CHEMISTRY COMMUNITY

Gian Boco 2G

You can use the strength of bonds and bond energies to determine if a reaction is endothermic or exothermic by comparing the sum of the bond energies on the reactants side and that of the products side and then comparing the two. If the reactants have a higher total bond energy or have stronger bond...

Gian Boco 2G

I think diatomic molecules are just molecules that involve two atoms in total while the diatomic elements allude to the elements that pretty much only exist in a homogenous diatomic arrangement such as O2 and H2.

Gian Boco 2G

Bonded atoms are stable so it requires energy to break these bonds and move them into a less stable state. Atoms forming bonds go into a more stable state, therefore releasing energy.

Gian Boco 2G

Molar heat capacity alludes to the amount of energy required to change the temperature of one mole of the substance while specific heat alludes to the same concept but for one gram of the substance.

Gian Boco 2G

Temperature is commonly defined as the average kinetic energy of the molecules. This derives from the increase of heat and temperature as the observed molecules move fast and increase their speed.

Gian Boco 2G

H2O is in excess amounts on both sides of the equation and can therefore be neglected as its minimal changes would not have any considerable effects

Gian Boco 2G

As said above, I think using molarity (mol/L) is the most convenient that volume is already taken care of. However, it can be used with moles and other units of pressure as long as the units throughout the table are consistent.

Gian Boco 2G

According to Lavelle, the safest would be below 10^-4

Gian Boco 2G

Yep, just use the ice box and the molar ratios for the "change" column. Then use the equilibrium value given to solve for those x's and then plug them back in to get the actual mole values at equilibrium. Then divide by the volume to get M, and then plug it in to the eq. expression.

Gian Boco 2G

On the right side of the equation, both the numerator and the denominator have a square (once you combine the like terms in the denominator), meaning the entire fraction can be squared. That way, you can square root both sides of the equation and solve with algebra from there.

Gian Boco 2G

Make sure the units for pressure and volume match(:

Gian Boco 2G

Their volume does not change and they do not dissolve in solution

Gian Boco 2G

Yep just temp in K

Gian Boco 2G

Yep, Q is just the expression anytime during the expression wheres as K is the value for when it is in equilibrium.

Gian Boco 2G

bars and atm, so make sure you have the right constant (:

Gian Boco 2G

You can't know simply from that. You would need the other bonded atoms as well as the shape of the molecule in order to see if the dipole moments remain or cancel out.

Gian Boco 2G

Carbon has 4 valence electrons that can form bonds. In CO2, there are 2 regions of electron density in which the sp hybrid orbitals participate. There are still two more electrons, therefore there are two non hybridized orbitals.

Gian Boco 2G

I think you would look at the number of regions of electron density vs the number of valence electrons available to that atom. For example, a carbon that has three regions of electron density would have 3 sp3 hybridized orbitals and 1 sp that isn't. Carbon has 4 valence electrons.

Gian Boco 2G

In a pi bond orbitals are overlapping outside of the intermolecular axis. Thus, rotating the bond about that axis would break the pi bond. On the other hand, sigma bonds overlap along this axis and thus allowing rotation.

Gian Boco 2G

Anything that is additional to the single bond would be the pi bond, so you would have to look at the lewis structure.

Gian Boco 2G

I cement the visualization of the shape and then the associated bond angles using flashcards and I'd say it helped a lot

Gian Boco 2G

2 lone pairs and 2 bonding domains on the central atom in O for H2O forms a "bent" shape

Gian Boco 2G

Nope because hydrogen bonding denotes an intermolecular force

Gian Boco 2G

Phosphate has access to the 3d state, which means it can add extra valence electrons

Gian Boco 2G

yep it seems like it, probably synonymous as long as the specie has one unpaired electron making it unstable and existing only for a short amount of time

Gian Boco 2G

Dipole-dipole interactions occur between the partial positive and partial negative ends of two polar molecules. London-Dispersion Forces occur as a result of temporary and arbitrary electron position within its density, which means it occurs in all molecules.

Gian Boco 2G

Yep, thus affecting the shape of the molecule. Their repulsion is stronger than bonding pairs.

Gian Boco 2G

Keep in mind that though some molecules have polar bonds, these molecules can be non polar still based on their shape.

Gian Boco 2G

I think it would be safe to know the bond angles with the common shapes and geomteries.

Gian Boco 2G

Yep because Lewis Bases are defined to donate electrons.

Gian Boco 2G

Izamary Marquez 2L wrote:Would this make cations and anions more likely to be acids or donors? Is there a correlation?

Cations donate electrons, resulting in their positive charge, making them Lewis bases. Anions accept electrons and then become negatively charged, making them Lewis acids.

Gian Boco 2G

Electronegativity, atomic radius, and either being a single, double, triple bond contributes to bond length.

Gian Boco 2G

Having a single unpaired electron would form a radical.

Gian Boco 2G

Resonance structures are just different Lewis structures of the same molecule. The most stable resonance structures would have the least number of atoms with a formal charge and most especially a sum closest to 0 for formal charges of the whole molecule.

