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Chemistry 17 Online
OpenStudy (anonymous):

Which equivalence factor should you use to convert from 6.26 moles of O2 to molecules of O2? (Points : 3) (1 molecule/6.02 x 1023 mol O2) (6.02 x 1023 molecules O2/1 mol O2) (6.02 x 1023 molecules O2/6.26 mol O2) (6.26 mol O2/6.02 x 1023 molecules O2) Again, I believe it is the 4th!

OpenStudy (anonymous):

Molecules /6.02x10^23=moles moles*6.02x10^23=molecules

OpenStudy (anonymous):

The third one? And I did work on it, but that response was totally helpful!

OpenStudy (anonymous):

it should be 6.26 *6.02x10^23

OpenStudy (anonymous):

That isn't an option...

OpenStudy (anonymous):

Weird !

OpenStudy (anonymous):

I mean you seem like you know what you are doing... could it be the teacher made a mistake?

OpenStudy (anonymous):

Its a basic rule .. ! I'am sure of this rule though

OpenStudy (anonymous):

Well,It is a basic rule as i said Maybe there is a trick somewhere ,ugh i'm kinda tired to notice xD

OpenStudy (anonymous):

I think a chemistry TEACHER should know the basic rules of what they are TEACHING.

OpenStudy (unklerhaukus):

\(6.26 \text{ mol } \text O_2\) \[=6.26 \text{ mol } \text O_2\times6.02\times10^{23}\frac{\text{molecules}}{\text{mol}}\]\[=(6.26\times 6.02)\times10^{23} \text{molecules } \text O_2\]\[\approx 3.8\times10^{24} \text{molecules } \text O_2\]

OpenStudy (unklerhaukus):

we have multiplied the quantity of gas by the second option

OpenStudy (anonymous):

Your answer doesn't match any of the choices given... it matches what the other person said

OpenStudy (unklerhaukus):

option 2

OpenStudy (unklerhaukus):

Which equivalence factor should you use to convert from \(\color{teal}{6.26\text{ moles of O}_2}\) to \(\color{brown}{\text{molecules of O}_2}\)? \[\color{red}{(6.02 \times 10^{23} \text{molecules O}_2/1 \text{ mol O}_2)}\] \(\color{teal}{6.26 \text{ mol O}_2}\) x \(\color{red}{(6.02 \times 10^{23} \text{molecules O}_2/\text{mol O}_2)}\) = 3.8e24\(\color{brown}{\text{ molecules of O}_2}\)

OpenStudy (anonymous):

OHHH! That makes total sense now! I was looking at it like what "factor set" instead of the proper set-up.

OpenStudy (preetha):

Unkle is right again!!!

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