CHEMISTRY COMMUNITY

Sidharth D 1E

To expand on that, reactions that give a straight line when lnk_r is plotted against 1/T are said to show Arrhenius behavior

Sidharth D 1E

Sue Bin Park 2I wrote:do we know if the final is going to be timed? or will he post it and give us more than the 1.5 hours he said it would take?

I am pretty sure he mentioned allotting from 3-4 hours for the actual final exam, even though you should be able to finish it in half the time.

Sidharth D 1E

I'm pretty sure it will be posted on his class website, so thats where to look for it when Lavelle sends an email saying he posted it.

Sidharth D 1E

I don't think there will by any derivations on the exam, so not much/basic calculus might be needed.

Sidharth D 1E

For example when dealing with hydrogen gas in the galvanic cell, you would need to include Pt(s)

Sidharth D 1E

Are you referring to what to do given a concentration cell situation or in a galvanic cell?

Sidharth D 1E

No, he mentioned at least on the midterm that there will not be any derivations so I think that same principle applies to the final.

Sidharth D 1E

I believe you should be prepared to be given the one that is on Lavelle's website

Sidharth D 1E

N should be the number of electrons transferred in that scenario.

Sidharth D 1E

Also, traditionally in a galvanic cell diagram, the anode will be on the left side while the cathode will be on the right side.

Sidharth D 1E

My TA mentioned that Dr. Lavelle will most likely keep it simple with the anode on the left side and cathode on the right side.

Sidharth D 1E

Yes, you separate sides of a reaction with a double line.

Sidharth D 1E

No, there is no equation or formula we need to write down for the salt bridge. It's just important to draw it on the diagram.

Sidharth D 1E

A salt bridge is meant to maintain charge as the electrons are transferred from the anode to cathode

Sidharth D 1E

The purpose of a salt bridge is to maintain charge as the electrons are transferred from the anode to cathode

Sidharth D 1E

A salt bridge is meant to maintain charge as the electrons are transferred from the anode to cathode

Sidharth D 1E

Salt bridges are meant to maintain charge as the electrons are transferred from the anode to cathode

Sidharth D 1E

Normally the electrode potentials are given as standard reduction potentials.

Sidharth D 1E

Most metals that are given in the question without a charge are in the solid state and those which have a charge normally are from a dissociated compound, so they will be aqueous.

Sidharth D 1E

As a good rule of thumb, remember that oxygen in a compound other than O2 will have an oxidation number of -2 and hydrogen in a compound other than H2 will have an oxidation number of +1.

Sidharth D 1E

Oxygen will have an oxidation number of -2, unless it is in O2, or other exceptions like hydrogen peroxide. I believe the same goes with hydrogen being +1.

Sidharth D 1E

Lavelle suggested to split the redox reaction into two half reactions which separate the oxidation and the reduction reactions. You can use it to find out which elements are oxidized and reduced, and also balance redox reactions (since the number of electrons transferred should cancel out in the two...

Sidharth D 1E

Half reactions separate the oxidation and the reduction reactions into two half reactions. You can use the individual two half reactions to find out which elements are oxidized and reduced, and also balance redox reactions (since the number of electrons transferred should end up cancelling out.

Sidharth D 1E

Another acronym to remember it by is: OIL RIG = Oxidation is the Loss of electrons and Reduction is the Gains of electrons.

Sidharth D 1E

Yes, we can use Hess' law for Gibbs Free Energy because it's a state function.

Sidharth D 1E

Also, if a question mentions that a chemical reaction is favorable, it means that the reaction is spontaneous.

Sidharth D 1E

It means that a change in the enthalpy of a system is equal to the heat released or absorbed at constant pressure. Since the external pressure is equivalent to the pressure of the system:
deltaH = q - P(deltaV) + P(deltaV), so deltaH equals q.

Sidharth D 1E

Also, for nonlinear molecules, Cv is 3R and Cp is 4R

Sidharth D 1E

The change in entropy of the universe is always positive because some form of energy is always released (whether that may be from the system or the surroundings), even if it's endothermic.

