CHEMISTRY COMMUNITY

Shruti Kulkarni 2I

I know that we use the concentration equations, like E = Enaught - RT/nF ln Q for concentration cells. I was wondering if they can be used for Galvanic cells as well or are they exclusive to concentration cells?

Shruti Kulkarni 2I

I would definitely email a TA and see if they had worksheets or if it was just a general review. You could also search Chemistry Community or CCLE to see if they posted their worksheets on there.

Shruti Kulkarni 2I

Your answer seems to be right. It could be a rounding error, with you rounding up when sapling wants you to go to the first decimal place?

Shruti Kulkarni 2I

An intermediate is a compound that is produced in a reaction and then subsequently used up in a later step of the reaction. So, it is not present in the products or the reactants but is both created and used up over the course of the reaction.

Shruti Kulkarni 2I

You can tell if a reactant is part of the slow step if it is present in the overall rate law, as that only includes the reactants in the slow step.

Shruti Kulkarni 2I

In regards to equilibrium, a large k value will tell us that at equilibrium, there will be a larger amount of products formed in comparison to reactants. In the case of kinetics, a large k value tells us that the reaction is proceeding at a fast rate.

Shruti Kulkarni 2I

It does affect the rate constants as catalysts decrease the activation energy, increasing rate constant as seen through the Arrhenius equation. As Ea decreases, the overall value of k increases, meaning that a catalyst will increase the rate constant.

Shruti Kulkarni 2I

The n value represents the number of electrons transferred in the redox reaction! So, if you balance the two half-reactions to make sure that an equal number of electrons are being transferred, that number of electrons is the value you would input for n.

Shruti Kulkarni 2I

I believe we are done with new material after Wednesday's lecture and that Friday's lecture is a general review. From last quarter, I would agree that studying the information from previous midterms is very important.

Shruti Kulkarni 2I

It isn't a guarantee that there will be an intermediate, but it is highly likely for most reactions. You can tell if it is an intermediate if the compound does not show up in the overall balanced equation and is created in one step and used up in another step.

Shruti Kulkarni 2I

Usually, if the mechanism follows both of the steps outlined, it is most likely to be correct. However, it is not a tried and true method as there could be other mechanisms that also follow both steps and could equally be the correct mechanism for the reaction. So, to know if it is actually the reac...

Shruti Kulkarni 2I

Molecularity is defined as the number of individual molecules involved in an elementary step of a reaction. The rate law follows from molecularity, as the number of molecules involved in the step defines the step's rate law. So, for example, if you had a step that was A + A, it would have a molecula...

Shruti Kulkarni 2I

Yes! The oxidizing agent is the molecule that is reduced and is oxidizing another molecule. Vice versa, the reducing agent is the molecule being oxidized and reducing something else.

Shruti Kulkarni 2I

The moles, or n, refers to the number of electrons transferred in the balanced reaction.

Shruti Kulkarni 2I

The anode is on the left and is oxidized while the cathode is on the right and is reduced. So, find which molecules are being oxidized/reduced and you'll find the cathode and anode of the reaction.

Shruti Kulkarni 2I

I dont think the order of the reactants or products matters much, as long as each is on the correct side of the equation it should work.

Shruti Kulkarni 2I

Double-check your cell potentials to make sure you're using the right ones! Otherwise, your steps are right.

Shruti Kulkarni 2I

Adiabatic means no heat is exchanged between the system and its surroundings.

Shruti Kulkarni 2I

A state function only depends on the final and initial values, it doesn't care about the "path" taken to get there. Free energy is a state function because it follows these restrictions.

Shruti Kulkarni 2I

Delta G naught is the standard reaction Gibbs free energy. If the reaction is not occurring at standard conditions, then regular gibbs free energy is used, which takes into account variations in the temperature, pressure, and anything else.

Shruti Kulkarni 2I

It's referencing Gibbs Free Energy! This shows that delta G = delta H - T delta S

Shruti Kulkarni 2I

I'm going to be talking with my friends and catching up on all the hw for the week :D

Shruti Kulkarni 2I

Delta U = Delta H + w, since delta H = q at a constant pressure.

