Search found 24 matches

by Ramya Natarajan 1D
Tue Dec 05, 2017 10:06 pm
Forum: Bronsted Acids & Bases
Topic: 12.39
Replies: 2
Views: 179

Re: 12.39

For parts c and d, the molecules are actually +NH3OH and (CH3)2NH2+, so how would we find these values in the table that the problem tells us to look at? Are we supposed to set up some sort of equation from the conjugate bases that we do see in the table?
by Ramya Natarajan 1D
Tue Dec 05, 2017 1:32 pm
Forum: Amphoteric Compounds
Topic: Amphoteric and amphiprotic compounds
Replies: 4
Views: 259

Re: Amphoteric and amphiprotic compounds

To make sure you don't get them confused, I'd just look at the suffixes of each. In Amphiprotic molecules, the protons are able to be transferred and taken away to both a conjugate acid and base. An amphoteric molecule is simply able to react with both an acid and a base, but isn't necessarily one u...
by Ramya Natarajan 1D
Tue Dec 05, 2017 1:26 pm
Forum: Conjugate Acids & Bases
Topic: strong and weak acids and bases
Replies: 2
Views: 152

Re: strong and weak acids and bases

It can apply to any acid-base reaction, you're basically just looking for what gave and took away the proton in the same molecule.
by Ramya Natarajan 1D
Wed Nov 29, 2017 12:24 am
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Help understanding polydentate
Replies: 4
Views: 288

Re: Help understanding polydentate

you're right--the way to determine whether an atom is polydenate or not is to draw out the lewis structure and see whether multiple atoms have lone pairs available for bonding. These long pairs must be on different atoms, and they must make geometrical sense. For example, in the example we have in t...
by Ramya Natarajan 1D
Wed Nov 29, 2017 12:20 am
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Chelates
Replies: 2
Views: 161

Re: Chelates

So if you have two or more ligands that are bidentate, or can attach to more than one site, and these ligands are next to each other, then another central atom can create a bond with these two to form a chelate. A chelate will create a ring with the original coordinate compound. I recommend looking ...
by Ramya Natarajan 1D
Tue Nov 21, 2017 12:46 am
Forum: Determining Molecular Shape (VSEPR)
Topic: seesaw
Replies: 2
Views: 125

Re: seesaw

Additionally, remember that repulsion ranks from bond-bond<bond-lone<lone pair
by Ramya Natarajan 1D
Mon Nov 20, 2017 10:15 pm
Forum: Hybridization
Topic: Homework question #43
Replies: 2
Views: 160

Re: Homework question #43

Can you explain what they mean by "s-character" though? Besides equal electron density, what are the characteristics of a hybridized s bond? Would whether it increases or decreases the bond just depend on the ratio of s:p orbitals?
by Ramya Natarajan 1D
Mon Nov 13, 2017 11:16 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Sigma vs. Pi Bonds
Replies: 5
Views: 254

Re: Sigma vs. Pi Bonds

The way I think of it is that sigma bonds have electrons that can interact completely because they're directly overlapped, so all electrons are equally distributed and consequently can rotate. Pi bonds are more "stiff" because they can only interact peripherally since the electron repulsio...
by Ramya Natarajan 1D
Mon Nov 13, 2017 11:11 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: CO2 vs. H2O
Replies: 6
Views: 563

Re: CO2 vs. H2O

Keep in mind that in CO2, Oxygen is more electronegative, so the dipole moments are facing opposite directions (away from the central atom of Carbon), making the molecule polar. In H2O, Oxygen is more electronegative, so the dipole moments would face inwards towards oxygen and consequently cancel ou...
by Ramya Natarajan 1D
Wed Nov 08, 2017 12:01 am
Forum: Lewis Structures
Topic: 11/5 ch3 review
Replies: 2
Views: 171

Re: 11/5 ch3 review

A good thing to remember is that when you have extra Hydrogens or when the molecular formula is expanded like that (as it is in most organic formulas), hydrogen will often not be bonded to a central atom but one of the peripheral atoms like Oxygen.
by Ramya Natarajan 1D
Tue Nov 07, 2017 11:12 pm
Forum: Lewis Structures
Topic: 3.67B and Radicals
Replies: 2
Views: 147

3.67B and Radicals

In each of these compounds, an atom violates the octet rule. Identify the atom and explain the deviation from the octet rule: b) ClO2 I understand that Cl is the atom violating the octet rule, but I'm confused because there's an odd number of atoms. What would this cause the Lewis structure to look ...
by Ramya Natarajan 1D
Mon Oct 30, 2017 11:17 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Valence-shell configuration [ENDORSED]
Replies: 3
Views: 1656

Re: Valence-shell configuration [ENDORSED]

It's actually just the wording throwing you off! When they say group 5 transition metals, they mean the fifth group of transition metals, so you'd look at the group for Mn, not V (I made the same mistake). Therefore, the configuration would be d^5
by Ramya Natarajan 1D
Mon Oct 30, 2017 11:14 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Periodic Table differences
Replies: 3
Views: 197