Sidharth D 1E

A state function is something where the path doesn't matter, only the initial and final state. The reason why it's important is that since the path doesn't matter, you can subtract, add, multiply, and divid state functions (entropy in this case) as necessary.

Sidharth D 1E

Steam has more energy than boiling water does (40.7 kJ vs 5 kJ). So when the steam hits your skin you are immediately feeling 40.5 KJ of heat rather than 5KJ. Steam also absorbs heat so it will also increase in temperature.

Sidharth D 1E

Since the beaker is open and exposed to the atmosphere (which is very large in comparison to the beaker), the pressure is constant. The beaker can also equilibrate with the pressure of its surroundings.

Sidharth D 1E

Also, constant pressure can be assumed if you are dealing with an open system.

Sidharth D 1E

Also, the review sessions that are listed on Dr. Lavelle's website match the topics that we will be tested on.

Sidharth D 1E

When looking at the reversible, isothermal expansion of an ideal gas, if the external pressure of a system were to be increased even infinitesimally at any stage of the expansion, the piston would move in instead of out. Therefore, the work done during a reversible expansion of a gas is the maximum ...

Sidharth D 1E

Both of these are used to find enthalpy/delta H but also find change in internal energy which is q + w. Therefore, q can either be q_p or q_v depending on the given conditions.

Sidharth D 1E

Due to the autoprotolysis of water, there is already a concentration of 1 x 10^-7 M of [H₃O⁺], so that amount of HCl would not have any impact.

Sidharth D 1E

Increasing the pressure of the reactant will not impact the equilibrium constant, it will only shift the equilibrium to the right.

Sidharth D 1E

Only the first two topics: Chemical Equilibrium, and Acid and Base Equilibria are on the exam, so anything we are learning about thermodynamics will be on the midterm.

Sidharth D 1E

I think that Sig Figs do count on the exam, and refer to this link on how to use them: https://lavelle.chem.ucla.edu/wp-conten ... OUT_SF.pdf

Sidharth D 1E

Focus on the lecture material and the homework problems given on the syllabus, because they are meant to mimic the questions on the test.

Sidharth D 1E

You can approximate if the K value for the table is less than 10^-3.

Sidharth D 1E

You should just only need to know how to plug values of a quadratic equation into the quadratic formula.

Sidharth D 1E

When K is at or near 1, the energy of the reactants and the products are the same, which is extremely rare.

Sidharth D 1E

K_p is meant to be used when dealing with the partial pressures of various gases, and K_c is meant to be used for concentrations.

Sidharth D 1E

Yes, they are both calculated the same way, and the states of the reactants and products will identify whether to use K_c or K_p.

Sidharth D 1E

If Q < K, at some time during the reaction, then the forward reaction is favored.
If Q > K, at some time during the reaction, then the reverse reaction is favored.

Sidharth D 1E

The formula for Boyle's law is P₁V₁=P₂V₂, which illustrates the concept mentioned above.

Sidharth D 1E

Cl should not be able to accommodate for more than an octet because the principle quantum number is 2, and there is no 2d orbital that exists.

Sidharth D 1E

Resonance allows for the delocalization of electrons, so that reduces the strength of the H-O bond, which allows the acid to be stronger.

Sidharth D 1E

A heme complex attached to a protein is what makes up myoglobin (which can attach to a single O₂ molecule. Four myoglobin together make up a single hemoglobin.

Sidharth D 1E

There is a link on his website: https://lavelle.chem.ucla.edu/wp-conten ... pounds.pdf

Sidharth D 1E

They help chemically by binding directly to the molecule and they structurally affect the active site of the molecule. It makes the bond weaker of the molecule that they are trying to change, thereby reducing the activation energy required for the reaction.

Sidharth D 1E

The questions before that (6C.16) gives you more information regarding the the comparison, so I think that is what you use to help you figure out the question.

Sidharth D 1E

Identifying the subshell in which the electron exists in will give you the first two quantum numbers, and the orbital will give m_l and the spin of the electron will tell you if it is +1/2 or -1/2.

