Oh okay. Thank you! I found out what I did wrong. For this question would I use the equation with ln? For the 18 hour one I got 97.16% The 1 day I got 46.68%. The 5 half-lives I got 5.61%. But I got it wrong. Im not sure how to do it

Technetium-99m, an important radionuclide in medicine, has a half-life of 6.0 hours. What percentage of a sample of technetium-99m will remain undecayed after the following times?

Hi again! I need help with some chemistry questions. Sorry Im asking here again, something is wrong with my account and wont let me ask a new question.

There are two half-reactions: Cu(s) -> Cu^+ (aq) + e^- (oxydation, anode, reduction potential E anode = 0.52 V) and Ag^+ (aq) + e^- -> Ag (s) (reduction, cathode, reduction potential E cathode = 0.80 V. The cell will give E cathode - E anode = 0.80 - 0.52 = 0.28 V for a standard cell emf.

"Incorrect. You may have summed two reduction potentials. You need to flip the sign of the potential for the oxidation half-reaction before adding it to the potential for the reduction half-reaction."

This is what it said when i got it wrong

Apr 16th, 2015

Never mind I figured it out!

Apr 16th, 2015

For the following reactions, determine E°, ΔG°, and K, given the balanced half reactions and the standard reduction potentials. Also, determine if the reaction is spontaneous as written.

a.) 2Co^3+ + H3AsO3+H2O ------> 2Co^2+ +H3AsO4+2H^+

Co^3+ +e^- ----->Co^2+ E°=1.92 V

H3AsO4+2H^+ +2e^- ----->H3AsO3+H2O E°=0.575 V

b.) 4Fe^3+ +2H2O -----> 4Fe^2+ +O2+4H^+

Fe^3+ +e^- -----> Fe^2+ E°=0.771 V

(1/2)O2+2H^+ +2e^- -------> H2O E°=1.229 V

I keep getting this wrong and I have no idea what I am doing wrong.

Apr 16th, 2015

I still cant figure it out...

Apr 16th, 2015

For the question above I seem to keep getting it incorrect again. I tried to look at some help videos but still could not understand it.