CHEMISTRY COMMUNITY

EvanWang

Look out for Lyndon's study packets, they are very helpful and he usually has a big review session where he explains the answers. In my experience, they have helped a lot and are sometimes harder than the actual midterm/final, so they definitely prepare you well.

EvanWang

Because the reactions are the same on both sides, just going in opposite directions. This means that Ecathode=Eanode, and since Ecell=Ecathode-Eanode, Ecell=0.

EvanWang

For first order, the units are just s^-1

For second order, the units are L x mol^-1 x s^-1

For third order, the units are L^2 x mol^-2 x s^-2

The pattern continues in this way.

EvanWang

Katie Bart 1I wrote:Can the order keep increasing? Or do they stop at two?

Order can be higher than two, but they become less common the higher order you go. It is harder for three molecules to collide than for two molecules to collide. The same goes for higher orders.

EvanWang

are the units for rate different for reactions of different orders? The units for the rate are always going to be moles per liter per second, but the units for the rate constant k will change according to the overall order. k changes so that the proper units cancel so that the rate stays at moles p...

EvanWang

It's the other way around. Since Ecell = Ecathode-Eanode, you want the Eanode to be the most negative so that the overall cell potential is positive. At the anode, oxidation occurs.

EvanWang

$\Delta G^{\circ}$ just means that the reaction is happening at one atmosphere and all species concentrations are one molarity.

EvanWang

It is because you want the right units in the end when you multiply everything to find the rate using the differential or integrative rate law. In a reaction with an overall order of 4 for example, you will have the unit of mol/L multiplied by itself 4 times, so the units of k must be different so t...

EvanWang

Liquid mercury can conduct electricity, so it is one of the few exceptions were a solid electrode is not needed.

EvanWang

And also, maximum nonexpansion work is the same as Gibbs free energy, so emf directly relates to $\Delta G$

EvanWang

You can find the standard reduction potentials in Appendix 2B of the book.

EvanWang

C(gr) is used as an electrode because the anode has Hydrogen gas and hydrogen aqueous, so a solid must be included to allow electron transfer. Graphite and Platinum are common inert solids used as electrodes.

EvanWang

You should flip the reaction that makes the standard reduction potential of the overall cell positive.

EvanWang

Can concentration cells be made with a salt bridge? Or can they only be made with a porous disk?

EvanWang

We use the Nernst equation to find the cell reduction potential when we know the concentrations of the products and reactants, as well as the moles of electrons transferred and the standard reduction potential.

EvanWang

Yes, it is because they are both aqueous.

EvanWang

It means that it doesn't matter how many times the reaction occurs, E will remain the same as long as it is at standard conditions. Unlike in something like enthalpy or Gibb's free energy where the standard formation value gets multiplied by the moles of the reaction, E will remain the same for one ...

EvanWang

I also keep getting -2.7kJ. Not sure how it is supposed to be -27 honestly.

EvanWang

The salt bridge allows the charges between the anode and cathode to be the same. As the cathode gains electrons from the anode, the buildup of electrons on the cathode would eventually stop the flow of electrons.

EvanWang

The question asks you to balance the reaction of chlorine in water:
Cl2(g)-->HClO(aq) + Cl2(g)

How would you set up the half-reactions for this problem? Would one of them be just Cl2-->Cl2? I do not see how electrons can be transferred between two of the same species.

EvanWang

I don't think there is a way to calculate enthalpy of phase changes based on the heat capacity equation. The only way I know if calculating the enthalpy of phase changes is by adding and subtracting other enthalpies of phase changes because enthalpy is a state function, for example: Enthalpy of vapo...

EvanWang

Cv is the specific heat of a substance under constant pressure, while Cp is the specific heat of a substance under constant pressure. Generally, Cv is lower because a system at constant volume does no work, so all of the energy inputed into the system will be used toward increasing the temperature.

EvanWang

A bomb calorimeter has rigid walls that do not allow for expansion, therefore it is constant volume. A system with constant pressure would be open to the environment.

EvanWang

You cannot assume the concentration of a species is 1 if it is not given to you. If you are trying to solve for initial concentrations and you are given the change in concentration after reaching equilibrium, then you substitute that change into x in your ICE table and you can solve for concentratio...

EvanWang

The system will be under free expansion or it will be stated in the problem.

EvanWang

If I remember correctly, free expansion occurs when a gas is being heated in a vacuum. It is called free expansion because there is no external pressure, so the gas can expand without opposing a force (no work is done), so it is said to expand freely.

EvanWang

We can use Hess's law because enthalpy is a state property, so it doesn't matter the path you take to get from an initial condition to a final condition, the change in enthalpy will always be the same.

