Gold can be hammered into extremely thin sheets called golfleaf. If a 200-mg pie

Question

Gold can be hammered into extremely thin sheets called golfleaf. If a 200-mg piece of gold (density = 19.32 g/cm^3) ishammered into a sheet measuring 2.4 x 1.0 ft, what is the averagethickness of the sheet in meters? How might the thickness beexpressed without exponential notation, using an appropriate metricprefix?

How do I begin to approach this problem? Can you please take mestep by step through it. Thank you so much for your help http://answers.yahoo.com/question/index;_ylt=AgL8r5Ph.jUopPiZJA9Q6uDsy6IX;_ylv=3?qid=20090822130303AAhAcGi Gold can be hammered into extremely thin sheets called golfleaf. If a 200-mg piece of gold (density = 19.32 g/cm^3) ishammered into a sheet measuring 2.4 x 1.0 ft, what is the averagethickness of the sheet in meters? How might the thickness beexpressed without exponential notation, using an appropriate metricprefix?

How do I begin to approach this problem? Can you please take mestep by step through it. Thank you so much for your help http://answers.yahoo.com/question/index;_ylt=AgL8r5Ph.jUopPiZJA9Q6uDsy6IX;_ylv=3?qid=20090822130303AAhAcGi

Explanation / Answer

You are given the density of the metal and the mass. Now find thevolume of metal available V = mass    * (1/density) V = 200 mg *(1 cm3/ 19.32 g) *(1 g / 1000mg) = 0.0214592275 cm3 The total volume is shaped as a rectangular prism. Solve for theheight, if given the area. V = height*area area = 2.4 ft*1.0 ft *(12 in /ft)2 *(2.54cm/1in)2 = 2229.67296 cm2 area is in ft^2. So use 12 in = 1ft and 2.54 cm/ in forconversion height = V/area = (0.0214592275 cm3)/(2229.67296 cm2) = 9.62e-6 cm Or 9.62e-6 cm *(1x109 nm / 100 cm) = 96.2 nm (nanometers)

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