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Chemistry 52 Online
OpenStudy (anonymous):

l

OpenStudy (anonymous):

OpenStudy (anonymous):

@zepp

OpenStudy (anonymous):

@zepp here

OpenStudy (zepp):

Give me 5 minutes, I'll post the steps

OpenStudy (anonymous):

is it 14.4 mol CH2 ?

OpenStudy (anonymous):

don't need to post the steps man...im already kinda late for this :|

OpenStudy (zepp):

I got 10.79 mol

OpenStudy (anonymous):

yea it's correct

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.

OpenStudy (anonymous):

late for the due date lol

OpenStudy (zepp):

http://puu.sh/KAX4 There are the steps.

OpenStudy (anonymous):

when you get the x=474 do you just convert it into mol and get 10.79 ?

OpenStudy (zepp):

Yep

OpenStudy (zepp):

g <-molar mass of CO2 -> mol

OpenStudy (anonymous):

how did u get 162,324 g ?

OpenStudy (anonymous):

@zepp plz dont give up on me yet :(

OpenStudy (anonymous):

new problem above ^

OpenStudy (zepp):

162 is the mass of the 5.4 mol

OpenStudy (zepp):

uh sec

OpenStudy (anonymous):

i got it, i got it

OpenStudy (zepp):

Where's the new question? o.o

OpenStudy (anonymous):

hold on

OpenStudy (anonymous):

OpenStudy (anonymous):

i got the answer to the question i posted before so, here's the new one :D

OpenStudy (anonymous):

@zepp

OpenStudy (anonymous):

these question are really hard O_O

OpenStudy (zepp):

I love Chemistry, but not at 11pm >.> Hold on

OpenStudy (anonymous):

omg please...gotta finish it by 11pm T____T

OpenStudy (zepp):

That's exactly what I did in the other question :|

OpenStudy (anonymous):

this is a 3 steps problem....:(

OpenStudy (zepp):

Find the g of O2 required first, using stoichiometry

OpenStudy (anonymous):

51.2 mol of O2 <----- BAMZ!!!

OpenStudy (zepp):

lemme see

OpenStudy (zepp):

Wrong unit :(

OpenStudy (zepp):

51.2g*

OpenStudy (zepp):

http://puu.sh/KBtQ

OpenStudy (zepp):

and if you can do the first one, then you can do the second one :D

OpenStudy (anonymous):

im tired.....plz do for me zepp =.=

OpenStudy (anonymous):

i will greatly appreciated and will send u a basket of apples later :))

OpenStudy (zepp):

No

OpenStudy (zepp):

:(

OpenStudy (anonymous):

why r u so mean to me :(

OpenStudy (anonymous):

70.4 g CO2 <----- BAMZ

OpenStudy (zepp):

Not mean, if you can do the first, you can do the other one as well

OpenStudy (zepp):

See? you got it

OpenStudy (anonymous):

naw your mean :>

OpenStudy (anonymous):

no apple for u ! :<

OpenStudy (zepp):

lawl

OpenStudy (anonymous):

ok i have no idea wtf is the next question talking about so here...

OpenStudy (zepp):

Nitrogen atoms are bigger than Hydrogens

OpenStudy (anonymous):

when it talks about molecules that means its referring to Avogadro's number right ? or is it atom?

OpenStudy (anonymous):

oh wait u told me this earlier opps !

OpenStudy (zepp):

o.o this has nothing to do w/ avogadro

OpenStudy (anonymous):

yea xD

OpenStudy (zepp):

for 1 nitrogen, 3 hydrogen molecules will stick to it

OpenStudy (zepp):

For 3 for the first question

OpenStudy (anonymous):

3 is the answer for the first question you mean ?

OpenStudy (zepp):

yep

OpenStudy (zepp):

1 for 2nd 0 for 3rd and nitrogen for last.

OpenStudy (anonymous):

lol.. all 4 are wrongs.. :))

OpenStudy (zepp):

lies

OpenStudy (anonymous):

it's the truth.....and the truth sometime hurts!

OpenStudy (anonymous):

lol lol, it's all wrong zepp ! try again

OpenStudy (anonymous):

well at least we know it's hydrogen for the last one lol

OpenStudy (zepp):

Oh, 2NH3.... >.>

OpenStudy (zepp):

6 for 1st then

OpenStudy (anonymous):

46 minutes of desperate :(

OpenStudy (zepp):

6 0 1 Hydrogen I reversed the numbers in 2/3 >.> sorry about that

OpenStudy (anonymous):

yay correct :D

OpenStudy (zepp):

psh, don't underestimate me

OpenStudy (anonymous):

hehehehe :)))

OpenStudy (anonymous):

just a few more professor !! hang in there

OpenStudy (anonymous):

OpenStudy (anonymous):

next summer i will take vacation to Montreal yup :))

OpenStudy (zepp):

o.o

OpenStudy (zepp):

need some calculations, hold on

OpenStudy (anonymous):

sure sure...39 minutes left

OpenStudy (zepp):

how many questions left

OpenStudy (anonymous):

10 left lol =.=

OpenStudy (anonymous):

it's fine..take your time, its already too late anyway, at least got a passing grade on this :D

OpenStudy (zepp):

18.25 first answer

OpenStudy (zepp):

and 14.54g of CaCO3

OpenStudy (anonymous):

sorry for all this :(

OpenStudy (anonymous):

the problem u just did is correct thanks :d

OpenStudy (zepp):

o.O

OpenStudy (anonymous):

these shoulda be done by yesterday, but for some reason i got kicked out of OS and can't log back in :(

OpenStudy (zepp):

Os was down yesterday night

OpenStudy (anonymous):

really..no wonder

OpenStudy (anonymous):

i really need another place where i can contact you zepp...otherwise i dont know what to do when in need...

OpenStudy (zepp):

0.21mol

OpenStudy (anonymous):

u never add my yahoo o_0

OpenStudy (zepp):

thought I did D:

OpenStudy (anonymous):

mol of Iron ?

OpenStudy (zepp):

1st box yea

OpenStudy (anonymous):

@zepp having trouble ?

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