Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: nikita on May 27, 2008, 06:50:18 PM
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i dont even know what theyre asking for here, but i think if someone could tell me, i can solve it.
a roll of aluminum wire is used to make finishing nails. if the nails are 5.00 cm long and .300 cm in diameter and the roll has a mass of 225 kg, how many nails can be made from one roll?
now im thinking of a nail with no head and i picture a cylinder. apparently, i am to use the formula for the volume of a circle with length (eg. pi * r^2 * l). does anyone know why? a nail with no head is not a circle to me. anyway, i get this far, using the formula for a circle with length:
pi*r^2 * l = 3.14*.0225*5= .353cm
so i guess the volume is .353cm
so, then what? i need to find out a number rather than a weight
maybe if i knew why i was using the above formula, i wouldnt be so mystified.
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i dont even know what theyre asking for here, but i think if someone could tell me, i can solve it.
a roll of aluminum wire is used to make finishing nails. if the nails are 5.00 cm long and .300 cm in diameter and the roll has a mass of 225 kg, how many nails can be made from one roll?
now im thinking of a nail with no head and i picture a cylinder. apparently, i am to use the formula for the volume of a circle with length (eg. pi * r^2 * l). does anyone know why? a nail with no head is not a circle to me. anyway, i get this far, using the formula for a circle with length:
pi*r^2 * l = 3.14*.0225*5= .353cm
so i guess the volume is .353cm
so, then what? i need to find out a number rather than a weight
maybe if i knew why i was using the above formula, i wouldnt be so mystified.
Don't overcomplicate it. Ignore the nail premise. Read the question as 'You're making Al cylinders with a 5 cm length and a .3 cm diameter. You have 225 kg of Al stock to make these cylinders. How many cylinders can you make?' Try to answer it from there.
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ok, completely ignore my rantings about the formula for a cylinder. i am henceforth ignoring the back cover of my book which has led me astray.
i tend to overanalyze everything. sometimes this is good, but not today.
well, im officially an idiot because i have no idea how to apply mass and volume to a number of nails. i guess i have the volume of each separate nail, and now i need to find the mass of each nail? i know the total mass. then maybe each nails mass with the total mass, somehow will give me the number of nails?
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Are you given the density of Al in your book anywhere?
http://wiki.answers.com/Q/What_is_the_density_of_aluminum
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duh. this is how you can tell i havent taken chem in 10 years. i get it now.
m= dv
m= 2.7*.353
m = .953g
mass for each nail is .953g and total mass is 225kg
22500g/.935g = 2.25x10^5 nails.
thank you. i think i know what im missing in these probs now.
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didn't check your math, but looks good
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thanks.
the math was wrong because i transposed some numbers, but the solution is
m = dv
m= (2.698)(.353)
m= .952
m=225kg = 225000g
225000g/.952g = 2.36 x 10^5 nails.