CHEMISTRY COMMUNITY

Haochen He 3L

Hi! I am not sure how to determine and compare the acidic strength between H3PO3 and H3PO4.
Could someone explain it?

Haochen He 3L

Hi! I think in this question, C6H5NH3+ is a weak acid. So the equation is C6H5NH3+ + H2O --> C6H5NH2 + H3O+ Totally agree with Victor. The +1 charge is a hint for the weak acid.

Haochen He 3L

Hi! I think not every atom with lone pairs could accept protons. But N atom with three bonds usually can accept one proton. So in this case, 2 protons can be accepted by this base.

Haochen He 3L

Hi! All acids have conjugate bases. Strong acids usually have weak conjugate bases. Weak acids, on the other hand, have strong conjugate bases.

Haochen He 3L

Hi! To get the answer, you need to compare the pH and pKa. If the pKa is lower than pH, which means the HA is more acidic than the solution, HA will be ionized into H+ and A-. Thus this acid is negatively charged. Otherwise, (pKa is higher than pH), it'll remain neutral.

Haochen He 3L

Hi! en is shorthand for ethylenediamine. The formula of this ligand is NH2CH2CH2NH2. Hope it helps.

Haochen He 3L

Hi! In general, the oxidation state is the charge of that atom.

Haochen He 3L

Hi! I think as long as you put the transition metal at the beginning, you should be fine. The order of ligands is not so important.

Haochen He 3L

Hi! I think only the names are ordered alphabetically not including the prefixes.

Haochen He 3L

Hi! There are two Cl in the complex ion, so it should be dichloro which has -2 negetive charges. In addition there is a cl (-1) atom outside. Given that the complex compound is neutral, Co should have +3 charges, written as Cobalt (III).

Haochen He 3L

I believe there are hybridizations beyond sp3d2. However, I dont think we are going to use them, at least not in 14A lol.

Haochen He 3L

Yes! Every bond and lone pair is considered as one region of electron density.

Haochen He 3L

Hi! I think ethene has sp2 orbitals instead of sp3 orbitals. C has 3 electrons in 2sp2 orbitals and 1 electron in 2p electron. Hope it helps!

Haochen He 3L

According to Dr. Lavelle (in his latest lecture), both are correct. :)

Haochen He 3L

Both bonds and lone pairs are considered as regions of electron density and thus determine the hybridization. One bond (single, double, or trible) counts as one region of electron density. So does one lone pair. I dont think there is any exception.

Haochen He 3L

Yes. One lone pair is considered as one region of electron density. Hope it helps :)

Haochen He 3L

Yes, I think we should always consider the most stable structure when determining the shapes of molecules.

Haochen He 3L

I think your answer is correct. I'm not sure why it gives another answer instead.

Haochen He 3L

I think the bond angle depends on the exact atoms in the molecules. We can get the actual values through experiments data.

Haochen He 3L

Yes. I think it can tell you the shape of molecules since it represents the number of bonded atoms and lone pairs.

Haochen He 3L

I think MinjooPark_3F has explained very clearly. Basically, you should memorize those oxidation numbers.

Haochen He 3L

The formal charge of C should be -2. FC= valence electrons (4) - (long pairs (4) + sharing electrons/2 (2))

Haochen He 3L

I think the reason is that in resonance structures, electrons move around more (delocalized electrons), which requires less energy to stablize the structure

Haochen He 3L

Hi! I think you are right about getting each atom equal or close to 0 formal charge.

Haochen He 3L

Hi, I don't think we need to calculate the actual bond length. As long as you know some characters like the longer length the weaker strength, you should be fine.

Haochen He 3L

It would be -1 then :)

Haochen He 3L

To calculate the formal charge of each atom, you just need to use the formula formal charge = valence electrons - (shared electrons/2 + lone pairs)

Haochen He 3L

Hi!
I agree with what others have discussed. The answer is incorrect for having 26 electrons instead of 28 electrons.

Haochen He 3L

How do you know when it's best to use a double bond as opposed to using a single bond and a lone pair? I think it depends on the formal charge. It is better for atom to have 0 formal charge because it's more stable. So if a double bond can make the atom's formal charge equal to 0, then draw the dou...

Haochen He 3L

Hi!
I think you need to assign the formal charge when you draw the structure. It doesn't matter where you put the elements except the centreal one.

Haochen He 3L

According to the equation E(photon)-work function=Kinetic energy, the work funtion is a constant. So the higher frequency, the larger energy of Photo, and thus the larger kinetic energy.

Haochen He 3L

Hi！
I think you should have 5 posts per week. :)

Haochen He 3L

Hi!
You should use E of photo to minus the kinetic energy of the electron (mv^2/2) in order to get the work funtion.

Haochen He 3L

Yes, if it is Lyman series n1 should be 1

Haochen He 3L

The sulfate has four O atoms (6*4 electrons) so prof multiplied 4. In ammonium there are 4 H atoms, therefore he multiplied the number of electrons by 4

Haochen He 3L

I think both are OK, unless it is stated in the question.

Haochen He 3L

Hi! In part a, you get the energy emitted by an excited atom. For part b, you just need to find the number of sodium atoms in 5 mg sodium (using moalr mass of sodium and avogadro's number). Then multiply the result of part a and the number of sodium atoms.

Haochen He 3L

I just did all of them because it's said that some of midterms questions will come from the textbook. lol

Haochen He 3L

Hi! The number of electrons emitted is maximized when their kinetic energy equal to 0. So you just need to divide the total energy by the work funtion.

Haochen He 3L

Hi guys!

Does anyone how the phenomenon that cathode rays pass through metal foils can support wave model of electron?

Haochen He 3L

Delta X is +/- 1% of the hydrogen radius, whcih is (1.01-0.99)(0.05nm). Hope it helps!

Haochen He 3L

Th answer would be D, because the momentum of a sationary object is 0. The Heisenberg Indeterminacy is for moving object.

Haochen He 3L

Hi! The delta X as mentioned in the question is +/- 1% of the hydrogen radius (2*0.01*0.05nm). To get the delta V you need to first use the equation Delta p x Delta x >= h / 4 pi to get the value of Delta P. Then plug the values in delta P = M * delta V where M is the mass of an electron. Hope it he...

Haochen He 3L

yea, we should use the equation E(photon) - E (energy remove e-) = E (excess), where work function is the energy remove e-. the maximum kinetic energy is E (excess). Basically, you just need to plug in the values and do the calculation. Hope it helps !

Haochen He 3L

Hi! i think the prefix M (mage) in this question means 10^6. Hope it helps :)

Haochen He 3L

I believe the unit doesn't matter if you're just concerning about the calculation. It's ok to use milliliters for volume. However, if you are tring to solve M, the unit should be in liters. (mol/L)

Haochen He 3L

I think it's best to always check the equation first. Usually it's quite easy to tell whether it's balanced or not.

Haochen He 3L

Hi!! I usually use 4 sig figs. It seems to be enough based on the questions I've solved.

Haochen He 3L

there are four sig figs: 3204.

Haochen He 3L

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