CHEMISTRY COMMUNITY

Anna Goldberg 2I

What does it mean when it asks to calculate Ka? As well, why does the solutions manual choose the half reaction with the largest potential to be the anode? I thought the half reaction with the largest potential was the cathode, seeing as it has the most oxidizing power and is thus most likely to be ...

Anna Goldberg 2I

15.87 asks what conditions can be used to distinguish the two mechanisms-essentially, what mechanisms can be used to distinguish the two slow steps. If the reaction is carried out in a highly concentrated solution of sucrose, mechanism ii will behave as a pseudo-first order reaction, due to its depe...

Anna Goldberg 2I

Considering a B- is 80-83%, 500*.8 is 400 points, so 400/500 points is what would be needed to get an 80% in the class, assuming there is no curve.

Anna Goldberg 2I

This was covered in class on March 9th at the end of lecture. You graph the change in standard Gibbs free energy for the reaction, making sure to detail the changes in standard Gibbs free energy of activation.

Anna Goldberg 2I

How is the proposed structure for the activated complex determined?

Anna Goldberg 2I

Does anyone have any good tips on how to approach this problem and the diagram? I understand that you have to isolate the reactants and the products, and, by doing so, find the intermediates, but I'm not sure on the quickest way to do so.

Anna Goldberg 2I

Most likely, because zero order reactions don't contribute to reaction rate.

Anna Goldberg 2I

Yes, that is the formula that you use.
I would have to know the values for E cathode and E anode in order to tell you if your E cell is correct.

Anna Goldberg 2I

First, calculate E standard by E standard=E cathode-E anode
Then, use the equation E standard=(0.05916/n)*log(K) to solve for K.

Anna Goldberg 2I

Yes, that is correct. Even though C changes in concentration, the initial rate remains constant while A and B remain constant.

Anna Goldberg 2I

Will we be expected to derive by hand the integrated rate laws from the instantaneous rate laws on the Kinetics Test or Final Exam? Or will just knowing the instantaneous and integrated rate laws be sufficient?

Anna Goldberg 2I

ln(At)/(Ainitial)=-kt

When you do ln(A initial)/(At), the swap of the numerator and denominator makes ln(A initial)/(At)=kt

Anna Goldberg 2I

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

A one molar solution.

Anna Goldberg 2I

Considering 15.6 covers through Second-Order Integrated Rate Laws and 15.7 begins Reaction Mechanisms, I would assume the assigned problems between 15.1 and 15.44 would be good practice for test 3, considering the problem set for Reaction Mechanisms begins with problem 15.45.

Anna Goldberg 2I

In experiments 1 and 4, the initial rate stays the same while the concentrations of A and B stay the same between the two experiments-yet, C changes between the two experiments. Thus, C does not contribute to the initial rate, because a change in the initial rate does not result from a change in C.

Anna Goldberg 2I

Will we have to draw a physical diagram of the galvanic cell for the exam? Or just write out the galvanic cell diagram with the lines representing the phase changes and the salt bridge?

Anna Goldberg 2I

From examples Dr. Lavelle has done in class and from the textbook, I was under the assumption that, in a galvanic cell diagram, you wrote the reactants and the products for the anode and the cathode, with the anode and cathode separated by the double line representing the salt bridge. The reactants ...

Anna Goldberg 2I

As well, a more positive Ecell makes G more negative, and thus the reaction more spontaneous (from G=-nFE).

Anna Goldberg 2I

Metals that are reducing agents are in the anode, as they themselves are becoming oxidized, which occurs in the anode. Thus, you want these metals to have a more negative standard reducing potential, as this will make the Ecell more positive, from the relationship: Ecell=Ecathode-Eanode (you want to...

Anna Goldberg 2I

From what I've seen in the solutions, an inert electrode such as platinum is needed in the galvanic cell when the oxidized and reduced molecules are in the same solution (they are both aqueous, or both gas). Is an inert electrode needed when oxidized and reduced molecules are both solids? As well, I...

