How many molecules are in 4.75 moles H2SO4?
1 mole = 6.023x10^23 molecules
I know that, but I'm confused on how to convert it.
if you have 4.75 dozens of doughnuts, you have 4.75 x 12= if you have 4.75 moles of molecules, you have 4.75 x avogadros number
also do you know why it ends up as like 10^4? example: 4.84 * 10^4. Why does the power of 10 change?
it shouldn't go down
you should get 2.86x10^24
but 4.75*6.022= 28
it's not 6.023, it's 6.023x10^23 as in 602300000000000000000000
can you show me what variables to multiply because I'm still very confused.
you have 4.75 moles right? so multiply 4.75 x (6.023x10^23)=
don't we have to find the MM of H2SO4 and multiply it by that first?
molar mass
no they're asking you for the amount of molecules not grams
okay, but does (4.75 x 6.023) multiply with 10^23?
do you know what scientific notation is? avogadros number is in scientific notation 6.023x10^23, is one number not 2.
wouldn't 6.023 x 10^23 be some outrageous number
WAIT
I think I understand
yes it is, because molecules are really, like really small
say we multiply 4.75 x 602300000000000000000000 (something like that) how does it end up with 10^24?
lol pretend it wasn't 10^23 pretend it was 6023 (six thousand and twenty three) multiply it by 4.75 4.75*6023=28609.25 you went from 6.023x10^3 to 2.8x^10^4 one order of magnitude change same thing with the problem you're doing
oh alright that makes sense, but if we want to solve this equation again do we have to always multiply by 6023000000000000000000?
isn't there a simpler way?
if you're finding the number of atoms, molecules or ions, you have to use avogadros number
it's simpler if you use scientific notation 6.023*10^23
I mean you have to times by 6.023 x 10^23, but how do you use the scientific notation?
that is scientific notation 6.023*10^23, in most calculators you can input the number like that
Alright, thank you for your help. Will it be alright if I happen to message you again if I have any questions?
yeah, for sure. i have a crappy internet connection right now, so it might take me a bit, but il reply
alright thanks again for your help!
no problem !
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