Recent content by chemnoob

No you don't want constant current, you just need a somewhat stable voltage around 5 volts. That is, you need to go from the somewhat stable voltage of around 6 to 7 volts that your pack provides to a somewhat stable voltage around 5 volts that the phone's internal charging circuit requires...

Ok. So is it still correct to say that the sum of the oxidization numbers (of the individual atoms) in a molecule must equal the charge, -2 in the above example? To be honest I don't even know what the reaction products would be for that reaction, apart from the Cr^{\,3+} as that was given...

I've recently been trying to teach myself this stuff. Somebody let me know if this is way off base but this is a method that seems to work for me. A neutral molecule has an overall oxidization number of zero, and a charged radical or ion has an overall oxidization number equal to it's charge...

Ok thanks. That tends to confirm my suspicion that this distinction is not so widely used. Perhaps it's one of the authors own "thangs". BTW. He introduces this terminology while discussing the concept of conjugate acids and bases. He explains how the stronger the acid the weaker it's...

Yes, he gives Boric acid (HBH2O3) and Phenol (HOC6H5) as examples of "feeble" acids. I looked these up and they both have dissociation constants order of magnitude around 10^-10. By contrast the examples of weak acids he gives tend to have dissociation constants around 10^-4 and 10^-5.

Definition of a "Feeble" acid? I'm currently trying to re-familiarize myself with chemistry, after a very long absence of any study in this field. At the moment I'm working my way through the text "Fundamentals of Chemistry" by "David E Goldberg" and I notice the author introduces the concept...

I'm wondering what happens in a single displacement reaction if the displacing metal (the one that's going into the solution) reaches it's solubility limit before the other metal in the solution is fully displaced. Say for example you had a strong silver nitrate solution, and you placed enough...