CHEMISTRY COMMUNITY

Brian Nguyen 2I

Charlotte Chen 3B wrote:Do we still have to post this week if we already have 50 posts from previous weeks?

No, if you already have 50 posts, then you shouldn't have to post anymore.

Brian Nguyen 2I

Thank you so much for teaching this class! It really helped me gain confidence with concepts I had learned in chemistry that I had previously struggled with. Thanks for providing so many resources for us to use!

Brian Nguyen 2I

You put platinum on the side when the reaction doesn't contain a solid This makes sense, as there are no solids on the cathode right side of the diagram, but I was confused because on the anode left side of the diagram, there isn't a solid either. There's Hg in liquid form and Hg2 2+ in aqueous form.

Brian Nguyen 2I

Honestly, the review session really help you feel comfortable with old topics that you haven't covered in a while, so as long as you go to as many of those, you should be fine!

Brian Nguyen 2I

For part A of 6M.5, we are asked to write a cell diagram for the following reaction:
2 NO3-(aq) + 8 H+(aq) + 6 Hg(l) -> 3 Hg2 (2+) (aq) + 2NO (g) + 4H2O (l)

Why is that in the answer key, we put platinum at the end of the right side of the cell diagram but not on the left side?

Brian Nguyen 2I

To determine the unit for k, you just have to take into account that the unit for rate will always be M/sec. Thus, for first order, the unit would be 1/s for k. For second order, the unit would be 1/M s for k. For zero order, it would be M/s.

Brian Nguyen 2I

Take a look at the half reactions and balance out the electrons to make sure that there is the same amount of electrons on both sides of the redox equation.

Brian Nguyen 2I

An oxidizing agent is what facilitates reduction. Thus, the more likely something is to be reduced, the better of a oxidizing agent it is. Thus, it needs to have a high reduction potential.

Brian Nguyen 2I

The order describes how the concentration of a reactant will affect the overall rate. This is determined by the exponents in each concentration of the reactant in the rate law.

Brian Nguyen 2I

You can determine order based on different graphs. [A] vs. Time is linear for zero order, ln[A] vs. Time is linear for first order, and 1/[A] vs. Time is linear for second order.

Brian Nguyen 2I

K can indeed change, but only from a change in temperature.

Brian Nguyen 2I

Yes, this is possible. But given the high activation energy, it will take a long period of time to form such diamonds, and even so, the diamonds formed will most likely be minuscule in size.

Brian Nguyen 2I

The instantaneous rate will decrease over time, as there becomes less reactants to be used up.

Brian Nguyen 2I

Delta G of Gibbs' free energy determines whether something is thermodynamically favorable. This is basically spontaneity. A negative delta G is spontaneous.

Brian Nguyen 2I

Delta G of Gibbs' free energy determines whether something is thermodynamically favorable. This is basically spontaneity. A negative delta G is spontaneous.

Brian Nguyen 2I

Yes, the order matters. Always remember that the anode goes on the left while the cathode goes on the right of the diagram.

Brian Nguyen 2I

When you plug in n, you're plugging in the value of the number of electrons that are transferred in the redox reaction. This means that you use the number of electrons calculated when balancing the reaction itself.

Brian Nguyen 2I

Yes, electrons will always leave from the anode. Losing electrons is oxidation, so this is indeed the oxidized side.

Brian Nguyen 2I

Yes, these should be the same and you can use them interchangeably.

Brian Nguyen 2I

Yes, larger molecules have more complexity and thus have more entropy. Also, gases will have more entropy than liquids, and liquids will have more entropy than solids. Solids have the least entropy out of the phases.

Brian Nguyen 2I

Entropy is a state function, thus meaning that it doesn't matter which path is taken. You're asked to calculate the entropy of vaporization at 85 degrees, but you only are given the entropy values at 100 degrees, so you must create a path where 85 degrees is the start and end point. This is why you ...