Shruti Kulkarni 2I

Temperature can be measured in Kelvin if that is the K you are talking about. If you are talking about the equilibrium constant, k then changes in temperature will change the equilibrium constant for the reaction. If the reaction is endothermic, adding heat will shift the direction towards the produ...

Shruti Kulkarni 2I

I love to binge New Girl. I've watched it like 10 times!

Shruti Kulkarni 2I

The enthalpy for the reaction is given by the heater, which is running at 500 W (J/s). So, you would convert the W into J by multiplying by the 4.02 minutes over which each compound boils, as that would reference the time in which the compound is vaporized. This is the enthalpy in the reaction. Then...

Shruti Kulkarni 2I

I think it is important to note that two different heat capacities are given, and as we do not know how to use both in the same equation, it can be assumed that you would need to take it in steps. Also, as it is a heating and vaporization problem, the process of doing two things would probably indic...

Shruti Kulkarni 2I

The midterm covers up until Friday, 2/12's lecture. So, the lectures during the midterm week do not count towards the midterm and, as it is not cummulative, nothing covered in the previous midterm will show up as well.

Shruti Kulkarni 2I

I think Dr. Lavelle mentioned that he has to put in the points manually, so it isn't an accurate reflection until he updates them. Just wait it out!

Shruti Kulkarni 2I

I would assume that the same format would follow from 14A where the TA's could go over your exam questions with you. You would have to go to a TA's office hours and ask them to see if they'll go over your midterm with you.

Shruti Kulkarni 2I

U and E are extensive properties! They depend on the number of molecules are in the system.

Shruti Kulkarni 2I

When work is being done to a system, the work will be positive and if work is being done by the system, the work will be negative.

Shruti Kulkarni 2I

Since forming bonds requires energy and breaking bonds loses energy, you can determine if it's exo or endothermic based on if bonds are being created or destroyed. If a bond is being created, it probably is endothermic because making a bond requires energy, and if a bond is being broken, it is proba...

Shruti Kulkarni 2I

You can make x negligible for any equation where the k value is very small. It is not limited to just cubic equations.

Shruti Kulkarni 2I

I am definitely thinking about taking BL next quarter, simply because I wanted to learn all of the 14B material before diving straight in. I am wondering if taking it with Chem 14C is a good idea though, because of how hard Organic Chemistry is supposed to be. Also, having an online lab seems diffic...

Shruti Kulkarni 2I

Honestly, the only way to improve this is to slow down and double-check everything you are doing very thoroughly. You should also be thinking about what your calculator outputs and if it makes sense for the scenario as a whole, and if it doesn't then look through and see what mistake you could have ...

Shruti Kulkarni 2I

It means that the heat of the object's surroundings is equivalent to the heat of the object. So if the object lost energy to its surroundings, the surroundings would gain that same amount of energy.

Shruti Kulkarni 2I

Kelvin is a Celsius plus 273. So, for every one-unit increase in Celsius, there will be a one-unit increase in Kelvin as well.

Shruti Kulkarni 2I

Yes! When the steam touches something cooler, in this case, your skin, it condensates and therefore loses all the energy it gained in order to become and transferring it into your skin. This would cause you a much more severe burn in comparison to just water at 100 degrees, as it is also releasing a...

Shruti Kulkarni 2I

Yes! Since solids and liquids are not included in the k constant, changing their concentrations should not have an impact on the equilibrium of a system. However, remember to look out for aqueous solutions, because if the volume of the solvent is increased by a lot then it is possible that the conce...

Shruti Kulkarni 2I

We can assume that x is insignificant usually when Ka < 10^-4, or when x is less than 5% of the initial amount of the acid/base. We can assume x is insignificant because it is so small it will barely impact the overall concentration of the acid.

Shruti Kulkarni 2I

Salts can be acidic, basic, or neutral. If there is a salt with a cation from a strong base and an anion from a weak acid, the salt will be basic as the anion will draw a proton from water, forming OH- ions. If there is a salt with a cation from a weak base and an anion from a strong acid, the salt ...

Shruti Kulkarni 2I

I believe that any molecule with two atoms is considered diatomic. The diatomic elements are those that can form diatomic molecules with themselves.