Sidharth D 1E

Also, anything that is a group 1 or 2 metal cation that is attached to an oxide (o^2-) can also be a strong base. ex: CaO and Na₂O

Sidharth D 1E

I think we could be tested for their functions, so just memorize them to be on the safer side.

Sidharth D 1E

Atoms in a molecule hybridize because hybridized orbitals results in lower energy orbitals and they form a more stable compound.

Sidharth D 1E

Remember that each lone pair is considered as a separate area of electron density and that multiple bonds between the same two atoms is considered as one area of electron density.

Sidharth D 1E

An easy example to remember how bond length is affected by hybridization is this: the length of a sigma bond between a carbon atom and another atom is shorter for a carbon atom with a hybridization of sp² than a carbon atom with a hybridization of sp³.

Sidharth D 1E

I would say another way is to count how many regions of electron density there are that surround the atom (counting multiple bonds as one region of electron density and each lone pair as an area of electron density).

Sidharth D 1E

Yes, for example the sigma bond length between a carbon atom and another atom is shorter for a carbon atom with a hybridization of sp² than for a carbon atom with a hybridization of sp³.

Sidharth D 1E

Induced dipole-induced dipole interactions occur between all molecules. Dipole-dipole interactions occur between any molecules that have polar bonds, meaning there is an unequal sharing of electrons within the atoms of a molecule.

Sidharth D 1E

They mean that the dipole-dipole vectors from two opposite atoms when added together equal to zero. A nonpolar molecule has a net dipole movement of zero.

Sidharth D 1E

The test will probably focus on elements up until the 3p block.

Sidharth D 1E

In terms of hybridization, this allows an atom to have up to 5 or 6 areas of surrounding electron density (whether it may be lone pairs or bonds) and the hybridization for those central atoms would be sp³d and sp³d² respectively.

Sidharth D 1E

For the example you gave AX₂E₃, the lone pairs would be occupying the three planar areas of electron density and the two bonding locations would be on opposite poles, and that means the angle between them would be 180 degrees and the molecule as a whole would be linear.

Sidharth D 1E

The two electron pairs that are in a double bond stay together and repel other bonds or lone pairs together as a unit. The same goes for triple bonds and that is why they are considered as a single region of electron density in VSEPR.

Sidharth D 1E

The bigger the atoms are that are bonded together, the larger the bond length due to the electron-electron repulsions of the electron shells that distance the nucleus of an atom from the valence electrons that are involved in the bonding.

Sidharth D 1E

The additional bonding between the two atoms pulls the nuclei of the two atoms closer to each other, and for that reason double bonds are shorter than single bonds.

Sidharth D 1E

Van der Waals forces are the dipole-dipole interactions that exist between all molecules due to the natural movement of electrons within each molecule. They are relatively weak interactions, but are increase with a greater number of electrons interacting between the atoms of two separate molecules.

Sidharth D 1E

I dont think need to know anything other than trends for electronegativity

Sidharth D 1E

All molecules have some sort of dipole-dipole interaction (which is the van der Waals' forces that we discussed in class)

Sidharth D 1E

Atoms in that period have empty 3d-orbitals to accommodate for the expanded octet.

Sidharth D 1E

The next best option is to have the most elements with a formal charge of 0, with negative formal charges with the most electronegative elements of the molecule and positive formal charges with the most electropositive elements of the molecule.

Sidharth D 1E

Yes, we should still account for formal charge, because it allows for us to figure out the best structure that would work for that molecule.

Sidharth D 1E

Yes, it is more stable, because the formal charges for the molecule for the expanded octet result in the more dominant resonance structure.

Sidharth D 1E

Ionization energy is the energy needed to remove an electron from an atom and electron affinity is the energy released when an electron is added to a gas-phase atom. Ionization energy increases down a period because an atom that only needs to lose one electron to form an octet, would require less en...

Sidharth D 1E

Are those valence electrons also used in bonding when we draw out lewis structures?