EvanWang

In problems like these, it is important to recognize that whatever heat is released by the copper is absorbed by the water. -q copper =q water Also, it is important to know that the final temperature of the copper and water will be the same. We know this because once the water and copper reach the s...

EvanWang

The change in internal energy is deltaU=q+w, since there is no mention of heat, we assume q to be 0. Now, deltaU=w, so change in internal energy is just change in work you calculated previously.

EvanWang

No, because the definition of an isolated system is that no energy is exchanged with the environment. This includes heat and work. Consider it like a thermally-insulated and rigid bomb calorimeter.

EvanWang

Because they are not tailor-made to every type of bond in every molecule, they are an average of the energy required to break bonds across many different molecules. For example, the O-H bond in CCl3COOH is different than the O-H bond in CCH3COOH because we know that the Cl pulls electrons more than ...

EvanWang

I'm assuming you meant to say the Pkb of F- and that the reaction given to you was the dissociation of HF in water. To do this problem, you have to first convert the Pkb of F- into the Pka of HF. This can be done using 14-10.2 because F- and HF are conjugates. Once you find the Pka, you convert it i...

EvanWang

When q=0 and deltaU=q+w, then
deltaU=(0)+w
deltaU=w

d states deltaU=q or deltaU=0. For this to be true according to the derivation from before, w must equal 0.

EvanWang

Specifically, standard enthalpy of formation is the enthalpy required to form one mole of a substance from its elements in their most stable form. The standard enthalpy of formation for elements in their most stable form is 0.

EvanWang

If something is specified as being insulated and rigid, that means it doesn't exchange heat or work of expansion with the surroundings. This makes the system isolated. If these are not specified, but the system doesn't allow the exchange of matter with the surroundings, then the system of closed.

EvanWang

An increase in pressure by compression of the system will favor the side of the reaction with fewer moles of gas. This is because when you reduce the volume of the system (via compression) then the concentrations are increased, and the reaction quotient shifts above or below K. This shift in Q resul...

EvanWang

For most polyprotic acids except sulfuric acid, the second ionization contributes so few H₃O⁺ that its effect on pH is negligible.

EvanWang

He explained that if pressure is increased by decreasing volume (if a piston pushed down on the reaction, for example) then the concentration of each of the gases in the reaction would increase. We calculate concentration by using \frac{n}{V} where n is moles of gas and V is the volume of the contai...

EvanWang

Although adding an inert gas affects the total volume of the reaction, they do not interact with anything, thus causing there to be no change in concentration. I thought the whole point was that adding a inert gas doesn't affect the volume of the container, which was why it has no affect on concent...

EvanWang

Typically the Ka for the second deprotonation is extremely small. Apart from sulfuric acid, for most acids it doesn't have a meaningful effect on pH.

EvanWang

Yes, pH can be negative and it just means that the molarity of H+ is greater than 1. Google tells me that some acids have been observed at pH -18.

EvanWang

The values of the ICE table must be in concentration because they will be calculated using K_c, which deals with concentrations.

EvanWang

I have another similar question with 5J.3 - can someone explain why adding/removing a certain amount of a compound might affect its equilbrium? Adding or removing a certain amount of compound will not affect the equilibrium constant K because only temperature will do that. It will affect the equili...

EvanWang

When looking at problems like this, it is a good idea to keep the equation for K in mind. According to that equation, if one product increases the other products must decrease to keep the same ratio of K. Additionally, if one reactant increases then the products must increase to keep the same ratio ...

EvanWang

K B and K A are calculated the same way. From what I gathered from the lecture, K B has to do with the equilibrium constant of bases and has [OH - ] as a product. Conversely, K A has to do with the equilibrium constant of acids and has [H 3 O + ] as a product. That's the extent to how much they diff...

EvanWang

Temperature is the only thing that affects the equilibrium constant k. Factors such as concentrations of species, partial pressures of species, and volume of the container can all affect the reaction quotient, q, and drive the reaction in a certain direction in order to achieve equilibrium again.

EvanWang

You can find the solutions manual at Powell Library. It is on hold for this class, and every time I check it out it has been available. Just ask the help desk and have your Bruin Card on hand.

EvanWang

K_c is used for concentration in general, it applies to aqueous solutions and gases. K_p applies to partial pressures of gases.

EvanWang

A catalyst speeds up a reaction but is not used during it, it is neither a product nor a reactant. It doesn't affect the concentrations of product or reactant at equilibrium.

EvanWang

K is technically calculated using an attribute called the activity of a substance, and activity is unitless.

EvanWang

In a reaction where the only substance that can be inputed into K is on the reactant side (the rest of the reactants and products are either solids or pure liquids), is K calculated as just the concentration of that substance? Or is it the inverse of the concentration?