Anna Goldberg 2I

Also-in problems that have both volume and temperature changing, for example-use Cv in your delta S=nCln(T2/T1) temperature change calculation, as volume is constant while temperature is changing (even though volume changes in another part of the reaction). Conversely, in problems that have both del...

Anna Goldberg 2I

The first trend noticed is that the higher the bond dislocation energy, the higher the free energy of formation. This makes sense, as it should take more energy to form a molecule that requires more energy for separation. The second trend is that, as atomic number increases, the free energy of disso...

Anna Goldberg 2I

q reversible=-w
And delta S should equal zero for reversible.

Anna Goldberg 2I

A salt bridge facilitates counter ion transfer so the solutions stay neutral.

Anna Goldberg 2I

I'm noticing that the solutions manual uses different equations when solving for delta S when there is a temperature change. For example, for 9.13, the manual uses delta S=nRln(T2/T1). For 9.19, the manual uses delta S=Cp,mln(T2/T1). And then, for 9.43, the manual uses delta S=nCp,mln(T2/T1). How do...

Anna Goldberg 2I

Considering the readings posted on the syllabus do not include Box 8.1, and since Dr. Lavelle hasn't lectured on it, I assume we would not need to know the information.

Anna Goldberg 2I

What does Sm represent, as in question 9.89?

Anna Goldberg 2I

And then for b, just do a simple conversion of units.
For c, use delta S=-deltaH/T in order to make the comparison.

Anna Goldberg 2I

You use change in entropy/time = -qrev/time*T

Anna Goldberg 2I

When volume is constant, delta U = q. This is from the internal energy equation delta U = q -P*delta V (with -P*delta V substituted for work). When there is constant volume, delta V thus equals zero, resulting in delta U = q. You just substitute delta U in for q in the heat capacity equation. When p...

Anna Goldberg 2I

How are you able to determine the intermolecular forces for these molecules? Does it have to do with covalent/ionic bonding or hydrogen bonding?
As well, do liquids tend to have higher intermolecular forces?

Anna Goldberg 2I

If you were to have a question that asked you to determine which compound had a higher enthalpy of vaporization, how would you do so?
If, for example, a question asked you to rank H2O, NaCl, and N2 in order of increasing enthalpy of vaporization, how would you determine this?

Anna Goldberg 2I

After you do the above calculations with the bond enthalpies, you should have delta H = -242 kJ/mol. You then need to take into account that water is in it's gaseous phase here, and the question asks for you to find the enthalpy of formation at a liquid state. With this in mind, subtract the enthalp...

Anna Goldberg 2I

Keep in mind, as in question 8.73 c), the compounds may rearrange into different compounds, but the bonds between the elements remain the same. The reaction enthalpy is zero in situations such as this.

Anna Goldberg 2I

Question 8.1 offers good examples of open, closed, and isolated systems.

Anna Goldberg 2I

If you want to catch up on that, read only 17.5 and 17.6 and do problems 29, 31, 33, 35, and 37, which has you practice naming coordination compounds.

Anna Goldberg 2I

As well, what problems in the Chapter 8 problem set should we begin with?

Anna Goldberg 2I

Yes, I was also wondering which sections of the textbook would be advisable to read in order to follow along with Dr. Lavelle's lecture schedule.

Anna Goldberg 2I

Will we receive a list similar to this on the final? (I am assuming not).

Anna Goldberg 2I

A strong acid is an acid in which almost all of the molecules are deprotonated in solution. A weak acid is an acid in which only a small fraction of the molecules become deprotonated in solution.

Anna Goldberg 2I

Metals typically form basic oxides, nonmetals typically form acidic oxides, and substances that react with both acids and bases (see Figure 12.7-these are the elements in and close to the diagonal line of metalloids), are classified as amphoteric.
Reading 12.3 should also help!

Anna Goldberg 2I

Hybridization is linked to regions of electron density and shape. Linear is sp Trigonal planar is sp2 Tetrahedral is sp3 Trigonal bipyramidal is sp3d Octahedral is sp3d2 For compounds with lone pairs about the central atom, consider the electron arrangement. For example, a trigonal pyramidal compoun...

