CHEMISTRY COMMUNITY

Abby Soriano 1J

You should probably know that collisions need to have the appropriate amount of energy to break a bond and also be correctly oriented in order for a new molecule to form. This is also why we only cover up to second-order reactions, since it's much more difficult to get 3 or more molecules to all co...

Abby Soriano 1J

You'll most likely be asked to find the half-life of something you want to know the age of. By comparing how much of the reactant there was initially versus how much of it exists now, and by knowing the half-life, you should be able to calculate how old a substance is.

Abby Soriano 1J

The slowest step acts as a bottleneck so that, no matter how fast the rest of the reaction occurs, the overall reaction cannot be completed until the slowest step is done.

Abby Soriano 1J

The problem will most likely specify which of the steps is the slow step. You will probably be asked if a proposed mechanism matches the observed rate law, but you won't need to actually calculate a value for the slow step.

Abby Soriano 1J

I would assume that it would make it easier for the molecules to collide if they were bigger in size, so it is possible that the rate would increase, but I'm not too sure.

Abby Soriano 1J

The standard cell potential can be figured out through given values and occurs at standards conditions. The cell potential can occur at any condition and can be figured out if you know the standard cell potential and the value of Q.

Abby Soriano 1J

Platinum is added as an inert conductor for the movement of electrons in a galvanic cell when no solid is present to do this already.

Abby Soriano 1J

You can think of a galvanic cell as a battery. When the standard cell potential is negative, the cell's battery is essentially dead and would need an external source of power to keep it running. When it is positive, it doesn't need this source of power since the reaction is spontaneous (which is why...

Abby Soriano 1J

The electrodes are what allow for the movement of electrons from one side of a galvanic cell to the other. Platinum is used as an inert conductor (it doesn't affect the reaction) when there are no conducting solids already present in the reaction.

Abby Soriano 1J

Changing the mass of the electrode shouldn't affect the cell potential since cell potential is only affected by things like concentration, which solids don't play a part in.

Abby Soriano 1J

E standard = 0 in a concentration cell since you're essentially dealing with the same reaction occurring in both directions. The E standard of the reduction half-reaction is just the negative value of the E standard of the oxidation half-reaction, and so when you add them together, you get that the ...

Abby Soriano 1J

Reducing agents are the ones being oxidized because, in order for them to become oxidized, they must give up electrons. And so, by giving up those electrons, they are acting as a reducing agent for whatever is being reduced. The opposite goes for oxidizing agents being the ones who are reduced.

Abby Soriano 1J

It's important to look at the charge of the element you're working with rather than that of the entire compound when deciding if it is being oxidized or reduced. So although MnO4- has a 1- charge on the outside, that refers to the charge of the entire compound. Since oxygens always have a 2- charge ...

Abby Soriano 1J

If there is not already a conducting solid present in the reaction, then an inert conductor (like platinum) can be used as an electrode to transfer electrons between the half-reactions of the cell.

Abby Soriano 1J

I don't think Prof Lavelle would ask us to derive the equation on the final since his exams have to do more with applying the concepts we've learned rather than just reproducing the information he's given us. Instead, you should study what the difference between a differential rate law and integrate...

Abby Soriano 1J

We haven't talked much yet about properly balancing half-reactions, but I know that there are a lot more steps involved, such as adding H2O to balance any oxygens and adding H+ to balance any hydrogens. The number of electrons you add should be to properly balance the charges on both sides, and yes ...

Abby Soriano 1J

With the movement of electrons from the anode to the cathode, the charges of the solutions in a galvanic cell can buildup to be increasingly positive and negative. When this buildup of charge occurs, it can stop the movement of the electrons since they wouldn't want to continue moving towards an inc...

Abby Soriano 1J

A reaction is spontaneous if it has a negative delta G (meaning it also proceeds in the forward direction towards the products). When delta G is 0, the reaction is at equilibrium since the free energy of the products equals the free energy of the reactants and no more work can be done.

Abby Soriano 1J

Whenever you see any type of visual diagram for a galvanic cell, it's standard for the cathode (and thus reduction) to be represented on the right.

Abby Soriano 1J

It helps to separate the equation into its two half-reactions and balance both. When you do this, you're forced to think about how many electrons are transferred in order for the charges on both sides to be balanced, and from there you can plug in the number of moles of electrons into the equation y...

Abby Soriano 1J

We know that delta s = nRln(V2/V1). Since pressure and volume have an inverse relationship, delta S = nRln(P1/P2).

Abby Soriano 1J

Typically delta S refers to the delta S of the system while delta S total (delta S of the system + delta S of the surroundings) refers to the delta S of the universe.