Shruti Kulkarni 2I

I like listening to lowfi music or sometimes I use the instrumental versions of songs I like!

Shruti Kulkarni 2I

You have to look at the ions it is composed of! If the salt is made of a cation of a strong base and the anion of a weak acid, the salt itself would make the solution basic as the anion would steal a proton from water and leave OH- behind. In contrast, if it is made of a cation of a weak base and th...

Shruti Kulkarni 2I

You can get the equilibrium value of O2, as you know the moles of O2 at equilibrium and the volume of the container. Using the moles of O2 at equilibrium, you can use stoichiometry to find the moles of SO2 and SO3 present at equilibrium in the reaction. When you have those values, you can divide the...

Shruti Kulkarni 2I

You're right that solving for Kc and Kp is the same. Kp uses partial pressures while Kc uses concentrations. If you want to convert between the pressure to the concentration you can use PV = nRT, as n/V is concentration.

Shruti Kulkarni 2I

You can multiply both sides by (0.7 -x)^2 to remove the denominator and then expand each expression to create an equality. Afterward, you can move all the terms to one side and set the equation equal to zero. All that's left is to solve using the quadratic equation or even graphing out the equation ...

Shruti Kulkarni 2I

I think you need to remember that pH is just a measure of the H3O+ concentration in the solution, so when you dissolve the acids in the solution, it measures how many H3O+ ions are in the solution afterward. However, this is very relative with different concentrations of acids, weak or strong. The m...

Shruti Kulkarni 2I

Hi! So you're just supposed to play around with the equations and their k constants to find the k constant of the original equation. Personally, I inversed the N2(g)+3H2(g)↽−−⇀2NH3(g) equation, which would mean that you would inverse the k constant by doing 1/k. Then, I used I2(g)↽−−⇀2I(g) and multi...

Shruti Kulkarni 2I

Yes, that is correct. K is just Q when the products are at equilibrium, so you can use Q as a measure of determining which way the reaction will move. When K>Q, the reaction will shift towards the products, and if K< Q, the reaction will shift towards the reactants.

Shruti Kulkarni 2I

Concentrations cannot be negative because it is defined by moles/L. As it is impossible to have a negative number of moles or a negative number of L, the concentration must be positive.

Shruti Kulkarni 2I

I want to start working out a bit more as well as journaling my days so I remember the things that I did in the future!

Shruti Kulkarni 2I

I agree with what was said above, you cubed the 1.75 when it should be squared!

Shruti Kulkarni 2I

I think it follows the same format as 14A where the lectures were recorded and posted under the media tab on CCLE. I assume we would be informed via email if anything was synchronous!

Shruti Kulkarni 2I

I would agree with using chrome! It seems that sapling just operates better with chrome than any other browser so maybe use it instead.

Shruti Kulkarni 2I

Yep! This happens cuz weak acids have pH's lower than 7 and weak bases have pH's over 7, so they usually will be lowering or raising the pH.

Shruti Kulkarni 2I

I think you should know that if the pKa < pH, then the majority of atoms in the solution will be charged for acids. However, if the pKa > pH, then the majority of atoms in the solution will be neutral, for acids. The inverse is true for bases. Also, you should know that the pKa value being low indic...

Shruti Kulkarni 2I

Yes, you can just say less than or slightly less than to indicate the bond angles. I don't think we are expected to know exactly the angles of the bonds in the presence of lone pairs.

Shruti Kulkarni 2I

It just depends on the level of specification you want! If you wanna tell the energy level as well as the hybridization of the orbital, you would put the 2. But sp^3 is equally correct if you are not specifying the energy level.

Shruti Kulkarni 2I

You add the -ate to the TM when the charge of the coordination compound is negative to signal that the compound is negative. If the charge is neutral or positive, you will leave it as the metal, like cobalt.

Shruti Kulkarni 2I

You would name the ligands in alphabetical order, not including the prefixes, and then put the transition metal at the end, followed by whatever was not in the brackets.

Shruti Kulkarni 2I

I believe they'll be very similar otherwise we would have been informed. I would suspect it would once again be on a lockdown browser with a second device on a zoom call.

