CHEMISTRY COMMUNITY

mayasinha1B

How do you figure out how to draw a proposed structure for an activated complex? Will we be expected to know this for the test?

mayasinha1B

No but we do need to know how to analyze experimental data

mayasinha1B

I don’t know if there are any practice problems but I would know identifying factors for each group as well as be able to identify them in a molecule.

mayasinha1B

What was the purpose of the UV? Was it only to reach the activation energy?

mayasinha1B

I'm pretty sure the unique rate is the same for every reactant, so you just wouldn't take the stoichiometric coefficients into account.

mayasinha1B

The more negative the E standard for a given half reaction is, the less likely that reaction is to take place in the given reduction state. If you think about it, a more negative E standard yields a more positive delta G with a lowered likelihood of spontaneity. So the higher value will be reduced a...

mayasinha1B

If K>1 then that means the reaction favors the products, so the reaction is spontaneous.

mayasinha1B

You would look up the half reactions given in the table for Ag and Cu, and put together the half reactions to get the full balanced redox reaction. You would then use the given E standard values for the half reactions in order to solve for E standard for the full reaction. This can get you the value...

mayasinha1B

A negative rate would mean the reaction is occurring in the backwards direction.

mayasinha1B

Will a zero order reaction mean that changing the concentrations will result in no change?

mayasinha1B

The order of a reaction refers to the index, or exponent, to which its concentration term in the rate equation is raised.

mayasinha1B

Also how do you know to add KOH to the left side of the cell diagram. Why do you add Ni (s) on the right side of the cell diagram as the conducting electrode even though there is solid metal present? Help

mayasinha1B

Will we be provided the half reactions used in problem 15?

mayasinha1B

For homework problem 15 part a how do you get the half reactions displayed? Ag solid is not displayed in the skeleton reaction so how would you know to add this

mayasinha1B

You only use a comma if both are in the aqueous state and therefore in the same solution

mayasinha1B

Why is the Pt necessary on the left side. I thought it was only necessary when there is no solid metal present, but there is I2 solid present. The solutions manual explained that it is because both oxidized and reduced species are in the same solution, but how do we know that it wasn't given to us. ...

mayasinha1B

The more negative E is the one that is reduced. So, yes that is an accurate way to determine which side.

mayasinha1B

No there won't. If there are two negative reduction potentials, the more negative one will be reduced and the other will be oxidized, even if it isn't as favorable. The transferred electrons will always cancel out of the overall reaction, so one half reaction must gain electrons and the other must l...

mayasinha1B

We weren't given it for the midterm, but I would assume if we need it to do a problem (to calculate moles or something) we would be given it. If we aren't, there is likely another way to do the problem.

mayasinha1B

Anything he assigns us in problems is fair game for a test, so I would know how to calculate potentials with electrolysis.

mayasinha1B

Which delta S and delta G equations will be helpful to know for the midterm that are not given on the equation sheet?

mayasinha1B

It has to do with the degeneracy and total possible position orientations. Cis and trans are just two examples of possible orientations of a molecule.

mayasinha1B

The reversible expansion of an ideal gas has a pressure that is not constant, which causes us to need an integral to find the area under the curve which is w=-nRTln(V2/V1) The irreversible expansion is against a constant external pressure, so the calculation is easier and we can just use the area of...

mayasinha1B

The equipartition theorem states that energy is shared equally amongst all energetically accessible degrees of freedom of a system. The application that I can think of is calculating degeneracy and assuming energy is equally distributed. I don't think we learned any ways of using this in a calculati...

mayasinha1B

Will any of the calculations actually require using the integral to solve?

mayasinha1B

Vi = 1 L
Vf = 3 L

Then use the equation delta S = R ln(Vf/Vi)

mayasinha1B

Yes this equation is only used for ideal gases and calculated using properties of ideal gases.

mayasinha1B

Thanks for the tip!

mayasinha1B

Gibbs free energy is a term that defines whether a reaction will be spontaneous or not. It combines the entropy and enthalpy (delta H) to determine the favorability of a reaction.

mayasinha1B

Dr. Lavelle just said he didn't use the term disorder in his description. I think it should be fine to use it in a description.

mayasinha1B

No we don't need to know how to derive equations. Just plug in variables and solve.

mayasinha1B

We discussed this in class and there were several problems on it in the homework so I would say yes.

mayasinha1B

Anything discussed in class or given through homework questions I would say is fair to appear on a test. Dr. Lavelle did say that there are sometimes questions on the homework that are more difficult than test problems but I would still understand how to do every problem.

mayasinha1B

H = U +PV ΔH = Δ(U + PV) ΔH = ΔU = Δ(PV) At constant pressure, ΔH = ΔU + PΔV So if you are at constant pressure and constant volume, then ΔH = ΔU + P(0) and therefore ΔH = ΔU If we don't have ΔV but we have other info (still at constant P) PV = nRT V = nRT/P If we are at constant P and T then only n...

mayasinha1B

I've noticed that sometimes enthalpies are used in units of kJ and sometimes in kJ/mol. How do we know which is right to use when? I assumed that it was when an enthalpy is multiplied by a certain number of moles for a particular reaction but on problem 8.67 part b when it multiplies kJ/mol with mol...

mayasinha1B

A higher energy density will lead to a higher total amount of energy for a given volume, which will lead to a higher amount of energy overall. I.E. the car will drive further with a higher energy density given a set volume of the gas tank.

mayasinha1B

The only circumstance I can think of is if we are provided all the necessary information to calculate them.

mayasinha1B

In the problems we have been given, heat capacities have been provided, so I think it is safe to assume we will be given them on tests/exams.

