Search found 36 matches
- Sun Jun 10, 2018 7:58 pm
- Forum: Hybridization
- Topic: pi bond and sigma bond
- Replies: 3
- Views: 484
Re: pi bond and sigma bond
Sigma and pi bonds don't really correlate with s and p, respectively. Sigma bonds form when an s orbital overlaps with another s orbital, an s overlaps with a p, or a p overlaps with another p end-to-end. Pi bonds are formed when additional orbitals overlap side-by-side.
- Tue Jun 05, 2018 9:31 pm
- Forum: Properties & Structures of Inorganic & Organic Bases
- Topic: 12.49 Determining Base Strength
- Replies: 4
- Views: 964
Re: 12.49 Determining Base Strength
I think in general, since strong acids and bases are completely ionized in water, the more soluble an acid/base is in water, the stronger it is. So for this question, BrO- would be the stronger base.
- Tue Jun 05, 2018 9:13 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Ligand binding
- Replies: 4
- Views: 574
Re: Ligand binding
If you're asking about how to tell which part of the ligand binds to the metal, the atom (which is part of the ligand) with a lone pair is the one that's directly bonded to the central metal atom.
- Mon Jun 04, 2018 11:12 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Problem 17.33
- Replies: 2
- Views: 287
Re: Problem 17.33
Since the 2 lone pairs of H2O are on the same O atom, H2O can't bind to 2 different sites on a central metal atom. Therefore, H2O can bind to a maximum of only 1 site on a central metal atom, making it monodentate.
- Sat Jun 02, 2018 11:38 pm
- Forum: Bond Lengths & Energies
- Topic: Test 3 Q 8 [ENDORSED]
- Replies: 9
- Views: 1308
Re: Test 3 Q 8 [ENDORSED]
You can also think about it in terms of resonance. The first structure would have resonance (you can move the double bonds around and get the same structure), which means the double bonds are actually "spread out" throughout all the bonds. So, the actual (observed) bond of the circled bond...
- Wed May 30, 2018 9:50 pm
- Forum: Resonance Structures
- Topic: Resonance Review
- Replies: 3
- Views: 618
Re: Resonance Review
(In the context of the test question) A resonance structure is like a variation to the given structure, but the resonance structure has to be equivalent to the given structure. So, if there's 1 double bond in a structure and the rest of the bonds are single bonds, you can't just move the double bond...
- Wed May 30, 2018 9:29 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 4.13 D
- Replies: 1
- Views: 224
Re: 4.13 D
I don't think you have a lone pair on the middle N and have all 16 electrons accounted for and all the octets. If you put a lone pair in the middle, then either oxygen won't have its octet or the other nitrogen won't have its octet.
- Wed May 23, 2018 10:30 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Ions
- Replies: 4
- Views: 611
Re: Ions
The extra electron isn't really designated to a specific atom. You just have to account for the extra electron when you're adding up all the valence electrons in the molecule and then draw the structure the same way you would draw any other structure. Also, just put brackets around the structure wit...
- Wed May 23, 2018 10:22 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Formal charge with ions
- Replies: 4
- Views: 782
Re: Formal charge with ions
To calculate the formal charges of atoms in a molecule, you use the typical number of valence electrons for that atom. But if you added the formal charges of all the atoms in an ion, they should add up to the total charge of the ion. For example, in CH3(-), C has a formal charge of -1 (FC=4-2-3=-1) ...
- Wed May 23, 2018 10:13 pm
- Forum: Resonance Structures
- Topic: Worksheet 7 #2
- Replies: 1
- Views: 398
Re: Worksheet 7 #2
I drew one structure that had a double bond between the carbon and oxygen (all other bonds are single bonds) and then the other structure had a double bond between the nitrogen and carbon. The first structure would have 1 lone pair on N and the second structure would have a single bond between O and...
- Wed May 16, 2018 6:30 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: 3.49
- Replies: 3
- Views: 554
Re: 3.49
Oh wait is it because N wouldn't have its octet?
- Wed May 16, 2018 6:19 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: 3.49
- Replies: 3
- Views: 554
3.49
For part A, the question asks for the Lewis structure and formal charges of NO+. The back of the book says the correct Lewis structure is N and O attached by a triple bond, where N has 1 lone pair of e- and O has 1 lone pair of e-. So the formal charges would be 0 for N and +1 for O. I understand ho...
- Mon May 14, 2018 10:54 pm
- Forum: Lewis Structures
- Topic: Ionic Lewis Structures
- Replies: 2
- Views: 422
Ionic Lewis Structures
(Just wanted to clarify) When you draw the lewis structure for an ionic bond between a cation and an anion, do you just write the element of the cation with a positive charge next to the element of the anion with a negative charge? And you would draw out the electrons for only the anion but not the ...
