CHEMISTRY COMMUNITY

KristinaNguyen_1A

An ion is anything with a charge to it, so it can either be negative or positive. Ex: Ca^2+, Na+, Cl-

KristinaNguyen_1A

It is nonpolar because there is no electronegativity difference between the C-C, and because the dipole moments from partially positive H all goes towards partially negative C, so it cancels out in the end.

KristinaNguyen_1A

Sigma bond is the end to end overlap. All single bonds are sigma bonds and these bonds can usually rotate. Pi bonds occur when there are more than one bond (double or triple bonds) and these will break if it tried to rotate.

KristinaNguyen_1A

I think it will depend on the question. If it asks for lewis structure, you will not need to do the geometric shape, but if it specifically asks for the geometric shape you probably will have to so it is best to know them.

KristinaNguyen_1A

I am not sure about this but I think if you do have lone pairs on the top and on the bottom, it would cancel out and keep the geometry and shape at a linear.

KristinaNguyen_1A

HCN is linear because C will be the central atom, single bonded to H and triple bonded to N, with N have a lone pair of electrons but, I think, because the lone pair is on the N and not the central atom it will not cause the bonds to be pushed and so it remains as a linear. CH2N2 is a tetrahedral an...

KristinaNguyen_1A

How do you know when atoms can break the octet rule? or which atoms can break the octet rule? Elements from the 3rd period and below can have more than 8 valence electrons because you have the D orbital added to it. Some other exceptions are Boron and Aluminum which can have 6 valence electrons and...

KristinaNguyen_1A

It's better to use the formal charge equation given in class: FC=(valence electron)-(Lone pairs+1/2(bonds))

KristinaNguyen_1A

They are all interchangeable so I don't think it really matters. They are also considered interactions and not bonds because it will continuously interact and then break apart due to its weak interactions.

KristinaNguyen_1A

Energy decreases when wavelength increases because as wavelength increases, the wave frequency decreases (due to the fact that they are inversely proportional) because larger wavelength means fewer waves. Wave frequency gives off energy so if wavelength increases=frequency decreases, then energy als...

KristinaNguyen_1A

It was stated during class that the question will be specific on whether or not you would need to list all of the resonance structure or not. If it states for just one, any of the resonance structure will work as an answer. I think if it just asks for the Lewis structure then you would only need to ...

KristinaNguyen_1A

Octet rule is due to the atoms wanting to be stable. With 8 electrons, its shell will be full and that stabilizes the electrons. This is similar to how when we did orbitals, 3d^9 would take an electron from 4s^2, because it wants to fill up it's orbital to be full (3d^10). So in the case of electron...

KristinaNguyen_1A

n=energy level/shell (1,2,3,4...) l=orbital/subshell (0,1,2,3), (which is s,p,d,f) ml=orientation (so for example if you have the p orbital, you will have 3 subshells; __ __ __ and each one can hold 2 electrons, so you can use that to tell how many electrons there are and which/how many are unpaired...

KristinaNguyen_1A

I am not 100% sure, but I think the equation R(1/n^2-1/n^2) is used to find the frequency, where the first one is initial and the second one is final because when deriving it you would get a negative final plus a positive initial, therefore it can switch places to make it a little simpler. (This was...

KristinaNguyen_1A

For Cu, the electron configuration is an exception because the d orbital is nearly filled up, so it would want to get an e- to become stable. Instead of it being [Ar]3d^9 4s^2, it will be [Ar]3d^10 4s^1. Usually the orbitals goes s,p,d,f (increasing energy), but the d goes first in Cu because you ha...

KristinaNguyen_1A

Like someone said above, the order depends on the energy level. d starts with level 3 but is the same row as when s and p have the energy level of 4. Because of the difference in energy levels, d will go before s and p. The order of s and p does not change since they are in the same energy level.

KristinaNguyen_1A

I am not sure about this but I am assuming that, because d is the inner orbital, it will want to be filled up, while p would be the outer orbital if you think of it at 3d 4s 4p, so it does not have the need to fill up its orbital.

