5.00 lb of mercury metal is spilled onto a counter and forms a circular puddle...

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iammonster

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...with a diameter of 62.7 cm. ? What is the thickness of the pudddle of mercury? [ie. Volume=pie(d/2)^2h]

how do i do this problem?
 
1. You are basically asked to solve the volume of a cylinder as a circult puddle. V = pi*r^2 * h. From this you want to solve for h, the height. h = V / (pi * r^2) We are given r = 62.7 cm / 2 = 31.35 cm

2. You are given a mass of Hg, but you need to know really the volume of mecury. To get the volume of a liquid given its mass, you need to know its density (density = mass / volume). So you need to find this value (a constant) for mercury. Wikipedia is a decent source http://en.wikipedia.org/wiki/Mercury_(element), and it gives us the value of 13.53 g/ml = 13.53 g/cm^3.

3. The mass is not given in SI units of kg or grams, so we also need the conversion factor of 2.2 lb = 1 kg.

The volume of Hg is thus:

V = (5.00 lb) * (1 kg/2.2 lb) * (1000 g/.kg) * (1 cm^3 / 13.53 g) = 167.977 cm^3

Notice that I added conversion factors also and that all values are arranged so that units/dimensions cancel and leave us with a value in units of cubic centimeters, the volume. Once you get dimensions to properly cancel each other, you are just left with multiplying the values: 5 * 1/2.2 * 1000 * 1/13.53

And so:

h = 167.977 cm^3 / (pi * (31.35 cm)^2) = 5.44e-2 cm = 0.544 mm

Note that you were given THREE significant digits, and so your answer must have THREE significant digits.
 
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