EvanWang

It's tough to say but generally the bonding sites should be on the same side of the molecule. For example, cisplatin works because both of the Chlorines are on the same side and can bind to DNA at two points. If the Chlorines were in an orientation where they were opposite each other, the resulting ...

EvanWang

I would study the strong acids and bases, as well as the trends in what makes an acid/base strong or weak.

EvanWang

In the following reaction:
HCl(aq)+NH3(aq)→NH+4(aq)+Cl−(aq)

The NH3 accepts H+ ions donated from the HCl. That makes it a bronsted base.

EvanWang

I think it wouldn't hurt to specify the principle quantum number when you know what the central atom is. If you only know the areas of electron density around the central atom, then you can only discern the hybridization without the principle quantum number.

EvanWang

A photon is absorbed when the energy of an electron increases and its principle quantum number increases. The energy is positive when a photon is absorbed.

EvanWang

Yes, a molecule with hydrogen bonding must also have dipole-dipole interactions.

EvanWang

Only if the bonded atoms on either side of the central atom are the same will the molecule be nonpolar. If the bonded atoms are different, there will be a net dipole moment.

EvanWang

In most cases when there is a lone pair, the actual bond angles are less than the expected for the electron arrangement. This is because lone pairs are more diffuse than bonding atoms, in other words they take up more space. As a result, they push the bonding atoms away more than a bonding atom woul...

EvanWang

It does not mention that there are other ions around, therefore there are no ion-dipole forces. Also, the question does not mention that there are other nonpolar molecules around, so there are no dipole-induced dipole forces.

EvanWang

The molecular formulas of organic molecules will typically be written in sections. If a molecule has many sections, such as CH3(CH2)3CH3, then it will be elongated. A molecule written like C5H12, is spherical. Notice how they have the same amounts of C and H atoms, but they are written in a way that...

EvanWang

If all the equatorial atoms are the same, you can break the dipole vectors into their component vectors and you will see that they all cancel. The axial atoms are pointed 180 degrees apart, so their dipoles cancel.

EvanWang

The lewis structure where Se is surrounded by four oxygens allows for all the formal charges to be zero when two of the oxygens double-bond with Se and the other two oxygens single bond with Se and H. The lewis structure where the two hydrogens bond with the central Se atom doesn't allow for all for...

EvanWang

If you draw the molecule out, it will look like 2 C 2 H 5 groups bonded on either side of a central oxygen atom. The oxygen can only single bond with both of the carbons next to it because the carbons already have three bonds. This leaves the central oxygen atom with two bonding pairs and two lone p...

EvanWang

If a problem strictly mentions molar mass, or states that there are no dipole-dipole interactions or hydrogen bonding, then increasing molar mass increases the Van de Waal forces because larger molecules can have larger instantaneous dipole moments. Larger Van de Waal forces means more attraction be...

EvanWang

You determine the hybridization of orbitals by the number of areas of electron density around a central atom. Areas of electron density include bonding pairs and lone pairs. If there are n number of areas of electron density, you need n number of hybridized orbitals.

EvanWang

Look at the lewis structure to help figure out the hybridization of an atom. If a central atom has four areas of electron density around it, then you need a hybridization that corresponds to those four areas. In this case, sp 3 . If an atom has three areas of electron density, the hybridization is s...

EvanWang

The overall shape of the molecule is not not symmetrical, there is a net dipole moment because the dipoles are not equal in magnitude and in opposite directions.

EvanWang

According to my TA hybridization won't be on the test, but it wouldn't hurt to know since it explains the behavior of some of the topics mentioned on the test.

EvanWang

O₂ is nonpolar because it is two of the same atom, there is no net dipole moment because both oxygens "pull" on electrons the same amount in opposite directions.

EvanWang

Pi bonds form in cases where there is one or more p orbitals that have not been hybridized into an spⁿ orbital. These leftover p orbitals are able to form a pi bond by bonding perpendicular to the plane of the sigma bond.

EvanWang

I like to consider electronegativity as how much an atom will hog an electron when it is bonded. Since Neon already has an octet, it won't want any more electrons so, hypothetically, it will have very low electronegativity.

EvanWang

Although nonpolar molecules may have symmetrical electron distributions, there are instances when one area of the molecule is more negative than the other. This fluctuating electron distribution results in fluctuating dipoles. Van der Waals forces are the results of those instantaneous dipoles that ...

EvanWang

I think it is represented with an arrow with its tail beginning at the positive atom and pointing toward the negative atom.

EvanWang

Covalent bonds occur between two nonmentals, while ionic bonds occur between a metal and a nonmetal.

EvanWang

H bonding occurs with N, O, and F because those elements have high electronegativity. When they bond covalently with Hydrogen, they will hog the electrons more and make that part of the molecule partially negative and the hydrogen atom partially positive. While this can occur with other elements, th...

