CHEMISTRY COMMUNITY

604656370

You should look at the bond length. Longer bond length means weaker bonds which can be broken with lower energy (lower boiling point). Smaller bond lengths are stronger and thus require more energy (higher boiling point).

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Do you remember the exact wording of the question? What info was given about the gamma ray? Would it have been possible to find E(photon) using E=hv and then using that as the Ek for the snail?

604656370

Does this apply to HNO3 and HNO2 as well? I said HNO3 is stronger because the extra oygen in the molecule helps stabilize the charge from the lone pair.

604656370

LDF is found in every bond because its the most basic form of molecular attraction. Since electrons are constantly moving, at any given time, one atom will be more electronegative than the other atom. Thus, there will always be some degree of induced dipole-induced dipole (also known as LDF).

604656370

How many of these discussion posts do we need to have? And does this week count towards them?

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Its a Platinum atom bonded to two Cl atoms and 2 NH3 atoms. The cis also implies that the Cl are on the same side. At least that's what my TA said.

604656370

Do you know what topics will be covered? I know VSEPR, hybridization, bond angles & shapes will be on the test but am I missing anything else?

604656370

I've been using 3.0X10^8 m/s but just to be safe, use the constants he provides on the formula sheet.

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From what I remember, my TA said it shouldn't matter for this class. Just place the unpaired electron in a way that the formal charge is as close to 0 as possible.

604656370

I liked this Khan Academy video: https://www.khanacademy.org/science/organic-chemistry/gen-chem-review/hybrid-orbitals-jay/v/pi-bonds-and-sp2-hybridized-orbitals

This other video was also helpful and showed the bonds in 3D: https://www.youtube.com/watch?v=voIpywwj2Ks

604656370

Sigma bonds are the initial and strongest bonds formed between two atoms, whereas pi bonds are the subsequent bonds (2nd and 3rd) formed between the same atoms. For example, in O2 (double bond), the first bond created would be a sigma bond and the second one would be a pi bond. In N2 (triple bond), ...

604656370

Identifying the bonds allows you to determine bond strengths and lenghts (not the exact number but relative to other molecules/bonds).

604656370

Since sulfur is a period below oxygen, it is less electronegative than oxygen. Therefore, the difference in electronegativity between carbon and oxygen will be greater than that of carbon and sulfur.

604656370

Sigma bonds are the initial and strongest bonds formed between two atoms, whereas pi bonds are the subsequent bonds formed between the same atoms. For example, in O2 (double bond), the first bond created would be a sigma bond and the second one would be a pi bond. In N2 (triple bond), the first woul...

604656370

Draw the lewis structure for both molecules and look at what kind of bonds are created (single, double, triple). Based on this you can figure out which bonds are longer and weaker or shorter and stronger; remember, triple bonds are the shortest and strongest because they involve the sharing of more ...

604656370

How would you determine these trends for diagonal (non-adjacent) elements? For example, between Carbon and Sulfur.

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The Ms also helps with understanding Hund's Rule which states: every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied. This question popped up on the practice midterm too.

604656370

Since n = 5 and l = 1 this would be the 5p orbital. And the p orbitals can hold 6 electrons. Therefore, 6 electrons can have those quantum numbers.

604656370

Did you use this equation: $v = -R[1/n_{1}^{2} - 1/n_{2}^{2}]$ ? If you use this, which is derived from
$E_{n} = -hR/n^{2}$ you should get the right answer.

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Yes, that's correct. My TA mentioned that if we had the right answer but wrong sig figs, we would lose partial credit.

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Yes, that is correct! The uncertainty in velocity is the aggregate unknown velocity. For instance, if the velocity is stated to be 1 m/s +/- 0.55 m/s then the velocity could be between 0.45 m/s and 1.55 m/s. Thus, the uncertainty would be 1.55 - 0.45 = 1.10 m/s (or 2 X 0.55).

604656370

Based on the question, you can figure out what it's trying to ask and only use the necessary equation. A lot of questions are phrased similarly so doing enough will help you figure out which situations require which equations.

604656370

The answers shouldn't be different for a Hydrogen atom - (a)-(d) should still increase. It might just be a trick question to see if you understand the concept.

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My TA mentioned that it's safe to assume visibe light is between 400-700nm.

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You need to post 3 times per week (1pt per post) with either questions or repsonses to questions by Sunday 11:59 PM.

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The both are constants. The first number, h, is Planck's constant whereas the second number, c, is the speed of light. And for reference, the formula is $E = hc/\lambda$

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Exactly. Frequency and wavelength have an inverse relationship so as one increases, the other decreases. In this case, decreasing frequency means increasing wavelength. If you were to visualize this, it would mean the waves are getting elongated (less steep). Therefore, the slope decreases as well.

604656370

I was stuck on this one too but figured out that you can add the three reactions. The result will give you: Fe + 2Br2 + Na2CO3 ---> NaBr + CO2 + Fe3O4 Using this equation, you would need to: 1. Balance the equation 2. Calculate the moles of NaBr given you have 2.5t 3. Using the moles of NaBr and the...

604656370

If I'm understanding your question correctly, I think the way the professor did it (and in my opinion the more intuitive way) is to calculate the moles of the reactants and then compare them using the equation. This method is pretty quick and sets you up to answer other questions the problem may pos...

604656370

The first step woud be to find the number of moles of O2 and P4 in the reaction given that we have 5.77g of each. For example, since the molar mass of O2 is 32g/mol and we have a sample with 5.77g, there will be 5.77g / (32g/mol) = 0.180 mol of O2. Similarly, you can calculate that there are 0.0465 ...

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Re: Boiling Point

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