CHEMISTRY COMMUNITY

Ariel Davydov 1C

It probably will be in order to make sure no one is cheating or accessing any online resources during the test.

Ariel Davydov 1C

1) We know it will be Sn -> Sn2+ + 2e- because the charge of Sn in the salt mentioned is +2. 2) When given the standard reduction potentials of each half reaction for a spontaneous reaction, we know that the half reaction with the highest reduction potential will be reduced and thus will be at the c...

Ariel Davydov 1C

Hi! Since the cell possesses an aqueous solution, water is implied to be present in the cell, so there is no need to include it in the cell diagram. Hope this helps!

Ariel Davydov 1C

We are finding the cell potential of the anode since M is being oxidized in this reaction.

Ariel Davydov 1C

In this reaction, Br- is oxidized and Hg2+ is reduced, resulting in a cell potential of -0.17 V, which is nonspontaneous. Hope this helps!

Ariel Davydov 1C

This occurs in this particular example because the reaction begins with Ni2+ ions. Regardless of its standard reduction potential, the reaction begins with these ions since they are already present in the solution and thus is reduced to Ni(s).

Ariel Davydov 1C

Since all the species listed except for Co are ions, they more often than not will be in aqueous form, while those is solid form will have neutral charges. What helps me to remember this is picturing these ions as parts of a dissolved salt, i.e. aqueous ions. Hope this helps!

Ariel Davydov 1C

Both of these redox reactions can be found in the SRP table on page A16 in the back of the textbook. Since the standard reduction potential of Ag+ + e- -> Ag(s) (0.80 V) is higher than that of AgBr + e- -> Ag + Br- (0.07 V), it will be at the cathode while the latter reaction will be at the anode.

Ariel Davydov 1C

Since Ni is present in its charged aqueous forms, we can use solid nickel as a spectator that allows for current to flow. I believe we can still use C(graphite) or Pt(s) as the electrode.

Ariel Davydov 1C

You can also tell which one is the cathode and anode by calculating the E(not)cell through trial and error; the values you use to end up with an E(not)cell that is positive will tell you which species is reduced and which is oxidized.

Ariel Davydov 1C

I did, thank you so much for your help.

Ariel Davydov 1C

At a constant volume, deltaV is zero, so there is no expansion work occurring (w=0) and deltaU overall will only equal q. At a constant pressure, since there may be expansion work occurring, the heat given off can only be interpreted as deltaH or q, since the overall internal energy includes the en...

Ariel Davydov 1C

Thank you!

Ariel Davydov 1C

An acidified half reaction will be balanced using H2O(l) and H+(aq) to balance the oxygen and hydrogen atoms present in reaction.

Ariel Davydov 1C

I believe this is a typo, as Cl- is present in the answer shown in the answer key.

Ariel Davydov 1C

The only case in which this wouldn’t be specified is if hydrogen ions or hydroxide ions are present in reaction, which would indicate whether the solution is acidic or basic.

Ariel Davydov 1C

If the oxidation number of an element increases, that means the element is being oxidized (losing electrons) and thus is a reducing agent, since those electrons will be accepted by the element that is being reduced. If the oxidation number of an element decreases, the element is being reduced and th...

Ariel Davydov 1C

Because in order for the all charge of ozone to be zero, the oxidation numbers of all the individual oxygen atoms must add up to zero as well. Since there are three oxygen atoms, the only case in which adding all three oxidation numbers of the individual atoms, as well as all three atoms having iden...

Ariel Davydov 1C

Hi everyone, I can’t seem to find my Apple Pencil, and the last time I used it was during lecture today in CS50. It was Lecture 1 at 11 AM, and it has my full name, Ariel Davydov, on it. If anyone picked it up by any chance or turned it in somewhere could you please let me know? Thank you so much!

Ariel Davydov 1C

You can also calculate deltaG the same way.

Ariel Davydov 1C

All heat capacities are state functions since it does not matter how the heat was added or how the temperature was changed; you will still arrive at the same final temperature.

Ariel Davydov 1C

Delta S is entropy, not enthalpy. Since the reaction is endothermic (positive enthalpy), it requires heat input in order to occur, which indicates that this reaction is not typically spontaneous unless at a high temperature and with a high entropy.

Ariel Davydov 1C

If you are given the change in certain variables, you can use the individual ratios between that variable and one other variable to calculate the change. For example, if given the change in volume and asked to find the final pressure of the system, you can use P1V1=P2V2 and plug in the initial and f...