Gian Boco 2G

When only one of the atoms in a covalent bond donates both electrons that are shared, that bond is considered a coordinate covalent bond.

Gian Boco 2G

L is the number of unpaired electrons. In other words, count the number of electrons around an atom that is not participating in a bond.

Gian Boco 2G

Based on the group that the element falls under on the periodic table, you can determine the number of valance electrons. On the S block, Group 1 (alkali metals) have 1 valence electron and Group 2 (alkali earth metals) have 2 valence electrons. Skipping all the way to the p block, the Boron group h...

Gian Boco 2G

Keep track of the number of valence electrons it has by looking at what group it is under in the periodic table. The number of bonds can surpass that however if that central element is passed period 3 because these elements can use their 3d state to accommodate extra valence electrons past 8.

Gian Boco 2G

The formal charge of an atom in a molecule will tells you the charge associated with it. We use them to determine the best Lewis Structure for a molecule that has different possibilities. We want the lowest energy, and separating charges takes energy, therefore we want FC's of 0 ideally.

Gian Boco 2G

These atoms do not have up to 8 protons and therefore do not have enough to fullfill 8 valence electrons in its outer shell.

Gian Boco 2G

I think you can choose either. In the lecture, he clearly showed that using the empirical formula twice and then subtracting worked just as well as using the ryderberg equation but it helps teach the concepts. Just be careful with the Ryderberg constant

Gian Boco 2G

Cations would be smaller than their normal atoms because there are more protons than there are electrons, reducing electron repulsion as well allowing the protons to pull tighter on the lesser number of electrons, making that ionic radius smaller.

Gian Boco 2G

While they can be used to imply one another, they directly mean two different things. One has to do with energy changes once electrons are accepted while the other deals with the ability of an atom to attract other atoms.

Gian Boco 2G

The decrease of atomic radius as you travel across the periodic table horizontally is a result of increasing number of protons, pulling the electrons closer without the electrons jumping further away as they would down a group.

Gian Boco 2G

This post really helped me a lot! The way I tend to look at it is as if an electron isn't where it's supposed to be because in the configuration, it appears as if an energy level has been skipped, which is kind of what the electron does during its excited state.

Gian Boco 2G

I think the distinction is based mostly on context.

Gian Boco 2G

It depends on which equation you plug it into because like they said, same value but different units and the relation is based on the light energy equaiton.

Gian Boco 2G

Yes, but for future references, keep in mind that this is for Hydrogen atom only.

Gian Boco 2G

You can find the velocity based because you know the wavelength and mass values. So, isolate v in wavelength = h/mv and plug it into 1/2 mv^2.

Gian Boco 2G

Yep, and it will always equal wavelength * frequency when taken from the EM spectrum

Gian Boco 2G

My understanding is that it is only applicable to photon energy for electromagnetic waves.

Gian Boco 2G

Protons and neutrons have very similar mass. In fact, their masses together form the atomic mass of that element. Electrons however are so much smaller than them that the mass of electrons is neglected when dealing with atomic mass.

Gian Boco 2G

Yep reference sheets for constants are useful! I imagine that our assessments in the future, we will be provided with these needed constants along with our periodic table.

Gian Boco 2G

Like the previous replies said, e is used to mean "10 to the power of." You can even input your own "E" by pressing "2nd" and then "," (comma) on the TI-84 if you have one. This makes it a lot easier to input very low/high values without *10^x and adding all t...

Gian Boco 2G

I think that it's safe to assume that we would be given an image or reference of the spectrum. But, I think it would probably be really helpful to be familiar with some of the ranges such as that of visible light.

Gian Boco 2G

I'm in the same situation, and yes you can access the textbook. It should be on the right-hand side when you open Sapling as an ebook. I had to make an account to use the textbook. Omg how did I not see the giant textbook on the right hand side of the screen I'm so blind HAHA. Thank you so much thi...

Gian Boco 2G

I just googled the molar masses of the molecules ahaha. I feel like in the future we would be given the molar masses of molecules that we aren't familiar with such as this.

Gian Boco 2G

My textbook has yet to arrive along with my sapling code. However, I do have the free trial for sapling. Is it possible to access the book through the free trial of sapling?

Gian Boco 2G

How do we know that the ratio for the problem is 1 to 1 to 1? Are we just assuming this or do we have to figure that out? If so, how do we find the ratio? I assumed that it would be 1:1 because there's no way would be expected to balance it even if it wasn't. Also if you work out the problem using ...

Gian Boco 2G

I googled the molar masses as well hahaha! I think it's clear that having prior understanding of what these organic compounds are is unrealistic for us to already know. It's safe to assume that for future tests or quizzes, we are most likely to be given these molar masses as well other information w...

Gian Boco 2G

I initially thought that the number before the organic compound was the number of moles but I realized that that was part of the name hahaha. Because I don't see how we can be expected to be familiar with these compounds and balance the equation, I think we should be safe to assume that they are all...

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