Brian Nguyen 2I

Thermodynamically favorable is when the reaction is spontaneous, which would thus mean that the Gibbs' free energy is negative. When the Gibbs' free energy is positive, the reaction is nonspontaneous and thus thermodynamically unfavorable.

Brian Nguyen 2I

We won't have to use the integral equations. The integrals were just used as a way to prove the equations we have to memorize in class for entropy, such as deltaS= nCv(lnT2/T1) or deltaS = nR(lnV2/V1). Those equations are just for you to understand conceptually why we use the equations we use.

Brian Nguyen 2I

Try to remember the units of R in order to solve. Look at the context of the situation by looking at the units in the equation you're trying to solve, and make sure that the R you use properly cancels out the units.

Brian Nguyen 2I

This cannot occur. You cannot have a negative delta G. By taking a look at the equation ∆G = ∆H-T∆S, a negative entropy or deltaS will cancel out the minus sign in the equation, thus making T∆S a positive value. This positive value added with a positive enthalpy or positive delta H will give you a p...

Brian Nguyen 2I

Yes, that's correct. If delta G is negative, then the reaction is always considered spontaneous.

Brian Nguyen 2I

deltaS = q/T is used for when temperature is constant. deltaS = nR*ln(V2/V1), we use when there is a change in volume. Temperature is constant here. deltaS = nC*ln(T2/T1) is when temperature changes. C values are used depending on whether pressure or volume is constant. Cp is used when pressure is c...

Brian Nguyen 2I

Cp will be greater than Cv. Usually, the equation used will be Cp = Cv + R.

Brian Nguyen 2I

Entropy is how much disorder there is in a system. A positive entropy means there is more disorder, whereas negative means that there is less. The equation for entropy is S = kB(lnW).

Brian Nguyen 2I

It depends on what the kind of T is used. If it is delta T, it won't make a difference between Celsius and Kelvin. Since Kelvin is converted by adding 273.15 to the Celsius measurement, either unit does not matter with change in temperature. It only matters when it is T by itself. With just T, eithe...

Brian Nguyen 2I

Reversible reactions will stay in equilibrium, whereas irreversible reactions do not stay at equilibrium with the surroundings. Pressure is constant with irreversible reactions, hence the reason why one simply multiplies the negative pressure by the change in volume (-P x delta V) to calculate work....

Brian Nguyen 2I

Systems will perform work on the surroundings when the work is negative. Work is performed on the system when work is positive.

Brian Nguyen 2I

Salts are acidic if they contain a weak base anion and strong acid cation. They are basic if they contain a strong base anion and weak acid cation. They are neutral if they have a cation from a strong acid and an anion from a strong base.

Brian Nguyen 2I

To solve the problem, you use the equation PV=nRT and solve plugging in what's given to you in the equation. As for the pressure, you just need to remember that 760 torr is equivalent to 1 atm. The R value can be found on the equation sheet we're given. From here, you can solve the equation.

Brian Nguyen 2I

Other examples can include snow and ice. I also found online that iodine will sublime from solid to purple gas at 100 degrees Celsius. It also says that moth balls can also sublime.

Brian Nguyen 2I

Neutralization is just when an acid and base react to each other. In this case, the acid would be HCl and the base would be NaOH. These two react to form a salt and water. In this case, H2O and NaCl is formed.

Brian Nguyen 2I

Yes, I would assume that these diatomic molecules are in their most stable state. Thus, the enthalpy would be zero.

Brian Nguyen 2I

A molecule is considered diatomic so long that there are two atoms involved. O2, Br2, H2, etc are considered diatomic elements, since there are two atoms of the same element in a molecule.

Brian Nguyen 2I

State properties mean that you only need to focus on the final and initial values. In this case, for enthalpy, it is a state property since enthalpy is only calculated using the initial heat and final heat values.

Brian Nguyen 2I

Catalysts will lower the activation energy of a reaction. This does not affect the change in enthalpy, as enthalpy is a state reaction. Thus, this does not affect whether a reaction is exothermic and endothermic, which can be indicated from the change in enthalpy.