Ariel Davydov 1C

The purpose of the degree sign is to indicate that the relationship between enthalpy, entropy, and temperature in regards to spontaneity is deltaG = deltaH -T*deltaS only at standard conditions.

Ariel Davydov 1C

At a constant volume, deltaV is zero, so there is no expansion work occurring (w=0) and deltaU overall will only equal q. At a constant pressure, since there may be expansion work occurring, the heat given off can only be interpreted as deltaH or q, since the overall internal energy includes the ene...

Ariel Davydov 1C

Since we are calculating the enthalpy density solely for the combustion of TNT, it will be a positive value because it is only in terms of the energy absorbed in breaking the bonds of TNT, rather than the overall reaction as a whole.

Ariel Davydov 1C

In this question you don’t necessarily have to calculate the degeneracy and entropy of each separate molecule; you can compare them contextually. In the trans molecule, you can see that there are much fewer positions the blue atoms can take (only among the X axis) compared to in the cis molecule, wh...

Ariel Davydov 1C

Approach a problem in calculating the entropy of transition the same as you would in calculating the qtotal of a substance being raised from a non-boiling-point temperature to its boiling point and transition. Set up two separate components for the entropy for the system and add them: 1) the entropy...

Ariel Davydov 1C

The answer key says for that equation there is a decrease in entropy, not an increase.

Ariel Davydov 1C

Finding the heat capacity can allow you to find the final temperature of a system when a certain amount of heat is provided to it, how much heat is required to raise the temperature of a certain amount of grams or moles of a substance to a specific temperature, and more. The specific heat capacity a...

Ariel Davydov 1C

1) Bond enthalpy is the amount of energy needed to break a singular bond. For example, the energy needed to break a C-C bond (single bond between two carbon atoms) is +346 kJ/mol. If a bond is broken, it will release the same amount of energy and will be a negative term. 2) Standard enthalpy of form...

Ariel Davydov 1C

It is to accommodate for the sign of the change in volume. If work is being done on the system (meaning work is positive), the volume of the system will decrease (ex: piston being pushed in) and the change in volume will be negative. Without the negative in front of the equation to counterbalance th...

Ariel Davydov 1C

If the pressure is constant, this simply means that it is not changing. For example, the external pressure could be constant and held at 1 atm, while the internal pressure of a system could be constant and held at 2 atm. However, because these pressures differ, this system is not at equilibrium with...

Ariel Davydov 1C

These will typically be elements in their standard, lowest energy state. For elements found as diatomic molecules (Br, I, N, Cl, H, O, F), they will have an enthalpy of zero when found as diatomic gases, with the exception of Br2 being found in liquid form and I2 being in solid form. For other eleme...

Ariel Davydov 1C

In order to determine how a change in pressure affects the equilibrium of a system, it must be said how the pressure was changed in the first place. As the above comments said, if the pressure is increased by adding an inert gas, there will be no change in equilibrium, since the partial pressures of...

Ariel Davydov 1C

Since Ba(OH)2 is a strong base, it will dissociate almost entirely. Thus, if given the mass of Ba(OH)2 in this particular problem, you can calculate its molarity, set up a balanced equation of Ba(OH)2 dissociating, and use the ratios to find the concentrations of Ba2+ and OH-. From here, you can cal...

Ariel Davydov 1C

As the above comment says, delta H represents the change of energy in a reaction from its final state to its initial state, regardless of the path it took to get there. q is the heat given off by a system during a reaction, and its sign will tell us whether the reaction is exothermic (negative value...

Ariel Davydov 1C

Yes, and if the percent protonation of the base or percent deprotonation of the acid is less than 5%.

Ariel Davydov 1C

What the solutions manual is doing is converting the [OH-] concentration to pOH, finding pH by subtracting pOH from 14, and then calculating [H3O+] by using the pH value. This method will result in a different result if the concentration of H3O+ or OH- exceeds 1 M.

Ariel Davydov 1C

When calculating pH, if you end up with a negative value you must convert it to its positive form, which will be the answer. Since the pH scale ranges from 0 to 14, there cannot be a “true” negative pH value. This can be explained by the fact that strong acids at concentrations higher than 1.00 mol*...

Ariel Davydov 1C

Yes, but if asked to solve for partial pressures or Kp, you can use the Ideal Gas Law (PV=nRT) to solve for individual partial pressures and plug them into the Kp expression to solve for Kp.