Anna Goldberg 2I

For knowing which hybridized orbital to use, consider the regions of electron density in the compound. Linear: sp Trigonal planar: sp2 Tetrahedral: sp3 Trigonal bipyramidal: sp3d Octahedral: sp3d2 For compounds with lone pairs, consider the electron arrangement about the compound. For example, a com...

Anna Goldberg 2I

I was wondering this as well. Is it because the oxygen with two lone pairs and two bonded pairs has a zero formal charge? And that the other two oxygen both have a negative one formal charge?

Anna Goldberg 2I

1) Identify the Greek prefix that corresponds with the coordination number (how many ligands are bonded) 2) When naming, name the ligands in alphabetical order 3) Then after you have named all of the ligands in alphabetical order, then name the Transition Metal Cation and then it's Roman Numeral in ...

Anna Goldberg 2I

In question 73, though, how are you able to determine a more exact answer for the bond angle in a shape such as triangular pyramidal or bent? Since it asks you to rank the bond angles in increasing size in various molecules, these approximates are difficult to rank.

Anna Goldberg 2I

This is well illustrated in a couple of homework questions: 4.33, for example, asks you to just name the hybridization orbital. For 4.33 a, the answer is simply sp. On the other hand, 4.95 asks you to name the bonds between the atoms in the CH2=CHCHO molecule. Then, you use the coefficients in namin...

Anna Goldberg 2I

You can also use the equation given by Dr. Lavelle for hydrogen (En = -hR/n^2), as we seem to do examples generally with just hydrogen atoms. This rids of the issue of a negative answer with initial-nfinal

Anna Goldberg 2I

I also wouldn't be too worried about solving particle in a box problems, especially because that wasn't on our Chapter 2 test, so I doubt it would be on the midterm.

Anna Goldberg 2I

The double bond is between oxygen and nitrogen.

Anna Goldberg 2I

You can also look at how electrons are apportioned. If elements share electrons in the bond, it is a covalent bond. If electrons are completely "gained" and "lost" by elements bonded together, then the bond is ionic.

Anna Goldberg 2I

From the looks of this attached photo, the p orbitals seem to be straddling the axis. The p orbitals on each axis are the direct opposite of the other. Hope this helps
http://www.chem.ucla.edu/~harding/IGOC/P/p_orbital.html

Anna Goldberg 2I

Are we supposed to be able to perform calculations with the Schrodinger Equation?

Anna Goldberg 2I

In 1.35, assuming the velocities of the two people are similar, I think the problem is just trying to show the relationship between the de Broglie wavelengths of two objects of the same speed but different masses, that being that the object with the smaller mass will have the larger de Broglie wavel...

Anna Goldberg 2I

Calculating the de Broglie wavelength will not tell you if it is constructive or destructive. It just simply tells you how long the wavelength is.

Anna Goldberg 2I

I didn't take AP Chem in high school and took general chemistry my sophomore year. I've noticed people on this feed saying there are night study sessions at Hedrick-when and where are those at? Thank you.

Anna Goldberg 2I

I, like you, am uncertain as to whether it is okay to do the fundamental problems instead of the ones for Chapter 1. However, based on the syllabus and the material covered in class, I think your best bet would be to do four Chapter 1 problems for your homework. I was able to do four for my discussi...

Anna Goldberg 2I

I also keep track of the number of elements on each side of the equation. Also, I tend to balance the elements that are present in an even number on one side of the equation and an odd number on the opposite side of the equation first. It often makes subsequent calculations easier.

Anna Goldberg 2I

I first subtracted 26.45g (mass of crucible) from 28.35(the mass of the crucible and product), to get 1.9 g, which is the mass of the tin oxide. I then subtracted 1.5g (mass of the metallic tin) from 1.9g (the mass of the tin oxide) to get .4g, which is the mass of the oxygen in the air that the tin...

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