CHEMISTRY COMMUNITY

Astha Patel 2J

Adding on, a good thing to remember is the higher the activation energy is, the more dependent K is on temperature.

Astha Patel 2J

The way you solved it is correct. I would double check units.

Astha Patel 2J

After balancing the half-reactions, whatever the number of electrons is that get transferred from the oxidation reaction to the reduction reaction is n.

Astha Patel 2J

Because the axis is half live vs. concentration, and we know that the relationship in a zero order reaction for half life is [A]/2k, we know that as the concentration decreases, the half life will decrease as well. The relationship is directly related, not exponentially, so the graph would be linear...

Astha Patel 2J

Intermediates go from being a product of a previous step to a reactant in future steps. Catalysts start in the reactants and do not change and end up on the product side as well. Both cancel out but remember to check whether it is being used as a reactant or whether it's simply staying unchanged.

Astha Patel 2J

The way I like to think of it is that your team is only as good as your slowest player. So, even if three other people ran extremely fast, the majority of the time would be taken up by the slowest leg. So the rate at which your team ran is basically dependent and mostly accounted for by the slowest ...

Astha Patel 2J

If the atom within a molecule gets reduced or oxidized, then the oxidation state of the molecule would also have to change, so the molecule too would be considered either reduced or oxidized.

Astha Patel 2J

The first equation is for zero order and the second equation is for first order reactions.

Astha Patel 2J

Appendix 2B, which is linked in many textbook problems online has every half-reaction that appears in textbook problems.

Astha Patel 2J

Yes, the little circle means at standard conditions! Another thing to remember is that E naught and G naught are values that are static, as in they do not change even when the current conditions of the reaction change.

Astha Patel 2J

I had this same issue and it's a sapling formatting issue. Don't put parenthesis around your product and your answer should be correct.

Astha Patel 2J

Reaction quotients don't include solids and liquids and BCl3 is the only one in the gas phase so Q would equal 1/BCl3^s

Astha Patel 2J

In some text book problems, the cell diagrams include H+ (aq), but don't include H2O (l). When do we include H+ when writing cell diagrams? And why do we exclude water?

Astha Patel 2J

I had the same question as you and I got the same answer and it was correct. Maybe try rearranging the order of the products, because my products were listed PbO2 + Cl- + H2O + 2OH-

Astha Patel 2J

I don't think the order matters, but I remember one of the sapling questions marked me wrong for switching the order of two of the reactants so maybe there is some rule?

Astha Patel 2J

305405193 wrote:Is Wanda Vision that good? I LOVE MARVEL so I've been debating on seeing it, but everyone says its so GOODD. Please let me know if it's worth skipping studying time LOL

Literally go binge it rn. It's so good

Astha Patel 2J

Phase changes occur at equilibrium because all the energy is used for changing the states rather than heating up the compounds.

Astha Patel 2J

I think he was referring to the second law, which states that an isolated system's entropy will always increase over time. A system can have a negative change in entropy, but overall somewhere in the universe, the entropy will increase.

Astha Patel 2J

Salt bridges allow ions to go from the cathode solution to the anode solution to counter the increasing negative charge on the cathode side. If salt bridges didn't exist, the transfer of electrons would eventually stop. I'm not sure how exactly they move, but the driving force behind it may just be ...

Astha Patel 2J

Don't quote me on this but I think it has to do with how well a compound acts as an oxidizing agent, or how likely it is to be reduced. That explains why Flourine gas (fluorine has very good electron pulling power), has such a high standard potential.

Astha Patel 2J

If a compound is thermodynamically stable, then forming it would be thermodynamically spontaneous in terms of Gibbs free energy. Therefore, A negative Gibbs free energy for the formation of some compound indicates that it is stable. If the Gibbs free energy is positive, then it's unstable.

Astha Patel 2J

For both parts a and b, you'll use the equation deltaS = nCln(T2/T1). The difference between the two parts will be due to different heat capacity values, C. For the constant pressure, the heat capacity of a monatomic gas is (3/2)R, and for constant volume, the heat capacity is (5/2)R, where R is the...

Astha Patel 2J

I think another thing to remember is that in a situation with constant volume, the heat, q, is equal to the change in internal energy. Also, when there is no change in volume, gases do no work, so that's why the work term gets removed from deltaU = q + w

Astha Patel 2J

Nathan Lao 2I wrote:Wandavision is literally so good. I just don't like how each episode is a cliffhanger. They answer our questions about the previous episode but then introduce way more at the end of each one. The suspense kills me.

I get more and more confused the more episodes I watch

Astha Patel 2J

Basically, you have to find the change in entropy due to volume and then combine it with the change in entropy due to temperature change. Use nRln(v2/v1) and nCln(T2/T1) and then add the two together.

