:What does the single line represent then? I thought the single line was the salt bridge?

the single line | represents the interface between phases in contact with each other (including a porous disc/wall)

AnnaYan_1l

After you balance the half reactions, you can see how many electrons are being transferred (how many are lost in oxidation and how many are gained in reduction, which will be the same #) - that number = n

AnnaYan_1l

Jchellis 1I wrote:Wait do you put the OH- on the same or opposite side of H+?

Hi, you put the OH- on the same side as the H+ because OH- + H+ -> H2O. Hope that helps!

AnnaYan_1l

Hi! So, I am working on 14.11d, and I am a bit confused. The question is: Pt(s)|O2(g)|H+(aq) || OH-(aq)|O2(g)|Pt(s) and it asks to write the half-reactions and the balanced equation for the cell reaction... When I was writing the half-reaction for the anode side, I originally wrote it H+ -> O2, sinc...

AnnaYan_1l

I think another property of standard conditions is 25 degrees C

AnnaYan_1l

A state function - delta G can be added or subtracted - it doesn't matter what path is taken to get there, which is how we are able to use the three methods to calculate delta G.

AnnaYan_1l

When delta G is negative, it is spontaneous

AnnaYan_1l

Molecules that are perfectly ordered have no entropy at 0K (ex. CH4). For molecules that are not perfectly ordered, there is positional/residual entropy at 0K.

AnnaYan_1l

Venya Vaddi 1L wrote:The enthalpy of formation of O2 is zero since it is in its standard state.

Thank you!!!

AnnaYan_1l

For 4C, are we supposed to find/know the enthalpy of formation for the reactant O2, or should that be given?

Thanks!

AnnaYan_1l

I think the integrals are just used to understand why the equations were were given in class make sense!

AnnaYan_1l

Are UA sessions in Covel still happening today (February 10)? Thanks!

AnnaYan_1l

From my understanding, Kb is the Boltzmann constant, and is derived from the gas constant R divided by Avogadro's number. I don't think you need to know this though, as it is on the formula sheet!

Hope this helps!

AnnaYan_1l

I think these values will be given!

AnnaYan_1l

eden tefera 2B wrote:You can also cross out bonds if both the products and reactants both have the same number!

Would you be able to explain what you mean with an example? This seems like a helpful tip, but I'm not sure how to do this in practice. Thank you!

AnnaYan_1l

Would someone be able to explain how they do Hess's Law problems? I'm not able to follow the notes from class as clearly, and the problems in discussion confused me. How do you know when you need to reverse the reaction? Are you trying to get the same compound on both the reaction and products sides...

AnnaYan_1l

According to my notes: Internal energy, U, is a state property. So, changes in U are a function of initial and final states (delta U = Ufinal - U initial). The energy of a closed system can be changed by heating/cooling or compression/expansion. Delta U = energy transferred by heating (+q) + energy ...

AnnaYan_1l

According to the notes, a bomb calorimeter is an isolated system because nothing is exchanged with the surroundings!

AnnaYan_1l

The explanation above is really great! Just a quick correction, delta H = 2(188.32) - 541.66 + 92.22 = -72.90 not 72.90

:)

AnnaYan_1l

I believe they are always exothermic because the acids and bases are breaking bonds.

AnnaYan_1l

Molar heat capacity specifies that it is the amount of heat needed to raise the temperature of an object of 1 mole by 1 degrees celsius, while heat capacity does not specify an amount (i.e. 1 mole).

AnnaYan_1l

Thanks! This helps a lot. Follow up question...how do you determine how exactly the weak compounds will dissociate? ex. How would you know that NH2NH2 becomes NH2NH3+?

Thanks again!!!

AnnaYan_1l

The last time I took chemistry was a couple of quarters ago, so I am a bit lost with identifying acids and bases, and whether they are strong or weak. I am most worried with this test that I will be unable to identify how a compound might dissociate in water. Does anyone have any tips on how to do t...

AnnaYan_1l

An analogy that would be easy to understand that highlights this concept is as follows: If a millionaire gave someone who had no money $100, that loss of the $100 wouldn't affect the millionaire. In the same way, removing the -x from the denominator when doing the shortcut method doesn't change the ...

AnnaYan_1l

I think Kp is only used when the reaction only includes gases, so I would default to Kc

AnnaYan_1l

Changing temperature changes K.

In an endothermic reaction (heat is added), products are favored.
In an exothermic reaction (heat is released), reactions are favored.

