Search found 39 matches
- Sun Jun 10, 2018 1:01 am
- Forum: Conjugate Acids & Bases
- Topic: Conjugate Acids/Bases
- Replies: 8
- Views: 1161
Re: Conjugate Acids/Bases
A conjugate base is the leftover after an acid donates a proton. A conjugate acid is what is formed when a base accepts a proton. For example the conjugate base of HCO 3 - is CO 3 - because the acid HCO 3 - lost a proton. So that's how I point them out. I believe you are correct about the water part...
- Sat Jun 09, 2018 12:13 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 12.19
- Replies: 1
- Views: 412
Re: 12.19
Hello, we are trying to find the difference so we should subtract pH final - pH initial.
initial pH = -log (HCl)
final pH = -log(.12HCl)
final - initial = -log(.12HCl) - -log(HCl)
-(log(.12HCl)- log(HCl))
** log law: logA - logB = log (A/B)
-log (.12HCL/HCl)
-log(.12) = .92
initial pH = -log (HCl)
final pH = -log(.12HCl)
final - initial = -log(.12HCl) - -log(HCl)
-(log(.12HCl)- log(HCl))
** log law: logA - logB = log (A/B)
-log (.12HCL/HCl)
-log(.12) = .92
- Wed Jun 06, 2018 7:56 am
- Forum: Hybridization
- Topic: Hybridization
- Replies: 3
- Views: 412
Re: Hybridization
The sum of the subscripts should equal the total number of regions of electron density.
- Wed Jun 06, 2018 12:26 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Wording of Questions Regarding Molecular Shape
- Replies: 2
- Views: 393
Re: Wording of Questions Regarding Molecular Shape
You are correct in saying that an atom with two bonds and one lone pair would be called angular. No matter the wording of the question, we would never call a molecule with 2 bonds and one lone pair trigonal planar. Trigonal planar has 3 bonds while angular has 2 bonds and one lone pair. However, one...
- Mon Jun 04, 2018 10:13 am
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: 4 ligands shape?
- Replies: 6
- Views: 725
Re: 4 ligands shape?
I read in the book that square planar is usually formed when the central metal atom has d8 electron configuration. So that's why Pt2+ and Au3+ have the square planar complex.
- Tue May 29, 2018 8:51 pm
- Forum: Resonance Structures
- Topic: How to determine resonance
- Replies: 2
- Views: 466
Re: How to determine resonance
Molecules have resonance when the atoms stay in the same location but the locations of the lone pairs and the bonding pairs can be changed.
- Tue May 29, 2018 8:48 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Form vs. Shape
- Replies: 2
- Views: 305
Re: Form vs. Shape
I am not familiar with the post you are referring to but I am not sure if there is a formal distinction between form and shape. The person above me is perhaps correct in pointing out that water's bent shape is derived from a tetrahedral shape, but I am not aware of referring to the shape a lone pair...
- Tue May 29, 2018 8:28 pm
- Forum: Bond Lengths & Energies
- Topic: HW problem 6.99
- Replies: 1
- Views: 395
Re: HW problem 6.99
Hello I looked at this problem using Vsepr models. If we look at F - H ~ F (~ represents hydrogen bond) there are no lone pairs on H so we have AX2 and it is therefore linear and the bond angle is 180 degrees. Now looking at H - F ~ H we would have lone pairs. F already has 3 sets of lone pairs but ...
- Sun May 27, 2018 9:28 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Difference in angles
- Replies: 3
- Views: 405
Re: Difference in angles
I think you might have been confused because with certain Vsepr structures, such as trigonal bipyramidal, we can know the bond angles. However, when a lone pare is introduced it can effect the bond angles. For instance if a bond in a trigonal bipyramidal structure were replaced with a lone pair the ...
- Sun May 27, 2018 9:20 pm
- Forum: Ionic & Covalent Bonds
- Topic: Partial charge or full charge
- Replies: 4
- Views: 5570
Re: Partial charge or full charge
I believe fully positive or negative atoms occur in ionic bonds while partially positive or negative atoms occur in covalent bonds.
- Sun May 27, 2018 9:18 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lone pairs on non-central atoms
- Replies: 4
- Views: 774
Re: Lone pairs on non-central atoms
They affect the bond angle because they take up more space and generate stronger repulsion than a regular bond.