Astha Patel 2J

When you have a phase change, for example going from a solid to a liquid, delta H is positive because heat is going into the system to make this phase change happen. Going from a solid to a liquid is also a more disordered state, and therefore delta S would have to be positive.

Astha Patel 2J

qrev is heat from a reversible expansion/compression, but q is just a heat term. Make sure that you use qrev when solving for entropy changes.

Astha Patel 2J

Heat capacities talk about how much heat a system needs to gain, and thus will always be positive. However, when you have a loss of heat (negative q) that's usually because delta T is negative.

Astha Patel 2J

Do the textbook problems, they are your friends. I haven't gone to any UA workshops, but I've heard they're good practice. Also, I like reading over the outlines to make sure I know all of the bullet points.

Astha Patel 2J

Yeprem wrote:but for V in PV=nRT which V value do i use, V1 or V2?

Use V1, because the number of moles won't change when you expand a gas.

Astha Patel 2J

this did not age well

Astha Patel 2J

The best thing to do to be safe would be to check the units of whatever it is you're trying to solve or the gas constant given in the problem. Also, if you are using the change in temperature then celsius and kelvins are interchangeable because a change of +1 degrees celsius is the same as a change ...

Astha Patel 2J

When you form the bonds to make water and carbon dioxide, that releases more energy than it takes to break the bonds in the reactants. Also, think of combustion as producing heat, so you know that it's "releasing" heat, and would therefore be exothermic.

Astha Patel 2J

At constant pressure, work done by the system will be -delta n *RT. Therefore, for a system to DO work, the change in n would have to be a positive number, so the reaction should produce a greater number of moles of gas compared to the moles of gas used. For the equation A(s) + B(s) -> C(g), the del...

Astha Patel 2J

Heat depends on the path taken, so it's not a state function. However, enthalpy is.

Astha Patel 2J

The delta H given in the problem refers to the KJ per mole of S8. Therefore multiple the delta H with the number of moles of S8 (1.31).

Astha Patel 2J

You'll probably use it in some practice problems, if they give you moles of a substance and molar heat capacity. Otherwise, we usually use specific heat.

Astha Patel 2J

Delta H is the change in heat energy whereas delta U is the change in total energy

Astha Patel 2J

Maybe some upper divisions have calculus pre-reqs but I'm not sure.

Astha Patel 2J

Yes! Vapors are gases.

Astha Patel 2J

If pH<pKa, then it is in a more acidic environment. This means that the species will act as a base and will then accept a proton. Therefore in the solution, there is more HA present. If pH>pKa, then it's in a basic environment. This means that the species will act as an acid and donate its proton. T...

Astha Patel 2J

You'll have to test the assumption. So when you do your ice tables and solve for x (assuming that x is negligible in the denominator), then take x and divide it by the concentration of the initial solution. If this is greater than 5% then the assumption doesn't work and you have to use quadratic. To...

Astha Patel 2J

They both equal each other.

Astha Patel 2J

I'm pretty sure that's all you need to know because that is how you derive that equation.

Astha Patel 2J

Acids generally have a carboxyl group or an acidic hydrogen. If a compound is a group 1 hydroxide, or a group 1/2 oxide, it is a base. When you have salt compounds, what I usually do is split the compound into two ions. If the compound is strong, its conjugate acid/base will be weak so the solution ...

Astha Patel 2J

I would always plug it in to check if it's less than 5% because it really depends on the concentration of the initial solution. That's why I trust the 10^-4 limit rather than 10^-3.

Astha Patel 2J

Correct me if I'm wrong, but I believe that Kc is just a more specific way to describe K. However, if you want to convert between Kc and Kp, then I think the equation is Kp= Kc(RT)^delta(n), where delta n is the change in the number of moles (moles of products - moles of reactants). So calculating ...

Astha Patel 2J

So what you do is write the equation for Q, which is the same as Kc but the concentrations aren't equilibrium concentrations. So, Q = ([NH3]^2) / ([H2]^3 * [N2])
Then, just plug in the concentrations that they gave in the problem.

Astha Patel 2J

Since the problem gives you the equilibrium concentrations already, all you need to do is write the equation for Kc which would be Kc = [NO2]^2 / [N2O4]. The NO2 is squared because one mol of N2O4 yields two moles of NO2. Then just plug in the values.

Astha Patel 2J

For the problems in 6D, when they give you a compound, what I usually do is split the compound into two ions. For example, 6D.15 asks to find the pH of 0.19M NH4Cl. So looking at this compound you need to figure out whether its acidic or basic so you know whether to use Ka or Kb when solving for the...

Astha Patel 2J

Because the question says "immediately after the temperature change", the concentrations of the reactants or products haven't changed yet so Q is the same. However, because the reaction is endothermic, increasing the temperature will increase K (favoring the products).

