CHEMISTRY COMMUNITY

Annie Tong 2G

This is an acid-base reaction but not a neutralization reaction between there aren't any H+ and OH- ions to form water.

Annie Tong 2G

The partial charges of H2O (+ H and - O) are attracted to the partial charges of the ion (salt). For example, O - is attracted to Na + and H + is attracted to Cl -. These are ion dipole forces -- forces between the partial charge of a polar molecule (water) and the charge of an ion (Na or Cl). Ion d...

Annie Tong 2G

Because hard water contains more mineral ions like Ca2+ and Mg2+, it's more alkaline than soft water.

Annie Tong 2G

I'm not entirely sure but here's my guess: carbon's hybridization would be 2sp ↑ ↑ and 2p ↑ ↑. Nitrogen is 2sp ↑ ↑ and 2p ↑↓ ↑ ↑. The 2sp overlap forms a sigma bond and the two 2p overlaps form two pi bonds. The ↑↓ in nitrogen's 2p orbital is its lone pair.

Annie Tong 2G

I'm assuming the bond angles are based on each nitrogen as a central atom. Lone pair-bonding pair repulsions are stronger than bond pair-bond pair repulsions, so the lone pair on each nitrogen will take up a larger volume. This will cause the bond angles to be smaller than a predicted tetrahedral an...

Annie Tong 2G

Helium walks into a bar,
The bar tender says "We don't serve noble gases in here."
Helium doesn't react.

Annie Tong 2G

En is short for ethylenediamine C2H4(NH2)2, a bidentate molecule. So whenever you see en, it count for a coordination number of 2.

Annie Tong 2G

That would be the oxidation number of the metal. You'd have to figure it out based on the charges of the ligand and any cation/anion that are present. Take [Co(NH3)5(SO4)]+ for example. The overall charge is +1, so that's what you have to end up with. The charge of NH3 is zero and the charge of SO4 ...

Annie Tong 2G

By definition, strong acids are going to be fully/mostly deprotonated in a solution, meaning most of the acid molecules will lose their H+. You can also tell the strength of an acid by looking at the oxygens and hydrogens. If there are more than 2 oxygens for every hydrogen, it's likely that it's a ...

Annie Tong 2G

Lots of studying going on now that finals are approaching :/ I'm really glad that break is also approaching though.
It's nice to have someone ask; how are you?

Annie Tong 2G

Write the formula for the coordination compound:

9C.3 (d) sodium bosoxalato(diaqua)ferrate(III)

The answer is Na[Fe(OH2)2(C2O4)2] but I thought the ligands should be in alphabetical order?

Annie Tong 2G

Polarity depends on the electronegativity difference present between two bonded atoms in a molecule. A dipole will form and the molecule will gain partial positive and negative charges. If a molecule has certain areas with high dipole moments and other areas with low dipole moments, and those moment...

Annie Tong 2G

Localized electrons exhibit normal behavior, as in they remain close to one atom. Resonance hybrids contain some "abnormal" (delocalized) electrons. They will move between certain atoms in a molecule so they're "distributed" over them and can't be identified with only one atom pa...

Annie Tong 2G

Lone pairs (and also radicals) have their own density. However, they only influence the shape of a molecule in electron geometry, not molecular geometry.

Annie Tong 2G

2F.15 "Noting that the bond angle of an sp3 hybridized atom is 109.5 degrees and that of an sp2 hybridized atom is 120 degrees, do you expect the bond angle between two hybrid orbitals to increase or decrease as the s-character of the hybrids is increased?" The answer is that the bond angl...

Annie Tong 2G

For example, NH3 has three single N-H bonds and a lone pair located on the central N atom. Looking at the electron geometry, there are four bonding regions so the molecule is a tetrahedron. However because molecular geometry doesn't consider lone pairs, only the N-H bonds influence the shape. So the...

Annie Tong 2G

An unpaired electron is still an area of electron density although its repulsion strength is greatly reduced. I think we put E as usual but the subscript 1/2 is unnecessary.

