Jump to content

Chemistry Problem


kamy

Recommended Posts

Can someone explain to me how to answer this question, thanks

 

If all of the energy from the decay of 14C in your body could be captured and converted into electricity, how many "radioactive" people would be required to power a 100 W light bulb?:confused:

Link to comment
Share on other sites

okay, first you'll need to find out what the decay energy of C-14 is. google it or go on wikipedia. then, you'll need to know roughly how much C-14 is in the average human.

then, you need to work out how many decays per second go on in a human. then you can work out the power of C-14 decay in one human and its simple from there

Link to comment
Share on other sites

erm.... we All have Nitrogen and Potassium isotopes in us (as well as the Carbon mentioned).

 

the point I made in post #4 is that they also could be counted.

 

I deny all the rest as Pure Speculation, and plead the 5`th!!!

Link to comment
Share on other sites

I don't know about you but all the nitrogen isotopes in me are stable. The lonest lived radioisotope has a half life of under 10 minutes. I think my tritium, thorium, radium, 90strontium etc, loading will contribute many orders of magnitude more radiation than any radioisotopes of nitrogen.

Link to comment
Share on other sites

It's very very nearly true, at any one time I doubt there's more than a few atoms of radioactive nitrogen in me formed by cosmic ray interactions. Of course some theories say that protons decay given time so there are no stable isotopes.

I could use the ratio of the 2 stable nitrogen isotopes to look for patterns related to food etc in the same way that it can be done with carbon 13/ carbon 12 signatures. Why would they need to be radioactive?

 

In fact, I couldn't do it with the radioisotopes of nitrogen because they are absurdly rare. Anyway since they don't last long enough to be digested there wouldn't be any point trying.

I think you need to check some references.

Link to comment
Share on other sites

I think you might do better to avoid double posting and to start a new thread for a new topic. Since you are new, I'll answer anyway but please don't do it again.

As it happens the first equation is ballanced.

There might be lots of uses for that reaction.

Heating copper in air leaves it covered with a layer of oxide. That layer can be removed by washing with dilute H2SO4 to give CuSO4 which dissolves.

 

I'm not sure, but I think CuO might exist naturally as a copper ore. If it does then you might want to extract the ore with dilute acid to recover the copper for processing.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.