Jukelia Posted February 28, 2008 Share Posted February 28, 2008 I need help in starting and setting up this prolem. I do not want the answer. The Queastion is: Determine the mass in grams,of 6.02 X 10^23 silver atoms (sorry) Link to comment Share on other sites More sharing options...
iNow Posted February 28, 2008 Share Posted February 28, 2008 Shouldn't there be a unit of measure where you put the face? Something like kg, or picograms or something? As it stands now, you just move the decimal to right 23 times (adding zeros as you go) and you still don't know the unit of measure or what you're measuring. 602,000,000,000,000,000,000,000 Link to comment Share on other sites More sharing options...
MrSandman Posted February 28, 2008 Share Posted February 28, 2008 What is it in? In the first place the question is like saying you have 602,000,000,000,000,000,000,000 somethings. Doesn't make since. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted February 28, 2008 Share Posted February 28, 2008 6.02 x 1023 is one mole, correct? So you need to look up the molar mass of silver to see how much one mole of silver weighs. Link to comment Share on other sites More sharing options...
timo Posted February 28, 2008 Share Posted February 28, 2008 Shouldn't there be a unit of measure where you [=Jukelia] put the face?Nope.EDIT: vv I see, also explains the strange "sorry". vv Link to comment Share on other sites More sharing options...
iNow Posted February 28, 2008 Share Posted February 28, 2008 Nope. Notice their post was edited AFTER mine was submitted... Link to comment Share on other sites More sharing options...
CaptainPanic Posted February 28, 2008 Share Posted February 28, 2008 Sounds like someone is learning the concept of a "mole". A "mole" is indeed 6.02 x 10^23. It is just a number. Like a "dozen" (12) and a "gross" (144). This number is just a lot bigger than 12 or 144. So you don't need it to describe an amount of eggs or fruit (a mole of eggs would be a gigantic mountain). Instead it is used to describe amounts of the smallest particles: atoms and molecules. To answer the question: Determine the mass in grams,of 6.02 X 10^23 eggs (sorry) You first need the weight of 1 egg, no? So, I think you can figure out what kind of number you need first before you can solve the problem. Link to comment Share on other sites More sharing options...
thedarkshade Posted February 28, 2008 Share Posted February 28, 2008 I need help in starting and setting up this prolem. I do not want the answer.The Queastion is: Determine the mass in grams,of 6.02 X 10^23 silver atoms (sorry) I think this is pretty simple. At school we use: [math]\frac{m}{M}=\frac{N}{L}[/math] where m=mass M=molar mass N=number of atoms and L is a constant; and in this case m is what you are looking for. Having this equation, the rest is just math:D Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted February 28, 2008 Share Posted February 28, 2008 What exactly is L there? The number of atoms (in moles) times the molar mass of silver is the mass of the pile of atoms. I don't see the point of L unless you're dealing in inconsistent units. Link to comment Share on other sites More sharing options...
CaptainPanic Posted February 29, 2008 Share Posted February 29, 2008 This "L" will confuse people. L = 1 (dimensionless, i.e. no units)... skip it. It should not be in the formula, unless you intend to use pounds, ounces or some other idiot (non SI) unit for the weight. My advise: use grams. m = mass in [grams] M = molar mass (or atomic mass), in [grams/mole] n = number of moles, in [mole] (number of times you have 6.02*10^23). m = n*M (in words: mass = number of moles * molar mass) The molar mass, as I said before, is something you need to look up in a book or google (wikipedia, then search for "Silver" and read the right side of the page). Ok, I practically gave away the answer here... sorry. Link to comment Share on other sites More sharing options...
Recommended Posts
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 accountSign in
Already have an account? Sign in here.
Sign In Now