Abby Soriano 1J

delta s total refers to the universe and the delta s system + surr are self explanatory (your system and whatever you define as your surroundings). An example is a cup and a bench. The cup is my system and its on my surroundings (the bench). The key concept is that for a reversible system delta S t...

Abby Soriano 1J

Does anyone know how to calculate isothermal irreversible expansion? How does it differ from calculations for isothermal reversible expansion?

Abby Soriano 1J

Maya Beal Dis 1D wrote:Would this mean that the change in enthalpy for an adiabtic process is also zero?

I think that the change in enthalpy would also be zero for cases in which the pressure is constant since only at constant pressure is heat equal to the enthalpy.

Abby Soriano 1J

Things like work and heat should always be approached from the point of view of the system. Just as when heat is being applied to a system and delta H is a positive value, when work is being done on/to a system, the value of work is positive. When the system is doing work to its surroundings, work i...

Abby Soriano 1J

Entropy is more of an extensive property because as you increase the amount of the substances (take for example the number of gas particles in a container), you increase the possible number of positions (the entropy).

Abby Soriano 1J

Celine 1F wrote:If the expansion of the gas is isothermal, then this means the temperature remains constant and delta U = 0

Could you explain why delta U would equal 0? I'm a little confused about the connection between temperature and delta U.

Abby Soriano 1J

One important thing to note about the difference between reversible and irreversible processes is that reversible processes are slower (work is done in small increments) but the overall work done is more than that of irreversible processes (which are faster but less efficient). This has to do with t...

Abby Soriano 1J

Delta U denotes a change in internal energy while delta H is used for a change in enthalpy. At constant pressure, delta H = q (heat). Since delta U = q + w (heat + work), at constant pressure, delta U = delta H + w too.

Abby Soriano 1J

By definition, enthalpy is the amount of heat absorbed/released at a constant pressure. So that means that delta H only equals qp rather than qv.

Abby Soriano 1J

Delta H is equal to q only when the gas is at a constant pressure. By definition, enthalpy (deltaH) is the amount of heat released/absorbed at a CONSTANT PRESSURE, so the heat required to change temp of gas at a constant pressure is the same as enthalpy. So does that mean that the same does not hol...

Abby Soriano 1J

Do you know why or how delta H is equal to qp? What about the constant pressure makes the two equal?

Abby Soriano 1J

It's always a good idea to think of both work and heat from the perspective of the system. We know that endothermic reactions are categorized with a +delta H since they are taking in heat. With that same concept, we can think of positive work or heat as being applied to the system. If heat is added ...

Abby Soriano 1J

Heat and work are not state functions because the way to get to the final endpoint can be done so using different paths in which one expends more energy than the other. Getting from one place to another is not all that matters; it matters, however, how much heat and work you use up to get there.

Abby Soriano 1J

qp is enthalpy at constant pressure and qv is at constant volume. They are used when finding enthalpy/delta H but also find change in internal energy which is q + w so q can either be qp or qv depending on the given conditions I know that qp is another way to describe enthalpy, but can qv also be u...

Abby Soriano 1J

If delta H is positive, the reaction is endothermic and takes in heat on the reactant side. Therefore, if the temperature were to increase, there would be more heat coming in on the left, forcing the reaction to favor the right (the product side) to balance itself out again.

Abby Soriano 1J

We can add the final and initial values for state properties since we don't care about what goes on in the middle of the process. Work and energy are different in which we want to look at how much is being used and wasted on the journey.

Abby Soriano 1J

You should expect to know how to determine if a reaction is exothermic or endothermic based on the value of delta H (a negative value signifies an exothermic reaction), the reaction's response to added heat (does the addition of heat cause the reaction to favor the left or right), and the stability ...

Abby Soriano 1J

water is not included in the K constant when it is a solvent because it is viewed as pure. Liquids and solids are not included in the K constant since their concentrations do not significantly change. To add to this, since water is the solvent, it will be in excess and so its concentration will not...

Abby Soriano 1J

Whenever you do any calculations involving ICE tables, you can use the 5% rule to ensure that your approximation that x is small is valid.

Abby Soriano 1J

In acid and base equilibria problems, the concentration of water is in large excess and therefore left essentially unchanged, leading us to omit it from K expressions.

Abby Soriano 1J

You can make your calculations quicker without having to use the quadratic formula when you approximate x. When Ka or Kb is small enough, x can be assumed to be insignificantly small in comparison to the initial concentration of weak acid or base given. As a result, rather than writing out the initi...

