CHEMISTRY COMMUNITY

Astha Sahoo 3I

So first, you find the rate law of the slowest step, so that would be step 2, and the rate law would be rate = k[HClO][I-] since you use the reactants for the rate law. However, you need to find a substitute for HClO since it's an intermediate. So, you take the first reaction, find the rate laws for...

Astha Sahoo 3I

The intermediates aren't in the rate law, you would use another step to calculate the substitution of the intermediate. Also catalysts can be included in the rate law, and can be found when they're first part of the reactants in one step, then part of the products in a further step. Hope this helps ...

Astha Sahoo 3I

You would add the activation energy of the forward reaction with the positive delta H of the forward reaction, so you would get 97 kJ in total I believe.

Astha Sahoo 3I

If the solution is acidic, first balance extra Os with H20 and then balance out extra H+ with H+. If the solution is basic, balance out the extra Os with H20, then balance out extra Hs with OH-.

Astha Sahoo 3I

I think the half reactions were 2H+ -> H2 and 2Cl- -> Cl2. I'm not sure though why HCl is on both sides, I believe it might be because when you're conducting the experiment you would use HCl and not H+ and Cl- ions individually? That's just a wild guess though, and if someone has an actual answer th...

Astha Sahoo 3I

I think because Hg(l) is a liquid? It seems like the Pt(s) is only there when there are only aqueous elements or aqueous and gas elements.

Astha Sahoo 3I

So you take the rate law of the slowest step, so that would be step 2, and the rate law would be rate = k[HClO][I-].However, you need to find a substitute for HClO since it's an intermediate. So, you take the first reaction, find the rate laws for the reverse and forward reaction, and make them equa...

Astha Sahoo 3I

I had the same exact problem! I finally got it right, and this is what I did: I didn't put a exponent for 1, left out the [C], and didn't put a multiplication symbol in between [A] and [B], and it finally worked. my final answer was : k[A][B]^2. Hope this helps!

Astha Sahoo 3I

I have the same problem! If you do receive the Cal grant though, I think you can request to have it be applied to your summer classes through a petition. I found most of my information on this website: https://www.financialaid.ucla.edu/Undergraduate/Summer-Information#495121507-maintaining-summer-ai...

Astha Sahoo 3I

Basically, add H2O to balance out hydrogens/oxygens or OH- to balance out oxygens in the reactions, compared to adding H+ ions when balancing reactions in acidic conditions.

Astha Sahoo 3I

Since H2O is part of the "surroundings" and any changes in some water molecules is probably insignificant in the big picture, we don't include it. I feel like we would include H+ if it's significant(being oxidized from H2 or something along those lines)?

Astha Sahoo 3I

I'm not sure either, but since they're moles of electrons, I believe we're just dividing by the number of moles of electrons to get the value for one mole of electrons, since the value is for 2 moles of electrons transferred.

Astha Sahoo 3I

I was stuck on this problem for ages, it turns out I had the right equation, sapling just wanted the reactants in a certain order, so that may be the issue. Hope this helps :)

Astha Sahoo 3I

You would put the CL- near the double bar since they are close to the salt bridge, and then just write the other molecules in the order they are in the equations

Astha Sahoo 3I

When work is done on the system, the system gives off energy to the surroundings instead of expending energy to do work, thus work is negative.

Astha Sahoo 3I

Delta G not is at standard conditions, and delta G is at any temperature. You can use delta g not to find the delta g at other temperatures

Astha Sahoo 3I

I would take the constant inside of the ln (like 40 if it was ln40) and then raise e to that power!

Astha Sahoo 3I

It's basically how many states a molecule has with equal energy. In most problems it's given, but for some like NO it's easily figured out that there's only two states because it can only be orientated N-O or O-N

Astha Sahoo 3I

The Cv value is always less than the Cp value, the constant for Cp is always one more than the constant for Cv. There are different values for linear, non-linear, and monatomic molecules.

Astha Sahoo 3I

It's similar to delta H! When it's greater than 0, it's not gonna form spontaneously because it needs energy. But if less than 0, it does form spontaneously, since it releases energy and increases entropy!