Hope that helps! (:

AnnaYan_1l

Yes, the reaction continues until Q = K and equilibrium is reached.

AnnaYan_1l

Molar concentrations of a pure substance (solid or liquid) does not change in a reaction. Therefore, solids and liquids are not included in the expression.

AnnaYan_1l

Qc and Kc are calculate the same way, but Qc is used to determine which direction a reaction will proceed, while Kc is the equilibrium constant (the ratio of the concentrations of products and reactants when the reaction is at equilibrium). So, Qc could be = to Kc, but it may not be. When Q < K, the...

AnnaYan_1l

I don't think I understand how lone pairs change bond angles. Let's say there was a tetrahedral molecule. I know that those bond angles are 109.5, but when lone pairs start getting introduced, do the bond angles increase or decrease and why? Does this apply to all molecules that get lone pairs intr...

AnnaYan_1l

By "it," are you referring to the electron? If so, if the photon does not reach the threshold energy, no electron will be emitted.

AnnaYan_1l

What I did to solve this problem was: 5.00 g KMnO_{4} \cdot \frac{1 mol KMnO_{4}}{158.04 g KMnO_{4}} = 0.0316 mol KMnO_{4} M = \frac{n}{V} = \frac{0.0316 mol KMnO_{4}}{.150L} = 0.211 M KMnO_{4} n = MV = (0.211 M KMnO_{4})(0.020 L) = 0.00422 mol KMnO_{4} (0.00422 mol KMnO_{4})...

AnnaYan_1l

If the dipole moments do not cancel out, that would lead to a polar molecule.

AnnaYan_1l

Thank you! I really appreciate it.

AnnaYan_1l

Are there review sessions in Hedrick Hall this week? I was under the impression that there were from the website (image attached), but the UAs for 5-6pm and 7-9pm weren't here today, so I am unsure if I read something wrong or if there was a mix-up?

Thanks!

AnnaYan_1l

We made a similar chart in discussion too. I want to point out that I believe our TA mentioned something about the angles being less than 109.5 in some cases, I'm not exactly sure why or in what instances, but I think he mentioned lone pairs. I also think Lavelle said that we don't have to know the...

AnnaYan_1l

Yes, I believe that multiple atoms in the compound would have to have lone pairs for it to be polydentate. It wouldn't be polydentate if a single N had 3 lone pairs, for example. Instead, it would be polydentate if multiple N had lone pairs.

AnnaYan_1l

You would have to be given the molar mass (which is usually said to be done through mass spectrometry). In this example we did in class: Vitamin C sample of 8.00g is analyzed - the results are C 3.27g, H 0.366g, O 4.36g Let's jump down the steps and assume that you have calculated that the empirical...

AnnaYan_1l

I believe that KA stands for the equilibrium constant for an acid, and pKA provides the -log of the concentration of KA. pKA provides you a way to compare other pKAs if that makes sense.... Let's say a KA = 10^{-10} pKA = 10 and a KA = 10^{-8} pKA = 8 (so this is a stronger acid than the previous one)

AnnaYan_1l

Ligands are ions or neutral atoms that bond to central atoms/ions to create a TM complex/organometallic complex/coordination compound. Ligands act as Lewis bases [electron pair donors]. In the example we had in class, the first ligand we looked at was NH3. It donated its lone pair to Ni2+. Another l...

AnnaYan_1l

Are we expected to memorize these bonding angles? How do you know that the bonding angle is slightly less than 90 degrees? If there are lone pairs in a molecule, this creates repulsion, pushing certain atoms in the molecule "away," which slightly alters the bond angles that would exist if...

AnnaYan_1l

Hi Sammi! TeCl4 has a see saw shape. Professor Lavelle used a similar example in class using SF4 (I believe on Wednesday of Week 8 if you want to refer to your notes). The reason why TeCl4 and SF4 have a see saw shape is because they have the VSEPR notation of AX4E (4 bonding pairs and 1 lone pair)....

AnnaYan_1l

This helps so much, thank you!!! I was super worried about memorizing them.

AnnaYan_1l

Yes, bond angles would change, because of electron repulsion :)

AnnaYan_1l

Anions become highly polarized, while cations have polarizing power.

AnnaYan_1l

Hi! I think we are considering them as largely similar and interchangeable, but there might be a distinction that we are not focusing on.

