Search found 30 matches
- Mon Jun 04, 2018 11:57 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lone pairs
- Replies: 5
- Views: 669
Re: Lone pairs
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.
- Mon Jun 04, 2018 11:55 am
- Forum: Hybridization
- Topic: Hybrid orbitals
- Replies: 6
- Views: 785
Re: Hybrid orbitals
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.
- Mon Jun 04, 2018 11:52 am
- Forum: Hybridization
- Topic: Question 4.25 Part A
- Replies: 2
- Views: 364
Question 4.25 Part A
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...
- Mon May 28, 2018 5:54 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: E
- Replies: 3
- Views: 439
Re: E
It would be E subscript 3. So assuming there is one bonding pair (X) the formula would be AXE(subscript 3).
- Mon May 28, 2018 5:52 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: VSEPR confusion
- Replies: 8
- Views: 1319
Re: VSEPR confusion
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...
- Mon May 28, 2018 5:48 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: shapes
- Replies: 3
- Views: 517
Re: shapes
We learned about linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, square planar, and seesaw.
- Mon May 21, 2018 1:40 pm
- Forum: Dipole Moments
- Topic: lone pairs
- Replies: 3
- Views: 471
Re: lone pairs
Shared electrons between atoms will have dipole moments, not lone pairs.
- Mon May 21, 2018 1:39 pm
- Forum: Octet Exceptions
- Topic: Octet Rules
- Replies: 4
- Views: 693
Re: Octet Rules
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.
- Mon May 21, 2018 1:35 pm
- Forum: Electronegativity
- Topic: Electronegativiry vs electron affinity [ENDORSED]
- Replies: 4
- Views: 983
Re: Electronegativiry vs electron affinity [ENDORSED]
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...
- Mon May 14, 2018 12:23 pm
- Forum: Ionic & Covalent Bonds
- Topic: HW problem 3.21
- Replies: 4
- Views: 1371
Re: HW problem 3.21
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.
- Mon May 14, 2018 12:17 pm
- Forum: Ionic & Covalent Bonds
- Topic: Ionic Bonds
- Replies: 4
- Views: 532
Re: Ionic Bonds
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.
- Mon May 14, 2018 12:15 pm
- Forum: Ionic & Covalent Bonds
- Topic: VALENCE ELECTRONS [ENDORSED]
- Replies: 16
- Views: 2152
Re: VALENCE ELECTRONS [ENDORSED]
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.
- Mon May 07, 2018 1:45 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Special Cases [ENDORSED]
- Replies: 3
- Views: 501
Re: Special Cases [ENDORSED]
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.
- Mon May 07, 2018 1:42 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 3d before 4s
- Replies: 9
- Views: 1125
Re: 3d before 4s
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!
- Mon May 07, 2018 1:40 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: homework 2.43 part e [ENDORSED]
- Replies: 8
- Views: 1008
Re: homework 2.43 part e [ENDORSED]
I'm also slightly confused with this problem. Why does 4f^14 come before 5d^4 and 6s^2?
- Mon Apr 30, 2018 5:26 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Explanation to the answer of the question 2.31
- Replies: 4
- Views: 511
Re: Explanation to the answer of the question 2.31
6f would consist of the lanthanoids and 4d would be the row where yttrium starts and cadmium ends (when n = 5).
- Mon Apr 30, 2018 5:22 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: orbitals
- Replies: 3
- Views: 496
Re: orbitals
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.
- Mon Apr 30, 2018 5:10 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Energy, subshells, shells???
- Replies: 1
- Views: 322
Energy, subshells, shells???
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...
- Mon Apr 23, 2018 12:15 pm
- Forum: Properties of Light
- Topic: amplitude and intensity
- Replies: 11
- Views: 1111
Re: amplitude and intensity
Yes I believe we will be tested on that and everything else up until De Broglie's equation.
- Mon Apr 23, 2018 12:12 pm
- Forum: Properties of Light
- Topic: Homework Problem(s) [ENDORSED]
- Replies: 6
- Views: 655
Homework Problem(s) [ENDORSED]
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.)
- Mon Apr 23, 2018 12:08 pm
- Forum: Properties of Light
- Topic: Memorizing formulas test 2
- Replies: 16
- Views: 1773
Re: Memorizing formulas test 2
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!
- Mon Apr 16, 2018 9:33 pm
- Forum: Properties of Light
- Topic: Homework? [ENDORSED]
- Replies: 10
- Views: 1098
Re: Homework? [ENDORSED]
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.
- Mon Apr 16, 2018 9:25 pm
- Forum: Properties of Light
- Topic: HW 1.7 (using nm vs pm)
- Replies: 8
- Views: 769
Re: HW 1.7 (using nm vs pm)
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.
- Mon Apr 16, 2018 9:20 pm
- Forum: Properties of Light
- Topic: 1.13 (b)
- Replies: 6
- Views: 714
Re: 1.13 (b)
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.
- Mon Apr 09, 2018 11:34 am
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: What do the coefficients/subscripts represent [ENDORSED]
- Replies: 7
- Views: 1559
Re: What do the coefficients/subscripts represent [ENDORSED]
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...
- Mon Apr 09, 2018 11:24 am
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Accuracy vs Precision
- Replies: 23
- Views: 2743
Re: Accuracy vs Precision
Accuracy is how close your results are to the true value. Precision is how close your results are to each other.
- Mon Apr 09, 2018 11:12 am
- Forum: Administrative Questions and Class Announcements
- Topic: Homework for Week 2
- Replies: 2
- Views: 250
Homework for Week 2
Is the homework for week 2 the same topics we worked on in week 1? (Balancing chemical reactions, limiting reactants, etc.)
- Fri Apr 06, 2018 5:28 pm
- Forum: Balancing Chemical Reactions
- Topic: How to know the state of the molecules [ENDORSED]
- Replies: 6
- Views: 850
Re: How to know the state of the molecules [ENDORSED]
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
- Fri Apr 06, 2018 5:25 pm
- Forum: Balancing Chemical Reactions
- Topic: H.5.b [ENDORSED]
- Replies: 7
- Views: 919
Re: H.5.b [ENDORSED]
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.
- Fri Apr 06, 2018 5:19 pm
- Forum: Limiting Reactant Calculations
- Topic: Module 3: Limiting Reactant Calculations, Question 26 [ENDORSED]
- Replies: 4
- Views: 677
Re: Module 3: Limiting Reactant Calculations, Question 26 [ENDORSED]
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.