Chemistry
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OpenStudy (anonymous):
l
12 years ago
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OpenStudy (anonymous):
12 years ago
OpenStudy (anonymous):
@zepp
12 years ago
OpenStudy (anonymous):
@zepp here
12 years ago
OpenStudy (zepp):
Give me 5 minutes, I'll post the steps
12 years ago
OpenStudy (anonymous):
is it 14.4 mol CH2 ?
12 years ago
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OpenStudy (anonymous):
don't need to post the steps man...im already kinda late for this :|
12 years ago
OpenStudy (zepp):
I got 10.79 mol
12 years ago
OpenStudy (anonymous):
yea it's correct
12 years ago
OpenStudy (zepp):
Late for what? You are never late to understand the concept right now, I understood the topics, the concepts in maths, last year, and I did 3 of my homework, out of ~60, and I still aced the tests, as long as understand the concept, you are okay.
12 years ago
OpenStudy (anonymous):
late for the due date lol
12 years ago
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OpenStudy (anonymous):
when you get the x=474 do you just convert it into mol and get 10.79 ?
12 years ago
OpenStudy (zepp):
Yep
12 years ago
OpenStudy (zepp):
g <-molar mass of CO2 -> mol
12 years ago
OpenStudy (anonymous):
how did u get 162,324 g ?
12 years ago
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OpenStudy (anonymous):
@zepp plz dont give up on me yet :(
12 years ago
OpenStudy (anonymous):
new problem above ^
12 years ago
OpenStudy (zepp):
162 is the mass of the 5.4 mol
12 years ago
OpenStudy (zepp):
uh sec
12 years ago
OpenStudy (anonymous):
i got it, i got it
12 years ago
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OpenStudy (zepp):
Where's the new question? o.o
12 years ago
OpenStudy (anonymous):
hold on
12 years ago
OpenStudy (anonymous):
12 years ago
OpenStudy (anonymous):
i got the answer to the question i posted before so, here's the new one :D
12 years ago
OpenStudy (anonymous):
@zepp
12 years ago
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OpenStudy (anonymous):
these question are really hard O_O
12 years ago
OpenStudy (zepp):
I love Chemistry, but not at 11pm >.> Hold on
12 years ago
OpenStudy (anonymous):
omg please...gotta finish it by 11pm T____T
12 years ago
OpenStudy (zepp):
That's exactly what I did in the other question :|
12 years ago
OpenStudy (anonymous):
this is a 3 steps problem....:(
12 years ago
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OpenStudy (zepp):
Find the g of O2 required first, using stoichiometry
12 years ago
OpenStudy (anonymous):
51.2 mol of O2 <----- BAMZ!!!
12 years ago
OpenStudy (zepp):
lemme see
12 years ago
OpenStudy (zepp):
Wrong unit :(
12 years ago
OpenStudy (zepp):
51.2g*
12 years ago
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OpenStudy (zepp):
and if you can do the first one, then you can do the second one :D
12 years ago
OpenStudy (anonymous):
im tired.....plz do for me zepp =.=
12 years ago
OpenStudy (anonymous):
i will greatly appreciated and will send u a basket of apples later :))
12 years ago
OpenStudy (zepp):
No
12 years ago
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OpenStudy (zepp):
:(
12 years ago
OpenStudy (anonymous):
why r u so mean to me :(
12 years ago
OpenStudy (anonymous):
70.4 g CO2 <----- BAMZ
12 years ago
OpenStudy (zepp):
Not mean, if you can do the first, you can do the other one as well
12 years ago
OpenStudy (zepp):
See? you got it
12 years ago
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OpenStudy (anonymous):
naw your mean :>
12 years ago
OpenStudy (anonymous):
no apple for u ! :<
12 years ago
OpenStudy (zepp):
lawl
12 years ago
OpenStudy (anonymous):
ok i have no idea wtf is the next question talking about so here...
12 years ago
OpenStudy (zepp):
Nitrogen atoms are bigger than Hydrogens
12 years ago
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OpenStudy (anonymous):
when it talks about molecules that means its referring to Avogadro's number right ? or is it atom?
12 years ago
OpenStudy (anonymous):
oh wait u told me this earlier opps !
12 years ago
OpenStudy (zepp):
o.o this has nothing to do w/ avogadro
12 years ago
OpenStudy (anonymous):
yea xD
12 years ago
OpenStudy (zepp):
for 1 nitrogen, 3 hydrogen molecules will stick to it
12 years ago
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OpenStudy (zepp):
For 3 for the first question
12 years ago
OpenStudy (anonymous):
3 is the answer for the first question you mean ?
12 years ago
OpenStudy (zepp):
yep
12 years ago
OpenStudy (zepp):
1 for 2nd
0 for 3rd
and nitrogen for last.
12 years ago
OpenStudy (anonymous):
lol.. all 4 are wrongs.. :))
12 years ago
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OpenStudy (zepp):
lies
12 years ago
OpenStudy (anonymous):
it's the truth.....and the truth sometime hurts!
12 years ago
OpenStudy (anonymous):
lol lol, it's all wrong zepp ! try again
12 years ago
OpenStudy (anonymous):
well at least we know it's hydrogen for the last one lol
12 years ago
OpenStudy (zepp):
Oh, 2NH3.... >.>
12 years ago
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OpenStudy (zepp):
6 for 1st then
12 years ago
OpenStudy (anonymous):
46 minutes of desperate :(
12 years ago
OpenStudy (zepp):
6
0
1
Hydrogen
I reversed the numbers in 2/3 >.> sorry about that
12 years ago
OpenStudy (anonymous):
yay correct :D
12 years ago
OpenStudy (zepp):
psh, don't underestimate me
12 years ago
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OpenStudy (anonymous):
hehehehe :)))
12 years ago
OpenStudy (anonymous):
just a few more professor !! hang in there
12 years ago
OpenStudy (anonymous):
12 years ago
OpenStudy (anonymous):
next summer i will take vacation to Montreal yup :))
12 years ago
OpenStudy (zepp):
o.o
12 years ago
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OpenStudy (zepp):
need some calculations, hold on
12 years ago
OpenStudy (anonymous):
sure sure...39 minutes left
12 years ago
OpenStudy (zepp):
how many questions left
12 years ago
OpenStudy (anonymous):
10 left lol =.=
12 years ago
OpenStudy (anonymous):
it's fine..take your time, its already too late anyway, at least got a passing grade on this :D
12 years ago
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OpenStudy (zepp):
18.25 first answer
12 years ago
OpenStudy (zepp):
and 14.54g of CaCO3
12 years ago
OpenStudy (anonymous):
sorry for all this :(
12 years ago
OpenStudy (anonymous):
the problem u just did is correct thanks :d
12 years ago
OpenStudy (zepp):
o.O
12 years ago
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OpenStudy (anonymous):
these shoulda be done by yesterday, but for some reason i got kicked out of OS and can't log back in :(
12 years ago
OpenStudy (zepp):
Os was down yesterday night
12 years ago
OpenStudy (anonymous):
really..no wonder
12 years ago
OpenStudy (anonymous):
i really need another place where i can contact you zepp...otherwise i dont know what to do when in need...
12 years ago
OpenStudy (zepp):
0.21mol
12 years ago
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OpenStudy (anonymous):
u never add my yahoo o_0
12 years ago
OpenStudy (zepp):
thought I did D:
12 years ago
OpenStudy (anonymous):
mol of Iron ?
12 years ago
OpenStudy (zepp):
1st box yea
12 years ago
OpenStudy (anonymous):
@zepp having trouble ?
12 years ago