Abby Soriano 1J

Yea, you only need to find partial pressures if they ask you to find them. In equilibrium problems, they'll typically already provide you with concentrations which you can use instead of partial pressures (you just need to remember when to use Kc versus Kp).

Abby Soriano 1J

In this reaction, you're breaking a bond to get from X2 to 2X. Since energy is required to break a bond, the reaction is endothermic.

Abby Soriano 1J

You can approximate x only if the Ka or Kb value is small enough (less than 10^-3) since this will give you a very small x that would be an insignificant amount to remove from the bigger initial concentration of the weak acid or base they give you.

Abby Soriano 1J

By definition, the equilibrium constant has no units, as we're supposed to be using active masses instead of the molarity/ concentrations of the respective substances. During lecture, Dr. Lavelle talked about how in ideal situations, we assume that the activity equals the concentration, which is wh...

Abby Soriano 1J

K is a constant that only describes the [products]/[reactants] at equilibrium. Q can be calculated at any time during the reaction and is often compared to K in order to see if the reaction is moving in either the forward or reverse direction.

Abby Soriano 1J

In lecture, Lavelle said that if the value is between the two limits neither the reactant nor the product is favored and it is in a sort of intermediate state. Does that mean that we most likely won't be asked a question on whether the reactants or products are favored if the K value falls between ...

Abby Soriano 1J

If the reactants and products are in the aqueous state, then you should include them in your K expression (more specifically in regards to Kc).

Abby Soriano 1J

Whenever a question asks for the equilibrium constant, you have to figure out the phases of the reactants and products when you write out the reaction. If they are in the aqueous state, then use Kc. If they are in the gas phase, then use Kp. Writing out the entire reaction with all the phases will a...

Abby Soriano 1J

Yes, water is amphoteric because it can act as both a base and acid (depending on what it is initially reacting with). When acting as a base, it will accept a proton to produce H3O+; when acting as an acid, it will donate a proton to produce OH-.

Abby Soriano 1J

I'm pretty sure we only need to know about cisplatin since Dr. Lavelle discussed it in detail. It's good to know how it works rather than just knowing what it's used for. The reason why cisplatin is a useful chemotherapy drug is that its two Cl ligands (which are on the same side) get displaced by e...

Abby Soriano 1J

Anytime a Ka value is given, that alone is a good enough indication that you're being given a weak acid. For strong acids, the Ka value is too big to even be recognized. Ka = [A-][H+]/[AH] Due to the complete dissociation of strong acids, the Ka value moves towards infinity since the denominator wil...

Abby Soriano 1J

An easy way of recognizing something as an acid or base is seeing if it has a proton that it can donate. If it does, then, according to the Bronsted definition, it is an acid. You can always try drawing the Lewis structures to see if it has a lone pair it can donate. In this case, it would be a Lewi...

Abby Soriano 1J

As stated above, HCN will react with H2O to form CN- and H3O+. In this case, H2O is acting as a base since it is accepting a proton. The resulting H3O+ is the conjugate acid of H2O that can then act as an acid to donate a proton. However, it's important to remember that H2O is amphoteric and thus ca...

Abby Soriano 1J

A molecule that is a Lewis acid can also be considered a Bronsted acid (and vice versa) since their definitions do line up. In order for a molecule to accept an electron pair (be a Bronsted acid), it donates a proton (is a Lewis acid). Typically you can use both definitions to define an acid unless ...

Abby Soriano 1J

Reactions dealing with weak acids or bases use equilibrium arrows since they are not completely dissociated. On the other hand, in the case of strong acids and bases, complete dissociation allows for reactions to go in just one forward direction where the acid or base fully ionizes.

Abby Soriano 1J

Cisplatin is important because its two Cl ligands are on the same side. When it comes into contact with DNA, the two Cl's are displaced by the exposed N's on guanine, resulting in a strong bond that essentially stops all cell division and makes cisplatin useful as a chemotherapy drug.

Abby Soriano 1J

I would try to memorize and understand the biological importance of his examples (like cisplatin, myoglobin, and hemoglobin). It would be good to know the connection between the compounds' importance and how they are structured (such as why cisplatin being cis is important).

Abby Soriano 1J

I would try to memorize the most common acids and bases (or the ones that show up in the homework problems), especially the strong acids and bases included in the textbook.

Abby Soriano 1J

Lone pairs are not confined to a bond; the electrons exist in a cloud of probability that takes up more space and thus has a greater repulsion in order to maximize the amount of space between each other.

Abby Soriano 1J

You can make approximations (slightly less than __) but the only important thing is to know the concept behind why that is. The lone pairs have a greater repulsion that makes the bond angles slightly smaller than they should ideally be.

