anonymous
  • anonymous
A solid substance has a density of 12 g/mL. A cube made out of this substance has 1 mm edge length. If the atomic mass of the substance is 120.0 g/mol, calculate how many moles of the substance are in this cube.
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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aaronq
  • aaronq
first find the volume of the cube, then find the mass with the density, finally convert the mass to moles
anonymous
  • anonymous
How do you find the volume?
aaronq
  • aaronq
|dw:1442365426768:dw|

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anonymous
  • anonymous
So do you find the mass by using the density formula?
aaronq
  • aaronq
yep
aaronq
  • aaronq
you'll have to convert the volume (or the density) to match the units
anonymous
  • anonymous
I'm trying to work it out and getting confused.
Jhannybean
  • Jhannybean
\[\sf \rho = \frac{m}{v} \\ \rho = 120~\frac{g}{mL}~,~~v =(1mm)^3 \] \[\sf 1~mm^3 ~\times~ \left(\frac{1~cm}{1~mm}\right)^3~\times~\frac{1~mL}{1~cm^3} =~1~mL\] \[\sf \frac{120~g}{mL} =\frac{m}{1~mL}~\implies m = 120~g\]
Jhannybean
  • Jhannybean
When I see a value or conversion factor, I typically try to associate it with an appropriate formula
aaronq
  • aaronq
you made a mistake, 1 cm = 10 mm, so: \(\sf \large 1~mm^3 ~\times~ \left(\frac{1~cm}{10~mm}\right)^3~\times~\frac{1~mL}{1000~cm^3} =~0.001~mL\)
Jhannybean
  • Jhannybean
...it was a typo <_<
aaronq
  • aaronq
suuuuure it was :P
aaronq
  • aaronq
A+ for effort
anonymous
  • anonymous
Um...which way is it?
Jhannybean
  • Jhannybean
The way @aaronq wrote.
anonymous
  • anonymous
Thanks
anonymous
  • anonymous
Would it be 1000 or 100 for cm^3?
anonymous
  • anonymous
Alright. I was wondering why there was an extra 0
anonymous
  • anonymous
Thanks for the help!
Jhannybean
  • Jhannybean
Ok that was my fault. I had it backwards. \[\sf 1~mm^3 \times\left(\frac{1~cm}{10~mm}\right)^3~\times~ \frac{1~mL}{1~cm^3} =~1.0~\times~10^{-3}~mL \]Thats what it should be. There should be no 1000.
Jhannybean
  • Jhannybean
So once we have this, create a ratio: \[\sf \frac{120~g}{1~mL} = \frac{m}{1.0~\times~10^{-3}~mL} \implies m=~(120 \cdot 1.0~\times~10^{-3})g\]
aaronq
  • aaronq
i made a mistake too..it's supposed to be: \(\sf \large 1~mm^3 ~\times~ \left(\frac{1~cm}{10~mm}\right)^3~\times~\frac{1~cm^3}{1000~mm^3} =0.001~cm^3=~0.001~mL\)
aaronq
  • aaronq
it is 1000, 10^3= 1000 no?

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