Astha Patel 2J

Correct me if I'm wrong, but I believe that Kc is just a more specific way to describe K. However, if you want to convert between Kc and Kp, then I think the equation is Kp= Kc(RT)^delta(n), where delta n is the change in the number of moles (moles of products - moles of reactants).

Astha Patel 2J

Large equilibrium constants favor products, and a small equilibrium constant favors the reactants. At 1000K, F2 has a larger equilibrium constant than Cl2 (you'll probably solve for this in parts a and b). A large equilibrium constant in this example would favor the products of dissociation, therefo...

Astha Patel 2J

Large equilibrium constants favor products, and a small equilibrium constant favors the reactants. As an example, think about the dissociation of F2 and Cl2. At 1000K, F2 has a larger equilibrium constant than Cl2. A large equilibrium constant in this example would favor the products of dissociation...

Astha Patel 2J

You want to use equations that, when added together, give you the overall reaction. Sometimes, you'll have to flip one of the equations (use 1/K) or you may have to multiply all of the reactants and products of an equation by some constant (use K^constant). When you alter the individual equations, r...

Astha Patel 2J

The conjugate acid will gain a H+ (add 1 to the charge), and the conjugate base will lose a H+ (subtract 1 from the charge).

Astha Patel 2J

I would look for lone pairs on the lewis structure and then determine whether those pairs are in a favorable orientation, like with spacers. That's the only way to bind to the same central metal.

Astha Patel 2J

The structure doesn't have spacers making it impossible for two lone pairs to become bonding pairs simultaneously. It would try and bend the bond.

Astha Patel 2J

I would memorize the seven strong acids, because that rules out all the other acids as weak. For the relative strengths of the weaker acids it depends on the acid type: 1) binary acids are some HA acid, where A is some anion. With these acids, the strength increases as the electron affinity of A inc...

Astha Patel 2J

To reiterate what others have said, you take the charge of the element that interacts with the complex ion (which is what is in brackets). The total charge of the compound should be zero, so you know that the charge on the element(s) outside of the brackets should equal the negative of the total cha...

Astha Patel 2J

-ate is used when the complex ion has a negative charge. Usually, the coordination compound will contain some metal cation in front of the complex ion in the bracket.

Astha Patel 2J

I believe that oxalato, en, edta, dien, and carbonato are generally the ones we need to know as they come up in textbook problems. I haven't encountered any other ones.

Astha Patel 2J

The regions of electron density should equal the orbitals that are available from the hybridizations. sp = 2 orbitals, so, therefore, two regions of electron density. sp3 = 4 orbitals, so therefore four regions of electron density.

Astha Patel 2J

Ka is the acid dissociation constant and pKa is the negative log of the Ka value.

Astha Patel 2J

Coordination numbers refer to the number of bonds between the central metal and ligands. Regions of density could include lone pairs of electrons which does not pertain to coordination numbers.

Astha Patel 2J

Yes! Only prefixes that specify the number of ligand molecules is used. Anything outside of the bracket that pertains to the complex ion doesn't use a prefix.

Astha Patel 2J

The shape of the molecule can show whether dipoles are positioned to cancel each other out.

Astha Patel 2J

Oxygen gas can be assumed in excess during combustion reactions unless stated otherwise.

Astha Patel 2J

Since we will most likely be provided the molecular formal of compounds that we are unfamiliar with, you can probably search it up and still get good practice from doing the lewis structure. The past two midterms have always given us the molecular formulas so I wouldn't worry about it too much.

Astha Patel 2J

After drawing the Lewis structure, the first bonds (single) between any two atoms would be sigma bonds and any additional bonds between the atoms (making it a double or triple bond) would be pi bonds.

Astha Patel 2J

Chlorine and Oxygen gas are both diatomic molecules and will be written Cl2 and O2.

Astha Patel 2J

Molecules with resonance structures where the double or triple bond can be moved among different atoms consist of pi bonds, so depending on the lewis structures and whether resonance is possible, that can show whether the bonds are delocalized.

Astha Patel 2J

Electron density is the probability of finding an electron in a given region of an orbital. Greater electron density can indicate the more negative pole of a molecule.

Astha Patel 2J

CH3F is tetrahedral and I believe that the bond angles will decrease by a small amount.

Astha Patel 2J

To reiterate what others have said since bond angles are experimentally determined, I would remember the angles for the parent shapes where all bond angles are equal and then when larger areas of electron density are present then the bond angles would decrease.

Astha Patel 2J

I believe it would be linear.

Astha Patel 2J

The best way would be to compare electronegativities and whichever atom has the greater electronegativity (electron pulling power) would be the negative end of the pole. The relative differences in electronegativity would also tell us which ones have a greater dipole moment (greater difference in el...