Astha Sahoo 3I

Often delta S uses joules while delta H and most other terms use kilojoules, so I don't think there's a definite answer to that, I think you always need to look at the units and convert accordingly, but delta S is usually the only odd ball. Hope this helps :)

Astha Sahoo 3I

Enthalphy is the difference in heat, while entropy is the difference in disorder. Usually, spontaneous reactions involve the increase of entropy since the universe wants to always increase entropy.

Astha Sahoo 3I

Hi, can someone explain why we also have to calculate the entropy for cooling the water back down? Thanks You would use the same equation as when we're heating it up, delta S = nCvln(T2/T1), except you would use the Cv for the lower temperature and the lower temperature would be T2 and the higher t...

Astha Sahoo 3I

I used the gas constant of 8.31 and just converted pkA to atmospheres(bars would be a little more accurate actually)

Astha Sahoo 3I

We use the rule for logs. Ln(a) + ln(b) = ln(a*b). Thus, we multiply the values for concentrations, but for the pKa and pKb, since they are logs, they use the rule of logs.

Astha Sahoo 3I

An extensive property is one that changes as the amount of the substance changes. For example, when you add more water to a half-filled cup, the surface area, volume, and mass of the water change, because surface area, volume, and mass are extensive properties. Intensive properties would be density ...

Astha Sahoo 3I

If on both sides of the equation 4.184 is present, for example -qsurr = -qsys, and we're solving for final temperature, then we could cross out the specific heat capacity(4.184) on both sides. Hope this helps :)

Astha Sahoo 3I

No, since you always need to add heat to increase the temperature of a substance.

Astha Sahoo 3I

Just use the equation w = -P * delta V, and the first step equals 0 because the volume is constant. Hope this helps :)

Astha Sahoo 3I

Yes! In Justin's workshop today, he told us to think about the pH as the surrounding, and the pKa as the species, and the same applies to pOH and pKb!

Astha Sahoo 3I

I think we always do enthalpies of product - reactants. If we pretend that at all bonds are broken in reactants and all bonds are newly formed in products, we do subtract the released energy when breaking bonds from forming them, thus subtracting reactants from products. However, to make the math ea...

Astha Sahoo 3I

Yes! The 7 strong acids are HClO3, HBr, HCl, HI, HNO3, HClO4, H2SO4 :)

Astha Sahoo 3I

I believe that they usually open up more spots for second pass as they have done for most of the chem/math classes in the past! Not sure about Chem 14BL though. I would recommend getting Chem 14C since for a lot of majors it's a required course, and you can take 14BL anytime.

Astha Sahoo 3I

Funmi Baruwa wrote:oh okay so it is proton transfer, but it doesn't necessarily have to involve water molecules?

Yes! The proton transfer just has to happen between two of the same molecules.

Astha Sahoo 3I

The change is negative if the molecule is a reactant, the change is positive if the molecule is a product. For this rule to work all the time, make sure that the reaction is going forward (towards the right to the products).

Astha Sahoo 3I

He meant that the proton concentration was so low that it's irrelevant, and thus the pH of water, which is 7, is all that matters. The pH of an acid, no matter how weak, is never above 7, so if you do calculate that it's over 7, it means that the acid is irrelevant and the solution's pH is 7.

Astha Sahoo 3I

Since sublimation is a solid transforming into a gas, using Le Chatelier's principle, you can add the enthalpy of the reaction of solid going to a liquid(fusion) with the enthalpy of the reaction of a liquid going to gas (vaporization) to get the enthalpy of a solid going to a gas, or sublimation.

Astha Sahoo 3I

Yes. It's when a proton from two identical molecules ionizes one of the molecules. In water, the H+ from one H2O molecule ionizes the other one so H3O+ is formed, and the molecule that lost the proton is now OH-. Hope this helps!

Astha Sahoo 3I

Yes I believe by then the steam has diffused into the air, but if you were close enough to the steam you may get a minor burn or even a severe one.

Astha Sahoo 3I

I would ask your TA since every TA grades differently. Usually they are lenient from what I've heard and probably would be okay with you doing that, and perhaps post a couple more than usual for this week's posts to make up for being late? I was terrible at remembering to post on Chem community and ...