Shruti Kulkarni 2I

Hi! In this example, the second Cl is not directly bonded to the TM cation but instead bonded to the coordination compound that is formed. This would cause it to be outside the sphere of influence for the cation, and so its name wouldn't be changed in regards to being in the sphere of influence.

Shruti Kulkarni 2I

The most stable structure will have the lowest formal charge. If there must be a charge, it is most stable if the most electronegative atom holds the negative charge.

Shruti Kulkarni 2I

From what Dr. Lavelle said during the lecture, a Bronsted base and a lewis base are the same, it is just the way you observe them that's different. In terms of a Bronsted base, you would observe it from the actions of the protons, as it will receive a proton, in comparison to Lewis bases which will ...

Shruti Kulkarni 2I

From what I have seen, it appears that you would put the TM cation first and then put the ligands after it. I do not think the ligands need to be in alphabetical order but it couldn't hurt to follow that convention.

Shruti Kulkarni 2I

You look at the number of bonds for the atom. If it's a single bond, it is just one sigma bond. Double bonds will contain one sigma bond and one pi bond and triple bonds will contain one sigma bond and two pi bonds. Basically, there is only ever one sigma bond, the rest of them have to be pi bonds.

Shruti Kulkarni 2I

Hi! If the structure is symmetric, with the same kinds of atoms evenly spaced out then the molecule will be nonpolar. However, if the structure isn't symmetric and there are different atoms on each side, then the structure will be polar. This includes lone pairs, as if the structure is symmetric wit...

Shruti Kulkarni 2I

Yes! Every bond and lone pair is an area of electron density. However, double and triple bonds do not count as separate areas of electron density but instead are a single area of electron density.

Shruti Kulkarni 2I

Hi! Isoelectronic means that the atoms have the same amount of electrons. For example, O2-, F1- and Ne are isoelectric as they all have 10 electrons. From my understanding, isoelectronic atoms share similar properties, so knowing that they are isoelectronic can help you in understanding the properti...

Shruti Kulkarni 2I

In Dr. Lavelle's lecture today he mentioned that in the case of Benzene, the carbon will form three hybrid orbitals and keep one normal p orbital. He then said that the electron in the remaining p orbitals of each carbon will be delocalized and shared equally among the carbons. I was wondering if th...

Shruti Kulkarni 2I

Hi! From my understanding, I think that they contribute as they are a region of electron density around the central atom and contribute to the octet of the central atom, so they must be taken into account when forming hybrid orbitals as they impact the shape of the molecule.

Shruti Kulkarni 2I

Yes, they do! Lone pair electrons count as another region of electron density. In fact, they often have a larger electron density than a bond!

Shruti Kulkarni 2I

Yea! Just use the mass percentage to determine the empirical formula, and then divide by the molar mass to find the molecular formula. Then you can make a lewis structure of the molecular formula.

Shruti Kulkarni 2I

Yes, the answer key is most probably wrong. C has 4 electrons and each H has 1 electron, which will result in 7 electrons, implying a radical.

Shruti Kulkarni 2I

There's a lone pair in NO2- which changes its shape from linear to bent! The lone pair repels the bonds creating the lone pair shape.

Shruti Kulkarni 2I

Hi! If the shape is symmetric with the same atoms on each side, it won't be polar. If the same is asymmetric at all or has different atoms at different points, the bond will be polar.

Shruti Kulkarni 2I

That's right! The oxygen will have 2 lone pairs and will be double-bonded with C. The hydrogens will be single bonded with the C as well. Maybe double-check and see if the question asks for charges because that could be resulting in your error.

Shruti Kulkarni 2I

Hi! S actually has a larger atomic radius as it has more electrons, which would make the bond length between Si-S longer than that of Si-O. Atomic radii increase down the period! And as S has a larger radius, it will have a longer bond length in comparison to O.

Shruti Kulkarni 2I

That's right! Hydrogen bonds can only form when hydrogen is bonded to a highly electronegative ion, such as F. The difference in their electronegativities leads to the dipoles that are formed in the molecule which allows for the hydrogen bonds to occur. This is why hydrogen bonds only happen with N,...