KristinaNguyen_1A

This is because of stability. With 4d9 5s2, the d just needs one more electron in order to fill up the orbital. If it remains at 4d9 it is unstable, so it takes the electron from 5s2, the outer shell in order to fill its own. That causes it to become 4d10 5s1 instead.

KristinaNguyen_1A

The E=(-hR)/n^2, is to find the energy at that level. If you use this formula you will have to do it for the initial and for the final and then subtract it. The equation you put is the condensed way of doing it.

KristinaNguyen_1A

Because they are trying to keep the test doable timewise, there will be a lot of conceptual questions.

KristinaNguyen_1A

You will need to shorten the wavelength, which in turn increases the frequency and that will eject the electron. The relationship between photon and electron is a 1 to 1 relationship, so if 10 photons are shot at the metal, 10 electroms will be ejected.

KristinaNguyen_1A

He mentioned how if the value is less than 10^-18, it would be too small to detect. Anything greater than that is measurable.

KristinaNguyen_1A

In discussion it was mentioned that Lyman series is used whenever it asks for UV light (it starts off with n=1), while Balmer series is for visible light (it starts with n=2).

KristinaNguyen_1A

The molecular formula would be C5H5 because the model gives the actual amount. You will then try to figure out if you can simplify it and because both are divisible by 5, you divide it by five to get the empirical formula of CH.

KristinaNguyen_1A

I agree that Hz is the frequency of the waves, which is counted through complete wave cycles (one peak and one trough per cycle)

KristinaNguyen_1A

I went to a UA office hour and how she described it was that:
Precision= how CLOSE measurements are to ONE ANOTHER
Accuracy= closeness of measurements to its TRUE VALUES

KristinaNguyen_1A

I believe that since they give you the molar mass and the empirical formula, you will need to find the molar mass of the empirical formula: C=12.01 H2=2(1.008) CH2= 14.026 And then you find the ratio: 84g*mol^-1/14.026g*mol^-1 That will give you a value of 5.98888 so it is approximately 6 You will t...

KristinaNguyen_1A

1) Find the mass percentage (which is given) 2) You can assume that it is out of 100g (the mass percentage can be changed to grams) 3) Conver masses to molecular ratio (ex. mass ratio for carbon: 49.48g/12.01g*mol^-1) 4) Divide each by the smallest mol from the previous step and if needed, multiply ...

KristinaNguyen_1A

I agree that through the Hill System, H4 should go before Si, but I think it is disregarded because in either order it will not affect the balancing process, there will still be 4 H and 1Si in SiH4 and H4Si.

KristinaNguyen_1A

You will have to first convert 23.6g of PCl to moles (divide by the molar mass of PCl). Because in your balanced equation there is a 1:3 ratio for PCl:HCl, you will have to multiply 3 to the moles of PCl to find the moles of HCl. Afterward, you will have to convert HCl from moles to grams since it a...

CHEMISTRY COMMUNITY

Search found 30 matches

Re: Naming

Re: Problem 4.27

Re: Sigma Bonds

Re: VSEPR Drawing

Re: 4.1

Re: 4.3 Homework

Re: Octet rule

Re: Formal Charge

Re: Van der Waals [ENDORSED]

Re: Wavelength/frequency

Re: resonance structures on exams

Re: OCTET RULE [ENDORSED]

Re: 2.29

Re: E=R(1/n-1/n)?

Re: Order of Orbitals

Re: The order for s,p,d

Re: Cr and Cu exceptions

Re: 2.43

Re: Rydberg Equation

Re: test 2

Re: Frequency on Photoelectric Effect

Re: Measurable Wavelike properties

Re: Series Question

Re: Creating a molecular/empirical formula

Re: Clarification on Hz

Re: Accuracy vs Precision

Re: Empirical Formula

Re: Molecular Formula f.19

Re: H.3

Re: Video 3 Post- assessment [ENDORSED]