Astha Sahoo 3I

No, because solids don't really have concentration, so they have no impact on the K, and thus changing the amount of solid doesn't affect whether the reaction favors reactants or products. Hope this helps :)

Astha Sahoo 3I

What do you do to K when you need to divide the equation? I think you would raise it to 1/constant you're dividing by. For example if you're dividing by 2, it would be K^1/2. Because when you multiply an equation by a constant, you raise that constant to the exponent (e.g. multiplying by 2, it woul...

Astha Sahoo 3I

Yes! Like the other person above, you can change the PV = nRT, with n/V = [], to convert between the two. If you are switching between the two make sure to be careful because it's easy to make mistakes.

Astha Sahoo 3I

Since they form strong acids, they completely disassociate in water and don't alter the pH. Most(perhaps all) elements of strong acids and bases don't alter pH.

Astha Sahoo 3I

I'm assuming you're asking how to solve the problem. For the reaction given, you need to rearrange equations 1 and 3. You would multiply the first reaction by 3, and then flip the third equation around. By doing this, when you add the two equations, you get the reaction given initially. To find the ...

Astha Sahoo 3I

The correct answer would be A, because reactions never stop, but they do reach equilibrium when the concentrations of the products and reactants don't change. In order for this to happen, the forward and reverse reaction rates must be the same so there is no net change in the concentrations of react...

Astha Sahoo 3I

Siwa Hwang 3G wrote:Is it possible to use partial pressure for ICE table or no?

I haven't seen it done before yet, but I believe you can since the Kp equation is quite identical to the Kc one.

Astha Sahoo 3I

I definitely struggled with studying for the first couple of weeks too, but I found that going to at least one UA session every week really helped! Also being ahead with the textbook problems helped - I always did the Sapling ones last because they were slightly easier and didn't set as good of a fo...

Astha Sahoo 3I

When water is added, the concentration of the reactants increases, and thus to balance it out and make sure the reaction is at equilibrium (Q = Kc), the concentration of the products must increase. And yes, if CO2 partial pressure is decreased, in order to make more CO2 and restore equilibrium, O2 w...

Astha Sahoo 3I

Yes the only homework is the Sapling homework, and we get unlimited tries for each problem! Also, textbook problems are assigned as additional practice but not collected, so they are not mandatory, but definitely helpful for tests. Hopefully this helps you!

Astha Sahoo 3I

Yes I think just memorizing the strong acids and acetic acid should be enough, but I feel like Lavelle would include the chemical formula on the test since we haven't learned how to name acids yet.

Astha Sahoo 3I

Basically, you need to use the rule that if pH is less than pKa, the ion will be protonated, but if pH is greater than PKa, the ion will not be protonated.

Astha Sahoo 3I

Yes this is correct, because denticity depends on number of physically possible bonds.

Astha Sahoo 3I

In lecture Lavelle said to use Latin names only if it has negative charge (is an anion).

Astha Sahoo 3I

the seven strong acids are :HClO4, HClO3, H2SO4, HNO3, HBr, HCl, HI, and strong bases are the first column and heavy elements in the second column bonded with OH. All other acids and bases are weak. It's good to know that acids with COOH and CH3 and bases with nitrogen are weak.

Astha Sahoo 3I

Yes! Acidity of a compound is determined by concentration of H+ ions, so the lower the pH, the higher the concentration of H+ ions, and thus the more acidic.

Astha Sahoo 3I

So according to the solution manual, you just take the -log(0.12) and that gives you the answer!

Astha Sahoo 3I

After reading the textbook and listening to lecture, I'm still a bit confused on amphoteric compounds. I know that they are molecules that can either donate or accept a proton, but why would HSO4- not be amphoteric, but HCO3 isn't. Is it because if you add a proton to HSO4- it becomes a strong acid?...

Astha Sahoo 3I

In the textbook, it mentions that for binary acids especially, as you go down a group, the acid becomes stronger, since the bond enthalpy is lower. Thus, since F is at the top of the fluorine group, it's forms the weakest acid with H. All elements above Br form weak binary acids with hydrogen.

Astha Sahoo 3I

The structure of the molecule is in such a way that wherever you place the metal, only two oxygens can be bound to it. Gillian explained it really well, and I think the picture she provided is really helpful in explaining why only two oxygens can bind.

Astha Sahoo 3I

Since all the H atoms around the N central atom have the same electronegativity, even if the bonds were polar, since they all pull the same, the bonds would cancel each other out and thus there would be no significant partial charge on any of the atoms, making the molecule polar.