Annie Tong 2G

Nitrogen has a lone pair in addition to being bonded with the 2 oxygens. Because the repulsion strength between a lone pair and bonded pair is greater than that between two bonded pairs, the two bonds between N and O will be pushed down to become a bent shape.

Annie Tong 2G

For me I usually watch a video that's not related to the exam to distract myself from nerves right before taking it. Exercising or even just taking a walk around the neighborhood might also help if you've got the time.

Annie Tong 2G

^^I agree with everyone else. Just remember that bond angles are smaller if a lone pair is present because the repulsion strength between lone pairs is greater than that between a lone pair and bonding pair.

Annie Tong 2G

F would have more electrons than O, but F- and O2- have the same number of electrons (check the periodic table & their valence e-!). Because F has more protons than O, the former's effective nuclear charge is higher so F- is less polarizable than O2-.

Annie Tong 2G

One way to check if it's an ionic or covalent bond is through electronegativity. If the electronegativity values of the two atoms differ by more than 2, it has so much ionic character that it's considered an ionic bond. If the difference is less than 1.5, it has so much covalent character that it's ...

Annie Tong 2G

Hi, I think it's a policy that you can't take classes whose finals conflict. That's what I heard from my NSA, but you should check to make sure!

Annie Tong 2G

The structure is drawn as six fluorine atoms surrounding one As atom. Arsenic is capable of forming an expanded octet (not just 5 bonds) because it has passed the third row of the p-block. It's recommended to form 5 bonds with other atoms because it has 5 valence electrons, true, but that's to obtai...

Annie Tong 2G

Can anyone place these forces in order of increasing strength (weakest to strongest)? Ionic & Covalent (even though I know these two aren't intermolecular forces), Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding. I tried se...

Annie Tong 2G

There are many ways to draw a structure and we have to figure out which structure is best. The two ways I go by are checking if it's the simplest structure and its formal charge. Remember, the closer the formal charge is to zero, the more stable that molecule/compound is, and the more likely it is t...

Annie Tong 2G

Like a lot of other people said, writing helps a lot :) I usually take notes from his lectures, the textbooks, and write down little corrections by problems that I got wrong so I could drill it in my head. Practicing a huge amount of problems definitely helped me because I was able to take what I me...

Annie Tong 2G

Both water and salt compounds are polar, with positive and negative charges on opposite sides in the molecule. In water, the hydrogens are more positive while the oxygen is more negative. The positively-charged side of the water molecules are attracted to the negatively-charged chloride ions and the...

Annie Tong 2G

It would make sense if a larger electron cloud has a lower ionization energy because it'd be easier to remove an electron compared to a smaller electron cloud (relatively more "positive"). Not too sure if this applies to the bonded region like it does to atoms and molecules though.

Annie Tong 2G

I'm not super sure about this, so this is just what I think :) Electrons aren't actually "spinning" around, because as far as we know, electrons are infinitely small. There's not really anything to spin. But electrons do behave like they're "spinning" in experiments. They have &q...

Annie Tong 2G

If the problem says the atom is "excited", that means it's going to a higher energy level. So the change in energy for the atom is positive because it's absorbed. Because you already know it is part of the Lyman series, the initial energy level is n =1. You can find the frequency from the ...

Annie Tong 2G

The f-block has angular momentum number l = 3 (corresponds to f). The magnetic quantum number m sub l (which specifies the orbitals of the subshell) is equal to 2l + 1, so m sub l = 7. Each orbital has 2 electrons, so the f-block has a total of 14 electrons.

Annie Tong 2G

Do you want to send the question number and the actual question? I did all the Sapling hw so I might be able to walk you through it.

Annie Tong 2G

If n = 7, l = 2, m sub l = -1, the 7 tells us the energy level, l corresponds to the subshells (0 is s, 1 is p, 2 is d, 3 is f...), m sub l specifies the orbital of the subshell. Because m sub l = -1, we know it is talking about a specific orbital. Each orbital holds 2 electrons, so the answer is 2 ...

Annie Tong 2G

b) Use the equation hv = work function + 0 (KE). Substitute in v = 2.50 x 10^16 Hz to get work function = 1.66 x 10^-17. c) Use the equation E = 1/2mv^2 + work function. Use the answer from part b) to substitute work function and v (velocity) = 3.6 x 10^6 m/s (make sure to convert km to m!) and m = ...