Brian Nguyen 2I

Yes, this is definitely true. Exothermic reactions are always labeled by a negative delta H, whereas endothermic reactions are always labeled by a positive delta H. This is due to how exothermic reactions release heat, whereas endothermic reactions absorb heat.

Brian Nguyen 2I

If there's just one arrow, it means that the reaction usually proceeds in that direction most of the time. It doesn't necessarily mean that it will proceed only in that way, but just probably that for the majority of the time, it proceeds in such direction.

Brian Nguyen 2I

If the pressure increase did nothing to shift the reaction, chances are this is because the gas involved was an inert gases. Pressure increases and decreases can change concentrations in reactions, which shifts the reaction, but inert gases will not change volume. Thus, there is no change to concent...

Brian Nguyen 2I

Ka and Kb added together make Kw, which is always 1.0 times 10^14. Thus, as Kw always remains the same, as Ka increases, Kb will also decrease at the same time. This is also applied vice versa.

Brian Nguyen 2I

When you reverse a reaction, K is just equal to the reciprocal of K instead. This is 1 divided by K.

Brian Nguyen 2I

To use the ICE method, you have to know what ICE stands for. I stands for initial, C stands for change, and E stands for equilibrium. Using this method means you are trying to solve for E. First, you have to balance the equation given. Then for I, write out the initial concentrations given in the pr...

Brian Nguyen 2I

Whenever something is being produced, you use a positive sign for what is being made. For what we use to get to that end point, we use a negative sign. For example, if there is a shift to the right in the reaction, for the products you will use an add sign. You will use the subtract sign on the reac...

Brian Nguyen 2I

H2O is not included because solids and liquids are not included in equilibrium equations. The only exception would be if H2O was a gas.

Brian Nguyen 2I

K is affected by temperature based on whether the reaction is exothermic and endothermic. The reaction will favor product formation if the reaction is endothermic. The reaction will favor reactant formation is the reaction is exothermic.

Brian Nguyen 2I

As long as the gas isn't inert, then changing the pressure will change the reaction due to a change in volume. The change in volume changes the concentration of the products and reactants, which thus causes a reaction change.

Brian Nguyen 2I

Don't convert to pascals. We use atm or barr in these problems involving pressure, so there is no need for conversion.

Brian Nguyen 2I

When I use this website, I usually just scroll down to the topics on the home page and find the topic that we learned about for the week. Then I answer questions under certain posts. I usually use quick links a lot because it helps me check to see how many posts I've made so far. Sometimes, if you w...

Brian Nguyen 2I

The Bronsted definition involves protons, where as the Lewis definition involves electron pairs. Lewis acids are electron pairs acceptors, whereas Lewis bases are electron pair donors. Bronsted acids are proton donors, whereas Bronsted bases are proton acceptors.

Brian Nguyen 2I

To find the oxidation number, I just take the overall charge of the compound and then take into account the charges of other elements or compounds within the cation to find the charge of the metal ion.

Brian Nguyen 2I

Honestly, I just like to focus on solely textbook problems. The more practice I get, the more confident I feel about topics, since most of the exam questions are structured like the textbook problems.

Brian Nguyen 2I

Most of the concepts that we go over in 14A will be applied in 14B. This is probably especially seen with the material we covered with acids and bases, since we only touched on the surface of this subject before the final.

Brian Nguyen 2I

Reading the textbook is just helpful if I find that I need to find other ways to understand the material if lectures confuse me. Generally, the textbook is most useful as a source of practice, since I mainly do practice problems from there. The more practice problems I do, the more prepared I am for...

Brian Nguyen 2I

Dentate indicates that there is a binding site. The prefix indicates how many binding sites there are. For example, bidentate means there are two binding sites. In your case, tetradentate would have four binding sites.

