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Mathematics 10 Online
OpenStudy (anonymous):

What mass of lead (density 11.4 g/cm^3) would have an identical volume to 15.0 g of mercury(density 13.6 g/cm^3)

OpenStudy (anonymous):

say what?

OpenStudy (anonymous):

that aint nice

OpenStudy (anonymous):

umm. is this for chemistry?

OpenStudy (anonymous):

anyways, do you know how you go about solving this kinds of problems? like, do you know the setup?

OpenStudy (anonymous):

thats what im trying to get and it is for honors chem. i just dont get it

OpenStudy (anonymous):

oh ok i'm understanding this more... i think. i was thinking too hard. so first we have to get the volume of mercury to see what mass we need.

OpenStudy (anonymous):

do you guys use Dimensional Analysis (DA)?

OpenStudy (anonymous):

not yet - but what does that mean?

OpenStudy (anonymous):

nevermind then we can just use the d=m/v formula then.

OpenStudy (anonymous):

ohok

OpenStudy (anonymous):

Ok so let's play around with the formula so we can get the volume of 15.0g of mercury. |dw:1346380653011:dw|

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