Search found 32 matches
- Sat Dec 08, 2018 5:10 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: Strengths of H2S vs H2Se
- Replies: 1
- Views: 531
Strengths of H2S vs H2Se
In the final review session, it was said that H2Se has a greater boiling point (and therefore stronger bonds) because it has greater London dispersion forces than H2S does. However, in the chemistry textbook version 6 p.484, H2Se is expected to be a stronger acid than H2S because H2Se has weaker bon...
- Sat Dec 08, 2018 5:09 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: Strengths of H2S vs H2Se
- Replies: 4
- Views: 9189
Strengths of H2S vs H2Se
In the final review session, Lyndon said that H2Se has a greater boiling point (and therefore stronger bonds) because it has greater London dispersion forces than H2S does. However, in the chemistry textbook version 6 p.484, H2Se is expected to be a stronger acid than H2S because H2Se has weaker bon...
- Wed Dec 05, 2018 11:34 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: AX2E2
- Replies: 14
- Views: 4617
Re: AX2E2
A molecule with a VSEPR formula of AX2E2 would have a bent or v-shaped structure of atoms. A molecule with 4 things coming off of it (tetrahedral) would have bond angles of 109.5, but since two of these are lone pairs of electrons, the electrons will push other atoms away from it, lowering the bond ...
Re: Cyanide
Cyano and cyanido are the same when naming coordination compounds. In the "ligand names in coordination compounds" handout on Dr. Lavelle's website, it says cyanido is just the new IUPAC name convention.
- Wed Dec 05, 2018 11:22 am
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: ion-ion interactions
- Replies: 1
- Views: 1026
Re: ion-ion interactions
Ion-ion interactions are what take place during ionic bonds. The ion-ion interaction is what causes an ionic bond to occur. For example, Na+ and Cl- share ion-ion interactions and form an ionic bond.
- Wed Dec 05, 2018 11:19 am
- Forum: Sigma & Pi Bonds
- Topic: Identifying sigma & pi bonds
- Replies: 4
- Views: 917
Re: Identifying sigma & pi bonds
single bond - one sigma bond
double bond - one sigma, one pi bond
triple bond - one sigma, two bi bonds
I don't think we'd need to know how to draw it, but just need to be able to identify it.
double bond - one sigma, one pi bond
triple bond - one sigma, two bi bonds
I don't think we'd need to know how to draw it, but just need to be able to identify it.
- Wed Nov 28, 2018 1:24 pm
- Forum: Dipole Moments
- Topic: Lone pairs when determining hybridization
- Replies: 3
- Views: 335
Lone pairs when determining hybridization
When there are two lone pairs present on an atom, why are the lone pairs drawn opposite of eachother but when determining hybridization of the molecule the lone pairs are counted as just one region of electron density? Are two lone pairs on the central atom actually counted as only one region or two?
- Wed Nov 28, 2018 1:17 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond Angles
- Replies: 5
- Views: 621
Re: Bond Angles
There is no chart, but it helps me to just remember the main numbers and their possible variations
Linear = 180
Trigonal Planar = 120
Tetrahedral = 109.5
Trigonal Bipyramidal = 90 and 120
Octahedral = 90
and if lone pairs are present on the molecule, the angles will be less.
Linear = 180
Trigonal Planar = 120
Tetrahedral = 109.5
Trigonal Bipyramidal = 90 and 120
Octahedral = 90
and if lone pairs are present on the molecule, the angles will be less.
- Wed Nov 28, 2018 10:48 am
- Forum: Hybridization
- Topic: Drawing Hybrid Orbitals
- Replies: 3
- Views: 554
Re: Drawing Hybrid Orbitals
I don't think we'll need to know how to draw the energy shells of the hybrid orbitals. If anything, we might need to know general shape of the bonds that can arise from hybrid orbitals.
- Tue Nov 27, 2018 10:14 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: AX3E2
- Replies: 2
- Views: 296
Re: AX3E2
Equatorial lone pairs of electrons are more favorable than axial lone pairs because in equatorial pairs, the lone pairs are only near 2 atoms whereas in the axial position, the lone pairs would be near 3 atoms. The lewis structure with the least amount of atoms being pushed away is the lewis structu...