Brian Nguyen 2I

You order based on the actual ligand name, not by prefix. Let's say we had tetraammine and dichloro. Dichloro would not go before tetraammine because ammine comes before chloro alphabetically.

Brian Nguyen 2I

It's based on chloro and ammine. Thus, tetraammine would go first and then dichloro.

Brian Nguyen 2I

From what I remember silver and gold are some exceptions. However, I remember someone asked my TA if we were required to know any exceptions, and she said that you probably wouldn't have to remember them except probably the one for iron.

Brian Nguyen 2I

You name the ligands in alphabetical order. The prefixes are not taken into account when u order the ligands in alphabetical order. The ligand name is listed before the metal cation.

Brian Nguyen 2I

Boiling point is the point at which a liquid turns into a gas. Melting point is the point at which a solid turns into a liquid. The more intermolecular forces that are present, the higher the boiling and melting points of a substance will be.

Brian Nguyen 2I

Yes, it does happen when you have another compound with a lone pair of electrons. However, you also have to keep in mind that these form when the other compound has a hydrogen bonded to a fluorine, oxygen, or nitrogen atom.

Brian Nguyen 2I

Polar bonds are formed based on if there is a notable difference in electronegativity between two elements within a molecule. Polar molecules are formed when the dipole moments formed due to polar bonds don't cancel out. If they do cancel out, then these molecules are nonpolar.

Brian Nguyen 2I

Yes, London dispersion forces are always present. However, these forces are only significantly strong the larger the mass of the molecule.

Brian Nguyen 2I

Just look at the electronegativity. If there's a notable difference between the two electronegativities, then the bond is polar. If there isn't a notable difference, then the bond is non-polar.

Brian Nguyen 2I

Yes, they prefer the resonance structure with the lowest energy. This would be mean the formal charges are closest to zero on the atom.

Brian Nguyen 2I

You can have a nonpolar molecule even if there are polar bonds due to the fact that the dipoles formed from polar bonds are cancelled out from the overall shape of the molecule.

Brian Nguyen 2I

Higher boiling and melting points are indicated by if there are stronger intermolecular forces present. For example, if u have a molecule with dipole-dipole forces and London dispersion forces and a molecule with only London dispersion forces, then the molecule that has dipole-dipole forces has a hi...

Brian Nguyen 2I

You can identify a radical when there is the presence of a lone unpaired electron. One example is CH3.

Brian Nguyen 2I

Carbon and hydrogen's electronegativity difference is too small compared to that of hydrogen with the elements nitrogen, oxygen, and fluorine. Thus, it can't form partial charges for hydrogen bonds to occur.

Brian Nguyen 2I

Double bonds are stronger than single bonds. Thus, they are more stable than single bonds and less reactive than single bonds.

Brian Nguyen 2I

I don't think you have to memorize the actual values. It just matters if you know what the electronegativity trend is, as we've been taught in lecture.

Brian Nguyen 2I

Usually when it comes to helium, it's a noble gas, so we don't need to worry about it in terms of electronegativity. It has a filled outer energy level. It fulfills the duet rule.

Brian Nguyen 2I

I doubt that you will need to remember nomenclature. Lavelle focuses on mainly how well you know how to problem solve, so you shouldn't have to worry about nomenclature.

Brian Nguyen 2I

The test will be multiple choice for the midterm, so I doubt that we'll have to draw them out. If anything, we'll be given pictures of it on questions, but you probably won't be expected to draw them.

Brian Nguyen 2I

It decreases as you move across a period because the atomic number increases. When the atomic number increases, the amount of protons increase. There is more of a positive charge present, increasing the attraction between the electrons and protons, thus decreasing the space between the electrons and...

Brian Nguyen 2I

Metric conversions should be memorized. However, when it comes to any other conversions, they should be given on the test.

Brian Nguyen 2I

You have to post five times per week, since they check the time when there's a post. You can check your posts in quick links to see how many you have.

Brian Nguyen 2I

The elements on the left side of the periodic table are more likely to form cations due to the fact they have a small number of valence electrons. The smaller the number, the more likely you are to lose an electron instead of gain one.