- Sun May 20, 2018 2:23 pm
- Forum: Lewis Structures
- Topic: 3.41 c [ENDORSED]
- Replies: 4
- Views: 508
Re: 3.41 c [ENDORSED]
My question for this problem is more related to how to choose which atom is the central atom. At first, I made my structure with NH2 as the center atom, with CH2 and COOH on either side. However, the answer in the textbook has CH2 as the center atom. Why is this? Does it have to do with the formal ...
- Thu May 17, 2018 9:03 pm
- Forum: Lewis Structures
- Topic: Radicals
- Replies: 2
- Views: 356
Re: Radicals
I think radicals only becomes non-radical when it reacts with an antioxidant. However, I believe an antioxidant would neutralize the unpaired electron and it wouldn't be a radical anymore. So I agree with the person above.
- Thu May 17, 2018 8:38 pm
- Forum: Lewis Structures
- Topic: 3.41 c [ENDORSED]
- Replies: 4
- Views: 508
3.41 c [ENDORSED]
Hello, I am confused about the placement of the right most oxygen and hydrogen. Why does the book come to the conclusion it does? (The question is to complete the lewis structure for H2C(NH2)COOH)
Thanks!
- Wed May 16, 2018 9:04 am
- Forum: Ionic & Covalent Bonds
- Topic: Electron affinity [ENDORSED]
- Replies: 6
- Views: 781
Re: Electron affinity [ENDORSED]
Electron affinity is how likely an element is to gain an electron while electron ionization is the energy required to lose an electron. If we look at Fluorine we can see why these two concepts have the same trend. Fluorine has a high electron affinity because it wants to gain an electron to have an ...
- Thu May 10, 2018 10:23 pm
- Forum: Resonance Structures
- Topic: Help on 2.51
- Replies: 3
- Views: 469
Re: Help on 2.51
a. The electron configuration for Bi is [Xe] 4f^14 5d^10 6s^2 6p^3. All of the shell are full except for the p block so we want to look at that. I think it's helpful to draw out the orbitals for this. if you look at the 6p orbital ___ ___ ___ we would draw three parallel arrows (according to Hund's ...
- Thu May 10, 2018 10:15 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: H-Atom
- Replies: 3
- Views: 504
Re: H-Atom
The Rydberg constant was experimentally determined for Hydrogen only.
- Thu May 10, 2018 10:10 pm
- Forum: Ionic & Covalent Bonds
- Topic: Help on 2.45
- Replies: 2
- Views: 272
Re: Help on 2.45
c. Start on you periodic table at helium. Then move to the next row to Lithium to enter the 2s orbital. The 2 p block starts at Boron and the second element in the p block is carbon. The answer is carbon.
d. We were not assigned this part.
d. We were not assigned this part.
- Sun May 06, 2018 9:02 pm
- Forum: Trends in The Periodic Table
- Topic: 2.67
- Replies: 3
- Views: 315
Re: 2.67
Adding onto the previous post, ionization energies increase across a period AND decreases down a group. Ionization energy is the energy require to remove an electron while electron affinity is the energy that is released when an electron is added. More informally, electron affinity was explained to ...
- Thu May 03, 2018 3:11 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: f orbital
- Replies: 1
- Views: 214
Re: f orbital
Lanthanum is part of the d block. The electron configuration would [Xe} 5d^1 6s^2. The electron configuration for Lutetium is [Xe] 4f^14 5d^1 6s^2. I believe the f block has a its own set of exceptions but we do not have to know them. We are only being tested on the first row of the d block.
- Wed May 02, 2018 11:00 am
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Post Module Question #8/9 [ENDORSED]
- Replies: 3
- Views: 972
Re: Post Module Question #8/9 [ENDORSED]
I think the key here is realizing that Hydrogen has 1 electron so if you have a certain amount of electrons you have that amount of hydrogen atoms. I believe thats the why the hydrogen add on is in parentheses, signifying that the answer is the same for j/mol of electron and j/ mol of hydrogen atom....