- Tue Nov 27, 2018 10:10 pm
- Forum: Dipole Moments
- Topic: Dipole
- Replies: 2
- Views: 283
Re: Dipole
One way to tell if a molecule has dipole movement is if a molecule has lone pairs that do not cancel out. If the lone pairs on a molecule are symmetrical with one another, they will cancel out and not have dipole movement. However if a molecule has several lone pairs that do not cancel out, there wi...
- Sun Nov 25, 2018 11:15 pm
- Forum: Hybridization
- Topic: sigma/pi bonds
- Replies: 7
- Views: 714
Re: sigma/pi bonds
Atoms in sigma bonds can rotate and also overlap, whereas atoms in pi bonds can not rotate around the internuclear axis and are only end to end.
- Tue Nov 20, 2018 5:59 pm
- Forum: Bond Lengths & Energies
- Topic: H bonding
- Replies: 14
- Views: 1544
Re: H bonding
The presence of hydrogen bonds in molecules increases both melting and boiling points because greater intermolecular forces in the molecule will require more energy to be broken apart.
- Tue Nov 20, 2018 5:56 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lewis structure of CNO- with regards to formal charge
- Replies: 2
- Views: 3161
Lewis structure of CNO- with regards to formal charge
When drawing the Lewis structure of the CNO- (cyanate) ion why would there be a triple bond between N and C and a single bond between C and O instead of double bonds between both N, C and O? In both structures, the formal charges add to -1 but on the first structure mentioned, the -1 charge would be...
- Mon Nov 12, 2018 12:00 pm
- Forum: Ionic & Covalent Bonds
- Topic: Homework question 6.1
- Replies: 1
- Views: 293
Re: Homework question 6.1
Yes, a hydrogen bond is a dipole-dipole force because it is an attraction between a slightly positive hydrogen on one molecule and a slightly negative atom on another molecule, so you should consider it a dipole-dipole force.
- Tue Nov 06, 2018 11:41 pm
- Forum: Ionic & Covalent Bonds
- Topic: Covalent bonds
- Replies: 3
- Views: 420
Re: Covalent bonds
Yes, whenever two atoms of the same type are bonded together, it is covalent bonding. The type of bond will also depend on the group of the atom. For example, two hydrogen atoms will form one single bond to form H2, while two oxygen atoms will form one double bond between themselves to form O2, and ...
- Tue Nov 06, 2018 11:36 pm
- Forum: Bond Lengths & Energies
- Topic: Bond Lengths
- Replies: 6
- Views: 618
Re: Bond Lengths
The main things you need to know are that single bonds are longer than double bonds and double bonds are longer than triple bonds. For resonance structures, bond length will be an average of the possible bonds within the resonance structures of the molecule.
- Tue Nov 06, 2018 11:33 pm
- Forum: Resonance Structures
- Topic: Resonance Hybrids
- Replies: 3
- Views: 406
Re: Resonance Hybrids
I believe the octet rule has to be filled in all molecules unless it is an extended octet. And the central atom of a Lewis structure must have a complete octet. If it does not have enough bonds to fill the octet, that is when you draw lone pairs.
- Tue Nov 06, 2018 11:30 pm
- Forum: Lewis Structures
- Topic: Identifying the element
- Replies: 8
- Views: 956
Re: Identifying the element
Formal charge of the unknown atom, total electron numbers and how the unknown ion would add to that would be acceptable answers to this question I believe.
- Tue Nov 06, 2018 11:22 pm
- Forum: Lewis Structures
- Topic: Central Atom
- Replies: 3
- Views: 381
Re: Central Atom
Since the trend for ionization energy, electron affinity and electronegativity are the same, you can use any (least ionization energy, greatest electron affinity and greatest electronegativity) to determine the central atom in Lewis structures.
- Tue Oct 23, 2018 6:21 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Uncertainty in Kinetic Energy
- Replies: 1
- Views: 452
Uncertainty in Kinetic Energy
Use the above uncertainty in velocity to calculate the electron's uncertainty in kinetic energy. Then calculate the uncertainty in kinetic energy per mole of electrons (that is, per mole of hydrogen atoms). Comment on your value. I calculated the uncertainty in velocity as 2.3 x 105 m.s-1 How would ...
