CHEMISTRY COMMUNITY

Haison Nguyen 1I

I'm not sure about the first part of your question, but lone pairs will repel other lone pairs and bonded electrons so the angle would decrease compared to if there were no lone pairs affecting the bonding angles.

Haison Nguyen 1I

Regions of electron density are the amount of bonds (single, double, and triple all count as one) or lone pairs a central atom has. So if there are 5 regions of electron density for example, it would be sp3d.

Haison Nguyen 1I

For part A we were suppose to write the lewis structure of CCl2H2 and determine if it was nonpolar or polar. Based on the structure, it would make sense that it's polar since both Cls are next to each other and pulling on the electrons of C. But how are we suppose to know the structure of the molecu...

Haison Nguyen 1I

It would be E subscript 3. So assuming there is one bonding pair (X) the formula would be AXE(subscript 3).

Haison Nguyen 1I

A lewis structure just shows us how many valence electrons there are and what kind of bonds are used whereas the VSEPR model gives the actual shape of the valence electrons in a molecule. The shape is determined by the repulsion of the electrons which can create various shapes such as linear, tetrah...

Haison Nguyen 1I

We learned about linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, square planar, and seesaw.

Haison Nguyen 1I

Shared electrons between atoms will have dipole moments, not lone pairs.

Haison Nguyen 1I

I believe hydrogen, helium, lithium, beryllium, boron, and aluminum follow the same exception to the octet rule. Correct me if I am wrong though please.

Haison Nguyen 1I

The trends of electronegativity and electron affinity are the same. They increase to the right and increase when going up the periodic table. Electronegativity measures how strongly an atom of an element attracts bonding electrons and electron affinity is the amount of energy released when an electr...

Haison Nguyen 1I

The ground state configuration of AG is [Kr] 4d^10 5s^1 because of the half shell rule. Since the question is asking for the configuration of AG+, you would take away an electron leaving just [Kr] 4d^10.

Haison Nguyen 1I

Typically atoms will either gain or lose an electron until it form a noble gas configuration. It depends on which is easier to do for the element. Metals will more likely lose electrons and nonmetals will gain electrons.

Haison Nguyen 1I

You look at the group numbers or the rows of each element. However, I skip the d-block so where Boron is at would be group 3. For hydrogen, since it is in group 1, it has one valence electron. For another element like carbon, since it is group 4, it would have 4 valence electrons.

Haison Nguyen 1I

You just have to remember that the elements in d4 and d9 fall under the exceptions. It's because it's better to have all the orbitals half full than one full and one not have any at all.

Haison Nguyen 1I

When there are no electrons, 4s is lower in energy than 3d, but once 4s is filled, 3d becomes lower in energy. At least that is what I think happens, please correct me if I'm wrong!

Haison Nguyen 1I

I'm also slightly confused with this problem. Why does 4f^14 come before 5d^4 and 6s^2?

Haison Nguyen 1I

6f would consist of the lanthanoids and 4d would be the row where yttrium starts and cadmium ends (when n = 5).

Haison Nguyen 1I

Since Argon is a noble gas, the outer shell is filled and you can just start from the next row to get to bromine. So after you write Argon, you would start from the 4th row, but remember to use 3d before 4s. In this case, it would be 3d^10, 4s^2, and 4p^5 for bromine.

Haison Nguyen 1I

Hello can someone explain to me all the different terms and how they relate to each other? The math portion I understand, such as how to find l when you're given n, but the concepts itself confuse me. I know that n = size/shape and l corresponds to the orbitals, but I'm not sure exactly what they ar...

Haison Nguyen 1I

Yes I believe we will be tested on that and everything else up until De Broglie's equation.

Haison Nguyen 1I

I was having trouble with a few problems from Chapter 1 (such as 1.41. and 1.37). Both involved using the mass of either a proton, neutron, or both. Are we suppose to already know the mass of a proton or a neutron? (It wasn't given in the question.)

Haison Nguyen 1I

Equations will be given to us for the tests/exams, but practicing problems is a good way to help you memorize how to use each formula!

Haison Nguyen 1I

Homework usually depends on the day of your discussion, but as of right now you can do homework from Outline 1 or Outline 2 or both. Basically whatever Dr. Lavelle has taught us so far can be done for homework.

Haison Nguyen 1I

The problem itself says to use nm for part A and pm for part B. If the problem didn't tell you what units to use, you should probably use the units you were working with to solve the problem.

Haison Nguyen 1I

Are you sure your answer was 4.86x10^-17 m? My answer was 4.86x10^-7 m and then I converted it to nanometers because the chart itself was in nanometers.

Haison Nguyen 1I

When there is a coefficient in front of a molecule, you would multiply that number by the number of atoms in each element. So for H20, there are 2 hydrogen atoms and one oxygen atom. However, since there is a coefficient of 4 in front of it, you would multiply it by each element and get 8 hydrogen a...

Haison Nguyen 1I

Accuracy is how close your results are to the true value. Precision is how close your results are to each other.

Haison Nguyen 1I

Is the homework for week 2 the same topics we worked on in week 1? (Balancing chemical reactions, limiting reactants, etc.)

Haison Nguyen 1I

Usually the problems will give you the state of the molecules, but is not needed to actually solve the problem. (g) = gas, (s) = solid, (l) = liquid, and (aq) = aqueous

Haison Nguyen 1I

When there's a subscript following the parenthesis, you multiply that number by the number in the parenthesis. So yes Mg(N3)2 would be Mg(N6) and you ignore the MG since the subscript doesn't apply to that element. So since you have 6N on the left side, you try to get 6N on the right side as well.

Haison Nguyen 1I

Hello, you don't divide 0.20 mol by 81 g/mol because that does not lead to the theoretical yield. Instead of dividing, you would multiply 0.20 mols by 81 g/mol because that gives you the amount of grams in 0.20 mols (16.2 g). This should be the theoretical yield.

CHEMISTRY COMMUNITY

Search found 30 matches

Re: Lone pairs

Re: Hybrid orbitals

Question 4.25 Part A

Re: E

Re: VSEPR confusion

Re: shapes

Re: lone pairs

Re: Octet Rules

Re: Electronegativiry vs electron affinity [ENDORSED]

Re: HW problem 3.21

Re: Ionic Bonds

Re: VALENCE ELECTRONS [ENDORSED]

Re: Special Cases [ENDORSED]

Re: 3d before 4s

Re: homework 2.43 part e [ENDORSED]

Re: Explanation to the answer of the question 2.31

Re: orbitals

Energy, subshells, shells???

Re: amplitude and intensity

Homework Problem(s) [ENDORSED]

Re: Memorizing formulas test 2

Re: Homework? [ENDORSED]

Re: HW 1.7 (using nm vs pm)

Re: 1.13 (b)

Re: What do the coefficients/subscripts represent [ENDORSED]

Re: Accuracy vs Precision

Homework for Week 2

Re: How to know the state of the molecules [ENDORSED]

Re: H.5.b [ENDORSED]

Re: Module 3: Limiting Reactant Calculations, Question 26 [ENDORSED]