Brian Nguyen 2I

I usually just print out a periodic table and label it based on where the trends are. It helps for me to see it visually all in one place in order to memorize it.

Brian Nguyen 2I

I'm pretty sure that as long as you post something, it counts. As long as it is any interaction, your post should be counted as a point.

Brian Nguyen 2I

In my opinion, it will probably cover material starting from where we left off after we took Midterm 1 up until Focus 9 if I had to guess. I don't think it'll be a cumulative test, but there's probably bound to be mention of old topics at some point since most topics that we learn probably overlap w...

Brian Nguyen 2I

The spin state does not exactly mean that an electron points "up" or "down" but it differentiates between two different states itself. This can be labeled as a +1/2 or -1/2.

Brian Nguyen 2I

I initially had trouble with this question too. Some of the waves look continuous, but you need to check to make sure that the ends of each wave will meet perfectly. That's how you can differentiate between which waves are according to the model and which are not.

Brian Nguyen 2I

I heard that they're waiting until everyone has finished taking the exam. The last exam should've been taken on Friday, so anytime soon probably the scores will be up. You can probably check through CCLE to see your score soon.

Brian Nguyen 2I

I feel like you would have to know that conversion beforehand. However, I feel like we won't have to necessarily remember that kind of stuff for our upcoming midterm, so you probably don't have to worry about those kind of conversions.

Brian Nguyen 2I

Yes, I think you should be fine if you have some scratch paper as long as it is visibly blank, but I would check with the TA first just to be safe.

Brian Nguyen 2I

Intensity relates to amount of photons, whereas frequency relates to the quality of a photon. Frequency is the number of cycles per second.

Brian Nguyen 2I

Increasing amplitude doesn't have a correlation with the wavelength and frequency. However, it is important to note that the wavelength will decrease when frequency increases, and vice versa.

Brian Nguyen 2I

Kilograms is the SI Unit for mass, so you will have to keep in mind that when solving, your mass is converted in kilograms in order to get the right calculation.

Brian Nguyen 2I

I would try the practice problems in the textbook first, and then seek help at a step up session if you find yourself struggling to complete the practice problems. I hear that the step up sessions really help, so honestly it doesn't help to do both just for the extra practice.

Brian Nguyen 2I

I think it's just necessary that you know where things generally are on the spectrum. It might be helpful to keep in mind where visible light is on the spectrum (400-700 nm). You don't have to memorize the exact wavelengths, but it's necessary to know which has low and which has high frequency when ...

Brian Nguyen 2I

You need to use substitution in order to acquire this equation. You can solve for v in E = λv to get v = E/λ. Then you can substitute the v in E = hv to get E = hc/λ.

Brian Nguyen 2I

Usually you just look for whether it states in the problem itself. I also pay attention to units, so if they ask for Hertz, the problem is asking for frequency. If the problem asks for nanometers, they are looking for wavelength.

Brian Nguyen 2I

I try to make sure I take advantage of all the resources given. I watch all the modules and lectures and take notes. Then I try to divide evenly time to read the textbook and finish all the practice problems. It also helps for me to look at the the key concepts and try to write them on flashcards fo...

Brian Nguyen 2I

I always just don't think of rounding until I get to the answer that I solve for. Once I reach that point, I use significant figures to figure out where to round. I always just remember that leading zeros don't count and trailing zeros do count if there is a decimal present in the number when determ...

Brian Nguyen 2I

Hi! For me, it's just much easier for me to visualize which reactant is limiting using the first method. It is up to preference, but I just find the first method easier to understand since it shows exactly which reactant I will end up with excess.

Brian Nguyen 2I

I think after practicing a lot of problems, you start to pick up on the formulas of certain compounds, but typically as many people have said, you just need to focus on how to solve a problem. As long as you're good with the concepts taught in class and how to apply them, you should be fine.

Search found 102 matches