mayasinha1B

According to my understanding, at constant volume, all the heat energy is devoted to raising the temperature of the system, and no external work is done. When the reaction takes place at a constant pressure and not a constant volume, some of the heat energy does external work by expanding the system...

mayasinha1B

I think that is right that non-expansion work is specifically work where volume does not change

mayasinha1B

The work function has to do with the amount of work required to remove one electron from an atom.

mayasinha1B

As described above, ice melting is an endothermic reaction because it requires heat for the reaction to take place. As a result, surroundings, in Dr. Lavelle's example, one's hand, will cool down because the heat required for the reaction is absorbed from the surroundings (conservation of energy).

mayasinha1B

Both definitions of acids and bases refer to the same types of reactions. An acid is what likely loses a proton, which also means it will gain an electron pair to maintain it's octet. An acid will gain a proton but lose an electron pair to keep each atom at an octet. The test covers the problems and...

mayasinha1B

With changing temperature, Kw changes. For example, with water, pH will always be the same as pOH, but with increasing temperature, both values decrease due to an increasing Kw value (this relationship is a result of pH and pOH having a -log relationship to Kw).

mayasinha1B

The main distinction is definitely that strong acids dissociate more easily. It is also important to remember what Professor Lavelle discussed in class today (12/04/17) to help identify stronger vs. weak acids. Strong acids lose H+ easily and resulting anions are stable.

mayasinha1B

Proton: "I think I lost an electron"

Neutron: "Are you positive?"

mayasinha1B

Because you are given both moles and volume of each of the products and reactants except for NO, you can calculate each of their concentrations. Plugging in to the equilibrium equation using concentrations, you can solve for concentration of NO. Given it is in a 1 L container, you can use the concen...

mayasinha1B

If you calculate K (forwards) for the forward reaction, the backwards reaction has a K (backwards) value that is the inverse of the K (forwards). This is because you calculate using a ratio of the concentration of reactants to the power of stoichiometric coefficients and products to the power of the...

mayasinha1B

Lavelle says both ways will be accepted on a test and doesn't matter. Both ways are correct.

mayasinha1B

There are different ways of naming coordination compounds. Lavelle showed us the most common way, but he said he will accept both on a test. Both are correct ways of naming coordination compounds still.

mayasinha1B

Ionic bonds usually consist of a metal and a nonmental, or any compound with a difference of electronegativity greater than 2. Nonpolar molecules are ones with dipole moments (differences in electronegativity) that cancel out. Polar dipole moments do not cancel.

mayasinha1B

Since metals usually lose electrons in reactions, metals in the s-block have less electrons to lose and therefore are more likely to participate in a reaction. P-block metals have a few more electrons to lose to take part in a reaction and are less likely to form a compound or take part in a reaction.

mayasinha1B

h20 is not linear, and it's dipole moments do not cancel. If you draw dipole moments on co2 then you see that they cancel out, making co2 nonpolar.

mayasinha1B

When drawing Lewis structures, you count up total valence electrons for each atoms and draw bonds and lone pairs accordingly. If you follow the procedure established in lecture, you should get the diagram correctly.

mayasinha1B

Two reactants cannot limit each other. There is too little of one or the other, or they are present in the correct ratio, where they both would be used up completely and neither is limiting the other.

mayasinha1B

Electrons also fill each orbital in a subshell all with spin facing a certain direction (either +/- 1/2) before refilling the orbitals with the opposite spin.

mayasinha1B

Every covalent bond has some ionic character, so it is important to depict which atom pulls the electrons towards it slightly more (depicted with delta negative).

mayasinha1B

The book used the example that a diagnoal relationship can include similar properties in size, due to periodicity trends (increasing down a period and up a group)

mayasinha1B

Any element n=3+ can have an expanded octet by utilizing d-orbitals found in the third principal energy level and in higher principle energy levels. Common examples include PCl5 (discussed in class), SF6, ClF3, I3-, etc.

mayasinha1B

If the electron is going from n=4 to n=2, then it would be going from an excited state to a lower energy state, which would make sense that the resulting Energy from the Rydberg equation would be negative, as energy would be emitted in this scenario. In the equation, n1 is the initial and n2 is alwa...

mayasinha1B

I'm pretty sure my TA said we would only need to understand the conclusions and applications. I don't think we need to know the derivation.

mayasinha1B

Delta E is the change in energy, E(final)-E(initial), while just E would be the calculation of energy at a given energy state, n.

mayasinha1B

Diffraction also results from two or more waves interacting constructively or destructively.

mayasinha1B

When an electron is hit by a photon of light, it absorbs the energy, which can cause it to become excited and move up energy levels.

mayasinha1B

I am in the exact same boat. I haven't taken chemistry since sophomore year of high school and am now a first year in college. It's been helpful to just start doing problems in the book or online, and concepts start to come back to me. Working everything out is how I've been able to remember everyth...

mayasinha1B

If silver surfer and iron man teamed up, they'd be alloys.

mayasinha1B

What do you call a glass of water with a tooth in it?

1 molar solution.

mayasinha1B

If visualizing it and writing out all the steps helps you see it all better, definitely do it. The extra 15 seconds it takes would definitely be better than missing the points for the right answer. I also find that I tend to do that and it's saved me from many careless errors.

mayasinha1B

It should just be the section on review of chemical and physical principles, which is all we will have learned up until October 4. The syllabus said it will mainly cover problems from the homework.

mayasinha1B

You can calculate the grams of carbon using the given mass amount of CO2. You convert the mass of CO2 into moles using moles = mass/molar mass of CO2. Then, since CO2 and the compound are in a 1:1 ratio, you can use that same amount of moles to calculate the mass of Carbon using the equation moles=m...

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