Chemistry
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OpenStudy (anonymous):
What is the average mass of a single chlorine atom in grams?
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OpenStudy (anonymous):
@dpaInc here
OpenStudy (anonymous):
im having lunch right now....cuz waiting for u took forever D:
OpenStudy (anonymous):
How many atoms there are in a mole?
OpenStudy (anonymous):
6.02x10^23
OpenStudy (anonymous):
Cool, what's the molar mass of chlorine?
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OpenStudy (anonymous):
35.45
OpenStudy (anonymous):
Cool, so what's the mass of 1 atom of chlorine?
OpenStudy (anonymous):
take that multiply by that ?
OpenStudy (anonymous):
\[\large 1 \text{ mole}=6.023*10^{23} \text{atoms}\]
OpenStudy (anonymous):
\[\large 1 \text{ mole of chlorine}=34.45\text{g}\]
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OpenStudy (anonymous):
Transitivity property; \(\large 6.023*10^{23} \text{ atoms of chlorine}=34.45\text{g of chlorine}\)
Now you can do it.
OpenStudy (anonymous):
You are looking for the mass of 1 atom.
OpenStudy (anonymous):
34.45g/6.02x10^23 ??
OpenStudy (anonymous):
yep
OpenStudy (anonymous):
5.72x10^23
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OpenStudy (anonymous):
Should be \(\large 5.72*10^{-23}\)
OpenStudy (anonymous):
You probably forgot to put \(6.023∗10^{23}\) in parentheses.
OpenStudy (anonymous):
ohh so when you divide then you gotta put it in parenthesis? and when your multiply then don't need to ?
OpenStudy (anonymous):
uh oh..."You are close. Check for rounding errors, and keep four sigificant figures in your answer."
OpenStudy (anonymous):
why i put 5.723x10^-23 and it still wrong wtf o_0
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OpenStudy (anonymous):
@Blitzkrieg
OpenStudy (zepp):
Lemme see
OpenStudy (anonymous):
@zepp hello?
OpenStudy (zepp):
Internet must stop lagging for nothing ;[
OpenStudy (anonymous):
i got problem with this site yesterday too, otherwise this hw would've done :(
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OpenStudy (anonymous):
anyway, what to do with the 5.723x10^-23 ?
OpenStudy (zepp):
Try 5.73x10^-23
OpenStudy (anonymous):
still wrong...it said check for rounding errors and it should be 4 sig fig.
OpenStudy (anonymous):
omg chlorine is 35.45 not 34.45 zepp !! D:
OpenStudy (zepp):
zz :[
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OpenStudy (anonymous):
got 80% correct for that problem.... LOL after soo many fails
OpenStudy (zepp):
>.>
OpenStudy (anonymous):
i posted 35.45 and you copied it wrong, and then i copied you and ....wrong lmao :)))
OpenStudy (zepp):
@Blitzkrieg WRONG
OpenStudy (anonymous):
triple wrong... O__O
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OpenStudy (anonymous):
@Blitzkrieg bad!! Blitz!!
OpenStudy (anonymous):
What is the mass (in grams) of 9.52 × 1024 molecules of methanol (CH3OH)?
OpenStudy (anonymous):
Ahhh
OpenStudy (anonymous):
Try 5.8858 * 10^-23 then?
OpenStudy (anonymous):
i got it...already done...moving on!
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OpenStudy (anonymous):
What is the mass (in grams) of 9.52 × 1024 molecules of methanol (CH3OH)?
OpenStudy (anonymous):
that's 10^24 btw
OpenStudy (zepp):
Oh cool
OpenStudy (anonymous):
oh i know how to do this
OpenStudy (anonymous):
Now it's conversion mole <-> atoms
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OpenStudy (anonymous):
is it a 3 steps problem?
OpenStudy (anonymous):
first yu gotta convert from mole to mol and then from mol to grams right ?
OpenStudy (anonymous):
506 g of CH3OH ?
OpenStudy (zepp):
Mole = mol o.o
OpenStudy (zepp):
mol is just a unit for mole, abbreviation.
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OpenStudy (anonymous):
oohhhh lmao =.=
OpenStudy (anonymous):
well then i guess you just take that 9.52x10^24 x avogadro's number right ?
OpenStudy (zepp):
Divide
OpenStudy (zepp):
Looks like you have an enormous difficulty to distinguish when to divide or multiply :|
OpenStudy (anonymous):
why is it divide? =.=
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OpenStudy (zepp):
Number of Avagadro = # of atoms * (in) 1 mole of some substance.
OpenStudy (zepp):
\[\large 6.023*10^{23}=6.023*10^{23}*1\text{mol}\]
OpenStudy (anonymous):
please do the calculation for me.. i wanna see how u do it
OpenStudy (anonymous):
@zepp
OpenStudy (anonymous):
omg it is 506 g of CH3OH !! i got it right D:
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OpenStudy (anonymous):
forget u zepp, i want Blitz back :D
OpenStudy (zepp):
Dad suddenly wants to disturb me.
OpenStudy (anonymous):
:(
OpenStudy (anonymous):
next problem:
How many grams of Cl are in 285 g of CaCl2?
OpenStudy (zepp):
We did many of this kind of question, you should be able to do them easily now :(
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OpenStudy (anonymous):
ok let me try, you check my answer ok? :D
OpenStudy (zepp):
alright
OpenStudy (anonymous):
is it 140 g Cl?
OpenStudy (zepp):
Put it in the box and see if you got it right or not, be self-confident :)
OpenStudy (anonymous):
D: !
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OpenStudy (anonymous):
cuz i already got 4 wrong for that problem so...self confident is gone hahaha
OpenStudy (zepp):
Show steps pls
OpenStudy (anonymous):
5 wrong lol
OpenStudy (anonymous):
hold on..i got this
OpenStudy (anonymous):
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