Ariel Davydov 1C

If you look at the graph provided in the question, you will be able to determine the amount of A consumed and the amount of B and C formed in reaction in establishing equilibrium. If you set up an ICE table and plug in the initial and equilibrium partial pressure values, you can determine the change...

Ariel Davydov 1C

According to Lavelle’s slides, we can only do this if the equation is cubed AND the K value is <<< 10^-3.

Ariel Davydov 1C

When dealing with problems with shifts in equilibrium, try to approach problems with partial pressures as you would problems with molar concentrations. If you set up a K equation in terms of partial pressures, you will see that if you increase the partial pressure of a product, the system will shift...

Ariel Davydov 1C

If you balance the equation, use the values provided in the previous problem and convert them from bar to atm, you should get 2.96 x 10^3 (the inverse of K from #18). Hope this helps!

Ariel Davydov 1C

P is used to express partial pressures of gases in equilibrium equations.

Ariel Davydov 1C

In an experiment, 0.02 mol NO2 was introduced into a flask of volume 1.00 L and the reaction 2NO2 (g) ⇌ N2O4 (g) was allowed to come to equilibrium at 298 K. (a) Using information in Table 5E.2, calculate the equilibrium concentrations of the two gases. (b) The volume of the flask is reduced to hal...

Ariel Davydov 1C

A molecule is more thermodynamically stable if more of it exists post-reaction, since less of it breaks up into parts in order to form its product form. Thus, the molecule with the smallest K value (most reactant-favored reaction) will be the most stable. Since the reaction with Cl2 has a smaller K ...

Ariel Davydov 1C

Yes, the division of each factor by 100 is to convert from P/kPa to bar.

Ariel Davydov 1C

CaO is a Lewis base, and since it is a metal oxide, it will be a strong base that, when reacting with water, will form Ca(OH)2, a strong base that entirely dissociates into Ca2+ and hydroxide ions.

Ariel Davydov 1C

Draw the Lewis structure of boric acid, B(OH)3. (a) Is resonance important for its description? (b) The proton transfer equilibrium for boric acid is given in a footnote to Table 6C.1. In that reaction does boric acid act as a Lewis acid, a Lewis base, or neither? Justify your answer by using Lewis...

Ariel Davydov 1C

If an acid or base is strong, it will entirely ionize in solution. Thus, if the problem has a strong acid or base as a part of the equation, it must be written out in dissociated form (ex: HBr would be written as H+ and Br-). If the acid is weak (ex: CH3COOH, acetic acid), it does not entirely ioniz...

Ariel Davydov 1C

Finding the oxidation state of an element can be accomplished through finding the charges of other molecules in the compound and subtracting them from the overall charge of the complex. Since the oxidation state (aka charge of the element) + the charges of the other compounds in the complex = the ov...

Ariel Davydov 1C

HClO is a stronger acid than HBrO because Cl is a more electronegative element than Br, meaning it has a greater pull on electrons and thus can delocalize the negative charge of the acid's conjugate base (ClO^-) better than Br can in its conjugate base (BrO^-). This means that HClO's conjugate base ...

Ariel Davydov 1C

On the chart Dr. Lavelle sent us with the names of compounds in regard to coordination complex naming, there is a section with entirely neutral compounds. These include H2O, NH3, CO, NO, ethylenediamine (en), and diethylenetriamine (dien).

Ariel Davydov 1C

Based on Dr. Lavelle's lectures, it is safe to assume that metal oxides have basic character, nonmental oxides have acidic character, and certain elements that mirror the metalloid staircase on the periodic table are amphoteric, meaning they produce both an acid and a base in water. These elements a...

Ariel Davydov 1C

Exactly. Typically, if a Lewis structure omits lone pairs, we must fill them in ourselves. Since the oxygen has two bonds, we can assume that it has two lone pairs, since oxygen has 6 shareable valence electrons and in this molecule has a formal charge of zero. Thus, with a VSEPR equation of AX2E2 a...

Ariel Davydov 1C

AGulati_4A wrote:An unhybridized orbital

Why?

Ariel Davydov 1C

Will an electron prefer to occupy an unhybridized orbital or spin pair in hybridized orbital?

Ariel Davydov 1C

Bismuth was one of the exceptions mentioned in lecture today that acts as an amphoteric oxide.

