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Tarek Abushamma

Lone pairs have higher electron - electron repulsion and thus they take up more space than atoms, pushing the surrounding atoms closer together and lowing their bonding angle.

Tarek Abushamma

I think it could be monodentate or bidentate as well, it would depend on which would produce the most stable structure.

Tarek Abushamma

Lone pairs reduce bonding angles because they take up more space, resulting from their higher electron-electron repulsion, but I'm not exactly sure what you mean by something that has more than one pair having a specific name.

Tarek Abushamma

Keep in mind that the roman numerals indicate of the charge of the element. Ex: bismuth(III) has a +3 charge.

Tarek Abushamma

No it has to do with the shape. Since CH3(CH2)3CH3 is linear, it has a greater surface area, and thus more contact points for london forces to occur. The more interactions there are the stronger they are as a unit, which translates to the higher the boiling point.

Tarek Abushamma

Think of the bonds in a resonance structure with single and double bonds as a blend between the two. In reality, the molecule will not have discrete single and double bonds, rather all its bonds will be the same intermediate mix between single and double bonds. You could sort of think of it like a &...

Tarek Abushamma

The lone pair takes up space, so you could not arrange the fluorine atoms in a tetrahedral shape.

Tarek Abushamma

Whatever structure has the lowest energy state(such as the most favorable formal charge) contributes the most to resonance structure; so if all that separates equally low energy states is the position of a double bond, then showing all the alternate double bond positions would be the correct thing t...

Tarek Abushamma

In water the dipole moments are not linearly opposing each other, which is why they do not cancel each other out. For this reason the molecule is polar.

Tarek Abushamma

Are there any cases where all 10 extra are taken up?

Tarek Abushamma

Test three only covers chapter 3.

Tarek Abushamma

The cation of Hydrogen is just a proton, so there probably are not any electron shells associated with it.

Tarek Abushamma

Formal charge can be calculated using the formula: (# of valence electrons on an atom) - (non bonded electrons + number of bonds).

Tarek Abushamma

These trends have many of exceptions, however, so it is also important to take the element's electron configuration into consideration when determining how it would respond to additional electrons.

Tarek Abushamma

Also be mindful of exceptions, like oxygen, which has a lower ionization energy than nitrogen, yet is not in the center for the lewis dot of NO3.

Tarek Abushamma

I believe E=-hR/n^2 is only valid for hydrogen, yet this equation was not needed on the test.

Tarek Abushamma

With large everyday objects we can measure momentum and position with great precision, so the resulting uncertainty is negligible.

Tarek Abushamma

In order to solve this problem first convert the 2.93 eV to J. Then, convert both given wavelengths into J using E=hv to see which one corresponds to an energy greater than the 2.93 eV. After you decide which laser has the higher E, subtract the work function from this value in order to find the ene...

Tarek Abushamma

Nodal planes specify that there is 0 probability of finding an electron at that nodal plane, the don't really specify much about the way electrons are distributed in the areas that have a significant probability.

Tarek Abushamma

The equation to solve for the work function is E(Photon)=Work Function(Threshold Energy) + E(Kinetic)

Tarek Abushamma

I don't think we will have to directly apply the Shrodinger Equation to any problems in this course.

Tarek Abushamma

8.8nm would still be a part of the X-ray spectrum, I believe.

Tarek Abushamma

SI unit for distance is the meter and for time is the second, so you did the correct thing by converting it.

Tarek Abushamma

You can use this given frequency to calculate the threshold energy using the equation, E = (h)(v). From there you should be able to use the work function to calculate the energy of the photon needed to displace an electron the given 3.6x10^3 km/s.

Tarek Abushamma

I meant can you use a combustion reaction to find the empirical formula of a compound that does not contain oxygen? Or will oxygen always be present in the final empirical formula?

Tarek Abushamma

When using a combustion reaction to find the empirical formula of a compound, will oxygen always be one of the elements we are tying to solve for?

Tarek Abushamma

A discrete variable is something that has a defined set of possible values. A hydrogen atom is an example, as we can not have .33 or 1.7 hydrogen atoms. Instead, only discrete whole values (1 atom, 5 atoms, 456 atoms) are valid to quantify the variable.

Tarek Abushamma

Yes, I believe the Balmer series falls within the visible region of approximately 400nm - 700nm.

Tarek Abushamma

Yes, both of those would have two significant figures. Putting decimals less than one in scientific notation makes it very easy to tell which numbers are significant as only the digits that appear before the "x10" are.

Tarek Abushamma

The units of Avogadro's number is "things" per mole, which is why when the energy of a photon is given per mole, we have to multiply by Avogadro's number to determine energy per photon.

CHEMISTRY COMMUNITY

Search found 30 matches

Re: Atom vs lone pair

Re: HW 17.33

Re: Lone Pairs Affecting Bond Angles/Shape?

Re: Help on 3.25 [ENDORSED]

Re: 6.1 and 6.19 question

Re: Test Q.8 [ENDORSED]

Re: structure SF4

Re: Contributing to Resonance Structures

Re: dipole moments and polarization

Re: Expanded Octet

Re: Friday's Lecture [ENDORSED]

Re: 3.25 c)

Re: formal charge?

Re: Electron Affinity

Re: 3.33

Re: test 2 #4

Re: Everyday Objects and heisenberg's equation [ENDORSED]

Re: 1.69

Re: Nodal Planes [ENDORSED]

Re: What is work function? [ENDORSED]

Re: Shrodinger Equation

Re: HW 1.33

Re: DeBrogile Module

Re: HW 1.33

Re: Combustion Reaction [ENDORSED]

Combustion Reaction [ENDORSED]

Re: Quantized v. Discrete

Re: Balmer Series

Re: Sig fifs in 750 [ENDORSED]

Re: ENERGY PER MOLE OF PHOTONS [ENDORSED]