Astha Patel 2J

So basically the 4s subshell gets filled first before 3d. This is because, at the time where there are no electrons in the 3d subshell, the 4s subshell is lower in energy. However, once the 3d subshell starts getting filled, it has lower energy than the 4s subshell and therefore gets written before...

Astha Patel 2J

The formal charge on the carbon is a -2.

Astha Patel 2J

I think we just need to know that orbitals aren't pre-existing but rather describe the probabilities of finding electrons at a specific energy level.

Astha Patel 2J

Make sure you're using the pdf solutions of the textbook that's in the sapling website. It might just be an error in the solutions in the back of the ebook because in the pdf, the left structure is the favored one because of the minimized formal charges.

Astha Patel 2J

Yes, as soon as the 3d subshell starts getting filled, it is written before the 4s subshell because it is at a lower energy level.

Astha Patel 2J

GOSSIP GIRL !!!

and as soon as you finish start YOU

Astha Patel 2J

A lewis acid accpets electrons and therefore is usually a cation, a molecule where the central atom is electron deficient, or a molecule with polar double bonds. A lewis base donates the electrons and therefore is usually an anion, a molecule where atoms have excess electrons or a molecule with lone...

Astha Patel 2J

The ability for a molecule to be non-polar yet consist of polar bonds is due to the ability for dipoles to cancel out. This can be seen after looking at the shape of the molecule (for example, a symmetrical molecule will most likely have dipoles that cancel each other out).

Astha Patel 2J

So basically the 4s subshell gets filled first before 3d. This is because, at the time where there are no electrons in the 3d subshell, the 4s subshell is lower in energy. However, once the 3d subshell starts getting filled, it has lower energy than the 4s subshell and therefore gets written before ...

Astha Patel 2J

Electrons on the left side of the periodic table are more likely to form cations because of their low ionization energies and low electron affinities. They are more likely to lose their outermost electron as it isn't held as tightly by nuclear charge.

Astha Patel 2J

Your structure seems right to me. Did the question ask to put formal charges?

Astha Patel 2J

I think since the total formal charge of the ion is -1, it is considered more stable as opposed to a different lewis structure in which the oxygens would be given a formal charge value that isn't one even though they are more electronegative. I think when comparing lewis structures to see which is m...

Astha Patel 2J

How is it that Na+ is smaller than Cl-, but elemental Na is larger than elemental Cl?

Astha Patel 2J

So the first three quantum numbers specify an orbital. Each orbital can hold a maximum of 2 electrons. So the first three quantum numbers can specify at most 2 electrons. Once the fourth quantum number is given, then the four numbers specify a single electron.
Hope this helps :)

Astha Patel 2J

Someone had asked this a couple weeks ago and this was my reasoning: So the problem gives the energy requried to remove an electron. In this problem, the work function should equal the energy of the incoming photon, so hv= work function. Substitute frequency for c/lamda so that you can solve for wav...

Astha Patel 2J

I'm not 100% sure but I believe we just need to understand it conceptually and what it represents because it helps explain quantum numbers in orbitals, but I don't think we need to do calculations with wave functions because in the textbook many of the problems required integration.

Astha Patel 2J

For the quantum number l that specifies the subshell, you should probably just memorize that s = 0, p=1, and so on. For the Magnetic quantum number that gives you the specific orbital, i would remember that it depends on what l is : l, l-1, ...., -l

Astha Patel 2J

Use deBroglie's wavelength equation where wavelength = h/p, and p is momentum. p=mv. The problem gives you the velocity of the particle so all you have to do is plug it into the equation.

Astha Patel 2J

As long as the five posts are timestamped before 11:59pm of Sunday of every week, then you'll get the credit for all the posts.

Astha Patel 2J

Delta X refers to uncertainty in position. In a problem where they say that the electron is bound by an atom, the uncertainty of the position would be the length of the atom because the electron could be anywhere in that range. If the problem said something like the uncertainty was 10 nm +/- 2nm, th...

Astha Patel 2J

The mathematical representation of wave functions from Schrodinger's wave equation are essentially math models of specific orbitals. So each orbital has a unique wave function that specifically works for that orbital. So yes, it could be used as a form of identification, but I'm not too sure that it...

Astha Patel 2J

Hi guys :) So this is an even numbered problem from the textbook. I know how to do part a, but I'm confused on part b and c, specifically on the mass of the Li+ atom. The uncertainty principle has negligible consequences for macroscopic objects. However, the properties of nanoparticles, which have d...

Astha Patel 2J

It depends on the surface metal that has the electrons. Some metals require a greater frequency than others., so there's no exact range.

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