Ariel Davydov 1C

Typically, the VSEPR shapes of molecules will not allow for four lone pairs to bind to a transition metal. For example, if a tetrahedral molecule has shareable lone pairs on each attached atom, the shape will only allow for three lone pairs to bind to the transition metal, as the fourth lone pair wi...

Ariel Davydov 1C

If water is present and written as a part of the coordination complex, the transition metal will bond with it.

Ariel Davydov 1C

With the ligand's formula, you would generally be able to construct a Lewis structure and be able to analyze its shape and lone pairs available to share. If the ligand has numerous lone pairs that its geometry allows it to share simultaneously, it is a chelating complex.

Ariel Davydov 1C

A ligand can be polydentate if it has multiple locations where it can give up a lone pair AND its geometry allows for it to do so. An example of a ligand with multiple shareable lone pairs but that is not a polydentate is H2O, whose bent shape only allows for it to share one lone pair. Examples of p...

Ariel Davydov 1C

Hi! So on this graph, we were supposed to illustrate the number of electrons emitted once the photons reached a certain frequency. In order to approach this problem, you must remember that all the equations we are given (E=hv, E - work function = KE) is calculating the energy of ONE PHOTON or ONE EL...

Ariel Davydov 1C

I've heard Chem 14BL also picks up where Chem 14B left off, which would make taking them simultaneously kind of difficult.

Ariel Davydov 1C

Moleuclar geometry and molecular shape are one and the same. What differs, however, is electron geometry, which takes into account all areas of electron density. Molecular shape/geometry considers electron repulsion between lone pairs and lone pairs and atoms. For example, NH3 will have a molecular ...

Ariel Davydov 1C

From what I have heard, we will only be tested on sigma and pi bonds. Studying hybridization can help you understand sigma and pi bonds more, though!

Ariel Davydov 1C

I believe the only compounds we will be asked to name (or come up with when given the name) are the coordinate covalent complexes, which Lavelle will be lecturing on during Friday's lecture.

Ariel Davydov 1C

The coordinate covalent bond between a nonmetal and a TM metal is "provided" by the nonmetal in the form of an extra lone pair. For example, if we are discussing a coordinate covalent bond between Cl- and Pb2+, the bond between one chlorine atom and one lead atom would be provided by one v...

Ariel Davydov 1C

For seesaw shaped molecules, would the angles be less than 120 and less than 90 degrees, or flat out 120 and 90 degrees? I keep seeing different answers everywhere.

Ariel Davydov 1C

How would we treat a molecule with a radical in regards to determining its VSEPR shape? For example, in part B of question 2E.5, the molecule OClO has an uneven number of electrons. Would you treat that one additional electron as another shell of electron density?

Ariel Davydov 1C

Yes! Hydrogen bonding is a special type of dipole-dipole attraction between molecules where hydrogen (a very not-electronegative element) displays dipole-dipole attraction with very electronegative elements, specifically N, O, and F.

Ariel Davydov 1C

There are also some really great Youtube videos that help you visualize the shapes of these molecules and understand them rather than depending on sole memorization. I highly recommend Tyler DeWitt's video on VSEPR. Hope this helps!

Ariel Davydov 1C

Sure! Listed from strongest to weakest, the intermolecular forces responsible for interaction between molecules are ion-ion, hydrogen bonding, ion-dipole, dipole-dipole, dipole-induced dipole, and London/dispersion forces. Overall, the major theme in strength of intermolecular forces is that the str...

Ariel Davydov 1C

You are not expected to memorize the boiling and melting points of substances, but you are expected to be able to determine the strength of the intermolecular forces between like molecules and tell which substances will have higher or lower boiling and melting points based on their strength.

Ariel Davydov 1C

We will be learning how to determine bond order with hybridization at a later time this quarter.

Ariel Davydov 1C

Also make sure the amount of electrons in the central atom allows for the formal charge to be equal to zero! Since the central element in a molecule is almost always the least electronegative of the molecule, it will usually have a formal charge of zero or a positive value.

Ariel Davydov 1C

In addition, with elements with expanded octets, the best way to determine the amount of extra electrons they typically store is through calculating its formal charge. For example, with S, if it forms four bonds with other elements, it will typically have one additional lone pair of electrons in ord...

Ariel Davydov 1C

And thus conversely, a molecule displaying low covalent character (and high ionic character) would be more soluble in water than those of high covalent character. The character of a molecule can be determined either through the differences in electronegativity between the cation and anion of the mol...