Shruti Kulkarni 2I

Dipole-dipole and LDF forces are similar as they occur from the dipoles of molecules. However, dipole-dipole forces occur in polar atoms/molecules while LDF forces occur spontaneously in any atom/molecule as a result of a shift in its electron density.

Shruti Kulkarni 2I

I don't think there are any more modules for the subjects we are covering and Dr. Lavelle said that he probably won't upload anymore.

Shruti Kulkarni 2I

Hi! I'm pretty sure that the length of the bonds in your Lewis Structures don't matter as long as you conceptually understand that double and triple bonds are shorter and the reasoning behind this.

Shruti Kulkarni 2I

Noble gases are usually left out of periodic table trends as they have a full octet and therefore do not want to gain any or lose any electrons. So, they are not used in periodic table trends such as electronegativity.

Shruti Kulkarni 2I

Hi! Isoelectric atoms have the same number of electrons (ex. F- and Ne). However, as you can see they will have different charges and different properties overall.

Shruti Kulkarni 2I

Hi! So the charge on the outside of the brackets represents the overall charge of the atom or molecule. Usually, you can figure out the charge of the molecule based on the properties of the atoms that make it up. As Dr.Lavelle said in the lecture, you would use the formula (FC = ( S/2 + LP)) to find...

Shruti Kulkarni 2I

I don't think it matters at all! As long as your structure is accurate the way you put the atoms should not influence anything.

Shruti Kulkarni 2I

Hi! I agree with the above statements. If you want to find the number of total electrons of the element, you look at the atomic number. And if you want to find the number of valence electrons, you would look at the number of electrons in the s and p orbitals for rows 1-3.

Shruti Kulkarni 2I

That does look like the correct answer. I agree with what is said above about removing the commas! The only other thing I can think of is if you are using the carrots (^) to show the exponents instead of putting the actual exponents. Make sure to use the exponent feature on sapling to actually put e...

Shruti Kulkarni 2I

Hi! Generically, the corresponding orbitals do follow the pattern l = 0 is s, l = 1 is p, l = 2 is d, and l = 3 is f. So, for the 7p^5 example you mentioned, it would have quantum numbers of n = 7 and l = 1, as you are in the p-orbital.

Shruti Kulkarni 2I

I have always been told ionic bonds are the strongest, then covalent bonds, then London forces. I would assume that ionic would be the strongest because the atoms actually give up or gain electrons, they don't share them as is the case with covalent bonds.

Shruti Kulkarni 2I

Yes! Lone pairs indicate a higher reactivity as the atom either wants to lose that electron or gain another one in order to have a full orbital or subshell.

Shruti Kulkarni 2I

Hi! So what helps me in understanding orbitals is to think of them as addresses! You will need to memorize a few properties of the orbitals, for example, that p orbitals have 6 electrons, but when you think of them as addresses it becomes clearer. Think of the energy level, n, as a country. So, anyo...

Shruti Kulkarni 2I

I believe that Dr. Lavelle has stated that objects with a wavelength of 10^-15 m or larger have wavelike properties, and objects with a smaller wavelength do not have discernable wavelike properties as their wavelength is so small it is imperceptible.

Shruti Kulkarni 2I

De Broglie's formula is used to find the velocity of an electron. You cannot immediately use E = hv, as the kinetic energy is a portion of this energy, not all of it. Due to this, you would have to isolate the electron's velocity and then solve for the kinetic energy using the kinetic energy equatio...

Shruti Kulkarni 2I

Hey! An electron in the 5s state is said to be at the 5th energy level in the s orbital. So, n= 5 and l = 0.

Shruti Kulkarni 2I

Wavelength is measured in m, or some variety of meters, like nanometers, which will probably be given to us in the problem itself if we need to change the length to nm or do any other conversions. Frequency is measured in Hertz (Hz), which is 1/s. This too can be converted into MHz and similar value...

Shruti Kulkarni 2I

Hi! What I tend to do is attempt to practice problems so that I can understand the steps that should be taken to start each solution. By practicing lots of problems, I am able to grasp the overall flow of how to solve them quicker. For more conceptual topics, I take notes on them and then proceed to...

Shruti Kulkarni 2I

He mentioned during lecture that black body radiation is not a concept we need to know!

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