- Mon May 14, 2018 10:49 pm
- Forum: Octet Exceptions
- Topic: Radicals [ENDORSED]
- Replies: 6
- Views: 885
Re: Radicals [ENDORSED]
If the total # of electrons is an odd number, then there is a radical.
- Sat May 12, 2018 10:55 am
- Forum: Ionic & Covalent Bonds
- Topic: Main group metals
- Replies: 2
- Views: 405
Re: Main group metals
I think the main group metals are the metals in the s block and p block (metals in groups 1, 2, 13, 14, 15, 16, 17, and 18). For elements like Ga, In, Sn, Tl and Pb, I don't think they have to lose all of their electrons in their d orbital, because their valence electrons reside in only the s and p ...
- Wed May 09, 2018 5:26 pm
- Forum: Trends in The Periodic Table
- Topic: question
- Replies: 6
- Views: 768
Re: question
Electron affinity is how likely an atom will take an electron or how much it "wants" an electron. Electronegativity is how strongly an atom will attract shared electrons to itself (more related to bonding). Both electron affinity and electronegativity show similar trends on the periodic t...
- Wed May 09, 2018 5:22 pm
- Forum: Photoelectric Effect
- Topic: Kinetic Energy
- Replies: 2
- Views: 414
Re: Kinetic Energy
It's because this isn't always true, kinetic energy also depends on what the threshold energy or work function is.
- Wed May 09, 2018 5:09 pm
- Forum: Trends in The Periodic Table
- Topic: question
- Replies: 6
- Views: 768
Re: question
Electron affinity is how likely an atom will take an electron or how much it "wants" an electron. Electronegativity is how strongly an atom will attract shared electrons to itself (more related to bonding). Both electron affinity and electronegativity show similar trends on the periodic ta...
- Sat May 05, 2018 1:50 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration Exceptions
- Replies: 3
- Views: 413
Re: Electron Configuration Exceptions
Oh thanks for letting me know, I didn't know that! So I guess tungsten is the exception to the exception?
From high school chem I know that the rules get more complicated as you move down the periodic table, but I wouldn't worry too much about it since we should really just focus on the first 4 rows.
From high school chem I know that the rules get more complicated as you move down the periodic table, but I wouldn't worry too much about it since we should really just focus on the first 4 rows.
- Sat May 05, 2018 1:41 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Midterm Review Q5 part f [ENDORSED]
- Replies: 2
- Views: 375
Re: Midterm Review Q5 part f [ENDORSED]
If l=1, this means that the possible ml values are 1, 0, and -1. Since there are 3 possible ml values, there are 3 different orbital states in which electrons can exist. Since 1 orbital can hold up to 2 electrons, 3 orbitals can hold up to 6 electrons. Another way to think about it is by looking at ...
- Sat May 05, 2018 1:30 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration Exceptions
- Replies: 3
- Views: 413
Re: Electron Configuration Exceptions
You shouldn't have to memorize any electron configurations, you should just know how to figure out the electron configurations of an element when asked. So you should know how to read the periodic table (s-, p-, and d-blocks) and be able to write down the corresponding e- configuration using the rul...
- Fri May 04, 2018 10:45 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Knowing the Balmer series HW q1.57
- Replies: 3
- Views: 500
Re: Knowing the Balmer series HW q1.57
I figured this problem out by first solving for n2 when the wavelength is 656.3 nm (first line in the Balmer series). It's a fact that the Balmer series starts at n=2, so you know that n1=2. Then using hc/lambda = -(hR/n^2)+(hr/n^2) where lambda=656.3, the first n=2, and the second n is 3. So now yo...
- Thu Apr 26, 2018 5:46 pm
- Forum: DeBroglie Equation
- Topic: Calculating wavelength of an electron
- Replies: 4
- Views: 596
Re: Calculating wavelength of an electron
You could use De Broglie's equation: lambda=h/mV, where h is Planck's constant, m is the mass of the electron (which is a constant that would be given to you or you could just look it up) in kg, and V is the velocity in m/s.
- Thu Apr 26, 2018 5:43 pm
- Forum: Photoelectric Effect
- Topic: Question 1.33, part b
- Replies: 1
- Views: 235
Re: Question 1.33, part b
I think you can just use E=hv to solve for the threshold energy. So E=(6.626*10^-34)(2.50*10^16) should give you the correct answer.
- Thu Apr 26, 2018 5:35 pm
- Forum: DeBroglie Equation
- Topic: 1.33 [ENDORSED]
- Replies: 2
- Views: 426
Re: 1.33 [ENDORSED]
There's a kg "hidden" in h. The units for h are J*s and a Joule is the same as kg*(m^2)*(s^-2). Taking this into account, the kg from h will cancel with the kg from mass, and the other units also cancel until you're left with meters.