EvanWang

I would look thoroughly through the topics discussed in the syllabus and try to do a problem pertaining to each one. Bring up questions you have with your TA and spread the studying out over several days.

EvanWang

I asked my TA and he said that we aren't expected to know nomenclature right now, but I would try to memorize the important polyatomic ions like phosphate.

EvanWang

It is energetically favorable for Cu to have a ground state electron configuration of [Ar]3d 10 4s 1 . This is because when the 4s shell is half-filled and the 3d shell is completely filled, the atom is more stable. When you take away an electron, you take it away from the 4s shell because it is fur...

EvanWang

Resonance allows for delocalization, in which one or more electrons are spread out evenly across all bonds. This lowers the overall energy of the molecule.

EvanWang

No, resonance happens when you have the same arrangement of atoms, but different arrangements of electrons. For example, in ClO- the single bond could be in four different locations.

EvanWang

The prefix Mega (M) in SI units always means 10^6. So 1 MHz is 1*10^6 Hz.

EvanWang

Elements that are closer to the left side of the periodic table generally give up electrons and elements on the right side of the periodic table generally take electrons. This is reflected in the increasing ionization energy going right to left across a period.

EvanWang

In addition to the half-filled 4s orbital lending more stability, a completely full 3d orbital is more stable than one that is partially filled. If I remember correctly, Dr. Lavalle also said something about the 3d orbital having slightly less energy than the 4s orbital, but I'm not sure if that has...

EvanWang

In addition to what everyone else said, I think it's because the noble gasses don't react much and are very stable, much like the innermost electrons that they represent.

EvanWang

When a wavelength of light is absorbed in the UV (Lyman) region, then can we assume that n i is 1? And for wavelengths of light emitted that correspond to the Lyman series, can we assume that n f is 1? Or do we always assume that n f is 1 for regardless of whether the electron emits or absorbs a pho...

EvanWang

An orbital must have 3 quantum numbers, and no two orbitals can have the same quantum numbers. Knowing this, b and d both indicate only 1 orbital because they list 3 quantum numbers. For a and c, since all 3 quantum numbers are not given we must list the other possible numbers to find the number of ...

EvanWang

I think it just means to use the De Broglie equation because the De Broglie equation is derived from the equations E=pc, c= wavelength*frequency, and E=hv which we have learned before. However, this equation only works for any particle with a rest mass, momentum, and wavelength so if it asks for th...

EvanWang

This is a multistep reaction. You have to first balance both reaction equations. Next, you have to convert the amounts of phosphorus and oxygen into moles. Then, you use the mole ratio to find out which reactant will run out first. Phosphorus is the limiting reactant because it will run out first. O...

EvanWang

When a question asks whether a certain orbital can exist, it is asking if the quantum numbers are possible. If we look at 6f as an example, Because: n=6 The possible values of l are: l=(0 to n-1) or 0,1,2,3,4,5 We know that the f subshell occurs when l=3. Since 3 is listed in the possible values of ...

EvanWang

We assume the speed of light to be constant, so any changes in frequency won't impact the speed of the wave. Also, $\frac{h}{mv}$ only applies to particles with resting mass like electrons, not electromagnetic radiation.

EvanWang

I think you just need to memorize the fact that visible light occurs between 400 and 700 nm.

EvanWang

The questions for 1A begin on page 9. The last question for 1F is on page 63.

EvanWang

1.0003 has 5 significant figures because the zeroes come between nonzero digits. 0.0003 only has 1 significant figure. This is because you can rewrite it as 3x10^-4 (which only has one sigfig). The leading zeroes are not significant even if they come after a decimal point.

EvanWang

Generally, the order the elements are given in the problem will be the same order you can write them in your answer. In organic molecules, that normally means C, H, then everything else.

EvanWang

Different elements have different threshold energies. If it's not given, you can always determine the threshold energy by using
Threshold energy=Energy of photon-kinetic energy of ejected electron

EvanWang

You use the molar masses of oxygen and hydrogen to find the mass of one mole of water molecules. Next, divide that number by avogadro's constant to find the mass of one water molecule.

EvanWang

Calcium is in the second group of the periodic table so calcium ions are Ca2+. Hydroxide ions have a -1 charge, so you need 2 OH to make the overall charge neutral. So Calcium hydroxide is Ca(OH)2.

EvanWang

For sig figs, do we round to the correct number of sig figs at the end or do we constantly round to the correct number as we go? Generally, it is better to try to use exact values throughout the calculations and then round at the end. If your calculator has it, it could be helpful to save the exact...

EvanWang

I don't think we need to calculate the level of precision and accuracy, we just need to identify situations and experimental results that are precise, accurate, both and neither.

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