- Tue May 01, 2018 9:22 am
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Homework 2.55 part c
- Replies: 3
- Views: 433
Re: Homework 2.55 part c
Gianna Graziano 1A wrote:Hi Natalie,
2.55(c) actually has a typo, as stated in Dr. Lavelle's correction to the textbook problems the correct answer is actually (n-1) d^3 ns^2. You were right!
Thank you! I didn’t know such a thing existed
- Mon Apr 30, 2018 3:49 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Homework 2.55 part c
- Replies: 3
- Views: 433
Homework 2.55 part c
2.55 Give the notation for the valence-shell configuration (including the outermost d electrons) of (a) the alkali metals; (b) Group 15 elements; (c) Group 5 transition metals; (d) "coinage" metals (Cu, Ag, Au). Hello, I got parts a,b, and d of this problem correct. However the answer to p...
- Wed Apr 25, 2018 10:18 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Homework problem 2.1 (d) [ENDORSED]
- Replies: 2
- Views: 342
Re: Homework problem 2.1 (d) [ENDORSED]
I answered that the radius increases because when we move from 1s to 2p we are increasing the size of the probability function. The books says this "the average distance of an electron from the nucleus when it occupies any of the orbitals in the shell with n = 2 is greater than when n = 1, and ...
- Wed Apr 25, 2018 12:47 pm
- Forum: Properties of Light
- Topic: Homework Problem(s) [ENDORSED]
- Replies: 6
- Views: 616
Re: Homework Problem(s) [ENDORSED]
I was needing help on question 1.13 part B in which it asks "What is the name given to the spectroscopic series to which this transition belongs?" The answer being 2.8x10^-7 m or 2.8x10^9 nm. Can someone answer me to what series it is and why? For 1.13 part A I got 486 nm. For part B, a w...
- Mon Apr 23, 2018 12:21 pm
- Forum: Properties of Light
- Topic: Homework Problem(s) [ENDORSED]
- Replies: 6
- Views: 616
Re: Homework Problem(s) [ENDORSED]
Those masses would be given to us on a test.
- Mon Apr 16, 2018 10:15 pm
- Forum: Properties of Light
- Topic: HW 1.15
- Replies: 4
- Views: 373
Re: HW 1.15
The book does say that the Lyman series is defined by n1 = 1but why is that? Couldn't we still be in the Lyman series and start at, for instance, n =2 and go to n = 3?
- Mon Apr 16, 2018 9:53 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: #42 In Post-Module Assessment [ENDORSED]
- Replies: 5
- Views: 382
Re: #42 In Post-Module Assessment [ENDORSED]
You're right it should be E4 - Ex. Since, I used a positive energy I was able to switch them. Really, the 7.55 X 10^-20 should be negative since we are losing energy. If you don't switch them is look like this: -7.55 X 10^-20 = E4 - Ex We also can calculate that E4 = (-hr/16) = -1.36 X 10^-19 Using ...
- Mon Apr 16, 2018 8:21 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: #42 In Post-Module Assessment [ENDORSED]
- Replies: 5
- Views: 382
Re: #42 In Post-Module Assessment [ENDORSED]
Update: I figure it out!
I made the mistakes of not converting the frequency to energy and subtracting in the wrong order.
E = hv
E = h (1.14 X 10^14) = 7.55 X 10^-20
7.55 X 10^-20 = Ex - E4
Ex = (-hr)/x^2 = -6.06 X 10^-20
x^2 = 35.9
x = n = 6
I made the mistakes of not converting the frequency to energy and subtracting in the wrong order.
E = hv
E = h (1.14 X 10^14) = 7.55 X 10^-20
7.55 X 10^-20 = Ex - E4
Ex = (-hr)/x^2 = -6.06 X 10^-20
x^2 = 35.9
x = n = 6
- Mon Apr 16, 2018 8:04 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: #42 In Post-Module Assessment [ENDORSED]
- Replies: 5
- Views: 382
Re: #42 In Post-Module Assessment [ENDORSED]
I am also stuck on this one but I do not think you plug in every quantum level. I began by using the formula E n = (-hR/n 2 ). If we consider 1.14 X 10 14 to be the delta E, I think the formula delta E = e final - E initial should work. My progress so far looks like this: 1.14 X 10 14 = E 4 - E x 1....
