Anyone good at chemistry????

Pllllllllllllllllllllllllllllllleas e I'm hoping there's someone who's good at chem on here, I have my midterm tomorrow and I need help with a problemmmm!!! Please please please lol Please. I beg of you. I know there has to be someone who can help.

 

:bigcry:

 

well perhaps iffin ya stated yer problem someone could help????

 

Ok, I'll just write the problem out.. here it is.

The approximate diameter of the nucleus of a lead atom is 7.8x10^-5 Angstrom and its mass is approximately 207 amu. The diamete of the earth is 7826 miles. Assuming that all nuclei are spherical and that there is no space between touching nuclei and that the earth consists entirely of such nuclei, how many suns of mass of 2.0x10^20 kg would be needed to equal the mass of one earth of nuclei?

Maybe someone can glance at it and help in some way or another...

 

Are you doing dimensional analysis?

 

Yeah.. I can help you with that a little later, i'm high and running late right now...

 

*blank stare*
*starts turning blue*
(oops! almost forgot to breathe!)

 

don't do that silly!
breathing's important!



but yeah i have no fucking clue, i like science when it's mythbusters science

 

Duno, I'm so lost in this class.

 

No this is actually kind of simple, you just need to know the conversions like amu --> kg, then set it up in fractions and cancel the other stuff with cross multiplication. in this case it looks like everything will cancel out leaving only a number - the number of suns

 

Okay good.. cause right now my answer is.... 4.667x10^-5 suns

Which is wrong. Very wrong.

 

What class is it? general or college chem?

I dont know how much of this stuff I remember, but will help you out where I can.

 

Yeah its probably going to be at least one sun haha

 

heres what you do...

convert *7.8x10^-5* angstroms to miles, then multiply by 7826 miles and you get the number of nuclei needed to match earth's diameter. Then you multiply that number by 207 amu and you get the mass of one Earth of nuclei.

Then you divide the mass of one Earth of nuclei by 2.0x10^20 kg (the mass of suns) and thats your answer.

Dont get no easier than that.

 

1 angstroms = 6.21371192 × 10-14 miles

you can google the conversions.

I never went to either of my chemistry classes unless it was a test day.

As long as you have google and a ti-83 your fine.

 

College, Inorganic Chem 1.

=) You're my friend, cause you're smart.

I know it calls for conversions... I just have problems with conversions for some reason, I just caaaaan't figure them out.

Thank you.

 

Yeah, the only problem is when I get into the class for the test, and I have a conversion problems, and no google.

I dunno, conversions like this just really confuse me, like I can't even figure out what I'm trying to get, the amount of nuclei or the suns of mass shit?

Ugh, I have like 4 years of chemistryyyyy its just so confusing.

 

they dont let you bring a cheat sheet with formulas on it?

If you have a chem teacher who expects you to memorize every fucking formula then your teacher is a dick...you need to drop the class and take it again next semester with a different instructor.

 

These are just stiochiometric exercises. It's best to write down all the relations given and all the conversions known first, determine what units the answer is expected to be in, and piece together them so that everything cancels out but the desired unit.

 

Yeah ozone is doing it in a way thats harder to understand if you are confused, posthumous is right too and using the method I was talking about. If you still need help maybe i can write it out and scan it or something. Its called dimensional analysis, this might help http://en.wikipedia.org/wiki/Dimensional_analysis