Astha Sahoo 3I

I think so? I usually base angles off of the hybridization of the atom in question and/or shape of the molecule, and there's a lot of textbook questions on it as well so I would expect it on the final.

Astha Sahoo 3I

I'm pretty sure it's on December 13th from 11:30 to 1pm? At least that's what it says on the test and exam schedule on the chem 14a website.

Astha Sahoo 3I

An easy way I remember is that in one bonding region, the first bond is always a sigma bond, and then the rest are pi bonds. For example, in a tripe bond, there are two pi bonds and one sigma bond.

Astha Sahoo 3I

If there's dipole moments in the molecule that do not cancel each other out, you can usually assume the molecule is polar.

Astha Sahoo 3I

I think for many molecules that just have c-c and c-h bonds, there's a ring or triangle form of the molecule. I'm not sure if there's a rule for it, but I always just experiment to see if there's a ring structure.

Astha Sahoo 3I

Br- is one whole shell larger than Na, making it's electrons(especially valence electrons) farther apart from the nucleus, thus giving it a larger electron cloud.

Astha Sahoo 3I

Would an ionic bond have a greater boiling/melting point than a covalent molecule? Like would NaCl have a greater melting point than CH4? But if the covalent molecule was really big for example C3H8 or something similar, would the covalent molecule have a higher boiling point because of size? I gues...

Astha Sahoo 3I

Yes you are right! To add on, a similar equation is the equation to find the number of lone pairs in a molecule: (# of valence electrons - # of shared electrons)/2, so for N^2, that would be (10 - 6)/2, so that would equal two lone pairs on the molecule N^2, one on each nitrogen atom. Hope this helps!

Astha Sahoo 3I

How do vapor pressure, solubility, and boiling point all correlate? Would NaF, which has a high solubility, have a high boiling point(due to strong IMF) and a low vapor pressure(also due to IMF)? So does a high solubility correlate with a high boiling point and low vapor pressure, and a low solubili...

Astha Sahoo 3I

I hope this is the right category to post my question in. I was just hoping someone could explain the solubility trends in the periodic table and the solubility rules that we should know for this class? An example that's been debated a lot is why NaF is more soluble than NaI, but AgI is more soluble...

Astha Sahoo 3I

I believe they are more stable because the electrons are moving around more, stabilizing charge, and thus there is less energy used to stabilize charge and bond the atoms.

Astha Sahoo 3I

I agree with everything said above! I just wanted to add on a couple of points. One proton has much more of a significant impact on the nucleus's charge than an electron, for it is so much bigger than an electron. Also in a period, the electrons are all in one subshell, and thus one proton added has...

Astha Sahoo 3I

I believe this is more of a theory, and no current atoms have a 6f subshell. However, they perhaps could get excited to the 6f subshell?

Astha Sahoo 3I

Dispersion forces somewhat rely on electronegativity. The way I think of it is that every molecule has a slight positive and slight negative charges at some place, and these slight opposite charges on different molecules leads to slight attraction, namely dispersion forces.

Astha Sahoo 3I

The g orbital is more of a theory currently, but I believe as we find more elements, there is potential for a g subshell. The theory is that the g subshell would have 9 orbitals, and from this information you can derive that l = 4 and that mi has nine possibilities for this subshell.

Astha Sahoo 3I

Basically if the electron is closer to the nucleus, it has greater effective charge since it's attracted to the protons in the nucleus. So when n increases, effective nuclear charge decreases. Also as you move along a period in the periodic table, the effective nuclear charge also increases because ...

Astha Sahoo 3I

I agree with all the previous answers! N is the energy level and shell, l is the subshell, and mi is the spin of the electron, so basically which direction the electron is spinning. One thing I remember from lecture is that if given n, l can equal any number n-1.

Astha Sahoo 3I

I think electrons always occupy orbitals shell by shell. By that I mean that electrons will try to fill up a whole shell before moving on to the next one, since electrons like to be closer to the nucleus. Thus, an electron would prefer to fill up the 3rd shell rather than move from the 3d subshell t...