AnnaYan_1l

This is what I have in the notes: - All ionic bonds have some covalent character (this can be seen in Na+ and Cl-, because the anion's electrons are pulled into the bonding region, giving Na+ and Cl- some covalent character [in other words, the positive cation pulls electrons from negative anions to...

AnnaYan_1l

As the people above have stated, the octet rule is basically a rule in which atoms combine with other atoms so that they have 8 valence electrons. However, there are exceptions to this rule, so keep this in mind.

AnnaYan_1l

How do we determine which atom or ion can be classified as the radical? I think this would depend on the molecule given in a problem...If it had an odd number of electrons (when adding up valence electrons), it would be safe to assume one of the atoms would be considered a radical. This can be seen...

AnnaYan_1l

Hi Betty! I was told in my discussion that anything below this line (see attached image) could potentially have more than 8 electrons.

AnnaYan_1l

One way to think of it is this: Lewis structures are used to show how a molecule may look. However, when you draw a single Lewis structure, you may not be showing all of the possibilities (a double bond could be somewhere else, etc.). So, the Lewis structure you drew would be just a "generalize...

AnnaYan_1l

Jacob is correct! This is how you are easily able to determine how much of a product was made as a result of a limiting reactant.

AnnaYan_1l

You can use the wavelength to determine the energy released (where an electron went from a higher energy level to a lower energy level). This energy is the difference in energy between the two states. Hope this helps! Try to work out the problem, and if you are still having trouble I can write out t...

AnnaYan_1l

The "rule" is that 2nd round ionization energy always requires more energy than 1st round. This is because it requires energy to remove the first electron (which makes the element a positive ion). It takes even more energy to remove an electron from the now positive ion than it would have ...

AnnaYan_1l

Electron affinity is a measure of energy change when an electron is added while electronegativity is how attracted an atom is to electrons. While electron affinity is increasing, electronegativity is decreasing since as an atom is less attractive to electrons, it requires more energy to add an elec...

AnnaYan_1l

I didn't use M(initial)V(initial)=M(final)V(final) because more solution wasn't being added to the aqueous solution. The idea of M(initial)V(initial)=M(final)V(final) is that you would be adding more solution with NiCl2 (for example), and the original solution would still be in the beaker (if that m...

AnnaYan_1l

Hi Maya! So you can basically determine whether an electron is in a ground state or an excited state by seeing whether the electron configurations violate Aufbau principle, Hund's rule, or the Pauli's exclusion principle. If an electron configuration violates any principle/rule, then it is represent...

AnnaYan_1l

Hi, Anna! I had the same issue with this problem originally. It is important to remember that Balmer series means that the electron transitions from a higher energy level to n=2. So, we can assume that: 656.3 nm is a transition from n=3 to n=2 486.1 nm is a transition from n=4 to n=2 434.0 nm is a t...

AnnaYan_1l

I did not have the same version of the test, but here is what I think are the correct steps: So, it says that 50.0 mL of .200M of NiCl2 aqueous solution is prepared (called electrolyte). An electrode surface is placed into the solution, and after the electrodeposition process, some of the Ni is depo...

AnnaYan_1l

I am a little confused as to how an orbital can be a negative number. Say n=3 so you are in the 3d subshell. Th orbitals it says can be: -2,-1,0,1,2 but I don't understand what these numbers correspond to? For a 3d subshell, n = 3 l = 2 (because a d-orbital responds to an l of 2) (the allowed value...

AnnaYan_1l

The question is referring to ml, so: For part a, when l = 0 (it is an s orbital), there is only one possible value for ml (which is 0), so the answer is 1. When l = 2, this means ml can be: -2, -1, 0, 1, 2 (5 values) When l = 1, this means that ml can be: -1, 0, 1 (3 values) When l = 3, this means t...

AnnaYan_1l

Yes, I agree with the people above. Additionally, in the video module, Professor Lavelle showed the correct atomic model: The electron is not confined to the nucleus and we now know, from experimental observation that the size of an atom is determined by its electrons outside of the nucleus. For a H...

AnnaYan_1l

It depends on what you're talking about, but yes, they have a definite mass that can be important in some contexts.

AnnaYan_1l

I agree with the people above! p (which means momentum) is = to (mass)(velocity) which is usually given to you in some shape or form in the question. It is for a moving object (not light)

AnnaYan_1l

When memorizing the general order of wavelengths of the rays from gamma ray to radio waves, do we need to know the general trend of the wavelengths within the visible light as well as the numerical range for the waves? You just need to know generally the electromagnetic spectrum so like the differe...