Re: Periodic Table differences

I was confused on this also when I had to write the electron configuration for elements farther down periods in the d block. For reference, I recommend counting group number to determine what position the element is in instead of the actual period, because we don't know if there's a gap where the la...
by Ramya Natarajan 1D
Mon Oct 30, 2017 11:09 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Ions
Replies: 2
Views: 189

Re: Ions

I'm fairly sure that at this point of the class, the molecular formula would be given to us like it was during the first two tests since we haven't yet been taught nomenclature, or at least that's what Professor Lavelle said during the first week. Hope this helps!
by Ramya Natarajan 1D
Tue Oct 24, 2017 6:41 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Sequen of Orbitals
Replies: 4
Views: 230

Re: Sequen of Orbitals

I'm fairly sure it's just a different way of ordering it but with the same logic. So yes, the 4s comes first, but in our final answer we rearrange the sequence in numerically ascending order?
by Ramya Natarajan 1D
Tue Oct 24, 2017 6:35 pm
Forum: Trends in The Periodic Table
Topic: Size of Atomic Radii
Replies: 3
Views: 244

Re: Size of Atomic Radii

The electron interactions are also very important. Down a period, more and more shells of electrons effectively shield the outer electrons from the nucleus, which means those negative electrons are being less attracted by the protons' positive charge and repelled by the negative charge of the other ...
by Ramya Natarajan 1D
Tue Oct 24, 2017 6:29 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Test 3 [ENDORSED]
Replies: 7
Views: 444

Re: Test 3 [ENDORSED]

Test 3 covers whatever was left off from Chapter 1 (So from 1.6 onwards) and all of Chapter 2!
by Ramya Natarajan 1D
Mon Oct 16, 2017 10:16 pm
Forum: DeBroglie Equation
Topic: Fig 1.19
Replies: 2
Views: 215

Re: Fig 1.19

Hi! So this picture is an illustration of the dual slit experiment, which proved the duality of light as both a wave and particle (there are really good videos on youtube for it!). Picture a box with two tiny slits in it. Light is entering from both slits, but if light is a particle, you only expect...
by Ramya Natarajan 1D
Mon Oct 16, 2017 10:10 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Rydberg Equation! [ENDORSED]
Replies: 8
Views: 643

Rydberg Equation! [ENDORSED]

Hi, so I'm really confused with the Rydberg equation. In the book, the official equation is v = R(1/n^2 - 1/n^2), but when we derived it in class, a negative was applied to the equation (-R). I know we use E = -Rv/n on the actual equations sheet, but I'm still confused about the negatives as it's me...
by Ramya Natarajan 1D
Mon Oct 09, 2017 11:55 pm
Forum: *Black Body Radiation
Topic: Stefan-Boltzmann Law
Replies: 3
Views: 392

Stefan-Boltzmann Law

Hi all!
I'm confused on the significance of the Boltzmann Law. I understand that it states that the intensity is quantized since it is in linear proportion of a constant, but how is intensity even measured? I know it's amplitude squared, but it doesn't relate to energy, so what's it's significance?
by Ramya Natarajan 1D
Mon Oct 09, 2017 11:51 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Balmer and Rydberg series
Replies: 3
Views: 180

Balmer and Rydberg series

Hello all!
How exactly do the Balmer and Rydberg series differ from each other? Is the Balmer series only proven for H atoms, and if so, when would we even use it? Also, what does "n" signify in both these equations, spectral lines? Would "n" always be given in that case?
by Ramya Natarajan 1D
Thu Oct 05, 2017 4:08 pm
Forum: Molarity, Solutions, Dilutions
Topic: G25
Replies: 4
Views: 329

Re: G25

Hi! There are actually multiple other ways to do this problem. So the first way is to do it manually, which is what we did in a TA office hour section I attended. Obviously you wouldn't do all 90 doubles manually, but after the first 2 or 3 you start to see that each subsequent double decreases by h...
by Ramya Natarajan 1D
Mon Oct 02, 2017 9:37 pm
Forum: Empirical & Molecular Formulas
Topic: Self-Test M.4A (p.F114)
Replies: 3
Views: 362

Re: Self-Test M.4A (p.F114)

Hi there! I think I actually had the same problem, but the difference should be very insignificant, in most cases less than 0.05 I'd guess. Also, I seem to notice that in all these problems, we need to round up. Can anyone confirm?
by Ramya Natarajan 1D
Mon Oct 02, 2017 9:28 pm
Forum: Molarity, Solutions, Dilutions
Topic: G25 homework problem
Replies: 7
Views: 457

Re: G25 homework problem

Hey there! You know how many molecules you have, so that's a great start, and you should multiply that by (1/2)^90 because you doubled the number you had all the way to 90 dilutions: 6.022e20 (1/2)^90 = 0.000000048 molecules, or essentially, so insignificant that there can be said to be no molecules...

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