Astha Sahoo 3I

I believe valence electrons have more energy since they have a weaker bond towards the nucleus compared to electrons near the nucleus, based off of effective nuclear charge. Since weaker bonds usually have more energy, I would say valence electrons have more energy.

Astha Sahoo 3I

Yes I believe the textbook forgot to put two electrons at the bottom of the fluorine so it would have six valence electrons.

Astha Sahoo 3I

You would need to fill the 3rd shell before going onto the fourth one. An electron would be pulled into the 3rd subshell if there was space in the 3rd subshell. Electrons are pulled towards the nucleus due to effective nuclear charge, so if there is a space in the 3rd subshell, an electron will move...

Astha Sahoo 3I

In one of the UA sessions I think someone said that electronic transitions in molecules are more complicated and complex than electronic transitions in atoms, because there's more subshells. We haven't gone into a lot of depth on the topic though so I'm assuming we won't be tested on it on this firs...

Astha Sahoo 3I

Since the wavelength is more than 400nm, it is part of the Balmer series, meaning that the n initial will be 2. Using this information, use the Rydberg's equation to find n final, and you can substitute v with c/wavelength. Hope this helps!

Astha Sahoo 3I

Usually you would compute the answer using L as units since that's the SI unit, but once you get the answer you can multiply it by 1000 if you want the answer in mL!

Astha Sahoo 3I

You can derive the Rydberg's equation from the Bohr's formula! Also if you multiply both sides by h in the Rydberg's equation, you'll get Bohr's formula.

Astha Sahoo 3I

Thanks everyone for your help! I had 7 as the answer before, it turns out I misread the answer sheet. So sorry!

Astha Sahoo 3I

So the energy may increase due to more photons, but the frequency and wavelength of each photon would stay the same! It's like if you had 100 cars crashing into a building at the same speed - the impact would be 100x that of one car, but the speed for all of the cars would still be the same.

Astha Sahoo 3I

I agree with what was said above! Also make sure you have all the signs(+/-) the right way since that messed me up the first couple of times I did the problem.

Astha Sahoo 3I

Waves can pass through material, while electrons cannot, thus this would support the wave model. Electrons would bounce off and not pass through, waves can.

Astha Sahoo 3I

In the problem 1E.13 part c, it asks for the electron configuration of silver. I put down [Kr] 4d^9, 5s^2, but the answer says the correct answer would be [Kr]4d^105s^1. If the atom isn't excited, why would one of the electrons from the s orbital go to the d orbital? I thought electrons only moved f...

Astha Sahoo 3I

For part a in 1D.13, it asks how many values of the quantum number 1 are possible when n=7. The answer is 2, but I'm confused on why this would be the answer?

Astha Sahoo 3I

In lecture Professor Lavelle said that a negative sign in the delta E for an ejected electron means that the bound electron has less energy than a free electron. Does that mean the electron is more stable when it's bound?

Astha Sahoo 3I

Why exactly is delta E negative when electrons emits light? Is it because it's releasing energy like an exothermic reaction?

Astha Sahoo 3I

The UV spectrum of the hydrogen atom corresponds to the Lyman series, which represents a jump to the principle quantum energy level n=1. first, you would find the frequency of the 102.6 nm wave using the equation: frequency= c/lambda. Then, you can solve for the initial energy level (from which the...

Astha Sahoo 3I

I think you would usually put 3d after 4s because when you follow the lines on the electron configuration chart, 3d comes after 4s.

Astha Sahoo 3I

During lecture, the professor said that if the energy of the photon was greater than the energy needed to remove an electron, the electron would not be emitted. I'm confused on why this would occur?

Astha Sahoo 3I

For question 1A.11, it talks about the lower energy levels for different series - what does that exactly mean? For example, the Lyman series has a lower energy level of n = 1.

Astha Sahoo 3I

Do you mean reacts with 2.00g NO2? If so, then yes you're correct because both have molar coefficients of 1, and thus whichever reactant is present in the least amount of moles will be the limiting reactant, because the reactants have coefficients in a 1:1 ratio. However if they had different coeffi...

Astha Sahoo 3I

Divide the grams of 2-butanone by its molar mass to calculate the number of moles.Multiply the number of moles of 2-butanone by the molar mass of the product, since there are no molar coefficients in this problem. This will give you the theoretical yield (should be 0.78g). Then, divide the actual ma...

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