- Mon Oct 22, 2018 7:58 pm
- Forum: Properties of Electrons
- Topic: Atomic Spectra
- Replies: 1
- Views: 320
Atomic Spectra
The meter was defined in 1963 as 1,650,763.73 wavelengths of radiation emitted by krypton-86 (it has since been redefined). What is the wavelength of this krypton-86 radiation? To what region of the electromagnetic spectrum does this wavelength correspond (i.e. infrared, ultraviolet, x-ray, etc.)? W...
- Sun Oct 21, 2018 2:15 am
- Forum: Photoelectric Effect
- Topic: spectral lines
- Replies: 6
- Views: 635
Re: spectral lines
In what type of problems will we need to use this equation and does anyone know if we will be tested on this on the 2nd test?
- Sun Oct 21, 2018 2:00 am
- Forum: Einstein Equation
- Topic: Numbers to memorize [ENDORSED]
- Replies: 37
- Views: 4256
Re: Numbers to memorize [ENDORSED]
While these numbers will be given on a formula sheet, I have learned that learning the numbers and when to use them gives me a greater understanding of the concept. Plus, by the time I know the numbers, I have become familiar with the process used in the types of problems.
- Sun Oct 21, 2018 1:56 am
- Forum: Photoelectric Effect
- Topic: units
- Replies: 12
- Views: 1147
Re: units
A helpful tip to also know is that kg.m^2.s^-2 is equal to 1 Joule. You can use this when using the Ek = 1/2mv^2 equation. Example 1.5 in the Chemistry Principles Textbook version 6 explains this well.
- Sun Oct 14, 2018 6:44 pm
- Forum: Photoelectric Effect
- Topic: Sodium metal surface
- Replies: 4
- Views: 513
Re: Sodium metal surface
For the work function, I got 1.506 x 10^5 J, but on the post assessment survey for photoelectric effect, the answer showed up as wrong.
Would the frequency be (1.99 x 10^-19)/(6.626 x 10^-34) = 3.01 x 10^-14 Hz?
Would the frequency be (1.99 x 10^-19)/(6.626 x 10^-34) = 3.01 x 10^-14 Hz?
- Sun Oct 14, 2018 6:09 pm
- Forum: Photoelectric Effect
- Topic: Sodium metal surface
- Replies: 4
- Views: 513
Sodium metal surface
Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1. I got the KE of the ejected electron as 1.99 x 10^-19. How much energy is required to remove an electron from one sodium atom? and What is the frequency of ...
- Sun Oct 14, 2018 5:56 pm
- Forum: Photoelectric Effect
- Topic: Maximum possible energy of emitted electrons
- Replies: 1
- Views: 232
Maximum possible energy of emitted electrons
Molybdenum metal must absorb radiation with a minimum frequency of 1.09 x 1015 s-1 before it can emit an electron from its surface. The minimum energy needed for this is 7.22 x 10^-19. If molybdenum is irradiated with 194 nm light, what is the maximum possible kinetic energy of the emitted electrons...
- Sun Oct 14, 2018 5:50 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric effect
- Replies: 5
- Views: 452
Photoelectric effect
In photoelectric experiments, typically what part of the electromagnetic spectrum is the incoming light?
I thought the incoming light was visible light. Could someone explain why this is not the case/what part of the EM the incoming light is from?
I thought the incoming light was visible light. Could someone explain why this is not the case/what part of the EM the incoming light is from?
- Fri Oct 05, 2018 12:01 pm
- Forum: Significant Figures
- Topic: Scientific Notation
- Replies: 7
- Views: 423
Re: Scientific Notation
Use scientific notation in order to end the problem with the correct number of sigfigs.
- Fri Oct 05, 2018 11:59 am
- Forum: Significant Figures
- Topic: Number of Sig Fig
- Replies: 8
- Views: 635
Re: Number of Sig Fig
When calculating throughout the problem, only round to the proper amount of sigfigs at the very end of the problem, otherwise your answer will be significantly off than what it should be.
- Fri Oct 05, 2018 11:52 am
- Forum: Empirical & Molecular Formulas
- Topic: When finding empirical formula
- Replies: 4
- Views: 495
Re: When finding empirical formula
Yes, usually they give you the percent composition, but if it is not given, divide the mass given by the mass of the total compound to find percent composition.