- Sat Apr 14, 2018 3:38 pm
- Forum: Photoelectric Effect
- Topic: Photons in a light source [ENDORSED]
- Replies: 4
- Views: 221
Re: Photons in a light source [ENDORSED]
I reviewed my notes and cannot find where that was said. are you sure it said photons? you could be mistaking that for electrons. if so no electrons are ejected if the energy of the photon is not greater than or equal to the work function aka the energy needed to eject an electron. so back to the s...
- Sat Apr 14, 2018 3:25 pm
- Forum: Photoelectric Effect
- Topic: E(Photon) greater than/equal to (Energy Remove e-) [ENDORSED]
- Replies: 2
- Views: 158
Re: E(Photon) greater than/equal to (Energy Remove e-) [ENDORSED]
The statement from the book is correct. I think you may have accidentally wrote it down backwards.
If E(photon) < threshold energy then the electron will not eject.
If E(photon) > threshold energy then the electron will eject.
If E(photon) < threshold energy then the electron will not eject.
If E(photon) > threshold energy then the electron will eject.
- Sat Apr 14, 2018 11:23 am
- Forum: Photoelectric Effect
- Topic: Released electrons [ENDORSED]
- Replies: 1
- Views: 175
Released electrons [ENDORSED]
What happens after an electron is released from a metal in the photoelectric experiment. Does it eventually return to the cation? In the case of the threshold energy equaling the energy of the photon, does the electron release and immediately rejoin the cation? I'm trying to picture the experiment i...
- Sat Apr 14, 2018 11:03 am
- Forum: Photoelectric Effect
- Topic: Photons in a light source [ENDORSED]
- Replies: 4
- Views: 221
Photons in a light source [ENDORSED]
Assuming you had some light source, are all of the photons in that light source of the same energy level? The formula E = hv makes me think the answer would be yes but in lecture, one of the slides said "low frequency light has no photons with efficient energy" which makes me think some li...
- Thu Apr 12, 2018 8:57 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Fundamental practice problem L.7 part b
- Replies: 5
- Views: 2299
Re: Fundamental practice problem L.7 part b
I did it like this:
454 g fat (1mol fat/ 891.48 g fat) = .509 mol fat
.509 mol fat (163 mol O2/ 2 mol fat) = 41.507 mol O2
41.507 mol O2 (32 g O2/ 1 mol O2) = 1328.16g O2 = 1.33 x 103 g O2
454 g fat (1mol fat/ 891.48 g fat) = .509 mol fat
.509 mol fat (163 mol O2/ 2 mol fat) = 41.507 mol O2
41.507 mol O2 (32 g O2/ 1 mol O2) = 1328.16g O2 = 1.33 x 103 g O2
- Wed Apr 11, 2018 12:22 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Parent acids
- Replies: 1
- Views: 469
Parent acids
I have not taken chemistry in 3.5 years so I am trying to figure out the correct naming of cations and anions. In section F.D.2 the book has a chart of common anions and their parent acids but doesn't say anything about the parent acids. It appears that the parent acids just have an H added to the a...
- Sun Apr 08, 2018 1:21 pm
- Forum: Limiting Reactant Calculations
- Topic: M.15
- Replies: 5
- Views: 5826
Re: M.15
How do you come up with the equation Al + Cl2 --> AlCl3. I get that Cl is a diatomic molecule so it makes sense that we should use Cl2 but how do we know to put Cl3 in the product?
- Sun Apr 08, 2018 11:00 am
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: E.15 [ENDORSED]
- Replies: 10
- Views: 1565
Re: E.15 [ENDORSED]
This may be a bit obvious but I did not realize that M is not an element. It just stands for metal. I was confused about why people were saying to treat it as a variable but it is literally nothing more than a variable. And a sulfide is a compound with sulfur so it makes sense to combine the molar m...
- Sun Apr 08, 2018 10:52 am
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Knowing molecular formulas
- Replies: 1
- Views: 139
Knowing molecular formulas
For question E.9 in the homework it asks for the formula of magnesium sulfate heptahydrate. However, the question does not provide information (such as number of moles of each element used or percent composition) to find the molecular formula. Should I know from the chemical name what the molecular ...