Ariel Davydov 1C

I also struggled with this problem, and after searching it up, I believe your structure is the actual correct structure and that there is an error in the solutions manual, as using the formal charge of the elements would tell us that your structure is the lowest energy state of the molecule.

Ariel Davydov 1C

It would most likely be limited to the number of electrons available in the element's d-orbital. Typically though, you would use the formal charge of the element to determine how many electrons its expanded octet would typically hold at its lowest energy state.

Ariel Davydov 1C

Hi, For part C of this question, I'm a little confused as to how there is resonance with the structure for HClO3. The structure you end up with (with the lowest formal charge) doesn't have any bonds that can be switched without changing the number of double/single bonds in the molecule, yet the answ...

Ariel Davydov 1C

As for writing out electron configurations, Lavelle said that he prefers us to write 3d before 4s, and the same goes for the f-block and other d-orbitals.

Ariel Davydov 1C

Because potassium has a positive charge (+1) and phosphorus has a negative charge (-3), it is safe to assume that they will bond ionically, especially since the molecule is K3P.

Ariel Davydov 1C

It helps if instead of using the frequency equation you derive it through this basic equation: deltaE = Efinal - Einitial. From here, you can sub in En = -hr(1/n^2) for the final and initial energy values and solve from there. This will help prevent any future mixups with n values since they are ass...

Ariel Davydov 1C

He posted a pdf briefly with several locations listed based on last name and lecture but took it down due to the fires. It will probably be reposted shortly as a link on his class website.

Ariel Davydov 1C

The number of orbitals per angular momentum quantum number can be obtained by looking at the total number of potential m(l)-numbers for an l-number. For example, for l=3 (the f block), m(l) can be anything from 3 to -3 (3, 2, 1, 0, -1, -2, -3) meaning the total number of potential quantum numbers is...

Ariel Davydov 1C

Most transition metals have 2 valence electrons. Valence electrons are the sum total of all the electrons in the highest energy level (principal quantum number n). Most transition metals have an electron configuration that is [noble gas]ns2(n−1)d, so those ns2 electrons are the valence electrons. Ty...

Ariel Davydov 1C

If I'm thinking of the same question you're referring to (1B.27), there is an error in the solutions manual that Lavelle noted in his "Solution Manual Errors 7th Edition" PDF on his website. In this question, the textbook utilized that equation (1/2 h) instead of the one taught in class. I...

Ariel Davydov 1C

When losing an electron, the nucleus's pull on the electrons grows stronger, as with less negative energy the positive charge of the nucleus has a stronger pull on the electrons surrounding it, thus decreasing the size of the atomic radius. Because of this, it will require a greater amount of energy...

Ariel Davydov 1C

Because this equation provides us with the minimum uncertainty in position/momentum, this means that the actual uncertainty in position/momentum is equal to or greater than the value given or calculated, which means that, yes, the uncertainty can be greater than the value obtained. Hope this helps!

Ariel Davydov 1C

From what I have gathered, you will need to know and understand the basic concepts surrounding the practice problems to an extent great enough to apply it to more difficult questions. Your knowledge of these concepts will be tested through your ability to apply them to sometimes even never-before se...

Ariel Davydov 1C

We examine the photoelectric effect through this equation theoretically . Theoretically, if the energy of the photon equals that of the metal's work function, electrons will be emitted. While we know this in reality does not occur, we still utilize this equation in that sense, since the work functio...

Ariel Davydov 1C

The intensity of a light source is directly related to the number of photons and the rate at which they are projected onto the sample. The frequency is not related to the number of photons or intensity.

Ariel Davydov 1C

It will typically be given on the equation sheet provided during the test, but for personal reference, it is 9.1 x 10^-31 kg.

Ariel Davydov 1C

D, the photoelectric effect, best explains that electrons have particle-like properties. This experiment shows that, when light is shone onto a metal sample, electrons are emitted at a one-to-one ratio to the photons that are absorbed by the sample. This is also supported by the equation E=hv, which...

Ariel Davydov 1C

You would draw the atom as a circle and label its radius and diameter, since the diameter of the atom is the uncertainty in position for the electron in this problem.

Ariel Davydov 1C

All of the answers will be the same for a hydrogen atom. Even though the atoms' outermost electrons have different values of n (lithium's being n=2, while hydrogen's is n=1), the energy of the electron would increase in both a hydrogen and lithium atom. The value of n for the electron will also incr...

Ariel Davydov 1C

Yes, this is exactly what it means, because that is the range in which the speed could potentially fall into.

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