Abby Soriano 1J

A central atom can have multiple sigma bonds as long as they are all single bonds connecting the central atom to different atoms. Any double or triple bonds that are added in addition to those single bonds become pi bonds.

Abby Soriano 1J

Hydrogen bonds are a type of dipole-dipole interaction that are set apart simply because they are much stronger than regular ones. Their strength comes from the high electronegativity of the nitrogen, oxygen, and fluorine atoms they are bonded to.

Abby Soriano 1J

The dipole moments cancel each other out since they are all, essentially, the same (the only bond is a Br-F bond). If they were not all the same element (not all F or any other atom), then the dipole moments would not cancel, resulting in a net dipole moment.

Abby Soriano 1J

Yes, electron arrangement can be thought of as electron geometry. Electron geometry depends on the number of regions of electron density and is supposed to help you figure out molecular geometry. Molecular geometry is where looking only at bound atoms to name molecular shape comes into play.

Abby Soriano 1J

Unless they are the same element, all diatomic molecules are somewhat polar because every element has at least a slight difference in electronegativity that causes one atom to pull on the shared electrons more than the other.

Abby Soriano 1J

Temporary dipoles can occur even between two nonpolar molecules due to the constant movement of their valence electrons. As the electrons move, certain regions of the molecule become slightly more negative or slightly more positive, causing there to be a temporary dipole that can shift as quickly as...

Abby Soriano 1J

A central atom with only four bonding pairs will have a tetrahedral shape. Exceptions start to arise when lone pairs get introduced. For example, a central atom with four bonding pairs and one lone pair will have a molecular geometry of seesaw. We'll most likely go in depth on this topic on Friday.

Abby Soriano 1J

The presence of the two lone pairs on the oxygen makes it so that water has a bent (or angular) shape rather than linear. Lone pairs generally take up more space than bonded pairs do (since the electrons exist in a probability cloud rather than being confined in a bond) and so the repulsion of the l...

Abby Soriano 1J

Those are known as intermolecular forces because they are the forces that attract different molecules to one another. Polar and nonpolar bonding occurs between atoms in a molecule so we call them intramolecular (since they happen within the molecule).

Abby Soriano 1J

Dipole moments occur when electrons are not equally shared in a covalent bond. The more electronegative atom pulls on the electrons more, resulting in a slightly negative charge on the atom. Dipole moments are depicted with an arrow pointing towards the atom with the slightly negative charge.

Abby Soriano 1J

It would be [Ar]3d8 if Cu was just any other element; however, we have to remember that Cu is one of the exceptions. The electron configuration of neutral Cu is [Ar]3d10 4s1, and so the 2 electrons that are removed for Cu2+ is the 1 electron from the 4s orbital, then 1 electron from the 3d orbital l...

Abby Soriano 1J

Sometimes formal charges are impossible to get to 0 (for instance in compounds that have a charge on them, such as BrO3-) during which you need to draw a Lewis structure that has the lowest amount of charges and the lowest overall charges on each atom.

Abby Soriano 1J

We will most likely be given the covalent radii of different atoms, from which you can calculate the bond length by adding the radii together.

Abby Soriano 1J

E=hv is used only in regards to electromagnetic radiation since it uses constant c (the speed of light) while De Broglie is used for any other particle that has momentum since its velocity is variable.

Abby Soriano 1J

Whenever you draw Lewis structures, you want to follow 3 steps: 1) Ensure that you have the correct number of valence electrons 2) Try to satisfy the octet rule when you can (be mindful of exceptions) 3) Try to get the lowest formal charges possible on each element You start off everything with sing...

Abby Soriano 1J

They essentially refer to the same concept of ionic compounds having some covalent character; however, "highly polarizable" applies to anions (its electrons are being pulled by the cations) while "high polarizing power" applies to cations (whose positive charge pulls on the elect...

Abby Soriano 1J

The first five elements (H, He, Li, Be, B) can have incomplete octets, as can all group 13 elements. Elements in period 3 and higher can have expanded octet shells because of their d-orbitals.

Abby Soriano 1J

The cations in ionic compounds have a certain polarizing power: their positive charge makes them pull on the electrons of neighboring anions (that we describe as being polarizable to a certain extent). Because of this, there's a distortion in the electron cloud of the anion, making it so that there ...

Abby Soriano 1J

The question asks us to draw the Lewis structure of reactive species and to indicate which are radicals. I'm confused about letter a:
chlorine monoxide, ClO. Why is the radical on Cl rather than on O? Can you put it on O? What rules would apply to figuring out which element has the radical?