Annie Tong 2G

I don't think Sapling is fixed yet because some people are still saying their questions are in a really random order. The midterm will cover everything except the last 8 bullet points in Outline 2, so quantum numbers and orbitals won't be on the exam.

Annie Tong 2G

For the energy level n, the d-orbitals that belong to the (n−1) energy level are lower in energy than the s and p orbitals that belong to the n energy level.This means that you will have to switch the 3d orbitals with the 4s orbital to get 3d10 4s2 4p6 and 4d10 5s1. The 4d orbital is filled complete...

Annie Tong 2G

Each magnetic quantum number refers to a specific orbital of a subshell. So unless it's specified, we don't know which quantum number it is.

Annie Tong 2G

Hi! (So first this is question #5 by the way :) )

I found the energy of the photon by settling E = hc/λ and got 2.486 x 10^-20 J so I think your numbers were a bit off. Then you can divide 37.79 J by 2.486 x 10^-20 J to get the number of photons needed.

Annie Tong 2G

The original Rydberg equation is frequency v = R(1/n1^2 - 1/n2^2). The 1/λ is from when they substituted the λv = c equation into it. So basically they first set v = c/λ = R(1/n1^2 - 1/n2^2), which is c/λ = R(1/n1^2 - 1/n2^2), then divided both sides by c to get 1/λ = (1.0974 x 10^7 m^-1)(1/n1^2 - 1...

Annie Tong 2G

I used the formulas Kinetic energy = 1/2mv^2. The mass of an electron is given in the Fundamentals section of the book, 9.109 x 10^-31 kg. The velocity is given already, so you'd plug in everything to get Kinetic Energy = 1/2(9.109 x 10^-31 kg)(6.61 x 10^5 m/s)^2 = 1.99 x 10^-19 J.

Hope this helped!

Annie Tong 2G

Yep! If you see 2.222e23, that means it's 2.222 x 10^23. If it says 2.222e-23, that equals 2.222 x 10^-23.

Annie Tong 2G

n=1, n=2, n=3, n=4... and so on are the energy levels that an electron has, so you can't have anything in between. I think you're a bit confused about the energy of the incoming photon and electron, which are two different things. Between each electron energy level (n=...) there a specific amount of...

Annie Tong 2G

Kg, meters m, and and seconds s are all SI units so answers typically use them if the problem doesn't specify. Converting units also depends on the given formula and constants, so make sure to check those before you start solving anything!

Annie Tong 2G

I think you count the number of energy levels in between like you said. So it'd be 7-6, 7-5, 7-4, 7-3, 7-2, 7-1, one spectral line for each "level jump" for a total of 6.

Annie Tong 2G

Threshold energy is basically the energy per photon that is needed to remove an electron from a substance. For example, when a metal is illuminated by UV radiation, electrons are ejected ONLY when the frequency of the radiation is high enough (aka above a certain "threshold"). Because the ...

Annie Tong 2G

There's also answers for odd-numbered exercises at the back of the textbook.

Annie Tong 2G

For E23A you have 3.00 g of CuBr2 which equals 0.0134 mol of CuBr2 if you divide by its molar mass. Since Cu 2+ ions are in a 1:1 ratio with CuBr2 (it would be 2 mole Br : 1 mole CuBr2 for bromide ions), you'd multiply 0.0134 mol by (1 mol Cu 2+ / 1 mol CuBr2) and get 0.0134 mol Cu 2+ ions. For E23C...

Annie Tong 2G

Exactly what Cassidy and Anil said. I usually go by the rules of 1. what they ask for or 2. what seems reasonable given the context of the problem and the answer itself or 3. use SI units (which is probably the safest bet if the problem didn't specify the units).

Annie Tong 2G

Thanks guys! I got the answers to both :)

Annie Tong 2G

Here's the problem: "The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 x 10^3 km/s. (a) What is the wavelength of the ejected electron? (b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50 x 10^16 Hz. H...

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