Recent content by khanhhung2512

I still haven't understood all of what you're saying, but it's making sense to me now. Thank you very much. One more question, please. What kind of book do I need to read or study to have such a good knowledge of physical chemistry as yours, or at least to be able to fully comprehend this He2...

I mean, what's the reason the stable, low-energy excited state would spontaneously drop electron to transform to the unstable, high-energy ground state?

Suppose you're correct, then why would stabler excited He2 molecules spontaneously break apart as the electron(s) fall back to the ground state?

Because the ground state is lower in energy, and thus is stabler than the excited state, whatever bond orders they have. My explanation for this He2 question is as followed: In high-energy environments, some He atoms will exist in the excited state. Two He atoms might then form a He2 molecule...

I think the question didn't tell us anything about the configuration of the He2 molecule in high-energy environment.

Because with this explanation, excited-state configuration (σ1s)2(σ1s*)1(σ2s)1 with bond order 1 seems stabler than ground-state configuration (σ1s)2(σ1s*)2 with bond order 0, while it's not the case.

This is a question in "Chemical Principles, 6th Edition, Steven Zumdahl": Bond energy has been defined in the text as the amount of energy required to break a chemical bond, so we have come to think of the addition of energy as breaking bonds. However, in some cases the addition of energy can...

These are two answers I found on the net http://answers.yahoo.com/question/index?qid=20110915202418AA2v9AN While both answers seem wrong to me, one mentioned Advanced Inorganic Chemistry 6th ed (1999). p 868-869 (F. A. Cotton, G. Wilkinson, C. A. Murillo, M. Bochmann). If anyone has access to...

Well, the color we see is complementary to the color absorbed.

Copper (ii) chloride is a light brown solid, which slowly absorbs moisture to form a blue-green dihydrate. According to ligand field theory, water is a stronger field ligand than chloride. As a result, the dihydrate form should have a larger d orbital splitting than the anhydrous form. Thus...

The followings are the chemical reactions I learned at my university: K2S2O8(s) + H2SO4(concentrated) → H2S2O8 + K2SO4 H2S2O8 + 2H2O → H2O2 + 2H2SO4 H2O2 (heat)→ H2O + [O] [O] + O2 → O3 However, I'm wondering why I can't find these reactions anywhere on the Internet. What do you think...

Thanks. There's still one part in my questions. Is triple bond cylindrically symmetric?

I read in several websites that triple bonds cannot rotate freely. However, I've also read in the book "Chemical Principles" the following lines: "Various types of evidence suggest that the electron density around the two C-O bonds in CO2 is actually cylindrically symmetric—that is, the...

Yes, that's what I think. However, textbooks don't seem to incorporate that radiation in photoelectric effect. Is it because that amount of energy (according to Larmor formula) is too small and negligible?

Textbooks say that by measuring the stopping potential V0, we can determine the maximum kinetic energy with which electrons leave the cathode: eV0 = Kmax However, as I know, when we apply the stopping potential, the electrons will decelerate and radiate parts of their energies. Thus, the...