- Sat Apr 21, 2018 5:58 pm
- Forum: DeBroglie Equation
- Topic: Homework 1.33 Part C
- Replies: 3
- Views: 375
Re: Homework 1.33 Part C
So kinetic energy is this case would be \frac{1}{2}mV^{2} and we would just plug in the mass of an electron and the given velocity? Also, how would you be able to tell whether or not you should account for kinetic energy? Do we include kinetic E here, because we know that this problem is about the p...
- Thu Apr 19, 2018 5:10 pm
- Forum: DeBroglie Equation
- Topic: HW 1.33
- Replies: 12
- Views: 1210
Re: HW 1.33
Since you're given the velocity of an electron, to do part A, you can use the De Broglie equation lambda = h/mV where m is the mass of an election (you can just look this up, it's 9.11x10^-31) and V is the velocity and then solve for the wavelength.
- Thu Apr 19, 2018 2:16 pm
- Forum: Properties of Light
- Topic: HW 1.27
- Replies: 1
- Views: 271
Re: HW 1.27
32 W is the same as 32 J/s. Since you want to find the amount of photons generated after 2 seconds, multiply 32 J/s by 2 s (seconds cancel out) and then you're left with 64 J. 64 J is the total energy emitted (not sure if emitted is the right word) in 2 seconds. After using E=hc/lambda to find the e...
- Sat Apr 14, 2018 5:06 pm
- Forum: Empirical & Molecular Formulas
- Topic: Empirical and Molecular Formulas [ENDORSED]
- Replies: 4
- Views: 407
Re: Empirical and Molecular Formulas [ENDORSED]
Yup! The molecular formula tells you the actual number of atoms for each element, so you could just count them all up when given a diagram of a molecule. Then, like you said, you can simply the molecular formula to get the empirical formula (unless you can't simplify, then the molecular and empirica...
- Sat Apr 14, 2018 1:45 pm
- Forum: Photoelectric Effect
- Topic: Photons in a light source [ENDORSED]
- Replies: 4
- Views: 225
Re: Photons in a light source [ENDORSED]
E=hv calculates the amount of energy per photon of a light source with that particular frequency, so all photons of that light source would have the same amount of energy. I think the "low frequency light has no photons with sufficient energy" from lecture is referring to the 1 to 1 photon...
- Sat Apr 14, 2018 1:33 pm
- Forum: Properties of Light
- Topic: Units for Wavelengths [ENDORSED]
- Replies: 4
- Views: 340
Re: Units for Wavelengths [ENDORSED]
Yeah, your answer 0.000000522 (or 522 x 10^-9) is in meters and you could just leave it as meters unless the problem specifically asks for a wavelength in nanometers (or any other specific unit). If you want to convert your answer from m to nm, you can move the decimal 9 places to the right or multi...
- Thu Apr 12, 2018 5:30 pm
- Forum: SI Units, Unit Conversions
- Topic: Clarification on unit conversions
- Replies: 4
- Views: 288
Clarification on unit conversions
To convert from meters to nanometers, we just multiply the number of meters by 10^9 right? And then to convert meters to picometers, we just multiply by 10^12?
- Sun Apr 08, 2018 5:43 pm
- Forum: Empirical & Molecular Formulas
- Topic: Alternative way to solve for F.19
- Replies: 3
- Views: 215
- Sat Apr 07, 2018 9:18 pm
- Forum: Empirical & Molecular Formulas
- Topic: Why Divide by Smallest Number? [ENDORSED]
- Replies: 2
- Views: 2332
Re: Why Divide by Smallest Number? [ENDORSED]
When finding the empirical formula, you're ultimate goal is to have all the numbers as integers (no decimals), but when solving for the ratio you usually end up with decimals and it's hard to get rid of all of the decimals in one step. By dividing all of the numbers by the smallest number, there's a...
- Sat Apr 07, 2018 5:16 pm
- Forum: Empirical & Molecular Formulas
- Topic: Alternative way to solve for F.19
- Replies: 3
- Views: 215
Alternative way to solve for F.19
I solved for the molecular formula differently than the other students that replied in the other post about this homework question and I just wanted to make sure that my method of solving was ok. Instead of solving for the empirical formula first and then solving for the molecular formula, I just so...
- Sat Apr 07, 2018 4:53 pm
- Forum: Limiting Reactant Calculations
- Topic: In-Class Limiting Reactant and Theoretical Yield Problem [ENDORSED]
- Replies: 3
- Views: 340
Re: In-Class Limiting Reactant and Theoretical Yield Problem [ENDORSED]
In this particular problem, I think it's kind of confusing to think about the ratio of CaC2 to C2H2 as 1:2 instead of 1:1 because that would make the equation unbalanced. But if you had a different chemical equation where the ratio of the limiting reactant to the product the question referred to was...