AnnaYan_1l

Hi! The experiment requires a conductive material (metal), and since plastics and other materials are not conductive, the experiment wouldn't work with the use of a non-metal surface. Hope this clarifies things for you!

AnnaYan_1l

Why does a drop from n2 to n1 have more energy than a drop from n4 to n1? I thought I heard Professor Lavelle say on Monday, when answering someone's question that, a drop from n3 to n1 and then n2 to n1 emits 2 photons but has the same energy as a drop from n3 to n1? In this way, why wouldn't a dro...

AnnaYan_1l

I really like Komal's method! I tried to do something similar on Test 1.

AnnaYan_1l

NabilaNizam-1K wrote:Do we have to study the experiment in detail or do we just have to know the concepts and result of the experiment?

I don't think that we need to know the experiment in detail. Just understand the results/concepts!

AnnaYan_1l

In the equation E subscript n = (-hR)/(n^2), n = energy level. So in the worked example, at that point of the problem, we were calculating the energy at energy level 4 (n = 4):

E subscript 4 = - (hR)/(4^2)

In the next step:

E subscript 2 = (-hR)/(2^2)

I hope this was helpful.

AnnaYan_1l

Gabi Landes 1-H wrote:Is it expected of us to memorize the polyatomic ions?

I don't think it's expected to memorize the polyatomic ions yet! I think that will be expected later on in the course, but we haven't focused too much on it yet so I wouldn't worry about it until then (:

AnnaYan_1l

When you get the final answer for a problem, for example, #2. CHEM, an FM radio station, broadcasts at 99.5 MHz. What is the wavelength of the corresponding radio waves?" (from TA worksheet 2) would the resulting answer have be rounded to 3 sig figs because of the "99.5 MHz"? Yes, 3 ...

AnnaYan_1l

There is no difference between notations, so you can use either notation when writing out your problems!

AnnaYan_1l

What occurs to the kinetic energy that is the excess when the energy of the photon is greater than the threshold energy? I believe that in the photoelectric experiment, this kinetic energy is detected by a detector which shows the excess energy of the photon that surpassed the threshold energy (sin...

AnnaYan_1l

The Heisenberg Equation will not be tested in Test 2 (:

AnnaYan_1l

I believe it would be helpful to know, just in case. I think it'll help to have a general idea of how large each type of electromagnetic wave is (especially to know visible is between 400 and 700 nm, for example), if only to help you double check your work on the test. Hope that helps!

AnnaYan_1l

Hi, yes! You just refer to the electromagnetic spectrum to see where the wavelengths you found lie.

AnnaYan_1l

Jocelyn1B wrote:Sorry this is off topic but do we have to memorize these two equations or is it going to be given to us?

Hi Jocelyn! These equations as well as constants will be given on the test!

AnnaYan_1l

Yes, I believe this would be the best way to look at it.

AnnaYan_1l

I do not think this will be tested. :)

AnnaYan_1l

I'm super interested in this! Could you make a groupme?

AnnaYan_1l

I agree with Marisol! Frequency is shown as a s^-1 and wavelength is shown as m. When multiplied to get constant, this gives you m/s or m.s^-1. Hopefully this makes sense!

AnnaYan_1l

Yes, 1 Hz = 1 s-1 (one cycle per second).

AnnaYan_1l

Thank you!!! That makes so much sense now.

AnnaYan_1l

The question asks, "A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g mol 1. When 0.376 g of caffeine was burned, 0.682 g of carbon dioxide, 0.174 g of water, and 0.110 g of nitrogen were formed. Determine the empirical and molecular formulas of caffeine, and write the ...

AnnaYan_1l

There are different rules for different operations. Multiplication and division are pretty straight forward, but addition and subtraction are a little bit different. Multiplication and division: Round to the least number of significant figures of the numbers in the problem. (ex. 3.5 * 2.03 = 7.1) Ad...

AnnaYan_1l

I understand/agree with how Kuldeep got the answer to part A! I just have a couple quick questions...How does the portion of the question that says, "Some of this solution was transferred to a buret" relate to the question? Also, this might be a dumb question but how would you know the mol...

AnnaYan_1l

I usually do what works for the problem/how comfortable I am with the concept. So, for the first few problems I might break everything up into steps so I understand what I'm doing, and then when I feel more comfortable with the problem, I might write everything out on one line. Whatever helps you un...

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