Abby Soriano 1J

Also, be sure to remember that you can only use c=λv when dealing with photons since c represents the speed of light. Any other particles (such as a neutron) would use De Broglie's equation since its velocity will vary.

Abby Soriano 1J

My old high school teacher had a kind of silly yet helpful way of memorizing the electromagnetic spectrum:
"Remind Me If Visible Underwear eXcites Grandma" - Radio waves, Microwaves, Infrared, Visible, Ultraviolet, X-rays, Gamma rays
Hopefully, you find this as helpful as I did!

Abby Soriano 1J

The idea of frequency and energy being proportional is also demonstrated by Bohr's frequency condition (v = delta E/h) which states that only light with certain frequencies (and by extension certain energies) will be absorbed.

Abby Soriano 1J

The more reactive nature of s-block metals compared to p-block metals actually has to do with the lower ionization energies of s-block metals. Their lower first ionization energies make them more likely to lose an electron and thus be more reactive.

Abby Soriano 1J

The atomic radii decrease as you go across a period because the increased number of protons has a greater effect on the electrons around it (essentially the more protons there are, the more they pull on the electrons to move closer in, thereby decreasing the atomic radii).

Abby Soriano 1J

I don't know if it's required we memorize what's a metal, nonmetal, and metalloid, but I would recommend just knowing that boron, silicon, germanium, arsenic, antimony, tellurium, and polonium are metalloids. Everything to the right of those elements should be nonmetals and everything to the left sh...

Abby Soriano 1J

If you look on the class website, there's actually an error in the solutions manual for that problem. They made delta v = 5 m/s rather than 10 m/s as it should be (the velocity given was 5 m/s +- 5 m/s, so that would range from 0-10 m/s). The correct answer for delta x = 6.7 x 10-37 m (hopefully yo...

Abby Soriano 1J

Yes, the more you know about the momentum, the less you know about the position, and vice versa. This is due to the idea of how, at the atomic scale, methods of observing objects (i.e. knowing its position) with as small a mass as an electron's can actually disrupt other aspects of it such as its ve...

Abby Soriano 1J

If we're speaking on the subject of classical particles, they would not give rise to diffraction patterns when passing through the holes in the barrier since diffraction patterns result only from interacting waves. Since the particle is acting as a particle alone, it would not have diffraction patte...

Abby Soriano 1J

If you look on the class website, there's actually an error in the solutions manual for that problem. They made delta v = 5 m/s rather than 10 m/s as it should be (the velocity given was 5 m/s +- 5 m/s, so that would range from 0-10 m/s). The correct answer for delta x = 6.7 x 10-37 m (hopefully you...

Abby Soriano 1J

Your math was correct but the answer is C because, although energy can be expressed as a negative value for energy lost, when calculating things such as frequency or wavelength, you always use positive values since there is no such thing as a negative frequency, for example.

Abby Soriano 1J

It would be good just to know that the Balmer series occurs in the visible light region (and returns to an energy level of n=2) and that the Lyman series occurs in the UV region (and returns to an energy level of n=1).

Abby Soriano 1J

If you really wanted to use a constant that's a little more precise than 3.00x10^8 m/s but less tedious than 2.99792x10^8 m/s, I'd suggest using 2.998x10^8 m/s. But then again it shouldn't matter too much as long as your final sig figs are correct.

Abby Soriano 1J

In the De Broglie Equation, λ = h/p, where h is Planck's constant (6.626 x 10^-34 J s) and p is the momentum (mass x velocity). Since h is already a very small number, the momentum of very small and high velocity objects (such as electrons) are significant because they calculate a wavelength that is...

Abby Soriano 1J

Quite the opposite, when an electron moves up to a higher energy level, it is because it has absorbed energy that matches the energy difference between those two levels. An electron emits (loses) energy as a photon as it falls back down to a lower energy level.

Abby Soriano 1J

The question asks for the final and initial energy levels of the electron during the /emission/ of energy, so they're referring to the emission spectrum. As a result, n=3 would be the initial energy level while n=1 would be the final energy level since electrons only emit energy when falling from a ...

Abby Soriano 1J

There's never really a time when you should be wanting to use fractions when balancing equations since you want all of your stoichiometric coefficients to be whole numbers. Usually you end up needing to use fractions when there's no other way to balance something. For example, if you find that there...

Abby Soriano 1J

The 100 gram trick is used to make calculations involving finding an empirical formula easier! If you're given the percent composition of a sample, you might assume that you would need to work backwards in order to find the original number of grams in order to convert them to moles. However, with th...

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