Sidharth D 1E

in the book, they abbreviate their "h" as being h/2(pi) already.

Sidharth D 1E

They are different because to achieve maximum stability, those atoms take an electron from the 4s subshell and place it in the 3d subshell, so that they can have a full 3d subshell and half full 4s subshell (3d¹⁰ and 4s¹).

Sidharth D 1E

An atom is most stable when its subshells have complete orbitals (in this case d10) or when they have half complete orbitals (in this case d5). This is due to the symmetric distribution of electrons contributing to the atoms' stability

Sidharth D 1E

In class, he gave us those as the experimentally found values for the bond lengths, so those are the values for those bonds in specifically those molecules.

Sidharth D 1E

Yes, for the 3d orbitals: they are d_xy, d_yz, d_zx, d_x²-y², and d_z² respectively

Sidharth D 1E

Because the principle quantum number of the 4d subshell is less than the principle quantum number of the 5s subshell (4 < 5), the 4d state has lower energy.

Sidharth D 1E

So, what 1s 2 2s 2 2px 2py, means, is that carbon has 6 electrons, and it occupies these locations. First, there are 2 electrons located in the 1s orbital (which if you are following the periodic table follow the spots of Hydrogen and Helium). Next, there are another 2 electrons located in the 2s or...

Sidharth D 1E

how are orbitals, shells, and subshells related to one another? im not sure i understand Shells are composed of electrons that share the same principal quantum number (n), whereas subshells are composed of electrons that share the same angular momentum quantum number (l). Orbitals are composed of e...

Sidharth D 1E

Furthermore, it is written in that way, because if that element would lose an electron, it would lose it from the 4s² orbital, not the 3d orbital.

Sidharth D 1E

Specifically, the ultraviolet region corresponds with the Lyman series, the visible region with the Balmer series, and the infrared region with the Paschen series.

Sidharth D 1E

Make sure to watch his Audio-Visual Focus Topic on Wave Properties of Light: https://lavelle.chem.ucla.edu/wp-conten ... t.wmv.html

Sidharth D 1E

For example, for the problem we did with the 0.1 kg baseball moving at 35 m/s, the resulting wavelength that this mass would emit is 1.9 x 10^-34 m, which is a wavelength we cannot measure and, therefore, prove or disprove.

Sidharth D 1E

There is no exact time when you are supposed to do the Audio-Visual focus topics, as some people prefer to use them for revision on topics we learn in class, or to learn from them before class. Personally, I use them for revision.

Sidharth D 1E

In addition, it's probably noteworthy that visible light exists at around a wavelength of 500 nm and that the human eye detects EM radiation (visible light) within the 700 nm (red light) to 400 nm (violet light) region.

Sidharth D 1E

1st, I think the question is asking for the sulfide of the metal, not chloride (you mixed up E15 with E16) Step 1: Calculate the molar mass of the known parts of that compound [ the (OH) 2 ] molar mass of (OH) 2 is 34.02 g/mol Step 2: Subtract that known molar mass from the total to get the molar ma...

Sidharth D 1E

Step 1: Writing the base equation: C 7 H 16 (g) + O 2 (g) --> CO 2 (g) + H 2 O(g) Step 2: Balance the number of Carbons and Hydrogens on both sides because there are only one compound on either side that includes carbon, and one compound on either side that includes hydrogen: C 7 H 16 (g) + O 2 (g) ...

Sidharth D 1E

He means that the homework (5 problems) that we submit in our discussion sections, should be related to content that we have recently covered within our lectures. For the first week, our homework was due on a lecture day, but the next set of 5 problems are due next discussion class.

Sidharth D 1E

Our answer needs to have the same number of significant figures as the lowest number of significant numbers of any of the given values. For example, if we are given a sodium carbonate solution of 2.111 g to a 250. mL flask, the two givens have 4 significant figures and 3 significant figures respecti...

Sidharth D 1E

As far as what we have been practicing in our discussion sections, we do not need to worry about the states of the molecules until the very last step when stating a chemical equation or a net ionic equation.

Search found 98 matches