A 12.5 cm diameter piece of Whatman#1 filter paper has a thickness of 0.162 mm and a mass of 1.089 g. What is the density of the filter paper?
this is puzzling me, I know d=m/v but I am missing something because I am not getting the right answer. If you could, please also explain your logic.
2006-09-13 13:42:59 · 5 answers · asked by Rx 4 in Science & Mathematics ➔ Chemistry
wow, you guys all got it right! Thanks so much..I am not so stressed anymore. I knew I was missing one factor. DUH.
2006-09-13 14:15:17 · update #1
Take that back...some of you know the right answer! :)
2006-09-13 14:34:40 · update #2
d = 12.5 cm
r = 6.25 cm
Area (circle, I'm assuming) = pi*r^2
Volume of the filter paper = A*thickness
V = pi*(r^2)*(.162 mm)
V = pi*[(6.25 cm)^2]*(.0162 cm) (convert thickness to cm for dimensions to match)
V = 1.988 cm^3
Density = m/V
D = 1.089g/(1.988cm^3)
D = (approx.) 0.548 g/cm^3
2006-09-13 13:54:29 · answer #1 · answered by wheezer_april_4th_1966 7 · 0⤊ 0⤋
The paper volume is (pi x D^2)/4 x H = [(3.14 x 12,5^2)/4] x 0.162 = 19.87 cm^3.
If its mass is 1.089 g, its density is 1.089 g/19.87 cm^3 = 0.0548 g/cm^3.
I hope this is right.
2006-09-13 20:59:30 · answer #2 · answered by Xiquim 4 · 0⤊ 2⤋
volume = pi * radius squared * thickness
radius is 6.25 cm
thickness is .0162 cm
volume is therefore 1.988 cm cubed
density is 0.5478 grams per cubic centimeter
2006-09-13 20:51:26 · answer #3 · answered by piercesk1 4 · 0⤊ 0⤋
(6.25 cm)^2*pi*.0162cm= cm^3=v
1.089 g= m
So, just do 1.089/(6.25^2*pi*0.0162)
2006-09-13 20:52:15 · answer #4 · answered by jj 1 · 0⤊ 0⤋
67.222222g/cm3
2006-09-13 21:29:56 · answer #5 · answered by I Don't Know 2 · 0⤊ 2⤋